Financial Economics & Insurance
Albert Cohen
Actuarial Sciences ProgramDepartment of Mathematics
Department of Statistics and ProbabilityA336 Wells Hall
Michigan State UniversityEast Lansing MI
[email protected]@stt.msu.edu
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 1 / 203
Course Information
Syllabus to be posted on class page in first week of classes
Homework assignments will posted there as well
Page can be found athttps://mathdata.msu.edu/CP/RW/S458 001.html
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 2 / 203
Course Information
Many examples within these slides are used with kind permission ofProf. Dmitry Kramkov, Dept. of Mathematics, Carnegie MellonUniversity.
Book for course: Derivatives Markets (2e/3e) (Pearson-AddisonWesley) by Robert L.McDonald. Can be found in MSU bookstoresnow
Some examples here will be similar to those practice questionspublicly released by the SOA. Please note the SOA owns thecopyright to these questions.
This book will be our reference, and some questions for assignmentswill be chosen from it. Copyright for all questions used from this bookbelongs to Pearson.
From time to time, we will also follow the format of Marcel Finan’s ADiscussion of Financial Economics in Actuarial Models: A Preparationfor the Actuarial Exam MFE/3F. Some proofs from there will bereferenced as well. Please find these notes here
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 3 / 203
What are financial securities?
Traded Securities - price given by market.
For example:
StocksCommodities
Non-Traded Securities - price remains to be computed.
Is this always true?
We will focus on pricing non-traded securities.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 4 / 203
What are financial securities?
Traded Securities - price given by market.
For example:
StocksCommodities
Non-Traded Securities - price remains to be computed.
Is this always true?
We will focus on pricing non-traded securities.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 4 / 203
What are financial securities?
Traded Securities - price given by market.
For example:
StocksCommodities
Non-Traded Securities - price remains to be computed.
Is this always true?
We will focus on pricing non-traded securities.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 4 / 203
What are financial securities?
Traded Securities - price given by market.
For example:
StocksCommodities
Non-Traded Securities - price remains to be computed.
Is this always true?
We will focus on pricing non-traded securities.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 4 / 203
How does one fairly price non-traded securities?
By eliminating all unfair prices
Unfair prices arise from Arbitrage Strategies
Start with zero capitalEnd with non-zero wealth
We will search for arbitrage-free strategies to replicate the payoff of anon-traded security
This replication is at the heart of the engineering of financial products
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 5 / 203
How does one fairly price non-traded securities?
By eliminating all unfair prices
Unfair prices arise from Arbitrage Strategies
Start with zero capitalEnd with non-zero wealth
We will search for arbitrage-free strategies to replicate the payoff of anon-traded security
This replication is at the heart of the engineering of financial products
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 5 / 203
How does one fairly price non-traded securities?
By eliminating all unfair prices
Unfair prices arise from Arbitrage Strategies
Start with zero capitalEnd with non-zero wealth
We will search for arbitrage-free strategies to replicate the payoff of anon-traded security
This replication is at the heart of the engineering of financial products
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 5 / 203
How does one fairly price non-traded securities?
By eliminating all unfair prices
Unfair prices arise from Arbitrage Strategies
Start with zero capitalEnd with non-zero wealth
We will search for arbitrage-free strategies to replicate the payoff of anon-traded security
This replication is at the heart of the engineering of financial products
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 5 / 203
How does one fairly price non-traded securities?
By eliminating all unfair prices
Unfair prices arise from Arbitrage Strategies
Start with zero capitalEnd with non-zero wealth
We will search for arbitrage-free strategies to replicate the payoff of anon-traded security
This replication is at the heart of the engineering of financial products
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 5 / 203
More Questions
Existence - Does such a fair price always exist?
If not, what is needed of our financial model to guarantee at least onearbitrage-free price?
Uniqueness - are there conditions where exactly one arbitrage-freeprice exists?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 6 / 203
And What About...
Does the replicating strategy and price computed reflect uncertaintyin the market?
Mathematically, if P is a probabilty measure attached to a series ofprice movements in underlying asset, is P used in computing theprice?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 7 / 203
And What About...
Does the replicating strategy and price computed reflect uncertaintyin the market?
Mathematically, if P is a probabilty measure attached to a series ofprice movements in underlying asset, is P used in computing theprice?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 7 / 203
Notation
Forward Contract:
A financial instrument whose initial value is zero, and whose finalvalue is derived from another asset. Namely, the difference of thefinal asset price and forward price:
V (0) = 0,V (T ) = S(T )− F (1)
Value at end of term can be negative - buyer accepts this in exchangefor no premium up front
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 8 / 203
Notation
Forward Contract:
A financial instrument whose initial value is zero, and whose finalvalue is derived from another asset. Namely, the difference of thefinal asset price and forward price:
V (0) = 0,V (T ) = S(T )− F (1)
Value at end of term can be negative - buyer accepts this in exchangefor no premium up front
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 8 / 203
Notation
Forward Contract:
A financial instrument whose initial value is zero, and whose finalvalue is derived from another asset. Namely, the difference of thefinal asset price and forward price:
V (0) = 0,V (T ) = S(T )− F (1)
Value at end of term can be negative - buyer accepts this in exchangefor no premium up front
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 8 / 203
Notation
Interest Rate:
The rate r at which money grows. Also used to discount the valuetoday of one unit of currency one unit of time from the present
V (0) =1
1 + r,V (1) = 1 (2)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 9 / 203
Notation
Interest Rate:
The rate r at which money grows. Also used to discount the valuetoday of one unit of currency one unit of time from the present
V (0) =1
1 + r,V (1) = 1 (2)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 9 / 203
An Example of Replication
Forward Exchange Rate: There are two currencies, foreign anddomestic:
SBA = 4 is the spot exchange rate - one unit of B is worth SB
A of Atoday (time 0)
rA = 0.1 is the domestic borrow/lend rate
rB = 0.2 is the foreign borrow/lend rate
Compute the forward exchange rate FBA . This is the value of one unit
of B in terms of A at time 1.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 10 / 203
An Example of Replication
Forward Exchange Rate: There are two currencies, foreign anddomestic:
SBA = 4 is the spot exchange rate - one unit of B is worth SB
A of Atoday (time 0)
rA = 0.1 is the domestic borrow/lend rate
rB = 0.2 is the foreign borrow/lend rate
Compute the forward exchange rate FBA . This is the value of one unit
of B in terms of A at time 1.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 10 / 203
An Example of Replication
Forward Exchange Rate: There are two currencies, foreign anddomestic:
SBA = 4 is the spot exchange rate - one unit of B is worth SB
A of Atoday (time 0)
rA = 0.1 is the domestic borrow/lend rate
rB = 0.2 is the foreign borrow/lend rate
Compute the forward exchange rate FBA . This is the value of one unit
of B in terms of A at time 1.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 10 / 203
An Example of Replication
Forward Exchange Rate: There are two currencies, foreign anddomestic:
SBA = 4 is the spot exchange rate - one unit of B is worth SB
A of Atoday (time 0)
rA = 0.1 is the domestic borrow/lend rate
rB = 0.2 is the foreign borrow/lend rate
Compute the forward exchange rate FBA . This is the value of one unit
of B in terms of A at time 1.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 10 / 203
An Example of Replication
Forward Exchange Rate: There are two currencies, foreign anddomestic:
SBA = 4 is the spot exchange rate - one unit of B is worth SB
A of Atoday (time 0)
rA = 0.1 is the domestic borrow/lend rate
rB = 0.2 is the foreign borrow/lend rate
Compute the forward exchange rate FBA . This is the value of one unit
of B in terms of A at time 1.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 10 / 203
An Example of Replication: Solution
At time 1, we deliver 1 unit of B in exchange for FBA units of domestic
currency A.
This is a forward contract - we pay nothing up front to achieve this.
Initially borrow some amount foreign currency B, in foreign market togrow to one unit of B at time 1. This is achieved by the initial
amountSBA
1+rB(valued in domestic currency)
Invest the amountFBA
1+rAin domestic market (valued in domestic
currency)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 11 / 203
An Example of Replication: Solution
At time 1, we deliver 1 unit of B in exchange for FBA units of domestic
currency A.
This is a forward contract - we pay nothing up front to achieve this.
Initially borrow some amount foreign currency B, in foreign market togrow to one unit of B at time 1. This is achieved by the initial
amountSBA
1+rB(valued in domestic currency)
Invest the amountFBA
1+rAin domestic market (valued in domestic
currency)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 11 / 203
An Example of Replication: Solution
At time 1, we deliver 1 unit of B in exchange for FBA units of domestic
currency A.
This is a forward contract - we pay nothing up front to achieve this.
Initially borrow some amount foreign currency B, in foreign market togrow to one unit of B at time 1. This is achieved by the initial
amountSBA
1+rB(valued in domestic currency)
Invest the amountFBA
1+rAin domestic market (valued in domestic
currency)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 11 / 203
An Example of Replication: Solution
At time 1, we deliver 1 unit of B in exchange for FBA units of domestic
currency A.
This is a forward contract - we pay nothing up front to achieve this.
Initially borrow some amount foreign currency B, in foreign market togrow to one unit of B at time 1. This is achieved by the initial
amountSBA
1+rB(valued in domestic currency)
Invest the amountFBA
1+rAin domestic market (valued in domestic
currency)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 11 / 203
An Example of Replication: Solution
This results in the initial value
V (0) =FBA
1 + rA−
SBA
1 + rB(3)
Since the initial value is 0, this means
FBA = SB
A
1 + rA
1 + rB= 3.667 (4)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 12 / 203
An Example of Replication: Solution
This results in the initial value
V (0) =FBA
1 + rA−
SBA
1 + rB(3)
Since the initial value is 0, this means
FBA = SB
A
1 + rA
1 + rB= 3.667 (4)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 12 / 203
Outline1 Continuous Model-Probability
Expected ValuesApplication of Option Greeks
2 Continuous Model-Ito CalculusBrownian MotionBSMExamplesOptions on FuturesPath Dependent Options
3 Advanced TopicsHeat EquationGeneral Solution of Heat EquationApplication to B-S-M PDE
4 Discrete Multiperiod ModelArbitrageRisk Neutral ProbabilityAmerican OptionsExotic OptionsValuation via SimulationAlbert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 13 / 203
Black Scholes Pricing using Underlying Asset
In the next section, we will derive the following solutions to theBlack-Scholes PDE:
V C (S , t) = e−r(T−t)E [(ST − K )+ | St = S ]
= Se−δ(T−t)N(d1)− Ke−r(T−t)N(d2)
V P(S , t) = e−r(T−t)E [(K − ST )+ | St = S ]
= Ke−r(T−t)N(−d2)− Se−δ(T−t)N(−d1)
d1 =ln(
SK
)+ (r − δ + 1
2σ2)(T − t)
σ√T − t
d2 = d1 − σ√T − t
N(x) =1√2π
∫ x
−∞e−
z2
2 dz .
(5)
Notice that V C (S , t)− V P(S , t) = Se−δ(T−t) − Ke−r(T−t).Question: What underlying model of stock evolution leads to this value?How can we support such a probability measure?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 14 / 203
Normal Random Variables
We say that X is a Normal Random Variable with parameters µ, σ2 if
fX (x) =1√2πσ
e−(x−µ)2
2σ2 for −∞ < x <∞ (6)
Furthermore, we say that Z is a Standard Normal Random Variable if it isNormal with parameters µ = 0, σ = 1. Consider
z =x − µσ∫ ∞
−∞fX (x)dx =
∫ ∞−∞
1√2πσ
e−(x−µ)2
2σ2 dx
=
∫ ∞−∞
1√2π
e−(z)2
2 dz
=
∫ ∞−∞
fZ (z)dz
(7)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 15 / 203
Normal Random Variables
We user the notation X ∼ N(µ, σ2) and Z ∼ N(0, 1). By ourtransformation above, it can be seen that if Z ∼ N(0, 1), thenX = µ+ σ · Z ∼ N(µ, σ2). We can see this via
Φ(z) := FZ (z) = P[Z ≤ z ]
= P[X − µσ
≤ x − µσ
]
= P[X ≤ x ] =: FX (x)
1
σfZ
(x − µσ
)=
d
dxFZ (z)
=d
dxFX (x) = fX (x)
fZ (z) =1√2π
e−(z)2
2
(8)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 16 / 203
Normal Random Variables
We user the notation X ∼ N(µ, σ2) and Z ∼ N(0, 1). By ourtransformation above, it can be seen that if Z ∼ N(0, 1), thenX = µ+ σ · Z ∼ N(µ, σ2). We can see this via
Φ(z) := FZ (z) = P[Z ≤ z ]
= P[X − µσ
≤ x − µσ
]
= P[X ≤ x ] =: FX (x)
1
σfZ
(x − µσ
)=
d
dxFZ (z)
=d
dxFX (x) = fX (x)
fZ (z) =1√2π
e−(z)2
2
(8)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 16 / 203
Normal Random Variables
We user the notation X ∼ N(µ, σ2) and Z ∼ N(0, 1). By ourtransformation above, it can be seen that if Z ∼ N(0, 1), thenX = µ+ σ · Z ∼ N(µ, σ2). We can see this via
Φ(z) := FZ (z) = P[Z ≤ z ]
= P[X − µσ
≤ x − µσ
]
= P[X ≤ x ] =: FX (x)
1
σfZ
(x − µσ
)=
d
dxFZ (z)
=d
dxFX (x) = fX (x)
fZ (z) =1√2π
e−(z)2
2
(8)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 16 / 203
Convolutions and Sums of Independent Random Variables
fX+Y (a) =d
da
(∫ ∞−∞
FX (a− y)fY (y)dy
)=
∫ ∞−∞
d
daFX (a− y)fY (y)dy by cty of fX (Leibniz)
=
∫ ∞−∞
fX (a− y)fY (y)dy
(9)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 17 / 203
Sums of Normal Random Variables
For a sequence cini=1 of real numbers, we have for a sequence ofcorrelated normal random variables Xini=1, where the Xi ∼ N(µiσi ) withcovariance ρij that
n∑i=1
ciXi ∼ N
(n∑
i=1
µi ,
n∑i=1
n∑j=1
cicjρijσiσj
). (10)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 18 / 203
Lognormal Random Variables
We say that Y ∼ LN(µ, σ) is Lognormal if ln(Y ) ∼ N(µ, σ2).
As sums of normal random variables remain normal, products of lognormalrandom variables remain lognormal.
Recall that the moment-generating function ofX ∼ N(µ, σ2) ∼ µ+ σN(0, 1) is
MX (t) = E[etX ] = eµt+ 12σ2t2
(11)
If Y = eµ+σZ , then, it can be seen that
E[Y n] = E[enX ] = eµn+ 12σ2n2
(12)
and
fY (y) =1
σ√
2πyexp
(− (ln(y)− µ)2
2σ2
)(13)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 19 / 203
Lognormal Random Variables
We say that Y ∼ LN(µ, σ) is Lognormal if ln(Y ) ∼ N(µ, σ2).
As sums of normal random variables remain normal, products of lognormalrandom variables remain lognormal.
Recall that the moment-generating function ofX ∼ N(µ, σ2) ∼ µ+ σN(0, 1) is
MX (t) = E[etX ] = eµt+ 12σ2t2
(11)
If Y = eµ+σZ , then, it can be seen that
E[Y n] = E[enX ] = eµn+ 12σ2n2
(12)
and
fY (y) =1
σ√
2πyexp
(− (ln(y)− µ)2
2σ2
)(13)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 19 / 203
Stock Evolution and Lognormal Random Variables
One application of lognormal distributions is their use in modeling theevolution of asset prices S . If we assume a physical measure P with α theexpected return on the stock under the physical measure, then
ln
(StS0
)= N
((α− δ − 1
2σ2)t, σ2t
)⇒ St = S0e
(α−δ− 12σ2)t+σ
√tZ
(14)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 20 / 203
Stock Evolution and Lognormal Random Variables
We can use the previous facts to show
E[St ] = S0e(α−δ)t
P[St > K ] = N
(ln S0
K + (α− δ − 0.5σ2)t
σ√t
).
(15)
Note that under the risk-neutral measure P, we exchange α with r , therisk-free rate:
E[St ] = S0e(r−δ)t
P[St > K ] = N(d2).(16)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 21 / 203
Stock Evolution and Lognormal Random Variables
We can use the previous facts to show
E[St ] = S0e(α−δ)t
P[St > K ] = N
(ln S0
K + (α− δ − 0.5σ2)t
σ√t
).
(15)
Note that under the risk-neutral measure P, we exchange α with r , therisk-free rate:
E[St ] = S0e(r−δ)t
P[St > K ] = N(d2).(16)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 21 / 203
Stock Evolution and Lognormal Random Variables
The next challenge is to construct a process St that possesses the aboveproperties as well as continuity of paths.
Click here for a neat article relating actuarial reserving to option pricing!
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 22 / 203
Stock Evolution and Lognormal Random Variables
Risk managers are also interested in Conditional Tail Expectations (CTE’s)of random variables:
CTEX (k) := E[X | X > k] =E[X1X>k
]P[X > k]
. (17)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 23 / 203
Stock Evolution and Lognormal Random Variables
In our case,
E[St | St > K ] =
E
[S0e
(α−δ− 12σ2)t+σ
√tZ1
S0e(α−δ− 1
2σ2)t+σ
√tZ>K
]
P[S0e
(α−δ− 12σ2)t+σ
√tZ > K
]
= S0e(α−δ)t
N
(ln
S0K
+(α−δ+0.5σ2)t
σ√t
)
N
(ln
S0K
+(α−δ−0.5σ2)t
σ√t
)
⇒ E[St | St > K ] = S0e(r−δ)tN(d1)
N(d2)(18)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 24 / 203
Stock Evolution and Lognormal Random Variables
In fact, we can use this CTE framework to solve for the European Calloption price in the Black-Scholes framework, where P0[A] = P[A | S0 = S ]and
V C (S , 0) := e−rT E[(ST − K )+ | S0 = S
]= e−rT E0
[ST − K | ST > K
]· P0[ST > K ]
= e−rT E0
[ST | ST > K
]· P0[ST > K ]− Ke−rT P0[ST > K ]
= e−rTSe(r−δ)T N(d1)
N(d2)· N(d2)− Ke−rTN(d2)
= Se−δTN(d1)− Ke−rTN(d2).(19)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 25 / 203
Black Scholes Pricing using Prepaid Forwards
In order to apply the previous formulae to a myriad of underlying assets,we rewrite in terms of prepaid forwards:
d1(S , t) =ln(Se−δ(T−t)
Ke−r(T−t)
)+ 1
2σ2(T − t)
σ√T − t
=ln(F St,T
FKt,T
)+ 1
2σ2(T − t)
σ√T − t
d2 = d1 − σ√T − t.
(20)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 26 / 203
Black Scholes Analysis: Option Greeks
For any option price V (S , t), define its various sensitivities as follows:
∆ =∂V
∂S
Γ =∂∆
∂S=∂2V
∂S2
ν =∂V
∂σ
Θ =∂V
∂t
ρ =∂V
∂r
Ψ =∂V
∂δ.
(21)
These are known accordingly as the Option Greeks.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 27 / 203
Black Scholes Analysis: Option Greeks
Straightforward partial differentiation leads to
∆C = e−δ(T−t)N(d1)
∆P = −e−δ(T−t)N(−d1)
ΓC = ΓP =e−δ(T−t)N ′(d1)
σS√T − t
νC = νP = Se−δ(T−t)√T − tN ′(d1)
(22)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 28 / 203
Black Scholes Analysis: Option Greeks
as well as..
ρC = (T − t)Ke−r(T−t)N(d2)
ρP = −(T − t)Ke−r(T−t)N(−d2)
ΨC = −(T − t)Se−δ(T−t)N(d1)
ΨP = (T − t)Se−δ(T−t)N(−d1).
(23)
What do the signs of the Greeks tell us?
HW: Compute Θ for puts and calls.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 29 / 203
Black Scholes Analysis: Option Greeks
as well as..
ρC = (T − t)Ke−r(T−t)N(d2)
ρP = −(T − t)Ke−r(T−t)N(−d2)
ΨC = −(T − t)Se−δ(T−t)N(d1)
ΨP = (T − t)Se−δ(T−t)N(−d1).
(23)
What do the signs of the Greeks tell us?
HW: Compute Θ for puts and calls.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 29 / 203
Portfolio Sensitivity Analysis
For a portfolio of M options, each with weighting λi and∑M
i=1 λi = 1:
Greek(Portfolio) =M∑i=1
λiGreek(i thasset). (24)
If P(i) is the price of a portfolio of income streams: P(i) =∑n
k=1 Pk(i).
⇒ D = −(1 + i) · P′(i)
P(i)= −(1 + i) ·
∑nk=1 P
′k(i)
P(i)
= −(1 + i) ·n∑
k=1
P ′k(i)
P(i)= −(1 + i) ·
n∑k=1
Pk(i)
P(i)·P ′k(i)
Pk(i)
=n∑
k=1
Dk · qk
(25)
and so the portfolio duration is the weighted average of the individualdurations. What if the interest rate i is a random variable itself?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 30 / 203
Portfolio Sensitivity Analysis
For a portfolio of M options, each with weighting λi and∑M
i=1 λi = 1:
Greek(Portfolio) =M∑i=1
λiGreek(i thasset). (24)
If P(i) is the price of a portfolio of income streams: P(i) =∑n
k=1 Pk(i).
⇒ D = −(1 + i) · P′(i)
P(i)= −(1 + i) ·
∑nk=1 P
′k(i)
P(i)
= −(1 + i) ·n∑
k=1
P ′k(i)
P(i)= −(1 + i) ·
n∑k=1
Pk(i)
P(i)·P ′k(i)
Pk(i)
=n∑
k=1
Dk · qk
(25)
and so the portfolio duration is the weighted average of the individualdurations.
What if the interest rate i is a random variable itself?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 30 / 203
Portfolio Sensitivity Analysis
For a portfolio of M options, each with weighting λi and∑M
i=1 λi = 1:
Greek(Portfolio) =M∑i=1
λiGreek(i thasset). (24)
If P(i) is the price of a portfolio of income streams: P(i) =∑n
k=1 Pk(i).
⇒ D = −(1 + i) · P′(i)
P(i)= −(1 + i) ·
∑nk=1 P
′k(i)
P(i)
= −(1 + i) ·n∑
k=1
P ′k(i)
P(i)= −(1 + i) ·
n∑k=1
Pk(i)
P(i)·P ′k(i)
Pk(i)
=n∑
k=1
Dk · qk
(25)
and so the portfolio duration is the weighted average of the individualdurations. What if the interest rate i is a random variable itself?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 30 / 203
Option Elasticity
Define
Ω(S , t) := limε→0
V (S+ε,t)−V (s,t)V (S,t)
S+ε−SS
=S
V (S , t)limε→0
V (S + ε, t)− V (s, t)
S + ε− S
=∆ · SV (S , t)
.
(26)
Consequently,
ΩC (S , t) =∆C · SV C (S , t)
=Se−δ(T−t)
Se−δ(T−t) − Ke−r(T−t)N(d2)≥ 1
ΩP(S , t) =∆P · SV P(S , t)
≤ 0.
(27)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 31 / 203
Option Elasticity
Theorem
The volatility of an option is the option elasticity times the volatility of thestock:
σoption = σstock× | Ω | . (28)
The proof comes from Finan: Consider the strategy of hedging a portfolioof shorting an option and purchasing ∆ = ∂V
∂S shares.The initial and final values of this portfolio are
Initally: V (S(t), t)−∆(S(t), t) · S(t)
Finally: V (S(T ),T )−∆(S(t), t) · S(T )
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 32 / 203
Option Elasticity
Proof.
If this portfolio is self-financing and arbitrage-free requirement, then
er(T−t)(V (S(t), t)−∆(S(t), t) ·S(t)
)= V (S(T ),T )−∆(S(t), t) ·S(T ).
(29)It follows that for κ := er(T−t),
V (S(T ),T )− V (S(t), t)
V (S(t), t)= κ− 1 +
[S(T )− S(t)
S(t)+ 1− κ
]Ω
⇒ Var
[V (S(T ),T )− V (S(t), t)
V (S(t), t)
]= Ω2Var
[S(T )− S(t)
S(t)
]⇒ σoption = σstock× | Ω | .
(30)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 33 / 203
Option Elasticity
If γ is the expected rate of return on an option with value V , α is theexpected rate of return on the underlying stock, and r is of course the riskfree rate, then the following equation holds:
γ · V (S , t) = α ·∆(S , t) · S + r ·(V (S , t)−∆(S , t) · S
). (31)
In terms of elasticity Ω, this reduces to
Risk Premium(Option) := γ − r = (α− r)Ω. (32)
Furthermore, we have the Sharpe Ratio for an asset as the ratio of riskpremium to volatility:
Sharpe(Stock) =(α− r)
σ=
(α− r)Ω
σΩ= Sharpe(Call). (33)
HW Sharpe Ratio for a put? How about elasticity for a portfolio ofoptions? Now read about Calendar Spreads, Implied Volatility, andPerpetual American Options.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 34 / 203
Option Elasticity
If γ is the expected rate of return on an option with value V , α is theexpected rate of return on the underlying stock, and r is of course the riskfree rate, then the following equation holds:
γ · V (S , t) = α ·∆(S , t) · S + r ·(V (S , t)−∆(S , t) · S
). (31)
In terms of elasticity Ω, this reduces to
Risk Premium(Option) := γ − r = (α− r)Ω. (32)
Furthermore, we have the Sharpe Ratio for an asset as the ratio of riskpremium to volatility:
Sharpe(Stock) =(α− r)
σ=
(α− r)Ω
σΩ= Sharpe(Call). (33)
HW Sharpe Ratio for a put? How about elasticity for a portfolio ofoptions? Now read about Calendar Spreads, Implied Volatility, andPerpetual American Options.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 34 / 203
Option Elasticity
If γ is the expected rate of return on an option with value V , α is theexpected rate of return on the underlying stock, and r is of course the riskfree rate, then the following equation holds:
γ · V (S , t) = α ·∆(S , t) · S + r ·(V (S , t)−∆(S , t) · S
). (31)
In terms of elasticity Ω, this reduces to
Risk Premium(Option) := γ − r = (α− r)Ω. (32)
Furthermore, we have the Sharpe Ratio for an asset as the ratio of riskpremium to volatility:
Sharpe(Stock) =(α− r)
σ=
(α− r)Ω
σΩ= Sharpe(Call). (33)
HW Sharpe Ratio for a put? How about elasticity for a portfolio ofoptions? Now read about Calendar Spreads, Implied Volatility, andPerpetual American Options.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 34 / 203
Example: Hedging
Under a standard framework, assume you write a 4− yr European Calloption a non-dividend paying stock with the following:
S0 = 10 = K
σ = 0.2
r = 0.02.
(34)
Calculate the initial number of shares of the stock for your hedgingprogram.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 35 / 203
Example: Hedging
Recall
∆C = e−δ(T−t)N(d1)
d1 =ln(
SK
)+ (r − δ + 1
2σ2)(T − t)
σ√T − t
d2 = d1 − σ√T − t.
(35)
It follows that
∆C = N(
0.4)
= 0.6554. (36)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 36 / 203
Example: Risk Analysis
Assume that an option is written on an asset S with the followinginformation:
The expected rate of return on the underlying asset is 0.10.
The expected rate of return on a riskless asset is 0.05.
The volatility on the underlying asset is 0.20.
V (S , t) = e−0.05(10−t)(S2eS
)Compute Ω(S , t) and the Sharpe Ratio for this option.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 37 / 203
Example: Risk Analysis
By definition,
Ω(S , t) =∆ · SV (S , t)
=S · ∂V (S,t)
∂S
V (S , t)
=S d
dS (S2eS)
(S2eS)=
S · (2SeS + S2eS)
S2eS
= 2 + S .
(37)
Furthermore, since Ω = 2 + S ≥ 2, we have
Sharpe =0.10− 0.05
0.20= 0.25. (38)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 38 / 203
Example: Risk Analysis
By definition,
Ω(S , t) =∆ · SV (S , t)
=S · ∂V (S,t)
∂S
V (S , t)
=S d
dS (S2eS)
(S2eS)=
S · (2SeS + S2eS)
S2eS
= 2 + S .
(37)
Furthermore, since Ω = 2 + S ≥ 2, we have
Sharpe =0.10− 0.05
0.20= 0.25. (38)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 38 / 203
Example: Black Scholes Pricing
Consider a portfolio of options on a non-dividend paying stock S thatconsists of a put and a call, both with strike K = 5 = S0. What is the Γfor this option as well as the option value at time 0 if the time toexpiration is T = 4, r = 0.02, σ = 0.2.
In this case,
V = V C + V P
Γ =∂2
∂S2
(V C + V P
)= 2ΓC .
(39)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 39 / 203
Example: Black Scholes Pricing
Consider a portfolio of options on a non-dividend paying stock S thatconsists of a put and a call, both with strike K = 5 = S0. What is the Γfor this option as well as the option value at time 0 if the time toexpiration is T = 4, r = 0.02, σ = 0.2.
In this case,
V = V C + V P
Γ =∂2
∂S2
(V C + V P
)= 2ΓC .
(39)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 39 / 203
Example: Black Scholes Pricing
Consequently, d1 = 0.4 and d2 = 0.4− 0.2√
4 = 0, and so
V (5, 0) = V C (5, 0) + V P(5, 0)
= 5(N(d1) + e−4rN(−d2)− e−4rN(d2)− N(−d1)
)= 5(N(0.4) + e−4rN(0)− e−4rN(0)− N(−0.4)
)= 1.5542
Γ(5, 0) =2N ′(0.4)
0.2 · 5 ·√
4= N ′(0.4) =
1√2π
e−0.5·(0.4)2= 0.4322.
(40)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 40 / 203
Market Making
On a periodic basis, a Market Maker, services the option buyer byrebalancing the portfolio designed to replicate the payoff written into theoption contract.Define
Vi = Option Value i periods from inception
∆i = Delta required i periods from inception
∴ Pi = ∆iSi − Vi
(41)
Rebalancing at time i requires an extra (∆i+1 −∆i ) shares.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 41 / 203
Market Making
On a periodic basis, a Market Maker, services the option buyer byrebalancing the portfolio designed to replicate the payoff written into theoption contract.Define
Vi = Option Value i periods from inception
∆i = Delta required i periods from inception
∴ Pi = ∆iSi − Vi = Cost of Strategy
(42)
Rebalancing at time i requires an extra (∆i+1 −∆i ) shares.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 42 / 203
Market Making
Define
∂Si = Si+1 − Si
∂Pi = Pi+1 − Pi
∂∆i = ∆i+1 −∆i
(43)
Then
∂Pi = Net Cash Flow = ∆i∂Si − ∂Vi − rPi
= ∆i∂Si − ∂Vi − r(
∆iSi − Vi
) (44)
Under what conditions is the Net Flow = 0?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 43 / 203
Market Making
For a continuous rate r , we can see that if ∆ := ∂V∂S , Pt = ∆tSt − Vt ,
dV := V (St + dSt , t + dt)− V (St , t)
≈ Θdt + ∆ · dSt +1
2Γ · (dSt)2
⇒ dPt = ∆tdSt − dVt − rPtdt
≈ ∆tdSt −(
Θdt + ∆ · dSt +1
2Γ · (dSt)2
)− r (∆tSt − Vt) dt
≈ −
(Θdt + r(∆St − V (St , t))dt +
1
2Γ · [dSt ]2
).
(45)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 44 / 203
Market Making
If dt is small, but not infinitessimally small, then on a periodic basis giventhe evolution of St , the periodic jump in value from St → St + dSt may beknown exactly and correspond to a non-zero jump in Market Maker profitdPt .
If dSt · dSt = σ2S2t dt, then if we sample continuously and enforce a zero
net-flow, we retain the BSM PDE for all relevant (S , t):
∂V
∂t+ r(S∂V
∂S− V
)+
1
2σ2S2∂
2V
∂S2= 0
V (S ,T ) = G (S) for final time payoff G (S).
(46)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 45 / 203
Market Making
If dt is small, but not infinitessimally small, then on a periodic basis giventhe evolution of St , the periodic jump in value from St → St + dSt may beknown exactly and correspond to a non-zero jump in Market Maker profitdPt .
If dSt · dSt = σ2S2t dt, then if we sample continuously and enforce a zero
net-flow, we retain the BSM PDE for all relevant (S , t):
∂V
∂t+ r(S∂V
∂S− V
)+
1
2σ2S2∂
2V
∂S2= 0
V (S ,T ) = G (S) for final time payoff G (S).
(46)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 45 / 203
Note: Delta-Gamma Neutrality vs Bond Immunization
In an actuarial analysis of cashflow, a company may wish to immunizeits portfolio. This refers to the relationship between a non-zero valuefor the second derivative with respect to interest rate of the(deterministic) cashflow present value and the subsequent possibilityof a negative PV.
This is similar to the case of market maker with a non-zero Gamma.In the market makers cash flow, a move of dS in the stockcorresponds to a move 1
2 Γ(dS)2 in the portfolio value.
In order to protect against large swings in the stock causing non-lineareffects in the portfolio value, the market maker may choose to offsetpositions in her present holdings to maintain Gamma Neutrality or shewish to maintain Delta Neutrality, although this is only a linear effect.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 46 / 203
Option Greeks and Analysis - Some Final Comments
It is important to note the similarities between Market Making andActuarial Reserving. In engineering the portfolio to replicate thepayoff written into the contract, the market maker requires capital.
The idea of Black Scholes Merton pricing is that the portfolio shouldbe self-financing.
One should consider how this compares with the capital required byinsurers to maintain solvency as well as the possibility of obtainingreinsurance.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 47 / 203
Exam 1 Practice Solutions
Consider an economy where :
The current exchange rate is x0 = 0.011 $yen .
A four-year dollar-denominated European put option on yen with astrike price of 0.008$ sells for 0.0005$.
The continuously compounded risk-free interest rate on dollars is 3%.
The continuously compounded risk-free interest rate on yen is 1.5%.
Compute the price of a 4−year dollar-denominated European call optionon yens with a strike price of 0.008$.
ANSWER: By put call parity, and the Black Scholes formula, with theasset S as the exchange rate, and the foreign risk-free rate rf = δ,
V C (x0, 0) = V P(x0, 0) + x0e−rf T − Ke−rT
= 0.0005 + 0.011e−0.015·4 − 0.008e−0.03·4
= 0.003764.
(47)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 48 / 203
Exam 1 Practice Solutions
Consider an economy where :
The current exchange rate is x0 = 0.011 $yen .
A four-year dollar-denominated European put option on yen with astrike price of 0.008$ sells for 0.0005$.
The continuously compounded risk-free interest rate on dollars is 3%.
The continuously compounded risk-free interest rate on yen is 1.5%.
Compute the price of a 4−year dollar-denominated European call optionon yens with a strike price of 0.008$.ANSWER: By put call parity, and the Black Scholes formula, with theasset S as the exchange rate, and the foreign risk-free rate rf = δ,
V C (x0, 0) = V P(x0, 0) + x0e−rf T − Ke−rT
= 0.0005 + 0.011e−0.015·4 − 0.008e−0.03·4
= 0.003764.
(47)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 48 / 203
Exam 1 Practice Solutions
An investor purchases a 1−year, 50− strike European Call option on anon-dividend paying stock by borrowing at the risk-free rate r . Theinvestor paid V C (S0, 0) = 10. Six months later, the investor finds out thatthe Call option has increased in value by one: V C (S0.05, 0.5) = 11.Assuming (σ, r) = (0.2, 0.02). Should she close out her position after 6months?
ANSWER: Simply put, her profit if she closes out after 6 months is
11− 10e0.02 12 = 0.8995. (48)
So, yes, she should liquidate her position.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 49 / 203
Exam 1 Practice Solutions
An investor purchases a 1−year, 50− strike European Call option on anon-dividend paying stock by borrowing at the risk-free rate r . Theinvestor paid V C (S0, 0) = 10. Six months later, the investor finds out thatthe Call option has increased in value by one: V C (S0.05, 0.5) = 11.Assuming (σ, r) = (0.2, 0.02). Should she close out her position after 6months?ANSWER: Simply put, her profit if she closes out after 6 months is
11− 10e0.02 12 = 0.8995. (48)
So, yes, she should liquidate her position.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 49 / 203
Exam 1 Practice Solutions
Consider a 1−year at the money European Call option on a non-dividendpaying stock. If you are told that ∆C = 0.65, and the economy bears a1% rate, can you estimate the volatility σ?
ANSWER: By definition,
∆C = e−δTN(d1) = N( r + 1
2σ2
σ
)= N
(0.01 + 12σ
2
σ
)= 0.65
⇒0.01 + 1
2σ2
σ= 0.385
⇒ σ ∈ 0.0269, 0.7431 .
(49)
More information is needed to choose from the two roots computed above.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 50 / 203
Exam 1 Practice Solutions
Consider a 1−year at the money European Call option on a non-dividendpaying stock. If you are told that ∆C = 0.65, and the economy bears a1% rate, can you estimate the volatility σ?ANSWER: By definition,
∆C = e−δTN(d1) = N( r + 1
2σ2
σ
)= N
(0.01 + 12σ
2
σ
)= 0.65
⇒0.01 + 1
2σ2
σ= 0.385
⇒ σ ∈ 0.0269, 0.7431 .
(49)
More information is needed to choose from the two roots computed above.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 50 / 203
Exam 1 Practice Solutions
Consider a 1−year at the money European Call option on a non-dividendpaying stock. If you are told that ∆C = 0.65, and the economy bears a1% rate, can you estimate the volatility σ?ANSWER: By definition,
∆C = e−δTN(d1) = N( r + 1
2σ2
σ
)= N
(0.01 + 12σ
2
σ
)= 0.65
⇒0.01 + 1
2σ2
σ= 0.385
⇒ σ ∈ 0.0269, 0.7431 .
(49)
More information is needed to choose from the two roots computed above.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 50 / 203
Exam 1 Practice Solutions
Consider a 1−year at the money European Call option on a non-dividendpaying stock. If you are told that ∆C = 0.65, and the economy bears a1% rate, can you estimate the volatility σ?ANSWER: By definition,
∆C = e−δTN(d1) = N( r + 1
2σ2
σ
)= N
(0.01 + 12σ
2
σ
)= 0.65
⇒0.01 + 1
2σ2
σ= 0.385
⇒ σ ∈ 0.0269, 0.7431 .
(49)
More information is needed to choose from the two roots computed above.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 50 / 203
Exam 1 Review
When reviewing the material for exam 2, please note that the exam willfocus on parts of Chapters 11, 12, and 13. In your review, consider thefollowing milestones, examples, and HW questions (all references from 2ndedition):
The definition of the Black-Scholes pricing formulae for Europeanputs and calls.
What are the Greeks? Given a specific option, could you compute theGreeks?
What is the Option Elasticity? How is it useful? How about theSharpe ratio of an option? Can you compute the Elasticity andSharpe ration of a given option?
Q : 12.3, 12.5, 12.7, 12.9, 12.20
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 51 / 203
Exam 1 Review
What is Delta Hedging? Can you replicate the example on p.417?
If the Delta and Gamma values of an option are known, can youcalculate the change in option value given a small change in theunderlying asset value?
How does this correspond the Market Maker’s profit?
Q : 13.1, 13.3, 13.4
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 52 / 203
Brownian Motion
Consider a probability space (Ω,F ,P) and a process (Wt ,Ft) that lives onit, where Ft represents all the information about Wu0≤u≤t . Assumethat our pair satisfies, for s, t ≥ 0 and Aini=1 ⊆ F
P[W0 = 0] = 1
P[Wt ∈ dx ] = 1√2πt
e−x2
2t dx
P[limt→s Wt = Ws ] = 1
P[Wt+s −Ws ∈ A | Fs ] = P[Wt ∈ A] for all A ∈ FP[∩ni=1
Wti −Wti−1 ∈ Ai
] = Πn
i=1P[Wti −Wti−1 ∈ Ai ]
What’s so hard about that? Take a Z ∼ N(0, 1) and define
Xt =√tZ (50)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 53 / 203
Brownian Motion
Consider a probability space (Ω,F ,P) and a process (Wt ,Ft) that lives onit, where Ft represents all the information about Wu0≤u≤t . Assumethat our pair satisfies, for s, t ≥ 0 and Aini=1 ⊆ F
P[W0 = 0] = 1
P[Wt ∈ dx ] = 1√2πt
e−x2
2t dx
P[limt→s Wt = Ws ] = 1
P[Wt+s −Ws ∈ A | Fs ] = P[Wt ∈ A] for all A ∈ FP[∩ni=1
Wti −Wti−1 ∈ Ai
] = Πn
i=1P[Wti −Wti−1 ∈ Ai ]
What’s so hard about that? Take a Z ∼ N(0, 1) and define
Xt =√tZ (50)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 53 / 203
Brownian Motion
Consider a probability space (Ω,F ,P) and a process (Wt ,Ft) that lives onit, where Ft represents all the information about Wu0≤u≤t . Assumethat our pair satisfies, for s, t ≥ 0 and Aini=1 ⊆ F
P[W0 = 0] = 1
P[Wt ∈ dx ] = 1√2πt
e−x2
2t dx
P[limt→s Wt = Ws ] = 1
P[Wt+s −Ws ∈ A | Fs ] = P[Wt ∈ A] for all A ∈ FP[∩ni=1
Wti −Wti−1 ∈ Ai
] = Πn
i=1P[Wti −Wti−1 ∈ Ai ]
What’s so hard about that? Take a Z ∼ N(0, 1) and define
Xt =√tZ (50)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 53 / 203
Brownian Motion
Consider a probability space (Ω,F ,P) and a process (Wt ,Ft) that lives onit, where Ft represents all the information about Wu0≤u≤t . Assumethat our pair satisfies, for s, t ≥ 0 and Aini=1 ⊆ F
P[W0 = 0] = 1
P[Wt ∈ dx ] = 1√2πt
e−x2
2t dx
P[limt→s Wt = Ws ] = 1
P[Wt+s −Ws ∈ A | Fs ] = P[Wt ∈ A] for all A ∈ FP[∩ni=1
Wti −Wti−1 ∈ Ai
] = Πn
i=1P[Wti −Wti−1 ∈ Ai ]
What’s so hard about that? Take a Z ∼ N(0, 1) and define
Xt =√tZ (50)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 53 / 203
Quadratic Variation
Clearly, the last quantity, also known as independent increments, is whatmakes Brownian motion truly special. We can use this property to defineother, related properties. The first is the notion of quadratic variation.Simply put,
Wt+∆t −Wt ∼W∆t (51)
and so, for an i.i.d. N(0, 1) sequence Zini=1
n∑j=1
(Wtj+1 −Wtj
)2 ∼n∑
j=1
(Wtj+1−tj
)2
∼n∑
j=1
(√tj+1 − tjZj
)2
(52)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 54 / 203
Quadratic Variation
Assuming we have an even partition of [0,T ], with tj+1 − tj = Tn , then
n∑j=1
(Wtj+1 −Wtj
)2 ∼ T
n
n∑j=1
(Zj)2 = T · χ
2n
n→ T . (53)
To see why, recall that, for γ = 1√1−2t
,
Mχ2n(t) =
(E[et·Z
2])n
E[et·Z2] =
1√2π
∫ ∞−∞
etx2e−
x2
2 dx
=1√2π
∫ ∞−∞
e− x2
2γ2 dx
=1√
1− 2t
⇒ Mχ2nn
(t) =1√
(1− 2 tn )n→ et = E[et·1].
(54)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 55 / 203
Quadratic Variation
Assuming we have an even partition of [0,T ], with tj+1 − tj = Tn , then
n∑j=1
(Wtj+1 −Wtj
)2 ∼ T
n
n∑j=1
(Zj)2 = T · χ
2n
n→ T . (53)
To see why, recall that, for γ = 1√1−2t
,
Mχ2n(t) =
(E[et·Z
2])n
E[et·Z2] =
1√2π
∫ ∞−∞
etx2e−
x2
2 dx =1√2π
∫ ∞−∞
e− x2
2γ2 dx
=1√
1− 2t
⇒ Mχ2nn
(t) =1√
(1− 2 tn )n→ et = E[et·1].
(54)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 55 / 203
Quadratic Variation
Assuming we have an even partition of [0,T ], with tj+1 − tj = Tn , then
n∑j=1
(Wtj+1 −Wtj
)2 ∼ T
n
n∑j=1
(Zj)2 = T · χ
2n
n→ T . (53)
To see why, recall that, for γ = 1√1−2t
,
Mχ2n(t) =
(E[et·Z
2])n
E[et·Z2] =
1√2π
∫ ∞−∞
etx2e−
x2
2 dx =1√2π
∫ ∞−∞
e− x2
2γ2 dx
=1√
1− 2t
⇒ Mχ2nn
(t) =1√
(1− 2 tn )n→ et = E[et·1].
(54)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 55 / 203
Quadratic Variation
Assuming we have an even partition of [0,T ], with tj+1 − tj = Tn , then
n∑j=1
(Wtj+1 −Wtj
)2 ∼ T
n
n∑j=1
(Zj)2 = T · χ
2n
n→ T . (53)
To see why, recall that, for γ = 1√1−2t
,
Mχ2n(t) =
(E[et·Z
2])n
E[et·Z2] =
1√2π
∫ ∞−∞
etx2e−
x2
2 dx =1√2π
∫ ∞−∞
e− x2
2γ2 dx
=1√
1− 2t
⇒ Mχ2nn
(t) =1√
(1− 2 tn )n→ et = E[et·1].
(54)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 55 / 203
k th order Variation, where k > 2
Assuming we have an even partition of [0,T ], with tj+1 − tj = Tn , then
n∑j=1
(Wtj+1 −Wtj
)k ∼ (T
n
) k2
n∑j=1
(Zj)k
E
n∑j=1
(Wtj+1 −Wtj
)k ∝ Tk2 · n
nk2
→ 0
(55)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 56 / 203
Quadratic Variation
We can use this to define integration against Brownian Motion. This isdefined as∫ T
0f (Wt , t)dWt = lim
n→∞
n∑i=1
f (Wjh, jh) (Wjh+h −Wjh) (56)
In general, we can define the stochastic differential equation
dXt = µ(Xt , t)dt + σ(Xt , t)dWt (57)
and the accompanying Ito Equation for a new, stochastic calculus basedon the relationship
dWt · dWt = dt. (58)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 57 / 203
Black-Scholes-Merton Analysis
Independent of the model for evolution of underlying asset, pricingmust be arbitrage-free
Achieve this via replication of derivative up to end of term of contract
Mathematical tools we require are now more involved than linearalgebra; namely Ito Calculus and Partial Differential Equations
To enable this analysis, assume the following for the model
∃ (Ω,F ,P) for which Wt is a standard Brownian motion.
dSt = St (µdt + σdWt), a Geometric Brownian Motion models theasset evolution
For a function f (x , t) ∈ C 2,1(R2 × R
)and Yt ≡ f (Wt , t), Ito
Calculus gives us:
dYt =(ft(Wt , t) + 1
2 fxx(Wt , t))dt + fx(Wt , t)dWt
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 58 / 203
Black-Scholes-Merton Analysis
Independent of the model for evolution of underlying asset, pricingmust be arbitrage-free
Achieve this via replication of derivative up to end of term of contract
Mathematical tools we require are now more involved than linearalgebra; namely Ito Calculus and Partial Differential Equations
To enable this analysis, assume the following for the model
∃ (Ω,F ,P) for which Wt is a standard Brownian motion.
dSt = St (µdt + σdWt), a Geometric Brownian Motion models theasset evolution
For a function f (x , t) ∈ C 2,1(R2 × R
)and Yt ≡ f (Wt , t), Ito
Calculus gives us:
dYt =(ft(Wt , t) + 1
2 fxx(Wt , t))dt + fx(Wt , t)dWt
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 58 / 203
Black-Scholes-Merton Analysis
Independent of the model for evolution of underlying asset, pricingmust be arbitrage-free
Achieve this via replication of derivative up to end of term of contract
Mathematical tools we require are now more involved than linearalgebra; namely Ito Calculus and Partial Differential Equations
To enable this analysis, assume the following for the model
∃ (Ω,F ,P) for which Wt is a standard Brownian motion.
dSt = St (µdt + σdWt), a Geometric Brownian Motion models theasset evolution
For a function f (x , t) ∈ C 2,1(R2 × R
)and Yt ≡ f (Wt , t), Ito
Calculus gives us:
dYt =(ft(Wt , t) + 1
2 fxx(Wt , t))dt + fx(Wt , t)dWt
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 58 / 203
Black-Scholes-Merton Analysis
Independent of the model for evolution of underlying asset, pricingmust be arbitrage-free
Achieve this via replication of derivative up to end of term of contract
Mathematical tools we require are now more involved than linearalgebra; namely Ito Calculus and Partial Differential Equations
To enable this analysis, assume the following for the model
∃ (Ω,F ,P) for which Wt is a standard Brownian motion.
dSt = St (µdt + σdWt), a Geometric Brownian Motion models theasset evolution
For a function f (x , t) ∈ C 2,1(R2 × R
)and Yt ≡ f (Wt , t), Ito
Calculus gives us:
dYt =(ft(Wt , t) + 1
2 fxx(Wt , t))dt + fx(Wt , t)dWt
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 58 / 203
Black-Scholes-Merton Analysis
Independent of the model for evolution of underlying asset, pricingmust be arbitrage-free
Achieve this via replication of derivative up to end of term of contract
Mathematical tools we require are now more involved than linearalgebra; namely Ito Calculus and Partial Differential Equations
To enable this analysis, assume the following for the model
∃ (Ω,F ,P) for which Wt is a standard Brownian motion.
dSt = St (µdt + σdWt), a Geometric Brownian Motion models theasset evolution
For a function f (x , t) ∈ C 2,1(R2 × R
)and Yt ≡ f (Wt , t), Ito
Calculus gives us:
dYt =(ft(Wt , t) + 1
2 fxx(Wt , t))dt + fx(Wt , t)dWt
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 58 / 203
Black-Scholes-Merton Analysis
Independent of the model for evolution of underlying asset, pricingmust be arbitrage-free
Achieve this via replication of derivative up to end of term of contract
Mathematical tools we require are now more involved than linearalgebra; namely Ito Calculus and Partial Differential Equations
To enable this analysis, assume the following for the model
∃ (Ω,F ,P) for which Wt is a standard Brownian motion.
dSt = St (µdt + σdWt), a Geometric Brownian Motion models theasset evolution
For a function f (x , t) ∈ C 2,1(R2 × R
)and Yt ≡ f (Wt , t), Ito
Calculus gives us:
dYt =(ft(Wt , t) + 1
2 fxx(Wt , t))dt + fx(Wt , t)dWt
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 58 / 203
Black-Scholes-Merton Analysis
Independent of the model for evolution of underlying asset, pricingmust be arbitrage-free
Achieve this via replication of derivative up to end of term of contract
Mathematical tools we require are now more involved than linearalgebra; namely Ito Calculus and Partial Differential Equations
To enable this analysis, assume the following for the model
∃ (Ω,F ,P) for which Wt is a standard Brownian motion.
dSt = St (µdt + σdWt), a Geometric Brownian Motion models theasset evolution
For a function f (x , t) ∈ C 2,1(R2 × R
)and Yt ≡ f (Wt , t), Ito
Calculus gives us:
dYt =(ft(Wt , t) + 1
2 fxx(Wt , t))dt + fx(Wt , t)dWt
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 58 / 203
Black-Scholes-Merton Analysis
Independent of the model for evolution of underlying asset, pricingmust be arbitrage-free
Achieve this via replication of derivative up to end of term of contract
Mathematical tools we require are now more involved than linearalgebra; namely Ito Calculus and Partial Differential Equations
To enable this analysis, assume the following for the model
∃ (Ω,F ,P) for which Wt is a standard Brownian motion.
dSt = St (µdt + σdWt), a Geometric Brownian Motion models theasset evolution
For a function f (x , t) ∈ C 2,1(R2 × R
)and Yt ≡ f (Wt , t), Ito
Calculus gives us:
dYt =(ft(Wt , t) + 1
2 fxx(Wt , t))dt + fx(Wt , t)dWt
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 58 / 203
Black-Scholes-Merton Analysis
Independent of the model for evolution of underlying asset, pricingmust be arbitrage-free
Achieve this via replication of derivative up to end of term of contract
Mathematical tools we require are now more involved than linearalgebra; namely Ito Calculus and Partial Differential Equations
To enable this analysis, assume the following for the model
∃ (Ω,F ,P) for which Wt is a standard Brownian motion.
dSt = St (µdt + σdWt), a Geometric Brownian Motion models theasset evolution
For a function f (x , t) ∈ C 2,1(R2 × R
)and Yt ≡ f (Wt , t), Ito
Calculus gives us:
dYt =(ft(Wt , t) + 1
2 fxx(Wt , t))dt + fx(Wt , t)dWt
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 58 / 203
Black-Scholes-Merton Analysis
In general, for Zt = g(Xt , t), with dXt = µ(Xt , t)dt + σ(Xt , t)dWt , wehave
dZt = gt(Xt , t)dt +1
2gxx(Xt , t)dXtdXt + gxdXt
=
(gt(Xt , t) + µ(Xt , t)gx(Xt , t) +
1
2σ(Xt , t)2gxx(Xt , t)
)dt
+ σ(Xt , t)gx(Xt , t)dWt
(59)
Using the above assumptions, we arrive at the conclusion that we mustconstruct a portfolio Xt that matches the value Vt of the derivative wewish to price at all times t ≤ T , where T is the term of the contract.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 59 / 203
Black-Scholes-Merton Analysis
In general, for Zt = g(Xt , t), with dXt = µ(Xt , t)dt + σ(Xt , t)dWt , wehave
dZt = gt(Xt , t)dt +1
2gxx(Xt , t)dXtdXt + gxdXt
=
(gt(Xt , t) + µ(Xt , t)gx(Xt , t) +
1
2σ(Xt , t)2gxx(Xt , t)
)dt
+ σ(Xt , t)gx(Xt , t)dWt
(59)
Using the above assumptions, we arrive at the conclusion that we mustconstruct a portfolio Xt that matches the value Vt of the derivative wewish to price at all times t ≤ T , where T is the term of the contract.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 59 / 203
Black-Scholes-Merton Analysis
We match
dXt = dVt
XT = VT = G (ST )(60)
where G (S) is the payoff of the contract at time T if the value of theunderlying asset ST = S .
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 60 / 203
Black-Scholes-Merton Analysis
We match
dXt = dVt
XT = VT = G (ST )(60)
where G (S) is the payoff of the contract at time T if the value of theunderlying asset ST = S .
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 60 / 203
Black-Scholes-Merton Analysis
We match
dXt = dVt
XT = VT = G (ST )(60)
where G (S) is the payoff of the contract at time T if the value of theunderlying asset ST = S .
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 60 / 203
Black-Scholes-Merton Analysis
To achieve this, we recognize that Xt = Vt ⇒ dXt = dVt and by theassumption V = V (St , t), Ito provides:
dXt = r (Xt −∆tSt) dt + ∆tdSt = r (Vt −∆tSt) dt + ∆tdSt
dVt =
(∂V
∂t(St , t) +
1
2σ2S2
t
∂2V
∂S2(St , t)
)dt +
∂V
∂S(St , t)dSt .
(61)
This is due to the fact that dSt = µStdt + σStdWt ⇒ (dSt)2 = σ2S2
t dt.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 61 / 203
Black-Scholes-Merton Analysis
To achieve this, we recognize that Xt = Vt ⇒ dXt = dVt and by theassumption V = V (St , t), Ito provides:
dXt = r (Xt −∆tSt) dt + ∆tdSt = r (Vt −∆tSt) dt + ∆tdSt
dVt =
(∂V
∂t(St , t) +
1
2σ2S2
t
∂2V
∂S2(St , t)
)dt +
∂V
∂S(St , t)dSt .
(61)
This is due to the fact that dSt = µStdt + σStdWt ⇒ (dSt)2 = σ2S2
t dt.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 61 / 203
Black Scholes PDE
Matching terms implies
∆t =∂V
∂S(S , t)
∂V
∂t(S , t) +
1
2σ2S2∂
2V
∂S2(S , t) = r
(V (S , t)− S
∂V
∂S(S , t)
)V (S ,T ) = G (S)
(62)
This is the famous B-S-M PDE formulation for European option pricing,with payoff G (S). The question now - how do we solve it?!
Note:For dSt = µStdt + σStdWt , we have as the solution
St = Sue(µ− 1
2σ2)(t−u)+σ(Wt−Wu)
∼ Sue(µ− 1
2σ2)(t−u)+σWt−u
∼ Sue(µ− 1
2σ2)(t−u)+σ
√t−uZ .
(63)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 62 / 203
Black Scholes PDE
Matching terms implies
∆t =∂V
∂S(S , t)
∂V
∂t(S , t) +
1
2σ2S2∂
2V
∂S2(S , t) = r
(V (S , t)− S
∂V
∂S(S , t)
)V (S ,T ) = G (S)
(62)
This is the famous B-S-M PDE formulation for European option pricing,with payoff G (S). The question now - how do we solve it?!
Note:For dSt = µStdt + σStdWt , we have as the solution
St = Sue(µ− 1
2σ2)(t−u)+σ(Wt−Wu)
∼ Sue(µ− 1
2σ2)(t−u)+σWt−u
∼ Sue(µ− 1
2σ2)(t−u)+σ
√t−uZ .
(63)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 62 / 203
Our First Arbitrage Opportunity
Consider two perfectly correlated assets X ,Y whose evolution is modeledby
dXt = µ1Xtdt + σ1XtdWt
dYt = µ2Ytdt + σ2YtdWt
σ1 6= σ2
(64)
Assume wlog that 1σ1> 1
σ2. Design a portfolio that consists of
going long 1σ1Xt
units of X
short 1σ2Yt
units of Y ,
borrowing 1σ1− 1
σ2at continuous rate r .
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 63 / 203
Our First Arbitrage Opportunity
Borrowing the amount at time t means that the total net output at time tis nothing, but the evolution of our portfolio is
1
σ1XtdXt −
1
σ2YtdYt −
(1
σ1− 1
σ2
)rdt =
(µ1 − r
σ1− µ2 − r
σ2
)dt (65)
Unless the respective Sharpe Ratios are equivalent, ie unless
µ1 − r
σ1=µ2 − r
σ2(66)
one could make a deterministic profit with zero upfront capital.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 64 / 203
Spring 2009 Q10
Consider two perfectly correlated, non-dividend paying assets X ,Y whoseevolution is modeled by
dXt = 0.08Xtdt + 0.2XtdWt
dYt = 0.0925Ytdt − 0.25YtdWt(67)
An investor wishes to synthesize a risk-free asset by allocating 1000between X and Y . How much should she initially invest in the X?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 65 / 203
Spring 2009 Q10
If she goes long 5Xt
units of X for every 4Yt
units of Y she goes long, thenher portfolio has a deterministic growth rate. Symbolically,
dPt = ∆Xt dXt + ∆Y
t dYt
=5
Xt(0.08Xtdt + 0.2XtdWt) +
4
Yt(0.0925Ytdt − 0.25YtdWt)
= 0.77dt
(68)
So, for every 9 units she spends initially, she has 5 invested in X . It followsthat if she spends 1000 initially, 5
9 · 1000 is invested in X to obtain a riskfree asset.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 66 / 203
Spring 2009 Q10
If she goes long 5Xt
units of X for every 4Yt
units of Y she goes long, thenher portfolio has a deterministic growth rate. Symbolically,
dPt = ∆Xt dXt + ∆Y
t dYt
=5
Xt(0.08Xtdt + 0.2XtdWt) +
4
Yt(0.0925Ytdt − 0.25YtdWt)
= 0.77dt
(68)
So, for every 9 units she spends initially, she has 5 invested in X . It followsthat if she spends 1000 initially, 5
9 · 1000 is invested in X to obtain a riskfree asset.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 66 / 203
Spring 2009 Q10
If she goes long 5Xt
units of X for every 4Yt
units of Y she goes long, thenher portfolio has a deterministic growth rate. Symbolically,
dPt = ∆Xt dXt + ∆Y
t dYt
=5
Xt(0.08Xtdt + 0.2XtdWt) +
4
Yt(0.0925Ytdt − 0.25YtdWt)
= 0.77dt
(68)
So, for every 9 units she spends initially, she has 5 invested in X . It followsthat if she spends 1000 initially, 5
9 · 1000 is invested in X to obtain a riskfree asset.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 66 / 203
Spring 2009 Q18
Consider two perfectly correlated, non-dividend paying assets X ,Y whoseevolution is modeled by
Xt = X0e0.1t+0.2Wt
Yt = Y0e0.125t+0.3Wt
(69)
with a constant risk-free rate r for all t ≥ 0.
If you are constrained to a non-arbitrage market, solve for r .
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 67 / 203
Spring 2009 Q18
Recall that for constant µ, σ,
dSt = µStdt + σStdWt
⇒ St = S0e(µ− 1
2σ2)t+σWt .
(70)
For stock X , σ1 = 0.2 and µ1 = 0.1 + 12 · 0.2
2 = 0.12.For stock Y , σ2 = 0.3 and µ2 = 0.125 + 1
2 · 0.32 = 0.17.
Since the two assets are perfectly correlated, their Sharpe ratios are equal:
µ1 − r
σ1=µ2 − r
σ2
⇒ r = 0.02.(71)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 68 / 203
Spring 2009 Q18
Recall that for constant µ, σ,
dSt = µStdt + σStdWt
⇒ St = S0e(µ− 1
2σ2)t+σWt .
(70)
For stock X , σ1 = 0.2 and µ1 = 0.1 + 12 · 0.2
2 = 0.12.For stock Y , σ2 = 0.3 and µ2 = 0.125 + 1
2 · 0.32 = 0.17.
Since the two assets are perfectly correlated, their Sharpe ratios are equal:
µ1 − r
σ1=µ2 − r
σ2
⇒ r = 0.02.(71)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 68 / 203
Practice Aug 2010 Q13
Let Wt be a Brownian motion and define
Xt = 2Wt − 2
Yt = W 2t − t
Zt = t2Wt − 2
∫ t
0sWsds
(72)
Which of these has zero drift?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 69 / 203
Practice Aug 2010 Q24
Consider the SDE
dXt = λ (α− Xt) dt + σdWt (73)
where λ, α, σ > 0 and X0 are known. Solve for Xt .
Answer:Using integrating factor eλt , we compute
d(eλtXt
)= αλeλtdt + σeλtdWt
⇒ eλtXt − X0 = α(eλt − 1) + σ
∫ t
0eλsdWs
⇒ Xt = X0e−λt + α(1− e−λt) + σe−λt
∫ t
0eλsdWs .
(74)
Bonus: Can we compute limt→∞ Xt in some meaningful way?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 70 / 203
Practice Aug 2010 Q24
Consider the SDE
dXt = λ (α− Xt) dt + σdWt (73)
where λ, α, σ > 0 and X0 are known. Solve for Xt .Answer:Using integrating factor eλt , we compute
d(eλtXt
)= αλeλtdt + σeλtdWt
⇒ eλtXt − X0 = α(eλt − 1) + σ
∫ t
0eλsdWs
⇒ Xt = X0e−λt + α(1− e−λt) + σe−λt
∫ t
0eλsdWs .
(74)
Bonus: Can we compute limt→∞ Xt in some meaningful way?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 70 / 203
Practice Aug 2010 Q24
Consider the SDE
dXt = λ (α− Xt) dt + σdWt (73)
where λ, α, σ > 0 and X0 are known. Solve for Xt .Answer:Using integrating factor eλt , we compute
d(eλtXt
)= αλeλtdt + σeλtdWt
⇒ eλtXt − X0 = α(eλt − 1) + σ
∫ t
0eλsdWs
⇒ Xt = X0e−λt + α(1− e−λt) + σe−λt
∫ t
0eλsdWs .
(74)
Bonus: Can we compute limt→∞ Xt in some meaningful way?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 70 / 203
Practice: Brownian Bridge
Consider the SDE
dXt = − Xt
1− tdt + dWt
X0 = 0.(75)
Compute Xt .
Answer:Using integrating factor 1
1−t , we compute
d( Xt
1− t
)=
1
1− tdWt
⇒ Xt
1− t− X0 =
∫ t
0
1
1− sdWs
⇒ Xt = (1− t)
∫ t
0
1
1− sdWs .
(76)
Bonus: Can we compute limt→1 Xt in some meaningful way?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 71 / 203
Practice: Brownian Bridge
Consider the SDE
dXt = − Xt
1− tdt + dWt
X0 = 0.(75)
Compute Xt .Answer:Using integrating factor 1
1−t , we compute
d( Xt
1− t
)=
1
1− tdWt
⇒ Xt
1− t− X0 =
∫ t
0
1
1− sdWs
⇒ Xt = (1− t)
∫ t
0
1
1− sdWs .
(76)
Bonus: Can we compute limt→1 Xt in some meaningful way?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 71 / 203
Practice: Brownian Bridge
Consider the SDE
dXt = − Xt
1− tdt + dWt
X0 = 0.(75)
Compute Xt .Answer:Using integrating factor 1
1−t , we compute
d( Xt
1− t
)=
1
1− tdWt
⇒ Xt
1− t− X0 =
∫ t
0
1
1− sdWs
⇒ Xt = (1− t)
∫ t
0
1
1− sdWs .
(76)
Bonus: Can we compute limt→1 Xt in some meaningful way?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 71 / 203
Practice Aug 2010 Q32
Consider the non-dividend stock evolution
dSt = αStdt + σStdWt (77)
in a market that also provides a continuously compounded rate r onriskless asset. At time 0, Jane invests in a fund that employs aproportional investment strategy where at every point in time, 100kpercent is invested in S and 100(1− k) percent in the riskless asset. Solvefor Jane’s wealth Xt .
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 72 / 203
Practice Aug 2010 Q32
Assuming there are no transaction costs,
dXt
Xt= k
dStSt
+ (1− k)rdt
= k · (αdt + σdWt) + (1− k)rdt
= (kα + (1− k)r) dt + kσdWt
Xt = X0e(kα+(1−k)r− 1
2(kσ)2)t+kσWt
(78)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 73 / 203
Practice Aug 2010 Q32
Assuming there are no transaction costs,
dXt
Xt= k
dStSt
+ (1− k)rdt
= k · (αdt + σdWt) + (1− k)rdt
= (kα + (1− k)r) dt + kσdWt
Xt = X0e(kα+(1−k)r− 1
2(kσ)2)t+kσWt
(78)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 73 / 203
Practice Aug 2010 Q37
Consider the stock evolution
dSt = 0.03Stdt + 0.2StdWt (79)
If G = (S1 · S2 · S3)13 , calculate Var (ln (G ))
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 74 / 203
Bonds, Forwards,Barriers
In any market, we have two basic assets that we use to price derivativeswith - a stock and a bond. Let’s see how these carry through to the BSMframework
If G (S) = 1 then at time T , one unit of bond (or currency) ispresented. What is the price at time 0?
If G (S) = Sa then at time T , one unit of stock (or asset) ispresented. What is the price at time 0?
If S0 = 50, and we agree to pay 1 unit of currency if the stock everhits 100, otherwise we pay nothing, what is the value of this financialinstrument at time 0?
By the BSM technology, we must find a solution to a PDE that satisfiesall conditions imposed by the definition of the contract we wish to price.For the first two cases above, there are only final time conditions.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 75 / 203
Bonds, Forwards,Barriers
In any market, we have two basic assets that we use to price derivativeswith - a stock and a bond. Let’s see how these carry through to the BSMframework
If G (S) = 1 then at time T , one unit of bond (or currency) ispresented. What is the price at time 0?
If G (S) = Sa then at time T , one unit of stock (or asset) ispresented. What is the price at time 0?
If S0 = 50, and we agree to pay 1 unit of currency if the stock everhits 100, otherwise we pay nothing, what is the value of this financialinstrument at time 0?
By the BSM technology, we must find a solution to a PDE that satisfiesall conditions imposed by the definition of the contract we wish to price.For the first two cases above, there are only final time conditions.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 75 / 203
Bonds, Forwards,Barriers
In any market, we have two basic assets that we use to price derivativeswith - a stock and a bond. Let’s see how these carry through to the BSMframework
If G (S) = 1 then at time T , one unit of bond (or currency) ispresented. What is the price at time 0?
If G (S) = Sa then at time T , one unit of stock (or asset) ispresented. What is the price at time 0?
If S0 = 50, and we agree to pay 1 unit of currency if the stock everhits 100, otherwise we pay nothing, what is the value of this financialinstrument at time 0?
By the BSM technology, we must find a solution to a PDE that satisfiesall conditions imposed by the definition of the contract we wish to price.For the first two cases above, there are only final time conditions.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 75 / 203
Bonds, Forwards,Barriers
In any market, we have two basic assets that we use to price derivativeswith - a stock and a bond. Let’s see how these carry through to the BSMframework
If G (S) = 1 then at time T , one unit of bond (or currency) ispresented. What is the price at time 0?
If G (S) = Sa then at time T , one unit of stock (or asset) ispresented. What is the price at time 0?
If S0 = 50, and we agree to pay 1 unit of currency if the stock everhits 100, otherwise we pay nothing, what is the value of this financialinstrument at time 0?
By the BSM technology, we must find a solution to a PDE that satisfiesall conditions imposed by the definition of the contract we wish to price.For the first two cases above, there are only final time conditions.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 75 / 203
Bonds, Forwards,Barriers
In any market, we have two basic assets that we use to price derivativeswith - a stock and a bond. Let’s see how these carry through to the BSMframework
If G (S) = 1 then at time T , one unit of bond (or currency) ispresented. What is the price at time 0?
If G (S) = Sa then at time T , one unit of stock (or asset) ispresented. What is the price at time 0?
If S0 = 50, and we agree to pay 1 unit of currency if the stock everhits 100, otherwise we pay nothing, what is the value of this financialinstrument at time 0?
By the BSM technology, we must find a solution to a PDE that satisfiesall conditions imposed by the definition of the contract we wish to price.For the first two cases above, there are only final time conditions.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 75 / 203
Bonds, Forwards,Barriers
For the case of presenting a unit of currency at time T , it is a fairassumption that the value V (S , t) should be independent of S , as thepayoff G (S) = 1 is.
Our resulting PDE is in fact an ODE for V (t):
V ′(t) = rV (t)
V (T ) = 1(80)
The solution of course is V (t) = e−r(T−t), which is the discountedvalue of one unit of currence T − t units of time in the future.
The hedging strategy is ∆ = ∂V∂S = 0, which means no stocks are held
to replicate this, and all the money taken at time t is invested inbuying e−r(T−t) units of bond.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 76 / 203
Bonds, Forwards,Barriers
For the case of presenting a unit of currency at time T , it is a fairassumption that the value V (S , t) should be independent of S , as thepayoff G (S) = 1 is.
Our resulting PDE is in fact an ODE for V (t):
V ′(t) = rV (t)
V (T ) = 1(80)
The solution of course is V (t) = e−r(T−t), which is the discountedvalue of one unit of currence T − t units of time in the future.
The hedging strategy is ∆ = ∂V∂S = 0, which means no stocks are held
to replicate this, and all the money taken at time t is invested inbuying e−r(T−t) units of bond.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 76 / 203
Bonds, Forwards,Barriers
For the case of presenting a unit of currency at time T , it is a fairassumption that the value V (S , t) should be independent of S , as thepayoff G (S) = 1 is.
Our resulting PDE is in fact an ODE for V (t):
V ′(t) = rV (t)
V (T ) = 1(80)
The solution of course is V (t) = e−r(T−t), which is the discountedvalue of one unit of currence T − t units of time in the future.
The hedging strategy is ∆ = ∂V∂S = 0, which means no stocks are held
to replicate this, and all the money taken at time t is invested inbuying e−r(T−t) units of bond.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 76 / 203
Bonds, Forwards,Barriers
Consider the case V (S ,T ) = G (S) = Sa. Then if we assume the Ansatz
V (S , t) = SaV (t), (81)
substitution into the BSM PDE returns
V ′(t) +1
2σ2a(a− 1)V (t) = r(1− a)V (t)
V (T ) = 1.(82)
The solutiion of this ODE with a final time condition is
V (t) = exp([
r(1− a) +1
2σ2a(1− a)
](t − T )
)= e
−[r(1−a)+ 1
2σ2a(1−a)
](T−t)
(83)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 77 / 203
Bonds, Forwards,Barriers
Another way to see this is our solution is to again recall that under the riskneutral measure P, from u to time t, we have the evolution
St = Su exp([
r − 1
2σ2](t − u) + σWt−u
)
⇒ G (ST ) = SaT = Sa
t exp(a[r − 1
2σ2](T − t) + aσWT−t
)⇒ V (S , t) = e−r(T−t)E[G (ST ) | St = S ]
= Sae−r(T−t)ea
[r− 1
2σ2
](T−t)
· E[eaσWT−t ]
= Sae−r(T−t)ea
[r− 1
2σ2
](T−t)
· e12a2σ2(T−t)
= Sae
[−r+ar− 1
2σ2a+ 1
2a2σ2
](T−t)
= Sae
[−r(1−a)− 1
2σ2a(1−a)
](T−t)
(84)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 78 / 203
Bonds, Forwards,Barriers
Another way to see this is our solution is to again recall that under the riskneutral measure P, from u to time t, we have the evolution
St = Su exp([
r − 1
2σ2](t − u) + σWt−u
)⇒ G (ST ) = Sa
T = Sat exp
(a[r − 1
2σ2](T − t) + aσWT−t
)
⇒ V (S , t) = e−r(T−t)E[G (ST ) | St = S ]
= Sae−r(T−t)ea
[r− 1
2σ2
](T−t)
· E[eaσWT−t ]
= Sae−r(T−t)ea
[r− 1
2σ2
](T−t)
· e12a2σ2(T−t)
= Sae
[−r+ar− 1
2σ2a+ 1
2a2σ2
](T−t)
= Sae
[−r(1−a)− 1
2σ2a(1−a)
](T−t)
(84)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 78 / 203
Bonds, Forwards,Barriers
Another way to see this is our solution is to again recall that under the riskneutral measure P, from u to time t, we have the evolution
St = Su exp([
r − 1
2σ2](t − u) + σWt−u
)⇒ G (ST ) = Sa
T = Sat exp
(a[r − 1
2σ2](T − t) + aσWT−t
)⇒ V (S , t) = e−r(T−t)E[G (ST ) | St = S ]
= Sae−r(T−t)ea
[r− 1
2σ2
](T−t)
· E[eaσWT−t ]
= Sae−r(T−t)ea
[r− 1
2σ2
](T−t)
· e12a2σ2(T−t)
= Sae
[−r+ar− 1
2σ2a+ 1
2a2σ2
](T−t)
= Sae
[−r(1−a)− 1
2σ2a(1−a)
](T−t)
(84)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 78 / 203
Bonds, Forwards,Barriers
In the last case, time is not explicitly given as a condition, meaning thatthe option is perpetual - only events to do with the the underlying stockvalue S will factor into the price of the contract. Hence, we have again asimpler ODE for V (S), but now with boundary conditions:
1
2σ2S2V ′′(S) = r
(V (S)− SV ′(S)
)V (0) = 0
V (100) = 1
(85)
General solution of this ODE is
V (S) = c1S + c2S− 2rσ2 . (86)
Solution of this ODE with our boundary conditions leads to the closedform V (S) = 1
100S , and so the specific price asked for is V (50) = 0.50
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 79 / 203
Bonds, Forwards,Barriers
In the last case, time is not explicitly given as a condition, meaning thatthe option is perpetual - only events to do with the the underlying stockvalue S will factor into the price of the contract. Hence, we have again asimpler ODE for V (S), but now with boundary conditions:
1
2σ2S2V ′′(S) = r
(V (S)− SV ′(S)
)V (0) = 0
V (100) = 1
(85)
General solution of this ODE is
V (S) = c1S + c2S− 2rσ2 . (86)
Solution of this ODE with our boundary conditions leads to the closedform V (S) = 1
100S , and so the specific price asked for is V (50) = 0.50
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 79 / 203
Bonds, Forwards,Barriers
In the last case, time is not explicitly given as a condition, meaning thatthe option is perpetual - only events to do with the the underlying stockvalue S will factor into the price of the contract. Hence, we have again asimpler ODE for V (S), but now with boundary conditions:
1
2σ2S2V ′′(S) = r
(V (S)− SV ′(S)
)V (0) = 0
V (100) = 1
(85)
General solution of this ODE is
V (S) = c1S + c2S− 2rσ2 . (86)
Solution of this ODE with our boundary conditions leads to the closedform V (S) = 1
100S , and so the specific price asked for is V (50) = 0.50
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 79 / 203
Some Notes..
In our pricing examples, and in the BSM PDE, it turns out that theexpected return µ of the stock doesn’t factor directly into our calculations.Nevertheless, we note that
E[dStSt
]= E
[µdt + σdWt
]= µdt
E[dStSt
]= E
[rdt + σdWt
]= rdt
(87)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 80 / 203
Multivariable BSM PDE
Following in the same fashion as the above arguments, we can show for
dSi (t)
Si (t)= µidt +
m∑j=1
σijdW(j)t
αij =m∑
k=1
σikσjk
qi = the dividend rate of the i th asset
(88)
then
∂V
∂t+
1
2
n∑i ,j=1
αijSiSj∂2V
∂Si∂Sj+
n∑i=1
(r − qi )Si∂V
∂Si= rV
V (S1,S2, ..,Sn,T ) = G (S1, S2, ..,Sn)
(89)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 81 / 203
Forward Delivered in a Foreign Currency
Consider an asset with value S1 in yen and an exchange rate of S2 dollarsper yen. If the final payoff a contract is [S1(T )]a, then the risk-neutralprice of a contract that delivers this value in dollars is V (S1,S2, t), withpde
∂V
∂t+
1
2
2∑i ,j=1
αijSiSj∂2V
∂Si∂Sj+
2∑i=1
(r − qi )Si∂V
∂Si= rV
V (S1,S2,T ) = Sa1S2.
(90)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 82 / 203
Forward Delivered in a Foreign Currency
Once again, we assume the Ansatz V (S1, S2, t) = Sa1S2V (t) to obtain the
ODE with final condition:
V ′(t) + γV (t) = 0
V (T ) = 1,(91)
where γ := 12 (α11a(a− 1) + a(α12 + α21)) + a(r − q1) + (r − q2)− r .
It follows that
V (S1,S2, t) = Sa1S2e
γ(t−T ). (92)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 83 / 203
Forward Delivered in a Foreign Currency
Once again, we assume the Ansatz V (S1, S2, t) = Sa1S2V (t) to obtain the
ODE with final condition:
V ′(t) + γV (t) = 0
V (T ) = 1,(91)
where γ := 12 (α11a(a− 1) + a(α12 + α21)) + a(r − q1) + (r − q2)− r .
It follows that
V (S1,S2, t) = Sa1S2e
γ(t−T ). (92)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 83 / 203
Options on Futures
It may be that it would be to an investor’s advantage to hold an optioncontract on an asset associated with the underlying (ie commodity, stock,etc...). One that is more liquid and involves lower transaction costsanyway than the one involving the actual asset.
One way to enable this liquidity is to write the contract on the associatedFuture or Forward contract. Recall that
F = Ser(T−t) (93)
and so
∂
∂S=
∂
∂F
∂F
∂S= er(T−t) ∂
∂F∂2
∂S2=
∂
∂S
(er(T−t) ∂
∂F
)= e2r(T−t) ∂
2
∂F 2
∂
∂t=
∂
∂t+
∂
∂F
∂F
∂t=
∂
∂t− rF
∂
∂F
(94)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 84 / 203
Options on Futures
It may be that it would be to an investor’s advantage to hold an optioncontract on an asset associated with the underlying (ie commodity, stock,etc...). One that is more liquid and involves lower transaction costsanyway than the one involving the actual asset.
One way to enable this liquidity is to write the contract on the associatedFuture or Forward contract. Recall that
F = Ser(T−t) (93)
and so
∂
∂S=
∂
∂F
∂F
∂S= er(T−t) ∂
∂F∂2
∂S2=
∂
∂S
(er(T−t) ∂
∂F
)= e2r(T−t) ∂
2
∂F 2
∂
∂t=
∂
∂t+
∂
∂F
∂F
∂t=
∂
∂t− rF
∂
∂F
(94)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 84 / 203
Options on Futures
It may be that it would be to an investor’s advantage to hold an optioncontract on an asset associated with the underlying (ie commodity, stock,etc...). One that is more liquid and involves lower transaction costsanyway than the one involving the actual asset.
One way to enable this liquidity is to write the contract on the associatedFuture or Forward contract. Recall that
F = Ser(T−t) (93)
and so
∂
∂S=
∂
∂F
∂F
∂S= er(T−t) ∂
∂F∂2
∂S2=
∂
∂S
(er(T−t) ∂
∂F
)= e2r(T−t) ∂
2
∂F 2
∂
∂t=
∂
∂t+
∂
∂F
∂F
∂t=
∂
∂t− rF
∂
∂F
(94)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 84 / 203
Options on Futures
It may be that it would be to an investor’s advantage to hold an optioncontract on an asset associated with the underlying (ie commodity, stock,etc...). One that is more liquid and involves lower transaction costsanyway than the one involving the actual asset.
One way to enable this liquidity is to write the contract on the associatedFuture or Forward contract. Recall that
F = Ser(T−t) (93)
and so
∂
∂S=
∂
∂F
∂F
∂S= er(T−t) ∂
∂F∂2
∂S2=
∂
∂S
(er(T−t) ∂
∂F
)= e2r(T−t) ∂
2
∂F 2
∂
∂t=
∂
∂t+
∂
∂F
∂F
∂t=
∂
∂t− rF
∂
∂F
(94)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 84 / 203
Options on Futures
It may be that it would be to an investor’s advantage to hold an optioncontract on an asset associated with the underlying (ie commodity, stock,etc...). One that is more liquid and involves lower transaction costsanyway than the one involving the actual asset.
One way to enable this liquidity is to write the contract on the associatedFuture or Forward contract. Recall that
F = Ser(T−t) (93)
and so
∂
∂S=
∂
∂F
∂F
∂S= er(T−t) ∂
∂F∂2
∂S2=
∂
∂S
(er(T−t) ∂
∂F
)= e2r(T−t) ∂
2
∂F 2
∂
∂t=
∂
∂t+
∂
∂F
∂F
∂t=
∂
∂t− rF
∂
∂F
(94)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 84 / 203
Options on Futures
The resulting B-S-M equation is
Vt +1
2σ2F 2VFF − rV = 0
V (F ,T ) = G (F )(95)
Question: How do we solve for V (F , t) ? Certainly, it is much closer toHeat equation than the original form of the B-S-M PDE.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 85 / 203
Options on Futures
The resulting B-S-M equation is
Vt +1
2σ2F 2VFF − rV = 0
V (F ,T ) = G (F )(95)
Question: How do we solve for V (F , t) ? Certainly, it is much closer toHeat equation than the original form of the B-S-M PDE.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 85 / 203
Options on Futures
The resulting B-S-M equation is
Vt +1
2σ2F 2VFF − rV = 0
V (F ,T ) = G (F )(95)
Question: How do we solve for V (F , t) ? Certainly, it is much closer toHeat equation than the original form of the B-S-M PDE.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 85 / 203
Path Dependent Options: Types and Technology
So far, we have mainly described the methodology for fairly pricing whatare known as Vanilla options,namely Puts, Calls and Forwards. These areonly time-dependent, in that the only event that the price is contingent onis the payoff at the end of the term, T .
A more exotic option would provide extra security by enabling thepurchaser to receive a payment contingent on an event related to theunderlying asset the option is written on. This event is most often the firsttime a stock price hits a certain theshold. However, the event could berelated to the average or running maximum of the stock price over a giventime window.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 86 / 203
Barrier Options
One of the initial examples of the B-S-M equation we were able to pricewas a perpetual barrier option. In that case, if a stock hit an upper barrierin its price, one dollar was paid to the option purchaser, otherwise theoption continues on.
In most cases, a time limit is put on such an option. What happens if wewish to put an extra barrier condition on an otherwise plain vanillaEuropean Call option?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 87 / 203
Recall: European Barrier Options
Consider now the case where an option is activated or deactivated,depending on whether or not the asset has reached a certain price level(barrier) before the option expires.
Hence, we have another path dependent option with the need for anothertracking variable. In this case, we can add the maximum or minimum ofthe asset over its path. They are knocked in or knocked out of action ifthe barrier is reached before expiry, depending on the contract design.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 88 / 203
Recall: European Barrier Options
Consider now the case where an option is activated or deactivated,depending on whether or not the asset has reached a certain price level(barrier) before the option expires.
Hence, we have another path dependent option with the need for anothertracking variable. In this case, we can add the maximum or minimum ofthe asset over its path. They are knocked in or knocked out of action ifthe barrier is reached before expiry, depending on the contract design.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 88 / 203
Barrier Options: PDE Valuation
Consider a Down-and-Out Call Option:
∂V
∂t(S , t) +
1
2σ2S2∂
2V
∂S2(S , t) + r
(S∂V
∂S(S , t)− V (S , t)
)= 0
V (S ,T ) = (S − K )+
V (S , t) = 0 if S ≤ X
(96)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 89 / 203
Barrier Options
How do we solve such a PDE? Certainly, without this condition at X , wehave a solution of our plain Vanilla Call C (S , t). And, if X → 0, whatvalue should we should recover? Perhaps a modification of the solutionC (S , t) is a good place to start. With this in mind, we assume
V (S , t) = C (S , t)− B(S , t) (97)
where B(X , t) = C (X , t) and B(S , t) satisfies the same B-S-M PDE asthe European Call. This is reasonable assumption, as the B-S-M equationis linear.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 90 / 203
Barrier Options
Our next assumption is that we can rescale our function C (S , t) and stillhave it solve the B-S-M equation for Call options. Namely, try
B(S , t) =
(S
X
)αC (SβX γ , t)
∂B
∂S=
∂
∂S
((S
X
)αC (SβX γ , t)
)∂B
∂t=
∂
∂t
((S
X
)αC (SβX γ , t)
) (98)
Substitution into B-S-M equation leads us to α = 1− r2σ2 , β = −1, γ = 2
Hence,
V (S , t) = C (S , t)−(S
X
)1− 2rσ2
C
(X 2
S, t
)(99)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 91 / 203
General Barrier Options
Imagine the case now where we have an option that pays 1 if S ≥ X atany time t ≤ T , and pays 0 if S < X at expiration time T . The option isknocked out if St /∈ [L,U] for any t < T . In this case, our PDE is
∂V
∂t(S , t) +
1
2S2σ2∂
2V
∂S2(S , t) + r
(S∂V
∂S(S , t)− V (S , t)
)= 0
V (S ,T ) = G (S) if 0 ≤ S ≤ X
V (S , t) = 0 if S ≥ U
V (S , t) = 0 if S ≤ L
(100)
where G (S) is a general payoff function.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 92 / 203
Barrier Options
Once again, we can attempt a solution V (S , t) = A(t)B(S) and so weobtain:
A′(t)B(S) +1
2σ2S2A(t)B ′′(S) + r
(SA(t)B ′(S)− A(t)B(S)
)= 0
B(U) = 0
B(L) = 0
A(T )B(S) = G (S)
(101)
Modifications need to be made here if G (S) does not solve thecorresponding ODE for B(S).
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 93 / 203
Barrier Options
First, assume
1
2σ2S2B ′′(S) + rSB ′(S) = λB(S)
A′(t)− rA(t) = −λB(U) = 0
B(L) = 0
A(T ) = 1
(102)
We now have two separate ODE’s, one of which is easily solved:
A(t) = e(r−λ)(t−T ) (103)
but the question remains: what is λ?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 94 / 203
Barrier Options:Eigenvalue Approach
Again, we try the Ansatz B(S) = Sn. Solution of the ODE requires
1
2σ2n(n − 1) + rn = λ (104)
but neither boundary condition is satisfied. We need a more sophisticatedapproach, that of eigenfunction expansions. The solution of the boundaryvalue problem
1
2σ2S2B ′′(S) + rSB ′(S) = λB(S)
B(U) = 0
B(L) = 0
(105)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 95 / 203
Barrier Options:Eigenvalue Approach
can be shown to exist only for specific λ, namely a countable set of them:
ν =1
σ
(r − σ2
2
)Bn(S) =
σ√ln U
L
S−νσ sin
(πn ln X
L
ln UL
)
λn = r +ν2
2+σ2π2n2
2 ln2 UL
(106)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 96 / 203
Barrier Options:Eigenvalue Approach
Since each Bn(S) and e(r−λn)(t−T ) solve their respecitve ODE’s withboundary conditions, any linear combination of the products∑∞
n=0 cne(r−λn)(t−T )Bn(S) should solve the PDE. The only condition left
is to solve the final time condition
V (S ,T ) = G (S) (107)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 97 / 203
Barrier Options:Eigenvalue Approach
We can do this by taking the right sequence cn∞n=0 such that
G (S) = V (S ,T ) =∞∑n=0
cnBn(S) (108)
For an explanation of how to do this, proof of the form of the solutionsBn(S), and even more on Barrier options see Pricing Options on ScalarDiffusions: An Eigenfunction Expansion Approach by Dmitry Davydovand Vadim Linetsky
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 98 / 203
European Barrier Options
There are three types of barrier options:
Knock-out options: Options are deactivated when the asset pricereaches the barrier before option maturity. These are non-zero tobegin with.
down-and-out when the asset price has to decrease to reach thebarrier.up-and-out when the price has to increase to reach the barrier.
Knock-in options: Options are activated when asset price reaches thebarrier before option maturity. These are only activated if barrier isreached before option expiry.
down-and-in when the asset price has to decrease to reach the barrier.up-and-in when the price has to increase to reach the barrier.
Rebate options: Guaranteed level payment made to holder if the assetprice reaches the barrier before expiry. Payment can be made at thetime the barrier is reached or deferred to expiry time. Up Rebate ifbarrier is above the current spot price, Down Rebate if barrier isbelow the current spot price.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 99 / 203
Parity for Barrier Options
It is straightforward to see that
V knock-in0 + V knock-out
0 = V0 (109)
To see this, we have the explicit formulation usingτB := inf t <∞ : St = B:
V knock-out at B0 = e−rT E0
[G (ST )1τB>T
]V knock-in at B
0 = e−rT E0
[G (ST )1τB≤T
]V0 = e−rT E0 [G (ST )]
= e−rT E0
[G (ST )
(1τB>T + 1τB≤T
)]= V knock-out at B + V knock-in at B .
(110)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 100 / 203
Barrier Options: Example
Consider the following continuous economy with a non-dividend payingstock S , where S0 = K = 100, σ = 0.1 and r = 0.05. For a one-yearoption, we have the standard calculations
d1 =ln S0
K + (r − δ + 0.5σ2)T
σ√T
= 0.55
d2 = d1 − σ√T = 0.55− 0.1 = 0.45
N(d1) = P[Z ≤ d1] = 0.7088
N(d2) = P[Z ≤ d2] = 0.6736
⇒ V C0 = 100 · 0.7088− 100 · e−0.05 · 0.6736 = 6.805
⇒ V C ,knock-in at 990 = V C
0 = 6.805
⇒ V C ,knock-out at 990 = V C
0 − V C ,knock-in at 990 = 0.
(111)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 101 / 203
Barrier Options: Example
Consider the following continuous economy with a non-dividend payingstock S , where S0 = K = 100, σ = 0.1 and r = 0.05. For a one-yearoption, we have the standard calculations
d1 =ln S0
K + (r − δ + 0.5σ2)T
σ√T
= 0.55
d2 = d1 − σ√T = 0.55− 0.1 = 0.45
N(d1) = P[Z ≤ d1] = 0.7088
N(d2) = P[Z ≤ d2] = 0.6736
⇒ V C0 = 100 · 0.7088− 100 · e−0.05 · 0.6736 = 6.805
⇒ V C ,knock-in at 990 = V C
0 = 6.805
⇒ V C ,knock-out at 990 = V C
0 − V C ,knock-in at 990 = 0.
(111)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 101 / 203
Barrier Options: Example
Consider the following continuous economy with a non-dividend payingstock S , where S0 = K = 100, σ = 0.1 and r = 0.05. For a one-yearoption, we have the standard calculations
d1 =ln S0
K + (r − δ + 0.5σ2)T
σ√T
= 0.55
d2 = d1 − σ√T = 0.55− 0.1 = 0.45
N(d1) = P[Z ≤ d1] = 0.7088
N(d2) = P[Z ≤ d2] = 0.6736
⇒ V C0 = 100 · 0.7088− 100 · e−0.05 · 0.6736 = 6.805
⇒ V C ,knock-in at 990 = V C
0 = 6.805
⇒ V C ,knock-out at 990 = V C
0 − V C ,knock-in at 990 = 0.
(111)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 101 / 203
Barrier Options: Example
Consider the following continuous economy with a non-dividend payingstock S , where S0 = K = 100, σ = 0.1 and r = 0.05. For a one-yearoption, we have the standard calculations
d1 =ln S0
K + (r − δ + 0.5σ2)T
σ√T
= 0.55
d2 = d1 − σ√T = 0.55− 0.1 = 0.45
N(d1) = P[Z ≤ d1] = 0.7088
N(d2) = P[Z ≤ d2] = 0.6736
⇒ V C0 = 100 · 0.7088− 100 · e−0.05 · 0.6736 = 6.805
⇒ V C ,knock-in at 990 = V C
0 = 6.805
⇒ V C ,knock-out at 990 = V C
0 − V C ,knock-in at 990 = 0.
(111)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 101 / 203
Gap Options
In keeping with the extra freedom allowed in option design of the previoussections, consider a payoff
G (S) = (S − K1)1S>K2 = (S − K2)1S>K2 + (K2 − K1)1S>K2 (112)
where K1 is the familiar strike price and K2 is the trigger price thatactivates the payoff. It can be shown that
V C = Se−δTN(d1)− K1e−rTN(d2)
V P = K1e−rTN(−d2)− Se−δTN(−d1)
d1 =ln S
K2+ (r − δ + 0.5σ2)T
σ√T
d2 = d1 − σ√T
N(d1) = P[Z ≤ d1]
N(d2) = P[Z ≤ d2].
(113)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 102 / 203
Exchange Options
Consider now the case where one can exchange S for K , where K is nownon-constant and risky. Then
G (ST ,KT ) = max ST − KT , 0 (114)
and so
V C (S ,K , t) = e−r(T−t)E[G (ST ,KT ) | St = S ,Kt = K ]
= Se−δS (T−t)N(d1)− Ke−δK (T−t)N(d2)
d1 =ln Se−δS (T−t)
Ke−δK (T−t) + 0.5σ2(T − t)
σ√T − t
d2 = d1 − σ√T
σ =√σ2S + σ2
K − 2ρσSσK .
(115)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 103 / 203
Asian Options:Options on Averages
Another useful example of a path dependent option that is commonlyemployed is the Asian option. As an example, an airline may only buy afinite supply of jet fuel every year, but use it continuously during the year.To guard itself against large price movements, the company may choose touse financial instruments with a fixed purchase date to avoid having tocontinuously hedge against the commodity price movements. We assumethat the asset evolves according to
dSt = µ(St , t)dt + σ(St , t)dWt (116)
on some fixed probability space (Ω,F ,P)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 104 / 203
Asian Options:Options on Averages
Another useful example of a path dependent option that is commonlyemployed is the Asian option. As an example, an airline may only buy afinite supply of jet fuel every year, but use it continuously during the year.To guard itself against large price movements, the company may choose touse financial instruments with a fixed purchase date to avoid having tocontinuously hedge against the commodity price movements. We assumethat the asset evolves according to
dSt = µ(St , t)dt + σ(St , t)dWt (116)
on some fixed probability space (Ω,F ,P)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 104 / 203
Asian Options:Options on Averages
Another useful example of a path dependent option that is commonlyemployed is the Asian option. As an example, an airline may only buy afinite supply of jet fuel every year, but use it continuously during the year.To guard itself against large price movements, the company may choose touse financial instruments with a fixed purchase date to avoid having tocontinuously hedge against the commodity price movements. We assumethat the asset evolves according to
dSt = µ(St , t)dt + σ(St , t)dWt (116)
on some fixed probability space (Ω,F ,P)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 104 / 203
Asian Options:Options on Averages
A common instrument used is a forward contract. However, this fixes theprice once the contract is signed. Another choice companies have is topurchase an Asian option that pays, at time T ,
V (S ,T ) = max
(S − 1
T
∫ T
0Sudu, 0
)(117)
To be able to price this option at time t, two pieces of information arerequired:
St − the stock price
It :=1
t
∫ t
0Sudu − the running average stock price
(118)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 105 / 203
Asian Options:Options on Averages
A common instrument used is a forward contract. However, this fixes theprice once the contract is signed. Another choice companies have is topurchase an Asian option that pays, at time T ,
V (S ,T ) = max
(S − 1
T
∫ T
0Sudu, 0
)(117)
To be able to price this option at time t, two pieces of information arerequired:
St − the stock price
It :=1
t
∫ t
0Sudu − the running average stock price
(118)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 105 / 203
Asian Options:Options on Averages
A common instrument used is a forward contract. However, this fixes theprice once the contract is signed. Another choice companies have is topurchase an Asian option that pays, at time T ,
V (S ,T ) = max
(S − 1
T
∫ T
0Sudu, 0
)(117)
To be able to price this option at time t, two pieces of information arerequired:
St − the stock price
It :=1
t
∫ t
0Sudu − the running average stock price
(118)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 105 / 203
Asian Options:Options on Averages
A common instrument used is a forward contract. However, this fixes theprice once the contract is signed. Another choice companies have is topurchase an Asian option that pays, at time T ,
V (S ,T ) = max
(S − 1
T
∫ T
0Sudu, 0
)(117)
To be able to price this option at time t, two pieces of information arerequired:
St − the stock price
It :=1
t
∫ t
0Sudu − the running average stock price
(118)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 105 / 203
Asian Options:Options on Averages
Furthermore, we also require the differential form dIt :
∆I (t) ≡ I (t + ∆t)− I (t)
∆t
=1
∆t
∫ t+∆t
tSudu
→ Stdt ≡ dIt
as ∆t → 0
(119)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 106 / 203
Asian Options:Options on Averages
Furthermore, we also require the differential form dIt :
∆I (t) ≡ I (t + ∆t)− I (t)
∆t
=1
∆t
∫ t+∆t
tSudu
→ Stdt ≡ dIt
as ∆t → 0
(119)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 106 / 203
Asian Options:Options on Averages
With the triple of information (St , It , t), we can again construct a B-S-Mportfolio replication argument. On the one hand, we have theself-financing portfolio we construct, Xt using the underlying asset St andthe riskless bank account:
dXt = r (Xt −∆St) dt + ∆tdSt (120)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 107 / 203
Asian Options:Options on Averages
With the triple of information (St , It , t), we can again construct a B-S-Mportfolio replication argument. On the one hand, we have theself-financing portfolio we construct, Xt using the underlying asset St andthe riskless bank account:
dXt = r (Xt −∆St) dt + ∆tdSt (120)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 107 / 203
Asian Options:Options on Averages
And on the other hand, we have the price of the contract Vt ≡ V (St , It , t)that satisfies Ito’s equation
dVt =
(∂V
∂t+
1
2σ(St , t)2∂
2V
∂S2
)dt +
∂V
∂SdSt +
∂V
∂IdIt
=
(∂V
∂t+
1
2σ(St , t)2∂
2V
∂S2+ St
∂V
∂I
)dt +
∂V
∂SdSt
(121)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 108 / 203
Asian Options:Options on Averages
And on the other hand, we have the price of the contract Vt ≡ V (St , It , t)that satisfies Ito’s equation
dVt =
(∂V
∂t+
1
2σ(St , t)2∂
2V
∂S2
)dt +
∂V
∂SdSt +
∂V
∂IdIt
=
(∂V
∂t+
1
2σ(St , t)2∂
2V
∂S2+ St
∂V
∂I
)dt +
∂V
∂SdSt
(121)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 108 / 203
Asian Options:Options on Averages
Matching terms once again, with the constraint that Vt = Xt , we obtain
∂V
∂t+
1
2σ(S , t)2∂
2V
∂S2+ S
∂V
∂I+ r
(S∂V
∂S− V
)= 0
V (S , I ,T ) = max
(S − I
T, 0
)∆ =
∂V
∂S
(122)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 109 / 203
Asian Options:Options on Averages
Matching terms once again, with the constraint that Vt = Xt , we obtain
∂V
∂t+
1
2σ(S , t)2∂
2V
∂S2+ S
∂V
∂I+ r
(S∂V
∂S− V
)= 0
V (S , I ,T ) = max
(S − I
T, 0
)∆ =
∂V
∂S
(122)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 109 / 203
Asian Options:Similarity Solutions
This Asian option acts as a Call option contract, in that the strike K isreplaced by an average value over the path of the asset evolution. But, thedimension of the PDE has increased by 1. This makes the solution moredifficult to obtain. Still, we have the tools used in the solution of the heatequation and Barrier options, namely Similarity solutions.
We assume that
V (S , I , t) = SH(R, t)
R ≡ I
S
V (S , I ,T ) = max
(S − I
T, 0
)= S ·max
(1− I
ST, 0
)= S ·max
(1− R
T, 0
)= S · H(R,T )
(123)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 110 / 203
Asian Options:Similarity Solutions
This Asian option acts as a Call option contract, in that the strike K isreplaced by an average value over the path of the asset evolution. But, thedimension of the PDE has increased by 1. This makes the solution moredifficult to obtain. Still, we have the tools used in the solution of the heatequation and Barrier options, namely Similarity solutions.We assume that
V (S , I , t) = SH(R, t)
R ≡ I
S
V (S , I ,T ) = max
(S − I
T, 0
)= S ·max
(1− I
ST, 0
)= S ·max
(1− R
T, 0
)= S · H(R,T )
(123)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 110 / 203
Asian Options:Similarity Solutions
Substitution into the B-S-M equation derived above returns
∂H
∂t+
1
2σ2R2∂
2H
∂R2+ (1− rR)
∂H
∂R= 0
H(R,T ) = max
(1− R
T, 0
) (124)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 111 / 203
Asian Options:Similarity Solutions
Substitution into the B-S-M equation derived above returns
∂H
∂t+
1
2σ2R2∂
2H
∂R2+ (1− rR)
∂H
∂R= 0
H(R,T ) = max
(1− R
T, 0
) (124)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 111 / 203
Asian Options:Similarity Solutions
For boundary conditions similar to those imposed on the European call, wesee that if R →∞, then it must be that S → 0, and so the value of theoption should be worthless:
H(∞, t) = 0 (125)
If R → 0, then assuming that all derivatives of H with respect to R arefinite, it is reasonable to return to the BSM equation to obtain
∂H
∂t(0, t) +
∂H
∂R(0, t) = 0 (126)
Use tools similiar to those used in explicit solution of original EuropeanCall option C (S , t) to solve this PDE.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 112 / 203
Asian Options:Similarity Solutions
For boundary conditions similar to those imposed on the European call, wesee that if R →∞, then it must be that S → 0, and so the value of theoption should be worthless:
H(∞, t) = 0 (125)
If R → 0, then assuming that all derivatives of H with respect to R arefinite, it is reasonable to return to the BSM equation to obtain
∂H
∂t(0, t) +
∂H
∂R(0, t) = 0 (126)
Use tools similiar to those used in explicit solution of original EuropeanCall option C (S , t) to solve this PDE.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 112 / 203
Asian Options:Similarity Solutions
For boundary conditions similar to those imposed on the European call, wesee that if R →∞, then it must be that S → 0, and so the value of theoption should be worthless:
H(∞, t) = 0 (125)
If R → 0, then assuming that all derivatives of H with respect to R arefinite, it is reasonable to return to the BSM equation to obtain
∂H
∂t(0, t) +
∂H
∂R(0, t) = 0 (126)
Use tools similiar to those used in explicit solution of original EuropeanCall option C (S , t) to solve this PDE.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 112 / 203
Asian Options:Similarity Solutions
For boundary conditions similar to those imposed on the European call, wesee that if R →∞, then it must be that S → 0, and so the value of theoption should be worthless:
H(∞, t) = 0 (125)
If R → 0, then assuming that all derivatives of H with respect to R arefinite, it is reasonable to return to the BSM equation to obtain
∂H
∂t(0, t) +
∂H
∂R(0, t) = 0 (126)
Use tools similiar to those used in explicit solution of original EuropeanCall option C (S , t) to solve this PDE.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 112 / 203
Lookback Options
Another exotic, path-dependent option that is studied is the Lookbackoption. This contract agrees to pay the investor the value
max (JT − ST , 0) (127)
where Jt ≡ max0≤u≤tSu
Such an option is valuable, as it helps to reduce the regret of an investorwho wishes to sell his asset S at its highest value over a window of time[0,T ].
Of course, only the actual asset S is traded in, not its maximum Jt at timet. It follows that our replicating portfolio Xt is constructed using S andthe bank account only.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 113 / 203
Lookback Options
Another exotic, path-dependent option that is studied is the Lookbackoption. This contract agrees to pay the investor the value
max (JT − ST , 0) (127)
where Jt ≡ max0≤u≤tSu
Such an option is valuable, as it helps to reduce the regret of an investorwho wishes to sell his asset S at its highest value over a window of time[0,T ].
Of course, only the actual asset S is traded in, not its maximum Jt at timet. It follows that our replicating portfolio Xt is constructed using S andthe bank account only.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 113 / 203
Lookback Options
Another exotic, path-dependent option that is studied is the Lookbackoption. This contract agrees to pay the investor the value
max (JT − ST , 0) (127)
where Jt ≡ max0≤u≤tSu
Such an option is valuable, as it helps to reduce the regret of an investorwho wishes to sell his asset S at its highest value over a window of time[0,T ].
Of course, only the actual asset S is traded in, not its maximum Jt at timet. It follows that our replicating portfolio Xt is constructed using S andthe bank account only.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 113 / 203
Lookback Options
Another exotic, path-dependent option that is studied is the Lookbackoption. This contract agrees to pay the investor the value
max (JT − ST , 0) (127)
where Jt ≡ max0≤u≤tSu
Such an option is valuable, as it helps to reduce the regret of an investorwho wishes to sell his asset S at its highest value over a window of time[0,T ].
Of course, only the actual asset S is traded in, not its maximum Jt at timet. It follows that our replicating portfolio Xt is constructed using S andthe bank account only.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 113 / 203
Lookback Options
Another exotic, path-dependent option that is studied is the Lookbackoption. This contract agrees to pay the investor the value
max (JT − ST , 0) (127)
where Jt ≡ max0≤u≤tSu
Such an option is valuable, as it helps to reduce the regret of an investorwho wishes to sell his asset S at its highest value over a window of time[0,T ].
Of course, only the actual asset S is traded in, not its maximum Jt at timet. It follows that our replicating portfolio Xt is constructed using S andthe bank account only.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 113 / 203
Lookback Options
Once again, our B-S-M replication argument returns
dXt = r (Xt −∆tSt) dt + ∆tdSt = r (Vt −∆tSt) dt + ∆tdSt
dVt =
(∂V
∂t+
1
2σ(St , t)2∂
2V
∂S2
)dt
+∂V
∂SdSt +
∂V
∂JdJt
=
(∂V
∂t+
1
2σ(St , t)2∂
2V
∂S2
)dt +
∂V
∂SdSt +
∂V
∂J1St=JtdSt
(128)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 114 / 203
Lookback Options
Once again, our B-S-M replication argument returns
dXt = r (Xt −∆tSt) dt + ∆tdSt = r (Vt −∆tSt) dt + ∆tdSt
dVt =
(∂V
∂t+
1
2σ(St , t)2∂
2V
∂S2
)dt
+∂V
∂SdSt +
∂V
∂JdJt
=
(∂V
∂t+
1
2σ(St , t)2∂
2V
∂S2
)dt +
∂V
∂SdSt +
∂V
∂J1St=JtdSt
(128)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 114 / 203
Lookback Options
Once again, our B-S-M replication argument returns
dXt = r (Xt −∆tSt) dt + ∆tdSt = r (Vt −∆tSt) dt + ∆tdSt
dVt =
(∂V
∂t+
1
2σ(St , t)2∂
2V
∂S2
)dt
+∂V
∂SdSt +
∂V
∂JdJt
=
(∂V
∂t+
1
2σ(St , t)2∂
2V
∂S2
)dt +
∂V
∂SdSt +
∂V
∂J1St=JtdSt
(128)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 114 / 203
Lookback Options
Again, we can choose how we wish to hedge. It can be shown that∂V∂J (S ,S) = 0. This implies that ∆t = ∂V
∂S and we return our favouritePDE with an extra condition:
∂V
∂t+
1
2σ(S , t)2∂
2V
∂S2+ r
(S∂V
∂S− V
)= 0
V (S , J,T ) = max (J − S , 0)
V (0, J, t) = Je−r(T−t)
∂V
∂J(S ,S) = 0
∆ =∂V
∂S
(129)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 115 / 203
Lookback Options: Similarity Solutions
To enable a reduction in dimension, we again return to the notion of asimilarity solution. In the case of the Lookback option, we have
V (S , J, t) = JW (ζ, t)
ζ ≡ S
JV (S , I ,T ) = max (J − S , 0)
= J ·max
(1− S
J, 0
)= J ·max (1− ζ, 0)
= JW (ζ,T )
(130)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 116 / 203
Lookback Options: Similarity Solutions
The resulting PDE is now
Wt +1
2σ2ζ2Wζζ + rζWζ − rW = 0
W (0, t) = e−r(T−t)
W (ζ,T ) = max (1− ζ, 0)
Wζ(1, t) = W (1, t)
(131)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 117 / 203
Estimating Historical Volatility
Given an interval [0,T ] one can look at the returns over a partition of theinterval into n equal subintervals:
rt,h := ln(St+h
St
)(132)
and so for h = Tn ,
σ2 :=1
h· 1
n − 1·
n∑k=1
(r(k−1)h,h)2 (133)
Note that our estimator σ2 is stochastic.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 118 / 203
Estimating Historical Volatility
Given an interval [0,T ] and an assumed model
dStSt
= αdt + σdWt (134)
one can look at the returns over a partition of the interval into n equalsubintervals for h = T
n :
rt,h = ln(St+h
St
)∼ N
((α− 1
2σ2)· h, σ2h)
r =1
n
n∑k=1
r(k−1)h,h
σ2 =1
n − 1
n∑k=1
(r(k−1)h,h − r)2 =
∑nk=1(r(k−1)h,h)2
n − 1− nr2
n − 1.
(135)
It follows that n · r is our estimator for α and n · σ2 for σ2 over the period[0,T ]. See Q.51 in the SOA MFE practice for a nice worked example.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 119 / 203
What happens if σ 6= constant?
Returning to the continuous time setting, there are two models thataccount for a non-cosntant volatility.
Constant Elasticity of Variance:
dStSt
= αdt + σSβ−2
2 dWt
Heston Model (Stochastic Volatility):
dStSt
= αdt +√σtdWt
dσt = k · (σ(t)− σt)dt + σHdWt
dWt · dWt = ρdt.
(136)
Both lead to more mathematically complex solution techniques. However,empirical data suggests that they calibrate significantly better withobserved option data.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 120 / 203
What happens if σ 6= constant?
Returning to the continuous time setting, there are two models thataccount for a non-cosntant volatility.
Constant Elasticity of Variance:
dStSt
= αdt + σSβ−2
2 dWt
Heston Model (Stochastic Volatility):
dStSt
= αdt +√σtdWt
dσt = k · (σ(t)− σt)dt + σHdWt
dWt · dWt = ρdt.
(136)
Both lead to more mathematically complex solution techniques. However,empirical data suggests that they calibrate significantly better withobserved option data.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 120 / 203
What happens if σ 6= constant?
Returning to the continuous time setting, there are two models thataccount for a non-cosntant volatility.
Constant Elasticity of Variance:
dStSt
= αdt + σSβ−2
2 dWt
Heston Model (Stochastic Volatility):
dStSt
= αdt +√σtdWt
dσt = k · (σ(t)− σt)dt + σHdWt
dWt · dWt = ρdt.
(136)
Both lead to more mathematically complex solution techniques. However,empirical data suggests that they calibrate significantly better withobserved option data.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 120 / 203
Interest Rate Derivatives
Consider a zero-coupon, non-defaultable bond that pays V (T ,T ) = 1 atexpiry date T . The risk-free rate is a function of time r(t).
The corresponding boundary value problem is (holding T fixed)
dV
V= r(t)dt
V (T ,T ) = 1.(137)
The solution is
V (t,T ) = e−∫ Tt r(s)ds . (138)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 121 / 203
Interest Rate Derivatives
Consider a zero-coupon, non-defaultable bond that pays V (T ,T ) = 1 atexpiry date T . The risk-free rate is a function of time r(t).
The corresponding boundary value problem is (holding T fixed)
dV
V= r(t)dt
V (T ,T ) = 1.(137)
The solution is
V (t,T ) = e−∫ Tt r(s)ds . (138)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 121 / 203
Interest Rate Derivatives
Consider again a zero-coupon, non-defaultable bond that paysV (T ,T ) = 1 at expiry date T .
Now, the risk-free rate is a random variable indexed by time rt(ω) on a
probability space(
Ω,F ,P)
It follows that, if we have a related risk neutral measure P, then for astochastic process rt(ω), we expect the (martingale) solution
V (t,T ) = E[e−
∫ Tt rs(ω)ds
]. (139)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 122 / 203
Interest Rate Derivatives
Consider again a zero-coupon, non-defaultable bond that paysV (T ,T ) = 1 at expiry date T .
Now, the risk-free rate is a random variable indexed by time rt(ω) on a
probability space(
Ω,F ,P)
It follows that, if we have a related risk neutral measure P, then for astochastic process rt(ω), we expect the (martingale) solution
V (t,T ) = E[e−
∫ Tt rs(ω)ds
]. (139)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 122 / 203
Interest Rate Derivatives
Consider again a zero-coupon, non-defaultable bond that paysV (T ,T ) = 1 at expiry date T .
Now, the risk-free rate is a random variable indexed by time rt(ω) on a
probability space(
Ω,F ,P)
It follows that, if we have a related risk neutral measure P, then for astochastic process rt(ω), we expect the (martingale) solution
V (t,T ) = E[e−
∫ Tt rs(ω)ds
]. (139)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 122 / 203
Short Rate Model
For the evolution of interest rates r on(
Ω,F ,P)
, we can assume the
general model
drt = µ(rt , t)dt + σ(rt , t)dWt
dVt
Vt= α(rt , t)dt + σV (rt , t)dWt
(140)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 123 / 203
Short Rate Model
Consider now two bonds, one with expiry date T1 and the other withexpiry date T2. By the pricing formula
V (t,T1) = E[e−
∫ T1t rs(ω)ds
]V (t,T2) = E
[e−
∫ T2t rs(ω)ds
] (141)
where both are driven by the same Brownian motion W . It follows that,∀t, the Sharpe Ratios for bonds must be equal for any expiry date:
α(r , t)− r
σV (r , t)= λ(r , t) = Market Price of Risk . (142)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 124 / 203
Market Price of Risk
Recall from our initial discussion of equity models that under constant(µ, σ, r), the rate of return on a share of equity is
dStSt
= µdt + σdWt
= rdt + σdWt .
⇒ dWt = dWt +µ− r
σdt
= dWt + λdt
(143)
In words, the market price of risk, or Sharpe ratio, is the drift correctionthat allows us to transition from a Brownian motion under the physicalmeasure P to a Brownian motion under the risk-neutral measure P.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 125 / 203
Market Price of Risk
Recall from our initial discussion of equity models that under constant(µ, σ, r), the rate of return on a share of equity is
dStSt
= µdt + σdWt
= rdt + σdWt .
⇒ dWt = dWt +µ− r
σdt
= dWt + λdt
(143)
In words, the market price of risk, or Sharpe ratio, is the drift correctionthat allows us to transition from a Brownian motion under the physicalmeasure P to a Brownian motion under the risk-neutral measure P.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 125 / 203
Market Price of Risk
It follows that we can apply the same method/reasoning to interest ratemodeling:
drt = µ(rt , t)dt + σ(rt , t)dWt
= [µ(rt , t)− σ(rt , t)λ(rt , t)]dt + σ(rt , t)dWt .
dWt = dWt + λ(rt , t)dt.
(144)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 126 / 203
Market Price of Risk
It follows that we can apply the same method/reasoning to interest ratemodeling:
drt = µ(rt , t)dt + σ(rt , t)dWt
= [µ(rt , t)− σ(rt , t)λ(rt , t)]dt + σ(rt , t)dWt .
dWt = dWt + λ(rt , t)dt.
(144)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 126 / 203
Market Price of Risk
Once we are in the risk-neutral framework, then for
V (r , t,T ) = E[e−
∫ Tt rs(ω)ds | rt = r
]drt = [µ(rt , t)− σ(rt , t)λ(rt , t)]dt + σ(rt , t)dWt
(145)
it follows by Ito’s Lemma and the equivalence of all Sharpe ratios that Valso satisfies
0 =∂V
∂t+ [µ(r , t)− σ(r , t)λ(r , t)]
∂V
∂r+
1
2σ2(r , t)
∂2V
∂r2− rV
1 = V (r ,T ,T ).
(146)
Question: How do we determine the market price of risk λ?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 127 / 203
Market Price of Risk
Once we are in the risk-neutral framework, then for
V (r , t,T ) = E[e−
∫ Tt rs(ω)ds | rt = r
]drt = [µ(rt , t)− σ(rt , t)λ(rt , t)]dt + σ(rt , t)dWt
(145)
it follows by Ito’s Lemma and the equivalence of all Sharpe ratios that Valso satisfies
0 =∂V
∂t+ [µ(r , t)− σ(r , t)λ(r , t)]
∂V
∂r+
1
2σ2(r , t)
∂2V
∂r2− rV
1 = V (r ,T ,T ).
(146)
Question: How do we determine the market price of risk λ?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 127 / 203
Affine Models
As mentioned before, it is not clear what λ should be from just the shortrate evolution rt .
There are models, however, that make assumptions on the form of λ thatallow for easy computation of V (r , t,T ).
Assume the Ansatz
V (r , t,T ) = eA(t,T )+B(t,T )r . (147)
HW: What are the corresponding boundary value problems for A and B?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 128 / 203
Affine Models
Two important examples are
Vasicek with Constant λ :
drt = [γ(µ− r)]dt + σdWt
= [γ(µ− r)− σλ]dt + σdWt
CIR with λ(r , t) =λ√r
σ:
drt = [γ(µ− r)]dt + σ√rtdWt
= [γ(µ− r)− λr ]dt + σ√rtdWt .
(148)
HW: Solve the corresponding boundary value problems for A and B forboth the Vasicek and CIR models.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 129 / 203
Affine Models
Two important examples are
Vasicek with Constant λ :
drt = [γ(µ− r)]dt + σdWt
= [γ(µ− r)− σλ]dt + σdWt
CIR with λ(r , t) =λ√r
σ:
drt = [γ(µ− r)]dt + σ√rtdWt
= [γ(µ− r)− λr ]dt + σ√rtdWt .
(148)
HW: Solve the corresponding boundary value problems for A and B forboth the Vasicek and CIR models.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 129 / 203
Affine Models
Two important examples are
Vasicek with Constant λ :
drt = [γ(µ− r)]dt + σdWt
= [γ(µ− r)− σλ]dt + σdWt
CIR with λ(r , t) =λ√r
σ:
drt = [γ(µ− r)]dt + σ√rtdWt
= [γ(µ− r)− λr ]dt + σ√rtdWt .
(148)
HW: Solve the corresponding boundary value problems for A and B forboth the Vasicek and CIR models.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 129 / 203
Capital Structure
A company that has shareholders can be valued by the equation
At = Et + Bt
At := Value of Company’s Assets at time t.
Et := Value of Company’s Equity at time t.
Bt := Value of Company’s Debt at time t.
(149)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 130 / 203
Capital Structure
If we assume that the equity does not pay dividends, and that theoutstanding debt is of the form of a zero-coupon bond with face B,payable at time T , then
ET = maxAT − B, 0
BT = min
AT , B
= AT + min
B − AT , 0
= AT −max
AT − B, 0
= AT − ET
(150)
Here, default is implicitly assumed to coincide with the eventAT < B
.
This results in a turnover of the company’s assets to bondholders if assetsare worth less than the total value of bond outstanding.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 131 / 203
Capital Structure
It follows that, as BT = AT on the setAT < B
,
B0 = e−rT E0[BT ]
= e−rT E0[minAT , B
]
= e−rT E0[AT | AT < B] · P0[AT < B] + e−rT B · P0[AT > B]
= e−rT
[∫ B
0A · P0[AT ∈ dA] + B
∫ ∞B
P0[AT ∈ dA]
] (151)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 132 / 203
Capital Structure
Furthermore, we define
E[Recovery Rate] =E[BT | AT < B]
B
E[Loss Given Default] =B − E[BT | AT < B]
B
= 1− E[Recovery Rate]
(152)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 133 / 203
Credit Spread
We use the above to formulate
Credit Spread :=1
Tln
(B
B0
)− r
=1
Tln
(1
1− P0[AT < B] · E0[Loss Given Default]
)≈ 1
T· P0[AT < B] · E0[Loss Given Default].
(153)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 134 / 203
Merton Default Model
Recall that under the BSM framework
Pt [Default] = N
(−
ln At
B+ r − δ − 1
2σ2(T − t)
σ√T − t
)
Et [AT | Default] = Ate(r−δ)(T−t)
N
(− ln At
B+r−δ+ 1
2σ2(T−t)
σ√T−t
)
N
(− ln At
B+r−δ− 1
2σ2(T−t)
σ√T−t
) (154)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 135 / 203
Merton Default Model
Completing the analysis, under the assumption δ = 0,
B0 = e−rT[E0[AT | Default] · P0[Default] + B · P0[No Default]
]= A0N
(−
ln A0
B+ r + 1
2σ2T
σ√T
)+ Be−rT · N
(ln A0
B+ r − 1
2σ2T
σ√T
)= A0N(−d1) + Be−rTN(d2)
(155)This implies that
Credit SpreadMerton =1
Tln( B
B0
)− r =
1
Tln
(e−rT B
B0
)
= − 1
Tln
[N(d2) +
A0
Be−rT· N(−d1)
] (156)
Q: What happens to the credit spread as T → 0 ?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 136 / 203
Merton Default Model
Completing the analysis, under the assumption δ = 0,
B0 = e−rT[E0[AT | Default] · P0[Default] + B · P0[No Default]
]= A0N
(−
ln A0
B+ r + 1
2σ2T
σ√T
)+ Be−rT · N
(ln A0
B+ r − 1
2σ2T
σ√T
)= A0N(−d1) + Be−rTN(d2)
(155)This implies that
Credit SpreadMerton =1
Tln( B
B0
)− r =
1
Tln
(e−rT B
B0
)
= − 1
Tln
[N(d2) +
A0
Be−rT· N(−d1)
] (156)
Q: What happens to the credit spread as T → 0 ?Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 136 / 203
Calibration
Recall that under the BSM framework
Et [AT | Default] = Ate(α−δ)(T−t)
N
(− ln At
B+α−δ+ 1
2σ2(T−t)
σ√T−t
)
N
(− ln At
B+α−δ− 1
2σ2(T−t)
σ√T−t
)
Distance to Default :=E[ln (AT )]− B
σ√T − t
=ln (At) + α + 1
2σ2(T − t)− ln (B)
σ√T − t
(157)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 137 / 203
Calibration
Under this framework, we have also derived an important result forcalibration, where recovery upon default is priced via risk neutral measure,but may only be observed under a physical measure.
The two are related via
Et [AT | Default] = γ(t) · Et [AT | Default]
γ := e(r−α)(T−t)
N
(− ln At
B+r−δ+ 1
2σ2(T−t)
σ√T−t
)
N
(− ln At
B+r−δ− 1
2σ2(T−t)
σ√T−t
) N
(− ln At
B+α−δ− 1
2σ2(T−t)
σ√T−t
)
N
(− ln At
B+α−δ+ 1
2σ2(T−t)
σ√T−t
)(158)
Note that γ(t) = 1 for all 0 ≤ t ≤ T if α = r .
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 138 / 203
Calibration
Under this framework, we have also derived an important result forcalibration, where recovery upon default is priced via risk neutral measure,but may only be observed under a physical measure.
The two are related via
Et [AT | Default] = γ(t) · Et [AT | Default]
γ := e(r−α)(T−t)
N
(− ln At
B+r−δ+ 1
2σ2(T−t)
σ√T−t
)
N
(− ln At
B+r−δ− 1
2σ2(T−t)
σ√T−t
) N
(− ln At
B+α−δ− 1
2σ2(T−t)
σ√T−t
)
N
(− ln At
B+α−δ+ 1
2σ2(T−t)
σ√T−t
)(158)
Note that γ(t) = 1 for all 0 ≤ t ≤ T if α = r .
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 138 / 203
Solving the BSM PDE:
Lemma
Under a change of variables, solving the BSM PDE is equivalent to solvingthe Heat Equation
uτ = uxx
u(x , 0) = u0(x)(159)
ProofWe show this for a European Call - namely G (S) = (S − K )+ for somestrike K . Let us further write V (S , t) = C (S , t) to denote that we arelooking at a Call option. Our PDE is now
Ct +1
2σ2S2CSS + r (SCS − C ) = 0
C (0, t) = 0
limS→∞
C (S , t)
S= 1
C (S ,T ) = (S − K )+
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 139 / 203
Solving the BSM PDE:
Lemma
Under a change of variables, solving the BSM PDE is equivalent to solvingthe Heat Equation
uτ = uxx
u(x , 0) = u0(x)(159)
ProofWe show this for a European Call - namely G (S) = (S − K )+ for somestrike K . Let us further write V (S , t) = C (S , t) to denote that we arelooking at a Call option. Our PDE is now
Ct +1
2σ2S2CSS + r (SCS − C ) = 0
C (0, t) = 0
limS→∞
C (S , t)
S= 1
C (S ,T ) = (S − K )+
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 139 / 203
Solving the BSM PDE:
Lemma
Under a change of variables, solving the BSM PDE is equivalent to solvingthe Heat Equation
uτ = uxx
u(x , 0) = u0(x)(159)
ProofWe show this for a European Call - namely G (S) = (S − K )+ for somestrike K . Let us further write V (S , t) = C (S , t) to denote that we arelooking at a Call option. Our PDE is now
Ct +1
2σ2S2CSS + r (SCS − C ) = 0
C (0, t) = 0
limS→∞
C (S , t)
S= 1
C (S ,T ) = (S − K )+
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 139 / 203
Solving the BSM PDE:
Lemma
Under a change of variables, solving the BSM PDE is equivalent to solvingthe Heat Equation
uτ = uxx
u(x , 0) = u0(x)(159)
ProofWe show this for a European Call - namely G (S) = (S − K )+ for somestrike K . Let us further write V (S , t) = C (S , t) to denote that we arelooking at a Call option. Our PDE is now
Ct +1
2σ2S2CSS + r (SCS − C ) = 0
C (0, t) = 0
limS→∞
C (S , t)
S= 1
C (S ,T ) = (S − K )+Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 139 / 203
Solving the BSM PDE:
Assume the change of variables
S = Kex
t = T − 1
2σ2τ
C (S , t) = Kv(x , τ)
(160)
and so
vτ = vxx +
(2r
σ2− 1
)vx −
2r
σ2v
Kv(x , 0) = max(Kex − K , 0)⇒ v(x , 0) = max(ex − 1, 0)
(161)
Furthermore, assume v(x , τ) = eαx+βτu(x , τ).
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 140 / 203
Solving the BSM PDE:
Assume the change of variables
S = Kex
t = T − 1
2σ2τ
C (S , t) = Kv(x , τ)
(160)
and so
vτ = vxx +
(2r
σ2− 1
)vx −
2r
σ2v
Kv(x , 0) = max(Kex − K , 0)⇒ v(x , 0) = max(ex − 1, 0)
(161)
Furthermore, assume v(x , τ) = eαx+βτu(x , τ).
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 140 / 203
Solving the BSM PDE:
Assume the change of variables
S = Kex
t = T − 1
2σ2τ
C (S , t) = Kv(x , τ)
(160)
and so
vτ = vxx +
(2r
σ2− 1
)vx −
2r
σ2v
Kv(x , 0) = max(Kex − K , 0)⇒ v(x , 0) = max(ex − 1, 0)
(161)
Furthermore, assume v(x , τ) = eαx+βτu(x , τ).
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 140 / 203
Solving the BSM PDE:
Assume the change of variables
S = Kex
t = T − 1
2σ2τ
C (S , t) = Kv(x , τ)
(160)
and so
vτ = vxx +
(2r
σ2− 1
)vx −
2r
σ2v
Kv(x , 0) = max(Kex − K , 0)⇒ v(x , 0) = max(ex − 1, 0)
(161)
Furthermore, assume v(x , τ) = eαx+βτu(x , τ).
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 140 / 203
Solving the BSM PDE:
Assume the change of variables
S = Kex
t = T − 1
2σ2τ
C (S , t) = Kv(x , τ)
(160)
and so
vτ = vxx +
(2r
σ2− 1
)vx −
2r
σ2v
Kv(x , 0) = max(Kex − K , 0)⇒ v(x , 0) = max(ex − 1, 0)
(161)
Furthermore, assume v(x , τ) = eαx+βτu(x , τ).
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 140 / 203
Solving the BSM PDE:
Thus means that if we choose
α = −1
2
(2r
σ2− 1
)β = −1
4
(2r
σ2+ 1
)2
then
uτ = uxx , −∞ < x <∞u(x , 0) = e−αxmax(ex − 1, 0)
QEDAlbert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 141 / 203
Solving the BSM PDE:
Question - How do we solve for u(x , τ)?
Answer - depends!
Try solving the following basic problem on the half real line:
uτ = uxx , x ≥ 0
u(0, τ) = 1
with the basic guess that we can reduce to an ordinary differentialequation via the transformation
ζ :=x√τ
(162)
and the solution u(x , τ) = U(ζ)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 142 / 203
Solving the BSM PDE:
Question - How do we solve for u(x , τ)?
Answer - depends!
Try solving the following basic problem on the half real line:
uτ = uxx , x ≥ 0
u(0, τ) = 1
with the basic guess that we can reduce to an ordinary differentialequation via the transformation
ζ :=x√τ
(162)
and the solution u(x , τ) = U(ζ)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 142 / 203
Solving the BSM PDE:
Question - How do we solve for u(x , τ)?
Answer - depends!
Try solving the following basic problem on the half real line:
uτ = uxx , x ≥ 0
u(0, τ) = 1
with the basic guess that we can reduce to an ordinary differentialequation via the transformation
ζ :=x√τ
(162)
and the solution u(x , τ) = U(ζ)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 142 / 203
Solving the BSM PDE:
Question - How do we solve for u(x , τ)?
Answer - depends!
Try solving the following basic problem on the half real line:
uτ = uxx , x ≥ 0
u(0, τ) = 1
with the basic guess that we can reduce to an ordinary differentialequation via the transformation
ζ :=x√τ
(162)
and the solution u(x , τ) = U(ζ)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 142 / 203
General Solution of Heat Equation
The resulting ODE is
U ′′(ζ) = −1
2ζU ′(ζ)
U(0) = 1
U(∞) = 0
which has solution
U(ζ) = C
∫ ζ
0e−
s2
4 ds + D (163)
which, upon satisfaction of the boundary conditions at 0 and ∞ leads to
U(ζ) =1√π
∫ ∞ζ
e−s2
4 ds ⇒ u(x , τ) =1√π
∫ ∞x√τ
e−s2
4 ds (164)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 143 / 203
General Solution of Heat Equation
The resulting ODE is
U ′′(ζ) = −1
2ζU ′(ζ)
U(0) = 1
U(∞) = 0
which has solution
U(ζ) = C
∫ ζ
0e−
s2
4 ds + D (163)
which, upon satisfaction of the boundary conditions at 0 and ∞ leads to
U(ζ) =1√π
∫ ∞ζ
e−s2
4 ds ⇒ u(x , τ) =1√π
∫ ∞x√τ
e−s2
4 ds (164)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 143 / 203
General Solution of Heat Equation
Again for the basic equation
uτ = uxx ∀x ∈ (−∞,∞) (165)
we look for solutions of the form
u(x , τ) =1√τU(ζ)
ζ :=x√τ
which leads to the ODE of the form
d
dζ
[U ′(ζ) +
1
2ζU(ζ)
]= 0 (166)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 144 / 203
General Solution of Heat Equation
Again for the basic equation
uτ = uxx ∀x ∈ (−∞,∞) (165)
we look for solutions of the form
u(x , τ) =1√τU(ζ)
ζ :=x√τ
which leads to the ODE of the form
d
dζ
[U ′(ζ) +
1
2ζU(ζ)
]= 0 (166)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 144 / 203
General Solution of Heat Equation
Again for the basic equation
uτ = uxx ∀x ∈ (−∞,∞) (165)
we look for solutions of the form
u(x , τ) =1√τU(ζ)
ζ :=x√τ
which leads to the ODE of the form
d
dζ
[U ′(ζ) +
1
2ζU(ζ)
]= 0 (166)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 144 / 203
General Solution of Heat Equation
Again for the basic equation
uτ = uxx ∀x ∈ (−∞,∞) (165)
we look for solutions of the form
u(x , τ) =1√τU(ζ)
ζ :=x√τ
which leads to the ODE of the form
d
dζ
[U ′(ζ) +
1
2ζU(ζ)
]= 0 (166)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 144 / 203
General Solution of Heat Equation
Again for the basic equation
uτ = uxx ∀x ∈ (−∞,∞) (165)
we look for solutions of the form
u(x , τ) =1√τU(ζ)
ζ :=x√τ
which leads to the ODE of the form
d
dζ
[U ′(ζ) +
1
2ζU(ζ)
]= 0 (166)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 144 / 203
General Solution of Heat Equation
Again for the basic equation
uτ = uxx ∀x ∈ (−∞,∞) (165)
we look for solutions of the form
u(x , τ) =1√τU(ζ)
ζ :=x√τ
which leads to the ODE of the form
d
dζ
[U ′(ζ) +
1
2ζU(ζ)
]= 0 (166)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 144 / 203
General Solution of Heat Equation
from which we take the fundamental solution
U(ζ) =1
2√πe−
ζ2
4
u(x , τ) =1
2√πτ
e−x2
4τ =: Γ(x , τ)
(167)
Now, consider the family of functions δε(x), parametrized by ε > 0, with
δε(x) :=1
2ε1[−ε,ε](x) (168)
It follows that
limε→0
∫ ∞−∞
δε(x)φ(x)dx = φ(0) (169)
for any reasonable function φ(x)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 145 / 203
General Solution of Heat Equation
from which we take the fundamental solution
U(ζ) =1
2√πe−
ζ2
4
u(x , τ) =1
2√πτ
e−x2
4τ =: Γ(x , τ)
(167)
Now, consider the family of functions δε(x), parametrized by ε > 0, with
δε(x) :=1
2ε1[−ε,ε](x) (168)
It follows that
limε→0
∫ ∞−∞
δε(x)φ(x)dx = φ(0) (169)
for any reasonable function φ(x)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 145 / 203
General Solution of Heat Equation
from which we take the fundamental solution
U(ζ) =1
2√πe−
ζ2
4
u(x , τ) =1
2√πτ
e−x2
4τ =: Γ(x , τ)
(167)
Now, consider the family of functions δε(x), parametrized by ε > 0, with
δε(x) :=1
2ε1[−ε,ε](x) (168)
It follows that
limε→0
∫ ∞−∞
δε(x)φ(x)dx = φ(0) (169)
for any reasonable function φ(x)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 145 / 203
General Solution of Heat Equation
from which we take the fundamental solution
U(ζ) =1
2√πe−
ζ2
4
u(x , τ) =1
2√πτ
e−x2
4τ =: Γ(x , τ)
(167)
Now, consider the family of functions δε(x), parametrized by ε > 0, with
δε(x) :=1
2ε1[−ε,ε](x) (168)
It follows that
limε→0
∫ ∞−∞
δε(x)φ(x)dx = φ(0) (169)
for any reasonable function φ(x)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 145 / 203
General Solution of Heat Equation
from which we take the fundamental solution
U(ζ) =1
2√πe−
ζ2
4
u(x , τ) =1
2√πτ
e−x2
4τ =: Γ(x , τ)
(167)
Now, consider the family of functions δε(x), parametrized by ε > 0, with
δε(x) :=1
2ε1[−ε,ε](x) (168)
It follows that
limε→0
∫ ∞−∞
δε(x)φ(x)dx = φ(0) (169)
for any reasonable function φ(x)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 145 / 203
General Solution of Heat Equation
In general,
limε→0
∫ ∞−∞
δε(x − x0)φ(x)dx = φ(x0) (170)
We call the (weak) limit of δε(x) the Dirac delta δ(x), after the physicist.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 146 / 203
General Solution of Heat Equation
So, for any function u0(x), we have the relation
limε→0
∫ ∞−∞
δε(y − x)u0(y)dx = u0(x) (171)
It also follows that our fundamental solution Γ(x , τ)⇒ δ(x) as τ → 0.Since u(x , 0)” = ”δ(x) and Γ(x − s, τ) = Γ(s − x , τ) for fixed s, we have
∂
∂τ[u0(s)Γ(s − x , τ)] =
∂2
∂x2[u0(s)Γ(s − x , τ)]
u0(s)Γ(s − x , 0) = u0(s)δ(s − x)
(172)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 147 / 203
General Solution of Heat Equation
So, for any function u0(x), we have the relation
limε→0
∫ ∞−∞
δε(y − x)u0(y)dx = u0(x) (171)
It also follows that our fundamental solution Γ(x , τ)⇒ δ(x) as τ → 0.Since u(x , 0)” = ”δ(x) and Γ(x − s, τ) = Γ(s − x , τ) for fixed s, we have
∂
∂τ[u0(s)Γ(s − x , τ)] =
∂2
∂x2[u0(s)Γ(s − x , τ)]
u0(s)Γ(s − x , 0) = u0(s)δ(s − x)
(172)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 147 / 203
General Solution of Heat Equation
So, for any function u0(x), we have the relation
limε→0
∫ ∞−∞
δε(y − x)u0(y)dx = u0(x) (171)
It also follows that our fundamental solution Γ(x , τ)⇒ δ(x) as τ → 0.
Since u(x , 0)” = ”δ(x) and Γ(x − s, τ) = Γ(s − x , τ) for fixed s, we have
∂
∂τ[u0(s)Γ(s − x , τ)] =
∂2
∂x2[u0(s)Γ(s − x , τ)]
u0(s)Γ(s − x , 0) = u0(s)δ(s − x)
(172)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 147 / 203
General Solution of Heat Equation
So, for any function u0(x), we have the relation
limε→0
∫ ∞−∞
δε(y − x)u0(y)dx = u0(x) (171)
It also follows that our fundamental solution Γ(x , τ)⇒ δ(x) as τ → 0.Since u(x , 0)” = ”δ(x) and Γ(x − s, τ) = Γ(s − x , τ) for fixed s, we have
∂
∂τ[u0(s)Γ(s − x , τ)] =
∂2
∂x2[u0(s)Γ(s − x , τ)]
u0(s)Γ(s − x , 0) = u0(s)δ(s − x)
(172)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 147 / 203
General Solution of Heat Equation
So, for any function u0(x), we have the relation
limε→0
∫ ∞−∞
δε(y − x)u0(y)dx = u0(x) (171)
It also follows that our fundamental solution Γ(x , τ)⇒ δ(x) as τ → 0.Since u(x , 0)” = ”δ(x) and Γ(x − s, τ) = Γ(s − x , τ) for fixed s, we have
∂
∂τ[u0(s)Γ(s − x , τ)] =
∂2
∂x2[u0(s)Γ(s − x , τ)]
u0(s)Γ(s − x , 0) = u0(s)δ(s − x)
(172)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 147 / 203
General Solution of Heat Equation
Since
u0(x) =
∫ ∞−∞
u0(s)δ(s − x)ds (173)
and u0(x)u(s − x , τ) is a solution of the PDE for all s, we have that
v(x , τ) =
∫ ∞−∞
u0(s)u(s − x , τ)ds
=
∫ ∞−∞
u0(s)1
2√πτ
e−(s−x)2
4τ ds
(174)
is a solution of
vτ = vxx
v(x , 0) = u0(x)(175)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 148 / 203
General Solution of Heat Equation
Since
u0(x) =
∫ ∞−∞
u0(s)δ(s − x)ds (173)
and u0(x)u(s − x , τ) is a solution of the PDE for all s, we have that
v(x , τ) =
∫ ∞−∞
u0(s)u(s − x , τ)ds
=
∫ ∞−∞
u0(s)1
2√πτ
e−(s−x)2
4τ ds
(174)
is a solution of
vτ = vxx
v(x , 0) = u0(x)(175)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 148 / 203
General Solution of Heat Equation
Since
u0(x) =
∫ ∞−∞
u0(s)δ(s − x)ds (173)
and u0(x)u(s − x , τ) is a solution of the PDE for all s, we have that
v(x , τ) =
∫ ∞−∞
u0(s)u(s − x , τ)ds
=
∫ ∞−∞
u0(s)1
2√πτ
e−(s−x)2
4τ ds
(174)
is a solution of
vτ = vxx
v(x , 0) = u0(x)(175)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 148 / 203
General Solution of Heat Equation
The focus we have is on solving a specific PDE, namely the European Call.The previous slides have shown us that the solution can be derived fromthe fundamental solution of the Heat equation. After making thenecessary substitutions, and by using the notation
d1 :=ln S
K +(r + 1
2σ2)
(T − t)
σ√T − t
d2 :=ln S
K +(r − 1
2σ2)
(T − t)
σ√T − t
N(x) :=1√2π
∫ x
−∞e−
12s2ds
(176)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 149 / 203
General Solution of BSM PDE revisited
Then we have the closed form solution
C (S , t) = SN(d1)− Ke−r(T−t)N(d2) (177)
HW1: Can we use any theorems/ideas to use this work to find thecorresponding value of a European Put option without resorting to solvingthe associated PDE?
HW2: Prove that ∆C = ∂C∂S = N(d1)
HW3: Find the value of an option that pays 1 if the stock value is aboveK at time T , and 0 otherwise. Hint - use the above analysis!
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 150 / 203
General Solution of BSM PDE revisited
Then we have the closed form solution
C (S , t) = SN(d1)− Ke−r(T−t)N(d2) (177)
HW1: Can we use any theorems/ideas to use this work to find thecorresponding value of a European Put option without resorting to solvingthe associated PDE?
HW2: Prove that ∆C = ∂C∂S = N(d1)
HW3: Find the value of an option that pays 1 if the stock value is aboveK at time T , and 0 otherwise. Hint - use the above analysis!
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 150 / 203
General Solution of BSM PDE revisited
Then we have the closed form solution
C (S , t) = SN(d1)− Ke−r(T−t)N(d2) (177)
HW1: Can we use any theorems/ideas to use this work to find thecorresponding value of a European Put option without resorting to solvingthe associated PDE?
HW2: Prove that ∆C = ∂C∂S = N(d1)
HW3: Find the value of an option that pays 1 if the stock value is aboveK at time T , and 0 otherwise. Hint - use the above analysis!
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 150 / 203
General Solution of BSM PDE revisited
Then we have the closed form solution
C (S , t) = SN(d1)− Ke−r(T−t)N(d2) (177)
HW1: Can we use any theorems/ideas to use this work to find thecorresponding value of a European Put option without resorting to solvingthe associated PDE?
HW2: Prove that ∆C = ∂C∂S = N(d1)
HW3: Find the value of an option that pays 1 if the stock value is aboveK at time T , and 0 otherwise. Hint - use the above analysis!
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 150 / 203
General Solution of BSM PDE revisited
Then we have the closed form solution
C (S , t) = SN(d1)− Ke−r(T−t)N(d2) (177)
HW1: Can we use any theorems/ideas to use this work to find thecorresponding value of a European Put option without resorting to solvingthe associated PDE?
HW2: Prove that ∆C = ∂C∂S = N(d1)
HW3: Find the value of an option that pays 1 if the stock value is aboveK at time T , and 0 otherwise. Hint - use the above analysis!
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 150 / 203
Discrete Probability Space
Let us define an event as a point ω in the set of all possible outcomes Ω.This includes the events ”The stock doubled in price over two tradingperiods” or ”the average stock price over ten years was 10 dollars”.
In our initial case, we will consider the simple binary spaceΩ = H,T for a one-period asset evolution. So, given an initialvalue S0, we have the final value S1(ω), with
S1(H) = uS0,S1(T ) = dS0 (178)
with d < 1 < u. Hence, a stock increases or decreases in price,according to the flip of a coin.
Let P be the probability measure associated with these events:
P[H] = p = 1− P[T ] (179)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 151 / 203
Discrete Probability Space
Let us define an event as a point ω in the set of all possible outcomes Ω.This includes the events ”The stock doubled in price over two tradingperiods” or ”the average stock price over ten years was 10 dollars”.
In our initial case, we will consider the simple binary spaceΩ = H,T for a one-period asset evolution. So, given an initialvalue S0, we have the final value S1(ω), with
S1(H) = uS0,S1(T ) = dS0 (178)
with d < 1 < u. Hence, a stock increases or decreases in price,according to the flip of a coin.
Let P be the probability measure associated with these events:
P[H] = p = 1− P[T ] (179)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 151 / 203
Discrete Probability Space
Let us define an event as a point ω in the set of all possible outcomes Ω.This includes the events ”The stock doubled in price over two tradingperiods” or ”the average stock price over ten years was 10 dollars”.
In our initial case, we will consider the simple binary spaceΩ = H,T for a one-period asset evolution. So, given an initialvalue S0, we have the final value S1(ω), with
S1(H) = uS0,S1(T ) = dS0 (178)
with d < 1 < u. Hence, a stock increases or decreases in price,according to the flip of a coin.
Let P be the probability measure associated with these events:
P[H] = p = 1− P[T ] (179)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 151 / 203
Discrete Probability Space
Let us define an event as a point ω in the set of all possible outcomes Ω.This includes the events ”The stock doubled in price over two tradingperiods” or ”the average stock price over ten years was 10 dollars”.
In our initial case, we will consider the simple binary spaceΩ = H,T for a one-period asset evolution. So, given an initialvalue S0, we have the final value S1(ω), with
S1(H) = uS0,S1(T ) = dS0 (178)
with d < 1 < u. Hence, a stock increases or decreases in price,according to the flip of a coin.
Let P be the probability measure associated with these events:
P[H] = p = 1− P[T ] (179)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 151 / 203
Discrete Probability Space
Let us define an event as a point ω in the set of all possible outcomes Ω.This includes the events ”The stock doubled in price over two tradingperiods” or ”the average stock price over ten years was 10 dollars”.
In our initial case, we will consider the simple binary spaceΩ = H,T for a one-period asset evolution. So, given an initialvalue S0, we have the final value S1(ω), with
S1(H) = uS0,S1(T ) = dS0 (178)
with d < 1 < u. Hence, a stock increases or decreases in price,according to the flip of a coin.
Let P be the probability measure associated with these events:
P[H] = p = 1− P[T ] (179)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 151 / 203
Arbitrage
Assume that S0(1 + r) > uS0
Where is the risk involved with investing in the asset S ?
Assume that S0(1 + r) < dS0
Why would anyone hold a bank account (zero-coupon bond)?
Lemma Arbitrage free ⇒ d < 1 + r < u
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 152 / 203
Arbitrage
Assume that S0(1 + r) > uS0
Where is the risk involved with investing in the asset S ?
Assume that S0(1 + r) < dS0
Why would anyone hold a bank account (zero-coupon bond)?
Lemma Arbitrage free ⇒ d < 1 + r < u
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 152 / 203
Arbitrage
Assume that S0(1 + r) > uS0
Where is the risk involved with investing in the asset S ?
Assume that S0(1 + r) < dS0
Why would anyone hold a bank account (zero-coupon bond)?
Lemma Arbitrage free ⇒ d < 1 + r < u
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 152 / 203
Arbitrage
Assume that S0(1 + r) > uS0
Where is the risk involved with investing in the asset S ?
Assume that S0(1 + r) < dS0
Why would anyone hold a bank account (zero-coupon bond)?
Lemma Arbitrage free ⇒ d < 1 + r < u
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 152 / 203
Arbitrage
Assume that S0(1 + r) > uS0
Where is the risk involved with investing in the asset S ?
Assume that S0(1 + r) < dS0
Why would anyone hold a bank account (zero-coupon bond)?
Lemma Arbitrage free ⇒ d < 1 + r < u
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 152 / 203
Derivative Pricing
Let S1(ω) be the price of an underlying asset at time 1. Define thefollowing instruments:
Zero-Coupon Bond : V B0 = 1
1+r ,VB1 (ω) = 1
Forward Contract : V F0 = 0,V F
1 = S1(ω)− F
Call Option : V C1 (ω) = max(S1(ω)− K , 0)
Put Option : V P1 (ω) = max(K − S1(ω), 0)
In both the Call and Put option, K is known as the Strike.Once again, a Forward Contract is a deal that is locked in at time 0 forinitial price 0, but requires at time 1 the buyer to purchase the asset forprice F .
What is the value V0 of the above put and call options?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 153 / 203
Derivative Pricing
Let S1(ω) be the price of an underlying asset at time 1. Define thefollowing instruments:
Zero-Coupon Bond : V B0 = 1
1+r ,VB1 (ω) = 1
Forward Contract : V F0 = 0,V F
1 = S1(ω)− F
Call Option : V C1 (ω) = max(S1(ω)− K , 0)
Put Option : V P1 (ω) = max(K − S1(ω), 0)
In both the Call and Put option, K is known as the Strike.Once again, a Forward Contract is a deal that is locked in at time 0 forinitial price 0, but requires at time 1 the buyer to purchase the asset forprice F .
What is the value V0 of the above put and call options?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 153 / 203
Derivative Pricing
Let S1(ω) be the price of an underlying asset at time 1. Define thefollowing instruments:
Zero-Coupon Bond : V B0 = 1
1+r ,VB1 (ω) = 1
Forward Contract : V F0 = 0,V F
1 = S1(ω)− F
Call Option : V C1 (ω) = max(S1(ω)− K , 0)
Put Option : V P1 (ω) = max(K − S1(ω), 0)
In both the Call and Put option, K is known as the Strike.
Once again, a Forward Contract is a deal that is locked in at time 0 forinitial price 0, but requires at time 1 the buyer to purchase the asset forprice F .
What is the value V0 of the above put and call options?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 153 / 203
Derivative Pricing
Let S1(ω) be the price of an underlying asset at time 1. Define thefollowing instruments:
Zero-Coupon Bond : V B0 = 1
1+r ,VB1 (ω) = 1
Forward Contract : V F0 = 0,V F
1 = S1(ω)− F
Call Option : V C1 (ω) = max(S1(ω)− K , 0)
Put Option : V P1 (ω) = max(K − S1(ω), 0)
In both the Call and Put option, K is known as the Strike.Once again, a Forward Contract is a deal that is locked in at time 0 forinitial price 0, but requires at time 1 the buyer to purchase the asset forprice F .
What is the value V0 of the above put and call options?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 153 / 203
Derivative Pricing
Let S1(ω) be the price of an underlying asset at time 1. Define thefollowing instruments:
Zero-Coupon Bond : V B0 = 1
1+r ,VB1 (ω) = 1
Forward Contract : V F0 = 0,V F
1 = S1(ω)− F
Call Option : V C1 (ω) = max(S1(ω)− K , 0)
Put Option : V P1 (ω) = max(K − S1(ω), 0)
In both the Call and Put option, K is known as the Strike.Once again, a Forward Contract is a deal that is locked in at time 0 forinitial price 0, but requires at time 1 the buyer to purchase the asset forprice F .
What is the value V0 of the above put and call options?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 153 / 203
Put-Call Parity
Can we replicate a forward contract using zero coupon bonds and put andcall options?
Yes: The final value of a replicating strategy X has value
V C1 − V P
1 + (K − F ) = S1 − F = X1(ω) (180)
This is achieved (replicated) by
Purchasing one call option
Selling one put option
Purchasing K−F1+r zero coupon bonds
all at time 0.Since this strategy must have zero initial value, we obtain
V C0 − V P
0 =F − K
1 + r(181)
Question: How would this change in a multi-period model?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 154 / 203
Put-Call Parity
Can we replicate a forward contract using zero coupon bonds and put andcall options?Yes: The final value of a replicating strategy X has value
V C1 − V P
1 + (K − F ) = S1 − F = X1(ω) (180)
This is achieved (replicated) by
Purchasing one call option
Selling one put option
Purchasing K−F1+r zero coupon bonds
all at time 0.Since this strategy must have zero initial value, we obtain
V C0 − V P
0 =F − K
1 + r(181)
Question: How would this change in a multi-period model?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 154 / 203
Put-Call Parity
Can we replicate a forward contract using zero coupon bonds and put andcall options?Yes: The final value of a replicating strategy X has value
V C1 − V P
1 + (K − F ) = S1 − F = X1(ω) (180)
This is achieved (replicated) by
Purchasing one call option
Selling one put option
Purchasing K−F1+r zero coupon bonds
all at time 0.Since this strategy must have zero initial value, we obtain
V C0 − V P
0 =F − K
1 + r(181)
Question: How would this change in a multi-period model?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 154 / 203
Put-Call Parity
Can we replicate a forward contract using zero coupon bonds and put andcall options?Yes: The final value of a replicating strategy X has value
V C1 − V P
1 + (K − F ) = S1 − F = X1(ω) (180)
This is achieved (replicated) by
Purchasing one call option
Selling one put option
Purchasing K−F1+r zero coupon bonds
all at time 0.
Since this strategy must have zero initial value, we obtain
V C0 − V P
0 =F − K
1 + r(181)
Question: How would this change in a multi-period model?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 154 / 203
Put-Call Parity
Can we replicate a forward contract using zero coupon bonds and put andcall options?Yes: The final value of a replicating strategy X has value
V C1 − V P
1 + (K − F ) = S1 − F = X1(ω) (180)
This is achieved (replicated) by
Purchasing one call option
Selling one put option
Purchasing K−F1+r zero coupon bonds
all at time 0.Since this strategy must have zero initial value, we obtain
V C0 − V P
0 =F − K
1 + r(181)
Question: How would this change in a multi-period model?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 154 / 203
Put-Call Parity
Can we replicate a forward contract using zero coupon bonds and put andcall options?Yes: The final value of a replicating strategy X has value
V C1 − V P
1 + (K − F ) = S1 − F = X1(ω) (180)
This is achieved (replicated) by
Purchasing one call option
Selling one put option
Purchasing K−F1+r zero coupon bonds
all at time 0.Since this strategy must have zero initial value, we obtain
V C0 − V P
0 =F − K
1 + r(181)
Question: How would this change in a multi-period model?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 154 / 203
General Derivative Pricing -One period model
If we begin with some initial capital X0, then we end with X1(ω). To pricea derivative, we need to match
X1(ω) = V1(ω) ∀ ω ∈ Ω (182)
to have X0 = V0, the price of the derivative we seek.
A strategy by the pair (X0,∆0) wherein
X0 is the initial capital
∆0 is the initial number of shares (units of underlying asset.)
What does the sign of ∆0 indicate?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 155 / 203
Replicating Strategy
Initial holding in bond (bank account) is X0 −∆0S0
Value of portfolio at maturity is
X1(ω) = (X0 −∆0S0)(1 + r) + ∆0S1(ω) (183)
Pathwise, we compute
V1(H) = (X0 −∆0S0)(1 + r) + ∆0uS0
V1(T ) = (X0 −∆0S0)(1 + r) + ∆0dS0
Algebra yields
∆0 =V1(H)− V1(T )
(u − d)S0(184)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 156 / 203
Replicating Strategy
Initial holding in bond (bank account) is X0 −∆0S0
Value of portfolio at maturity is
X1(ω) = (X0 −∆0S0)(1 + r) + ∆0S1(ω) (183)
Pathwise, we compute
V1(H) = (X0 −∆0S0)(1 + r) + ∆0uS0
V1(T ) = (X0 −∆0S0)(1 + r) + ∆0dS0
Algebra yields
∆0 =V1(H)− V1(T )
(u − d)S0(184)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 156 / 203
Replicating Strategy
Initial holding in bond (bank account) is X0 −∆0S0
Value of portfolio at maturity is
X1(ω) = (X0 −∆0S0)(1 + r) + ∆0S1(ω) (183)
Pathwise, we compute
V1(H) = (X0 −∆0S0)(1 + r) + ∆0uS0
V1(T ) = (X0 −∆0S0)(1 + r) + ∆0dS0
Algebra yields
∆0 =V1(H)− V1(T )
(u − d)S0(184)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 156 / 203
Replicating Strategy
Initial holding in bond (bank account) is X0 −∆0S0
Value of portfolio at maturity is
X1(ω) = (X0 −∆0S0)(1 + r) + ∆0S1(ω) (183)
Pathwise, we compute
V1(H) = (X0 −∆0S0)(1 + r) + ∆0uS0
V1(T ) = (X0 −∆0S0)(1 + r) + ∆0dS0
Algebra yields
∆0 =V1(H)− V1(T )
(u − d)S0(184)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 156 / 203
Risk Neutral Probability
Let us assume the existence of a pair (p, q) of positive numbers, and usethese to multiply our pricing equation(s):
pV1(H) = p(X0 −∆0S0)(1 + r) + p∆0uS0
qV1(T ) = q(X0 −∆0S0)(1 + r) + q∆0dS0
Addition yields
X0(1 + r) + ∆0S0(pu + qd − (1 + r)) = pV1(H) + qV1(T ) (185)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 157 / 203
Risk Neutral Probability
Let us assume the existence of a pair (p, q) of positive numbers, and usethese to multiply our pricing equation(s):
pV1(H) = p(X0 −∆0S0)(1 + r) + p∆0uS0
qV1(T ) = q(X0 −∆0S0)(1 + r) + q∆0dS0
Addition yields
X0(1 + r) + ∆0S0(pu + qd − (1 + r)) = pV1(H) + qV1(T ) (185)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 157 / 203
If we constrain
0 = pu + qd − (1 + r)
1 = p + q
0 ≤ p
0 ≤ q
then we have a risk neutral probability P where
V0 = X0 =1
1 + rE[V1] =
pV1(H) + qV1(T )
1 + r
p = P[X1(ω) = H] =1 + r − d
u − d
q = P[X1(ω) = T ] =u − (1 + r)
u − d
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 158 / 203
If we constrain
0 = pu + qd − (1 + r)
1 = p + q
0 ≤ p
0 ≤ q
then we have a risk neutral probability P where
V0 = X0 =1
1 + rE[V1] =
pV1(H) + qV1(T )
1 + r
p = P[X1(ω) = H] =1 + r − d
u − d
q = P[X1(ω) = T ] =u − (1 + r)
u − d
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 158 / 203
Example: Pricing a forward contract
Consider the case of a stock with
S0 = 400
u = 1.25
d = 0.75
r = 0.05
Then the forward price is computed via
0 =1
1 + rE[S1 − F ]⇒ F = E[S1] (186)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 159 / 203
Example: Pricing a forward contract
Consider the case of a stock with
S0 = 400
u = 1.25
d = 0.75
r = 0.05
Then the forward price is computed via
0 =1
1 + rE[S1 − F ]⇒ F = E[S1] (186)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 159 / 203
This leads to the explicit price
F = puS0 + qdS0
= (p)(1.25)(400) + (1− p)(0.75)(400)
= 500p + 300− 300p = 300 + 200p
= 300 + 200 · 1 + 0.05− 0.75
1.25− 0.75= 300 + 200 · 3
5
= 420
Homework Question: What is the price of a call option in the caseabove,with strike K = 375?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 160 / 203
This leads to the explicit price
F = puS0 + qdS0
= (p)(1.25)(400) + (1− p)(0.75)(400)
= 500p + 300− 300p = 300 + 200p
= 300 + 200 · 1 + 0.05− 0.75
1.25− 0.75= 300 + 200 · 3
5
= 420
Homework Question: What is the price of a call option in the caseabove,with strike K = 375?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 160 / 203
General one period risk neutral measure
We define a finite set of outcomes Ω ≡ ω1, ω2, ..., ωn and anysubcollection of outcomes A ⊂ Ω an event.
Furthermore, we define a probability measure P, not necessarily thephysical measure P to be risk neutral if
P[ω] > 0 ∀ ω ∈ Ω
X0 = 11+r E[X1]
for all strategies X .
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 161 / 203
General one period risk neutral measure
We define a finite set of outcomes Ω ≡ ω1, ω2, ..., ωn and anysubcollection of outcomes A ⊂ Ω an event.Furthermore, we define a probability measure P, not necessarily thephysical measure P to be risk neutral if
P[ω] > 0 ∀ ω ∈ Ω
X0 = 11+r E[X1]
for all strategies X .
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 161 / 203
General one period risk neutral measure
The measure is indifferent to investing in a zero-coupon bond, or arisky asset X
The same initial capital X0 in both cases produces the same”‘average”’ return after one period.
Not the physical measure attached by observation, experts, etc..
In fact, physical measure has no impact on pricing
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 162 / 203
General one period risk neutral measure
The measure is indifferent to investing in a zero-coupon bond, or arisky asset X
The same initial capital X0 in both cases produces the same”‘average”’ return after one period.
Not the physical measure attached by observation, experts, etc..
In fact, physical measure has no impact on pricing
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 162 / 203
General one period risk neutral measure
The measure is indifferent to investing in a zero-coupon bond, or arisky asset X
The same initial capital X0 in both cases produces the same”‘average”’ return after one period.
Not the physical measure attached by observation, experts, etc..
In fact, physical measure has no impact on pricing
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 162 / 203
General one period risk neutral measure
The measure is indifferent to investing in a zero-coupon bond, or arisky asset X
The same initial capital X0 in both cases produces the same”‘average”’ return after one period.
Not the physical measure attached by observation, experts, etc..
In fact, physical measure has no impact on pricing
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 162 / 203
Example: Risk Neutral measure for trinomial case
Assume that Ω = ω1, ω2, ω3 with
S1(ω1) = uS0
S1(ω2) = S0
S1(ω3) = dS0
Given a payoff V1(ω) to replicate, are we assured that a replicatingstrategy exists?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 163 / 203
Example: Risk Neutral measure for trinomial case
Assume that Ω = ω1, ω2, ω3 with
S1(ω1) = uS0
S1(ω2) = S0
S1(ω3) = dS0
Given a payoff V1(ω) to replicate, are we assured that a replicatingstrategy exists?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 163 / 203
Example: Risk Neutral measure for trinomial case
Homework: Try our first example with
S0 = 400
r = 0.05, u = 1.25, d = 0.75
V digital1 (ω) = χ (S1(ω) > 450) .
Now, if you are given V digital0 = 1
1+r E[V digital1 ] = 0.25, price a call option
with strike K = 420.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 164 / 203
Solution: Risk Neutral measure for trinomial case
The above scenario is reduced to finding the risk-neutral measure(p1, p2, p3). This can be done by finding the rref of the matrix M:
M =
1 1 1 1500 400 300 420
1 0 0 0.25(1.05)
(187)
which results in
rref (M) =
1 0 0 0.26250 1 0 0.6750 0 1 0.0625
. (188)
It follows that (p1, p2, p3) = (0.2625, 0.675, 0.0625), and so
V C0 =
1
1.05E[S1 | S0 = 400] = 0.2625× (500− 420)
= 20.(189)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 165 / 203
Existence of Risk Neutral measure
Let P be a probability measure on a finite space Ω. The following areequivalent:
P is a risk neutral measure
For all traded securities S i , S i0 = 1
1+r E[S i
1
]Proof: Homework (Hint: One direction is much easier than others. Also,strategies are linear in the underlying asset.)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 166 / 203
Existence of Risk Neutral measure
Let P be a probability measure on a finite space Ω. The following areequivalent:
P is a risk neutral measure
For all traded securities S i , S i0 = 1
1+r E[S i
1
]Proof: Homework (Hint: One direction is much easier than others. Also,strategies are linear in the underlying asset.)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 166 / 203
Existence of Risk Neutral measure
Let P be a probability measure on a finite space Ω. The following areequivalent:
P is a risk neutral measure
For all traded securities S i , S i0 = 1
1+r E[S i
1
]
Proof: Homework (Hint: One direction is much easier than others. Also,strategies are linear in the underlying asset.)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 166 / 203
Existence of Risk Neutral measure
Let P be a probability measure on a finite space Ω. The following areequivalent:
P is a risk neutral measure
For all traded securities S i , S i0 = 1
1+r E[S i
1
]Proof: Homework (Hint: One direction is much easier than others. Also,strategies are linear in the underlying asset.)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 166 / 203
Complete Markets
A market is complete if it is arbitrage free and every non-traded asset canbe replicated.
Fundamental Theorem of Asset Pricing 1: A market is arbitrage freeiff there exists a risk neutral measure
Fundamental Theorem of Asset Pricing 2: A market is complete iffthere exists exactly one risk neutral measure
Proof(s): We will go over these in detail later!
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 167 / 203
Complete Markets
A market is complete if it is arbitrage free and every non-traded asset canbe replicated.
Fundamental Theorem of Asset Pricing 1: A market is arbitrage freeiff there exists a risk neutral measure
Fundamental Theorem of Asset Pricing 2: A market is complete iffthere exists exactly one risk neutral measure
Proof(s): We will go over these in detail later!
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 167 / 203
Complete Markets
A market is complete if it is arbitrage free and every non-traded asset canbe replicated.
Fundamental Theorem of Asset Pricing 1: A market is arbitrage freeiff there exists a risk neutral measure
Fundamental Theorem of Asset Pricing 2: A market is complete iffthere exists exactly one risk neutral measure
Proof(s): We will go over these in detail later!
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 167 / 203
Dividends
What about dividends? How do they affect the risk neutral pricing ofexchange and non-exchange traded assets? What if they are paid atdiscrete times? Continuously paid?
Recall that if dividends are paid continuously at rate δ, then 1 share attime 0 will accumulate to eδT shares upon reinvestment of dividends intothe stock until time T .
It follows that to deliver one share of stock S with initial price S0 at timeT , only e−δT shares are needed. Correspondingly,
Fprepaid = e−δTS0
F = erT e−δTS0 = e(r−δ)TS0.(190)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 168 / 203
Dividends
What about dividends? How do they affect the risk neutral pricing ofexchange and non-exchange traded assets? What if they are paid atdiscrete times? Continuously paid?
Recall that if dividends are paid continuously at rate δ, then 1 share attime 0 will accumulate to eδT shares upon reinvestment of dividends intothe stock until time T .
It follows that to deliver one share of stock S with initial price S0 at timeT , only e−δT shares are needed. Correspondingly,
Fprepaid = e−δTS0
F = erT e−δTS0 = e(r−δ)TS0.(190)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 168 / 203
Dividends
What about dividends? How do they affect the risk neutral pricing ofexchange and non-exchange traded assets? What if they are paid atdiscrete times? Continuously paid?
Recall that if dividends are paid continuously at rate δ, then 1 share attime 0 will accumulate to eδT shares upon reinvestment of dividends intothe stock until time T .
It follows that to deliver one share of stock S with initial price S0 at timeT , only e−δT shares are needed. Correspondingly,
Fprepaid = e−δTS0
F = erT e−δTS0 = e(r−δ)TS0.(190)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 168 / 203
Binomial Option Pricing w/ cts Dividends and Interest
Over a period of length h, interest increases the value of a bond by afactor erh and dividends the value of a stock by a factor of eδh.
Once again, we compute pathwise,
V1(H) = (X0 −∆0S0)erh + ∆0eδhuS0
V1(T ) = (X0 −∆0S0)erh + ∆0eδhdS0
and this results in the modified quantities
∆0 = e−δhV1(H)− V1(T )
(u − d)S0
p =e(r−δ)h − d
u − d
q =u − e(r−δ)h
u − d
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 169 / 203
Binomial Option Pricing w/ cts Dividends and Interest
Over a period of length h, interest increases the value of a bond by afactor erh and dividends the value of a stock by a factor of eδh.
Once again, we compute pathwise,
V1(H) = (X0 −∆0S0)erh + ∆0eδhuS0
V1(T ) = (X0 −∆0S0)erh + ∆0eδhdS0
and this results in the modified quantities
∆0 = e−δhV1(H)− V1(T )
(u − d)S0
p =e(r−δ)h − d
u − d
q =u − e(r−δ)h
u − dAlbert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 169 / 203
Binomial Models w/ cts Dividends and Interest
For σ, the annualized standard deviation of continuously compoundedstock return, the following models hold:
Futures - Cox (1979)
u = eσ√h
d = e−σ√h
General Stock Model
u = e(r−δ)h+σ√h
d = e(r−δ)h−σ√h
Currencies with rf the foreign interest rate, which acts as a dividend:
u = e(r−rf )h+σ√h
d = e(r−rf )h−σ√h
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 170 / 203
1- and 2-period pricing
Consider the case r = 0.10, δ = 0.05, h = 0.01, σ = 0.1,S0 = 10
Now price two digital options, using the
1 General Stock Model
2 Futures-Cox Model
with respective payoffs
V1 := χ(S1 ≥ 10)
V2 := χ(S2 ≥ 10)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 171 / 203
1- and 2-period pricing
Consider the case r = 0.10, δ = 0.05, h = 0.01, σ = 0.1,S0 = 10
Now price two digital options, using the
1 General Stock Model
2 Futures-Cox Model
with respective payoffs
V1 := χ(S1 ≥ 10)
V2 := χ(S2 ≥ 10)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 171 / 203
1- and 2-period pricing
We can solve for 2-period problems
on a case-by-case basis, or
by developing a general theory for multi-period asset pricing
In the latter method, we need a general framework to carry out ourcomputations
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 172 / 203
1- and 2-period pricing
We can solve for 2-period problems
on a case-by-case basis, or
by developing a general theory for multi-period asset pricing
In the latter method, we need a general framework to carry out ourcomputations
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 172 / 203
1- and 2-period pricing
We can solve for 2-period problems
on a case-by-case basis, or
by developing a general theory for multi-period asset pricing
In the latter method, we need a general framework to carry out ourcomputations
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 172 / 203
Risk Neutral Pricing Formula
Assume now that we have the ”regular assumptions” on our coin flipspace, and that at time N we are asked to deliver a path dependentderivative value VN . Then for times 0 ≤ n ≤ N, the value of thisderivative is computed via
Vn = e−rhEn [Vn+1] (191)
and so
V0 = e−NhE0 [VN ] (192)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 173 / 203
Risk Neutral Pricing Formula
Assume now that we have the ”regular assumptions” on our coin flipspace, and that at time N we are asked to deliver a path dependentderivative value VN . Then for times 0 ≤ n ≤ N, the value of thisderivative is computed via
Vn = e−rhEn [Vn+1] (191)
and so
V0 = e−NhE0 [VN ] (192)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 173 / 203
Risk Neutral Pricing Formula
Assume now that we have the ”regular assumptions” on our coin flipspace, and that at time N we are asked to deliver a path dependentderivative value VN . Then for times 0 ≤ n ≤ N, the value of thisderivative is computed via
Vn = e−rhEn [Vn+1] (191)
and so
V0 = e−NhE0 [VN ] (192)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 173 / 203
Risk Neutral Pricing Formula
Assume now that we have the ”regular assumptions” on our coin flipspace, and that at time N we are asked to deliver a path dependentderivative value VN . Then for times 0 ≤ n ≤ N, the value of thisderivative is computed via
Vn = e−rhEn [Vn+1] (191)
and so
V0 = e−NhE0 [VN ] (192)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 173 / 203
Computational Complexity
Consider the case
p = q =1
2
S0 = 4, u =4
3, d =
3
4
(193)
but now with term n = 3. There are 23 = 8 paths to consider. However,there are 3 + 1 = 4 unique final values of S3 to consider. In the generalterm N, there would be 2N paths to generate SN , but only N + 1 distinctvalues. At any node n units of time into the asset’s evolution, there aren + 1 distinct values.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 174 / 203
Computational Complexity
Also, at each value s for Sn, we know that in this example, Sn+1 = 43s or
Sn+1 = 34s. Using our multi-period risk-neutral pricing, we can generate
for vn(s) := Vn(Sn(ω1, ..., ωn)) on the node (event) Sn(ω1, ..., ωn) = s
vn(s) = e−rh[pvn+1
(4
3s)
+ qvn+1
(3
4s)]
(194)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 175 / 203
An Example:
Assume r , δ, and h are such that
p =1
2= q, e−rh =
9
10
S0 = 4, u = 2, d =1
2V3 := max 10− S3, 0
(195)
It follows that
v3(32) = 0
v3(8) = 2
v3(2) = 8
v3(0.50) = 9.50.
(196)
Compute V0.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 176 / 203
Markov Processes
If we use the above approach for a more exotic option, say a lookbackoption that pays the maximum over the term of a stock, then we find thisapproach lacking. There is not enough information in the tree or thedistinct values for S3 as stated. We need more. Consider our generalmulti-period binomial model under P
Definition We say that a process X is adapted if it depends only on theflips ω1, ..., ωn
Definition We say that an adapted process X is Markov if for every0 ≤ n ≤ N − 1 and every function f (x) there exists another function g(x)such that
En [f (Xn+1)] = g(Xn) (197)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 177 / 203
Markov Processes
This notion of Markovity is essential to our state-dependent pricingalgorithm. Indeed, since our stock process evolves from time n to timen + 1, using only the information in Sn, we can in fact say that for everyf (s) there exists a g(s) such that
g(s) = En [f (Sn+1) | Sn = s]
g(s) = e−rh[pf(4
3s)
+ qf(3
4s)] (198)
So, for any f (s) := VN(s), we can work our recursive algorithm backwardsto find the gn(s) := Vn(s) for all 0 ≤ n ≤ N − 1
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 178 / 203
Markov Processes
Returning to our example of a lookback option, we see that the problemwas that Mn := max0≤i≤nSi is not Markov by itself, but the pair (Mn,Sn)is. Why?
Let’s generate the tree!
Homework Can you think of any other processes that are not Markov?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 179 / 203
The Interview Process
Consider the following scenario: After graduating, you go on the jobmarket, and have 4 possible job interviews with 4 different companies. Sosure of your prospects that you know that each company will make anoffer, with an identically, independently distributed probability attached tothe 4 possible salary offers -
P [Salary Offer=50, 000] = 0.1
P [Salary Offer=70, 000] = 0.3
P [Salary Offer=80, 000] = 0.4
P [Salary Offer=100, 000] = 0.2
(199)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 180 / 203
The Interview Process
Consider the following scenario: After graduating, you go on the jobmarket, and have 4 possible job interviews with 4 different companies. Sosure of your prospects that you know that each company will make anoffer, with an identically, independently distributed probability attached tothe 4 possible salary offers -
P [Salary Offer=50, 000] = 0.1
P [Salary Offer=70, 000] = 0.3
P [Salary Offer=80, 000] = 0.4
P [Salary Offer=100, 000] = 0.2
(199)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 180 / 203
The Interview Process
How should you interview?
Specifically, when should you accept an offer and cancel theremaining interviews?
How does your strategy change if you can interview as many times asyou like, but the distribution of offers remains the same as above?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 181 / 203
The Interview Process
How should you interview?
Specifically, when should you accept an offer and cancel theremaining interviews?
How does your strategy change if you can interview as many times asyou like, but the distribution of offers remains the same as above?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 181 / 203
The Interview Process
How should you interview?
Specifically, when should you accept an offer and cancel theremaining interviews?
How does your strategy change if you can interview as many times asyou like, but the distribution of offers remains the same as above?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 181 / 203
The Interview Process: Strategy
At any time student will know only one offer, which she can either acceptor reject. Of course, if student rejects the first three offers, than she has toaccept the last one. Compute the maximal expected salary for the studentafter the graduation and the corresponding optimal strategy.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 182 / 203
The Interview Process: Optimal Strategy
The solution process Xk4k=1 follows an Optimal Stopping Strategy:
Xk(s) = maxs, E[Xk+1 | kth offer = s
]. (200)
At time 4, the value of this game is X4(s) = s, with s being the salaryoffered.
At time 3, the conditional expected value of this game is
E[X4 | 3rd offer = s
]= E[X4]
= 0.1× 50, 000 + 0.3× 70, 000
+ 0.4× 80, 000 + 0.2× 100, 000
= 78, 000.
(201)
Hence, one should accept an offer of 80, 000 or 100, 000, and rejectthe other two.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 183 / 203
The Interview Process: Optimal Strategy
This strategy leads to a valuation:
X3(50, 000) = 78, 000
X3(70, 000) = 78, 000
X3(80, 000) = 80, 000
X3(100, 000) = 100, 000.
(202)
At time 2, similar reasoning using E2[X3] leads to the valuation
X2(50, 000) = 83, 200
X2(70, 000) = 83, 200
X2(80, 000) = 83, 200
X2(100, 000) = 100, 000.
(203)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 184 / 203
The Interview Process: Optimal Strategy
At time 1,X1(50, 000) = 86, 560
X1(70, 000) = 86, 560
X1(80, 000) = 86, 560
X1(100, 000) = 100, 000.
(204)
Finally, at time 0, the value of this optimal strategy is
E0[X1] = E[X1] = 0.8× 86, 560 + 0.2× 100, 000 = 89, 248. (205)
So, the optimal strategy is, for the first two interviews, accept only anoffer of 100, 000. If after the third interview, and offer of 80, 000 or100, 000 is made, then accept. Otherwise continue to the last interviewwhere you should accept whatever is offered.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 185 / 203
Review
Let’s review the basic contracts we can write:
Forward Contract Initial Value is 0, because both buyer and sellermay have to pay a balance at maturity
(European) Put/Call Option Initial Value is > 0, because both onlyseller must pay balance at maturity.
(European) ”Exotic” Option Initial Value is > 0, because both onlyseller must pay balance at maturity.
During the term of the contract, can the value of the contract ever fallbelow the intrinsic value of the payoff? Symbolically, does it ever occurthat
vn(s) < g(s) (206)
where g(s) is of the form of g(S) := max S − K , 0, in the case of a Calloption, for example.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 186 / 203
Review
Let’s review the basic contracts we can write:
Forward Contract Initial Value is 0, because both buyer and sellermay have to pay a balance at maturity
(European) Put/Call Option Initial Value is > 0, because both onlyseller must pay balance at maturity.
(European) ”Exotic” Option Initial Value is > 0, because both onlyseller must pay balance at maturity.
During the term of the contract, can the value of the contract ever fallbelow the intrinsic value of the payoff? Symbolically, does it ever occurthat
vn(s) < g(s) (206)
where g(s) is of the form of g(S) := max S − K , 0, in the case of a Calloption, for example.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 186 / 203
Review
Let’s review the basic contracts we can write:
Forward Contract Initial Value is 0, because both buyer and sellermay have to pay a balance at maturity
(European) Put/Call Option Initial Value is > 0, because both onlyseller must pay balance at maturity.
(European) ”Exotic” Option Initial Value is > 0, because both onlyseller must pay balance at maturity.
During the term of the contract, can the value of the contract ever fallbelow the intrinsic value of the payoff? Symbolically, does it ever occurthat
vn(s) < g(s) (206)
where g(s) is of the form of g(S) := max S − K , 0, in the case of a Calloption, for example.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 186 / 203
Review
Let’s review the basic contracts we can write:
Forward Contract Initial Value is 0, because both buyer and sellermay have to pay a balance at maturity
(European) Put/Call Option Initial Value is > 0, because both onlyseller must pay balance at maturity.
(European) ”Exotic” Option Initial Value is > 0, because both onlyseller must pay balance at maturity.
During the term of the contract, can the value of the contract ever fallbelow the intrinsic value of the payoff? Symbolically, does it ever occurthat
vn(s) < g(s) (206)
where g(s) is of the form of g(S) := max S − K , 0, in the case of a Calloption, for example.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 186 / 203
Review
Let’s review the basic contracts we can write:
Forward Contract Initial Value is 0, because both buyer and sellermay have to pay a balance at maturity
(European) Put/Call Option Initial Value is > 0, because both onlyseller must pay balance at maturity.
(European) ”Exotic” Option Initial Value is > 0, because both onlyseller must pay balance at maturity.
During the term of the contract, can the value of the contract ever fallbelow the intrinsic value of the payoff? Symbolically, does it ever occurthat
vn(s) < g(s) (206)
where g(s) is of the form of g(S) := max S − K , 0, in the case of a Calloption, for example.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 186 / 203
Review
Let’s review the basic contracts we can write:
Forward Contract Initial Value is 0, because both buyer and sellermay have to pay a balance at maturity
(European) Put/Call Option Initial Value is > 0, because both onlyseller must pay balance at maturity.
(European) ”Exotic” Option Initial Value is > 0, because both onlyseller must pay balance at maturity.
During the term of the contract, can the value of the contract ever fallbelow the intrinsic value of the payoff? Symbolically, does it ever occurthat
vn(s) < g(s) (206)
where g(s) is of the form of g(S) := max S − K , 0, in the case of a Calloption, for example.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 186 / 203
Review
Let’s review the basic contracts we can write:
Forward Contract Initial Value is 0, because both buyer and sellermay have to pay a balance at maturity
(European) Put/Call Option Initial Value is > 0, because both onlyseller must pay balance at maturity.
(European) ”Exotic” Option Initial Value is > 0, because both onlyseller must pay balance at maturity.
During the term of the contract, can the value of the contract ever fallbelow the intrinsic value of the payoff? Symbolically, does it ever occurthat
vn(s) < g(s) (206)
where g(s) is of the form of g(S) := max S − K , 0, in the case of a Calloption, for example.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 186 / 203
Early Exercise
If σ = 0, and so uncertainty vanishes, then an investor would seek toexercise early if
rK > δS . (207)
If σ > 0, then the situation involves deeper analysis. Whether solving afree boundary problem or analyzing a binomial tree, it is likely that acomputer will be involved in helping the investor to determine the optimalexercise time.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 187 / 203
For Freedom! (we must charge extra...)
What happens if we write a contract that allows the purchaser to exercisethe contract whenever she feels it to be in her advantage? By allowing thisextra freedom, we must
Charge more than we would for a European contract that is exercisedonly at the term N
Hedge our replicating strategy X differently, to allow for thepossibility of early exercise
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 188 / 203
For Freedom! (we must charge extra...)
What happens if we write a contract that allows the purchaser to exercisethe contract whenever she feels it to be in her advantage? By allowing thisextra freedom, we must
Charge more than we would for a European contract that is exercisedonly at the term N
Hedge our replicating strategy X differently, to allow for thepossibility of early exercise
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 188 / 203
For Freedom! (we must charge extra...)
What happens if we write a contract that allows the purchaser to exercisethe contract whenever she feels it to be in her advantage? By allowing thisextra freedom, we must
Charge more than we would for a European contract that is exercisedonly at the term N
Hedge our replicating strategy X differently, to allow for thepossibility of early exercise
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 188 / 203
American Options
In the end, the option v is valued after the nth value of the stockSn(ω) = s is revealed via the recursive formula along each pathω := ω1, ω2, .., ωN:
vn(Sn(ω)) = maxg(Sn(ω)), e−rhE
[v(Sn+1(ω)) | Sn(ω)
]τ∗(ω) = inf k ∈ 0, 1, ..,N | vk(Sk(ω)) = g(Sk(ω)) .
(208)
Here, τ∗ is the optimal exercise time.
In the Binomial case, we reduce to
vn(s) = maxg(s), e−rh [pvn+1(us) + qvn+1(ds)]
τ∗ = inf k | vk(s) = g(s) .
(209)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 189 / 203
American Options
In the end, the option v is valued after the nth value of the stockSn(ω) = s is revealed via the recursive formula along each pathω := ω1, ω2, .., ωN:
vn(Sn(ω)) = maxg(Sn(ω)), e−rhE
[v(Sn+1(ω)) | Sn(ω)
]τ∗(ω) = inf k ∈ 0, 1, ..,N | vk(Sk(ω)) = g(Sk(ω)) .
(208)
Here, τ∗ is the optimal exercise time.
In the Binomial case, we reduce to
vn(s) = maxg(s), e−rh [pvn+1(us) + qvn+1(ds)]
τ∗ = inf k | vk(s) = g(s) .
(209)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 189 / 203
American Options
Some examples:
”American Bond:” g(s) = 1
”American Digital Option:” g(s) = 16≤s≤10
”American Square Option:” g(s) = S2.
Does an investor exercise any of these options early? Consider again thesetting
p =1
2= q, e−rh =
9
10
S0 = 4, u = 2, d =1
2.
(210)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 190 / 203
American Square Options
Consider the American Square Option. We know via Jensen’s Inequalitythat
e−rhE[g(Sn+1) | Sn
]= e−rhE
[S2n+1 | Sn
]≥ e−rh
(E[Sn+1 | Sn
]2)= e−rh
(erhSn
)2
= erhS2n > S2
n = g(Sn).
(211)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 191 / 203
American Square Options
It follows that vN(s) = s2 and
vN−1(s) = maxg(s), e−rhE[vN(SN) | SN−1 = s]
= max
s2, e−rhE[S2
N | SN−1 = s]
= e−rhE[S2N | SN−1 = s]
= e−rhE[vN(SN) | SN−1 = s]
(212)
In words, with one period to go, don’t exercise yet!! The American andEuropean option values coincide. Keep going. How about with twoperiods left before expiration?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 192 / 203
American Options
Let’s return to the previous European Put example, where
p =1
2= q, e−rh =
9
10
S0 = 4, u = 2, d =1
2V3 := max 10− S3, 0
(213)
It follows that S3(ω) ∈
12 , 2, 8, 32
. Use this to compute v3(s) and the
American Put recursively.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 193 / 203
American Options
Let’s return to the previous European Put example, where
p =1
2= q, e−rh =
9
10
S0 = 4, u = 2, d =1
2V3 := max 10− S3, 0
(213)
It follows that S3(ω) ∈
12 , 2, 8, 32
. Use this to compute v3(s) and the
American Put recursively.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 193 / 203
Lognormality
As in elementary credit and investment theory, we assume that on a
probability space(
Ω,F ,P)
, our asset St(ω) has an associated return over
any period (t, t + u) defined as
rt,u(ω) := ln
(St+u(ω)
St(ω)
)(214)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 194 / 203
Lognormality
If we partition the interval [t,T ] into n intervals of length h = T−tn , then
the return over the entire period can be taken as the sum of the returnsover each interval:
rt,T−t(ω) = ln
(ST (ω)
St(ω)
)=
n∑k=1
rtk ,h(ω)
tk = t + kh
(215)
We model the returns as being independent and possessing a binomialdistribution. Employing the Central Limit Theorem, it can be shown thatas n→∞, this distribution approaches normality.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 195 / 203
Binomial Tree and Discrete Dividends
Another issue encountered in elementary credit and investment theory isthe case of different compounding and deposit periods. This also occurs inthe financial setting where a dividend is not paid continuously, but ratherat specific times. It follows that the dividend can be modeled as deliveredin the middle of a binomial period, at time τ(ω) < T . This view is due toSchroder and can be summarized as viewing the inherent value of St(ω) asthe sum of a prepaid forward PF and the present value of the upcomingdividend payment D:
PFt(ω) = St(ω)− De−r(τ(ω)−t)
u = erh+σ√h
d = erh−σ√h
(216)
Now, the random process that we model as having up and down moves isPF instead of S .
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 196 / 203
Asian Options
In times of high volatility or frequent trading, a company may want toprotect against large price movements over an entire time period, using anaverage. For example, if a company is looking at foreign exchange marketsor markets that may be subject to stock pinning due to large actors.
As an input, the average of an asset is used as an input against a strike,instead of the spot price.
There are two possibilities for the input in the discrete case: h = TN and
Arithmetic Average: IA(T ) := 1N
∑Nk=1 Skh
Geometric Average: IG (T ) :=(
ΠNk=1Skh
) 1N
HW: Is there an ordering for IA, IG that is independent of T?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 197 / 203
Asian Options
In times of high volatility or frequent trading, a company may want toprotect against large price movements over an entire time period, using anaverage. For example, if a company is looking at foreign exchange marketsor markets that may be subject to stock pinning due to large actors.
As an input, the average of an asset is used as an input against a strike,instead of the spot price.
There are two possibilities for the input in the discrete case: h = TN and
Arithmetic Average: IA(T ) := 1N
∑Nk=1 Skh
Geometric Average: IG (T ) :=(
ΠNk=1Skh
) 1N
HW: Is there an ordering for IA, IG that is independent of T?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 197 / 203
Asian Options
In times of high volatility or frequent trading, a company may want toprotect against large price movements over an entire time period, using anaverage. For example, if a company is looking at foreign exchange marketsor markets that may be subject to stock pinning due to large actors.
As an input, the average of an asset is used as an input against a strike,instead of the spot price.
There are two possibilities for the input in the discrete case: h = TN and
Arithmetic Average: IA(T ) := 1N
∑Nk=1 Skh
Geometric Average: IG (T ) :=(
ΠNk=1Skh
) 1N
HW: Is there an ordering for IA, IG that is independent of T?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 197 / 203
Asian Options
In times of high volatility or frequent trading, a company may want toprotect against large price movements over an entire time period, using anaverage. For example, if a company is looking at foreign exchange marketsor markets that may be subject to stock pinning due to large actors.
As an input, the average of an asset is used as an input against a strike,instead of the spot price.
There are two possibilities for the input in the discrete case: h = TN and
Arithmetic Average: IA(T ) := 1N
∑Nk=1 Skh
Geometric Average: IG (T ) :=(
ΠNk=1Skh
) 1N
HW: Is there an ordering for IA, IG that is independent of T?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 197 / 203
Asian Options: An Example:
Notice that these are path-dependent options, unlike the put and calloptions that we have studied until now. Assume r , δ, and h are such that
p =1
2= q, e−rh =
9
10
S0 = 4, u = 2, d =1
2g(I ) = max I − 2.5, 0
(217)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 198 / 203
Asian Options: An Example:
Consider an arithmetic average with N = 2. Then
v2(HH) = max
8 + 16
2− 2.5, 0
= 9.5
v2(HT ) = max
8 + 4
2− 2.5, 0
= 3.5
v2(TH) = max
2 + 4
2− 2.5, 0
= 0.5
v2(TT ) = max
2 + 1
2− 2.5, 0
= 0
(218)
Compute v0, assuming a European structure. How about an Americanstructure?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 199 / 203
Asian Options: An Example:
Consider an arithmetic average with N = 2. Then
v2(HH) = max
8 + 16
2− 2.5, 0
= 9.5
v2(HT ) = max
8 + 4
2− 2.5, 0
= 3.5
v2(TH) = max
2 + 4
2− 2.5, 0
= 0.5
v2(TT ) = max
2 + 1
2− 2.5, 0
= 0
(218)
Compute v0, assuming a European structure.
How about an Americanstructure?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 199 / 203
Asian Options: An Example:
Consider an arithmetic average with N = 2. Then
v2(HH) = max
8 + 16
2− 2.5, 0
= 9.5
v2(HT ) = max
8 + 4
2− 2.5, 0
= 3.5
v2(TH) = max
2 + 4
2− 2.5, 0
= 0.5
v2(TT ) = max
2 + 1
2− 2.5, 0
= 0
(218)
Compute v0, assuming a European structure. How about an Americanstructure?
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 199 / 203
Back to the Continuous Time Case
Consider the case of a security whose binomial evolution is modeled as anup or down movement at the end of each day. Over the period of oneyear, this amounts to a tree with depth 365. If the tree is not recombining,then this amounts to 2365 branches. Clearly, this is too large to evaluatereasonably, and so an alternative is sought.
Whatever the alternative, the concept of replication must hold. This isthe reasoning behind the famous Black-Scholes-Merton PDE approach.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 200 / 203
Back to the Continuous Time Case
Consider the case of a security whose binomial evolution is modeled as anup or down movement at the end of each day. Over the period of oneyear, this amounts to a tree with depth 365. If the tree is not recombining,then this amounts to 2365 branches. Clearly, this is too large to evaluatereasonably, and so an alternative is sought.
Whatever the alternative, the concept of replication must hold. This isthe reasoning behind the famous Black-Scholes-Merton PDE approach.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 200 / 203
Monte Carlo Techniques
Recall our model for asset evolution
⇒ St = S0e(α−δ− 1
2σ2)t+σ
√tZ
Z ∼ N(0, 1).(219)
Consider now the possibility of simulating the stock evolution bysimulating the random variable Z , or in fact an i.i.d. sequence
Z (i)
ni=1
.
For a European option with time expiry T , we can simulate the expirytime payoff mulitple times:
V(S (i),T
)= G
(S (i))
= G(S0e
(α−δ− 12σ2)T+σ
√TZ (i)
)(220)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 201 / 203
Monte Carlo Techniques
Recall our model for asset evolution
⇒ St = S0e(α−δ− 1
2σ2)t+σ
√tZ
Z ∼ N(0, 1).(219)
Consider now the possibility of simulating the stock evolution bysimulating the random variable Z , or in fact an i.i.d. sequence
Z (i)
ni=1
.
For a European option with time expiry T , we can simulate the expirytime payoff mulitple times:
V(S (i),T
)= G
(S (i))
= G(S0e
(α−δ− 12σ2)T+σ
√TZ (i)
)(220)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 201 / 203
Monte Carlo Techniques
Recall our model for asset evolution
⇒ St = S0e(α−δ− 1
2σ2)t+σ
√tZ
Z ∼ N(0, 1).(219)
Consider now the possibility of simulating the stock evolution bysimulating the random variable Z , or in fact an i.i.d. sequence
Z (i)
ni=1
.
For a European option with time expiry T , we can simulate the expirytime payoff mulitple times:
V(S (i),T
)= G
(S (i))
= G(S0e
(α−δ− 12σ2)T+σ
√TZ (i)
)(220)
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 201 / 203
Monte Carlo Techniques
If we sample uniformly from our simulated valuesV(S (i),T
)n
i=1we
can appeal to a sampling-convergence theorem with the appoximation
V (S , 0) = e−rT1
n
n∑i=1
V(S (i),T
)(221)
The challenge now is to simulate our lognormally distributed assetevolution.
One can simulate the value ST directly by one random variable Z , or amultiple of them to simulate the path of the evolution until T .
The latter method is necessary for Asian options.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 202 / 203
Monte Carlo Techniques
If we sample uniformly from our simulated valuesV(S (i),T
)n
i=1we
can appeal to a sampling-convergence theorem with the appoximation
V (S , 0) = e−rT1
n
n∑i=1
V(S (i),T
)(221)
The challenge now is to simulate our lognormally distributed assetevolution.
One can simulate the value ST directly by one random variable Z , or amultiple of them to simulate the path of the evolution until T .
The latter method is necessary for Asian options.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 202 / 203
Monte Carlo Techniques
If we sample uniformly from our simulated valuesV(S (i),T
)n
i=1we
can appeal to a sampling-convergence theorem with the appoximation
V (S , 0) = e−rT1
n
n∑i=1
V(S (i),T
)(221)
The challenge now is to simulate our lognormally distributed assetevolution.
One can simulate the value ST directly by one random variable Z , or amultiple of them to simulate the path of the evolution until T .
The latter method is necessary for Asian options.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 202 / 203
Monte Carlo Techniques
If we sample uniformly from our simulated valuesV(S (i),T
)n
i=1we
can appeal to a sampling-convergence theorem with the appoximation
V (S , 0) = e−rT1
n
n∑i=1
V(S (i),T
)(221)
The challenge now is to simulate our lognormally distributed assetevolution.
One can simulate the value ST directly by one random variable Z , or amultiple of them to simulate the path of the evolution until T .
The latter method is necessary for Asian options.
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 202 / 203
Monte Carlo Techniques
There are multiple ways to simulate Z . One way is to find a randomnumber U taken from a uniform distribution U[0, 1].
It follows that one can now map U → Z via inversion of the Nornal cdf N:
Z = N−1(U). (222)
It can be shown that in a sample, the standard deviation of the sampleaverage σsample is related to the standard deviation of an individual drawvia
σsample =σdraw√
n. (223)
If σdraw = σ, then we can see that we must increase our sample size by22k if we wish to cut our σsample by a factor of 2k .
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 203 / 203
Monte Carlo Techniques
There are multiple ways to simulate Z . One way is to find a randomnumber U taken from a uniform distribution U[0, 1].
It follows that one can now map U → Z via inversion of the Nornal cdf N:
Z = N−1(U). (222)
It can be shown that in a sample, the standard deviation of the sampleaverage σsample is related to the standard deviation of an individual drawvia
σsample =σdraw√
n. (223)
If σdraw = σ, then we can see that we must increase our sample size by22k if we wish to cut our σsample by a factor of 2k .
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 203 / 203
Monte Carlo Techniques
There are multiple ways to simulate Z . One way is to find a randomnumber U taken from a uniform distribution U[0, 1].
It follows that one can now map U → Z via inversion of the Nornal cdf N:
Z = N−1(U). (222)
It can be shown that in a sample, the standard deviation of the sampleaverage σsample is related to the standard deviation of an individual drawvia
σsample =σdraw√
n. (223)
If σdraw = σ, then we can see that we must increase our sample size by22k if we wish to cut our σsample by a factor of 2k .
Albert Cohen (MSU) Math 458: Financial Economics & Insurance MSU 2015 203 / 203