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The Greeks: Derivatives of Option Prices MATH 472 Financial Mathematics J. Robert Buchanan 2018
The Greeks: Derivatives of Option PricesMATH 472 Financial Mathematics
J. Robert Buchanan
2018
Sensitivity Analysis
I In calculus a derivative gives you a measure of the rate ofchange of a dependent variable as an independentvariable is changed.
I In the world a finance an option is an example of aderivative, any financial instrument whose value is derivedfrom that of an underlying security.
I Here we will calculate the partial derivatives (in the senseof calculus) of option value formulas. These partialderivatives will allow us to determine how sensitive thevalues of options are to changes in independent variablesand parameters.
I In finance these partial derivatives are referred to as “theGreeks”.
I Unless otherwise specified, all results discussed are validonly for non-dividend paying securities.
Black-Scholes Option Pricing Formulas
w =ln(S/K ) + (r + σ2/2)(T − t)
σ√
T − t
Ce(S, t) = S Φ (w)− K e−r(T−t)Φ(
w − σ√
T − t)
Pe(S, t) = K e−r(T−t)Φ(σ√
T − t − w)− S Φ (−w)
Cumulative Distribution Function
The function Φ (w) is the cumulative distribution function
Φ (w) =1√2π
∫ w
−∞e−x2/2 dx
which by the Fundamental Theorem of Calculus has derivative
Φ′ (w) = φ (w) =1√2π
e−w2/2.
Partial Derivatives of w
w =ln(S/K ) + (r + σ2/2)(T − t)
σ√
T − t∂w∂t
=1
2σ√
T − t
(ln(S/K )
T − t− r − σ2
2
)∂w∂S
=1
σS√
T − t∂w∂r
=
√T − tσ
∂w∂σ
=√
T − t − wσ
Important Identity
Claim:
S φ (w) = Ke−r(T−t) φ(
w − σ√
T − t)
Theta Θ (1 of 3)
Theta Θ is the partial derivative with respect to time t .
Time is the only independent variable we are certain willchange before expiry. It is also the only deterministicindependent variable.
Theta Θ (2 of 3)
Ce = SΦ (w)− Ke−r(T−t)Φ(
w − σ√
T − t)
∂Ce
∂t= Sφ (w)
∂w∂t− rKe−r(T−t)Φ
(w − σ
√T − t
)− Ke−r(T−t)φ
(w − σ
√T − t
)[∂w∂t
+σ
2√
T − t
]
= Sφ (w)∂w∂t− rKe−r(T−t)Φ
(w − σ
√T − t
)− Sφ (w)
[∂w∂t
+σ
2√
T − t
]
= − σS2√
T − tφ (w)− r K e−r(T−t)Φ
(w − σ
√T − t
)< 0
Theta Θ (2 of 3)
Ce = SΦ (w)− Ke−r(T−t)Φ(
w − σ√
T − t)
∂Ce
∂t= Sφ (w)
∂w∂t− rKe−r(T−t)Φ
(w − σ
√T − t
)− Ke−r(T−t)φ
(w − σ
√T − t
)[∂w∂t
+σ
2√
T − t
]= Sφ (w)
∂w∂t− rKe−r(T−t)Φ
(w − σ
√T − t
)− Sφ (w)
[∂w∂t
+σ
2√
T − t
]
= − σS2√
T − tφ (w)− r K e−r(T−t)Φ
(w − σ
√T − t
)< 0
Theta Θ (2 of 3)
Ce = SΦ (w)− Ke−r(T−t)Φ(
w − σ√
T − t)
∂Ce
∂t= Sφ (w)
∂w∂t− rKe−r(T−t)Φ
(w − σ
√T − t
)− Ke−r(T−t)φ
(w − σ
√T − t
)[∂w∂t
+σ
2√
T − t
]= Sφ (w)
∂w∂t− rKe−r(T−t)Φ
(w − σ
√T − t
)− Sφ (w)
[∂w∂t
+σ
2√
T − t
]= − σS
2√
T − tφ (w)− r K e−r(T−t)Φ
(w − σ
√T − t
)< 0
Illustration
The value of a European Call decreases as expiry approaches(all other variables and parameters being constant).
Tt
Ce
Θ
Theta Θ (3 of 3)
For a European Put:
∂Pe
∂t= r K e−r(T−t)Φ
(σ√
T − t − w)
+ S φ (−w)∂w∂t
− K e−r(T−t)φ(σ√
T − t − w)[ σ
2√
T − t+∂w∂t
]= r K e−r(T−t)Φ
(σ√
T − t − w)− σS
2√
T − tφ (w)
Illustration
The value of a European Put decreases as expiry approaches(all other variables and parameters being constant).
Tt
Pe
Θ
Delta ∆ (1 of 2)
∆ was involved in the derivation of the Black-Scholes PDE andis defined to be the partial derivative with respect to the price ofthe security.
Ce = S Φ (w)− K e−r(T−t)Φ(
w − σ√
T − t)
∂Ce
∂S= Φ (w) + S φ (w)
∂w∂S− K e−r(T−t)φ
(w − σ
√T − t
)∂w∂S
= Φ (w) > 0
Question: what is the range of Delta for a European calloption?
Illustration
Consider a European Call with K = 50, T = 1, r = 0.10, andσ = 0.50.
20 40 60 80 100
0.
20.
40.
60.
0.
0.32
0.64
0.97
S
Ce Δ
Delta ∆ (2 of 2)
Recall the Put-Call Parity formula:
Pe + S = Ce + K e−r(T−t)
∂
∂S[Pe + S] =
∂
∂S[Ce + K e−r(T−t)]
∂Pe
∂S+ 1 =
∂Ce
∂S∂Pe
∂S= Φ (w)− 1 < 0
Question: what is the range of Delta for a European putoption?
Illustration
Consider a European Put with K = 50, T = 1, r = 0.10, andσ = 0.50.
20 40 60 80 100
0.
10.
20.
30.
40.
-1.
-0.75
-0.5
-0.25
-0.006
S
Pe Δ
Example (1 of 2)
The current price of a non-dividend-paying stock is $77 and itsvolatility is 35% per year. The risk-free interest rate is 3.25%per year. A portfolio is constructed consisting of one six-monthEuropean call option with a strike price of $80 and the cashobtained from shorting ∆ shares of the stock. The portfolio’svalue is non-random. What is ∆?
Example (2 of 2)The assumption the portfolio’s value is non-random is theassumption
(∆)S − Ce =
(∂Ce
∂S
)S − Ce = 0
made in deriving the Black-Scholes equation.
S = 77 σ = 0.35 T =6
12r = 0.0325 K = 80 t = 0
Using these values
w =ln(S/K ) + (r + σ2/2)(T − t)
σ√
T − t
≈ 0.0349666
∂Ce
∂S= ∆ = Φ (w)
≈ 0.513947
Example (2 of 2)The assumption the portfolio’s value is non-random is theassumption
(∆)S − Ce =
(∂Ce
∂S
)S − Ce = 0
made in deriving the Black-Scholes equation.
S = 77 σ = 0.35 T =6
12r = 0.0325 K = 80 t = 0
Using these values
w =ln(S/K ) + (r + σ2/2)(T − t)
σ√
T − t≈ 0.0349666
∂Ce
∂S= ∆ = Φ (w) ≈ 0.513947
Example (1 of 2)
Suppose a portfolio consists of a share of stock worth $75 anda European Put option on that stock with a strike price of $73and expiry in 3 months. Assume the risk-free interest rate is10% and the volatility of the stock price is 30%.
Find the Delta of the portfolio consisting of the stock and theoption.
Example (2 of 2)
The Delta of the portfolio is
∂
∂S[S + Pe] = 1 + Φ (w)− 1 = Φ (w)
calculated using the variables and parameters below.
S = 75 σ = 0.30 T = 3/12r = 0.10 K = 73 t = 0
w =ln(S/K ) + (r + σ2/2)(T − t)
σ√
T − t
≈ 0.421858
Φ (w) ≈ 0.663436
Example (2 of 2)
The Delta of the portfolio is
∂
∂S[S + Pe] = 1 + Φ (w)− 1 = Φ (w)
calculated using the variables and parameters below.
S = 75 σ = 0.30 T = 3/12r = 0.10 K = 73 t = 0
w =ln(S/K ) + (r + σ2/2)(T − t)
σ√
T − t≈ 0.421858
Φ (w) ≈ 0.663436
Gamma Γ
Gamma is the second partial derivative with respect to S, thus
Γ =∂
∂S[Φ (w)]
= φ (w)∂w∂S
∂2Ce
∂S2 =∂2Pe
∂S2 =e−w2/2
σS√
2π(T − t)> 0.
Gamma vs. S0
50 100 150 200S0
0.005
0.010
0.015
Γ
K = 100, σ = 0.25 T = 1 r = 0.0325
Gamma and Delta
For options far in-the-money or out-of-the-money, there is littlechange in ∆ and thus Γ is nearly zero.
50 100 150
0.
0.0025
0.005
0.0075
0.01
0.
0.21
0.42
0.63
0.85
S
Γ Δ
Gamma and At-the-Money Options
Consider an at-the-money option (S = K ), how does Gammabehave as expiry approaches?
0.0 0.2 0.4 0.6 0.8 1.0T
0.02
0.04
0.06
0.08
0.10Γ
S0<K
S0=K
S0>K
Solution
When S = K ,
limt→T−
Γ = limt→T−
e−((r+σ2/2)(T−t))2/(2σ2(T−t))
σK√
2π(T − t)
= limt→T−
e−(r+σ2/2)2(T−t)/(2σ2)
σK√
2π(T − t)= ∞.
Gamma vs. K and T
60 80 100 120 1400.0
0.2
0.4
0.6
0.8
1.0
K
T
S0 = 100, σ = 0.25 r = 0.0325
Relationships Between ∆, Θ, and Γ
Remember the Black-Scholes PDE:
r F = Ft + r S FS +12σ2S2FSS
Since Ft = Θ, ∆ = FS, and FSS = Γ then the Black-Scholesequation can be thought of as
r F = Θ + r S ∆ +12σ2S2Γ.
Changes in Option Values
Using differentials we can approximate changes in the prices ofoptions as underlying variables and parameters change.
Let F (S, t) be the value of an option at time t when the value ofthe underlying security is S, then we have the followingapproximations.
F (S, t + δt) ≈ F (S, t) + (Θ)δtF (S + δS, t) ≈ F (S, t) + (∆)δS
F (S + δS, t) ≈ F (S, t) + (∆)δS +12
(Γ)(δS)2
F (S + δS, t + δt) ≈ F (S, t) + (Θ)δt + (∆)δS +12
(Γ)(δS)2
Example (1 of 3)
A six-month call option with a strike price of $100 on a stockcurrently valued at $99 and having a volatility of σ = 0.40 costs$12.4911. The risk-free interest rate is r = 0.08. Estimate thevalue of the option at five months to expiry.
Ce(S, t + δt) ≈ Ce(S, t) + (Θ)δtCe(S, δt) ≈ Ce(S,0) + (Θ)δt
Ce(99,1/12) ≈ Ce(99,0) + (Θ)(1/12)
= 12.4911 +−14.5686
12= 11.2771
For comparison, the exact value is Ce(99,1/12) = $11.2322.
Example (1 of 3)
A six-month call option with a strike price of $100 on a stockcurrently valued at $99 and having a volatility of σ = 0.40 costs$12.4911. The risk-free interest rate is r = 0.08. Estimate thevalue of the option at five months to expiry.
Ce(S, t + δt) ≈ Ce(S, t) + (Θ)δtCe(S, δt) ≈ Ce(S,0) + (Θ)δt
Ce(99,1/12) ≈ Ce(99,0) + (Θ)(1/12)
= 12.4911 +−14.5686
12= 11.2771
For comparison, the exact value is Ce(99,1/12) = $11.2322.
Example (1 of 3)
A six-month call option with a strike price of $100 on a stockcurrently valued at $99 and having a volatility of σ = 0.40 costs$12.4911. The risk-free interest rate is r = 0.08. Estimate thevalue of the option at five months to expiry.
Ce(S, t + δt) ≈ Ce(S, t) + (Θ)δtCe(S, δt) ≈ Ce(S,0) + (Θ)δt
Ce(99,1/12) ≈ Ce(99,0) + (Θ)(1/12)
= 12.4911 +−14.5686
12= 11.2771
For comparison, the exact value is Ce(99,1/12) = $11.2322.
Example (2 of 3)
A six-month call option with a strike price of $100 on a stockcurrently valued at $99 and having a volatility of σ = 0.40 costs$12.4911. The risk-free interest rate is r = 0.08. Estimate thevalue of the option using Delta if the value of the stockincreases to $101.
Ce(S + δS, t) ≈ Ce(S, t) + (∆)δSCe(101,0) ≈ Ce(99,0) + (∆)(2)
= 12.4911 + (0.597666)(2)
= 13.6865
For comparison, the exact value is Ce(101,0) = $13.7137.
Example (2 of 3)
A six-month call option with a strike price of $100 on a stockcurrently valued at $99 and having a volatility of σ = 0.40 costs$12.4911. The risk-free interest rate is r = 0.08. Estimate thevalue of the option using Delta if the value of the stockincreases to $101.
Ce(S + δS, t) ≈ Ce(S, t) + (∆)δSCe(101,0) ≈ Ce(99,0) + (∆)(2)
= 12.4911 + (0.597666)(2)
= 13.6865
For comparison, the exact value is Ce(101,0) = $13.7137.
Example (2 of 3)
A six-month call option with a strike price of $100 on a stockcurrently valued at $99 and having a volatility of σ = 0.40 costs$12.4911. The risk-free interest rate is r = 0.08. Estimate thevalue of the option using Delta if the value of the stockincreases to $101.
Ce(S + δS, t) ≈ Ce(S, t) + (∆)δSCe(101,0) ≈ Ce(99,0) + (∆)(2)
= 12.4911 + (0.597666)(2)
= 13.6865
For comparison, the exact value is Ce(101,0) = $13.7137.
Example (3 of 3)
A six-month call option with a strike price of $100 on a stockcurrently valued at $99 and having a volatility of σ = 0.40 costs$12.4911. The risk-free interest rate is r = 0.08. Estimate thevalue of the option using Delta and Gamma if the value of thestock increases to $101.
Ce(S + δS, t) ≈ Ce(S, t) + (∆)δS +12
(Γ)(δS)2
Ce(101,0) ≈ Ce(99,0) + (∆)(2) +12
(Γ)(4)
= 12.4911 + (0.597666)(2) +12
(0.0138181)(4)
= 13.7141
For comparison, the exact value is Ce(101,0) = $13.7137.
Example (3 of 3)
A six-month call option with a strike price of $100 on a stockcurrently valued at $99 and having a volatility of σ = 0.40 costs$12.4911. The risk-free interest rate is r = 0.08. Estimate thevalue of the option using Delta and Gamma if the value of thestock increases to $101.
Ce(S + δS, t) ≈ Ce(S, t) + (∆)δS +12
(Γ)(δS)2
Ce(101,0) ≈ Ce(99,0) + (∆)(2) +12
(Γ)(4)
= 12.4911 + (0.597666)(2) +12
(0.0138181)(4)
= 13.7141
For comparison, the exact value is Ce(101,0) = $13.7137.
Example (3 of 3)
A six-month call option with a strike price of $100 on a stockcurrently valued at $99 and having a volatility of σ = 0.40 costs$12.4911. The risk-free interest rate is r = 0.08. Estimate thevalue of the option using Delta and Gamma if the value of thestock increases to $101.
Ce(S + δS, t) ≈ Ce(S, t) + (∆)δS +12
(Γ)(δS)2
Ce(101,0) ≈ Ce(99,0) + (∆)(2) +12
(Γ)(4)
= 12.4911 + (0.597666)(2) +12
(0.0138181)(4)
= 13.7141
For comparison, the exact value is Ce(101,0) = $13.7137.
Vega V (1 of 2)
Vega is the partial derivative with respect to volatility σ.
Ce = S Φ (w)− K e−r(T−t)Φ(
w − σ√
T − t)
∂Ce
∂σ= S φ (w)
∂w∂σ− K e−r(T−t)φ
(w − σ
√T − t
)[∂w∂σ−√
T − t]
= S φ (w)∂w∂σ− S φ (w)
[∂w∂σ−√
T − t]
=S√
T − t√2π
e−w2/2 > 0
Vega V (2 of 2)
According to the Put-Call Parity formula:
Pe + S = Ce + K e−r(T−t)
∂
∂σ[Pe + S] =
∂
∂σ
[Ce + K e−r(T−t)
]∂Pe
∂σ=
∂Ce
∂σ
∂Pe
∂σ=
S√
T − t√2π
e−w2/2
Remark: vega is identical for puts and calls.
Example (1 of 2)
Consider a three-month European put option on anon-dividend-paying stock whose current value is $200 andwhose volatility is 30%. The option has a strike price of $195and the risk-free interest rate is 6.25%.
1. Find the vega of the option.2. If the volatility of the stock increases to 31%, approximate
the change in the value of the put.
Example (2 of 2)
S = 200 σ = 0.30 T = 3/12K = 195 r = 0.0625 t = 0
Using these values
w =ln(S/K ) + (r + σ2/2)(T − t)
σ√
T − t
≈ 0.347952
V =S√
T − t√2π
e−w2/2
≈ 37.5509
Using the linear approximation,
dP = V dσ = (37.5509)(0.01) = 0.375509.
Example (2 of 2)
S = 200 σ = 0.30 T = 3/12K = 195 r = 0.0625 t = 0
Using these values
w =ln(S/K ) + (r + σ2/2)(T − t)
σ√
T − t≈ 0.347952
V =S√
T − t√2π
e−w2/2 ≈ 37.5509
Using the linear approximation,
dP = V dσ = (37.5509)(0.01) = 0.375509.
Example (2 of 2)
S = 200 σ = 0.30 T = 3/12K = 195 r = 0.0625 t = 0
Using these values
w =ln(S/K ) + (r + σ2/2)(T − t)
σ√
T − t≈ 0.347952
V =S√
T − t√2π
e−w2/2 ≈ 37.5509
Using the linear approximation,
dP = V dσ = (37.5509)(0.01) = 0.375509.
Rho ρ (1 of 2)
Rho is the partial derivative with respect to the risk-free interestrate r .
Ce = S Φ (w)− K e−r(T−t)Φ(
w − σ√
T − t)
∂Ce
∂r= S φ (w)
∂w∂r
+ K (T − t)e−r(T−t)Φ(
w − σ√
T − t)
− K e−r(T−t)φ(
w − σ√
T − t)∂w∂r
= K (T − t)e−r(T−t)Φ(
w − σ√
T − t)> 0
Rho ρ (2 of 2)
Starting with the Put-Call Parity formula:
Pe + S = Ce + K e−r(T−t)
∂
∂r[Pe + S] =
∂
∂r
[Ce + K e−r(T−t)
]∂Pe
∂r=
∂Ce
∂r− K (T − t)e−r(T−t)
∂Pe
∂r= −K (T − t)e−r(T−t)Φ
(σ√
T − t − w)< 0
Example (1 of 2)
Consider a three-month European put option on a stock whosecurrent value is $200 and whose volatility is 30%. The optionhas a strike price of $195 and the risk-free interest rate is6.25%.
1. Find the rho of the option.2. If the interest rate increases to 7.00%, approximate the
change in the value of the put.
Example (2 of 2)
S = 200 σ = 0.30 T = 3/12K = 195 r = 0.0625 t = 0
Using these values
w =ln(S/K ) + (r + σ2/2)(T − t)
σ√
T − t
≈ 0.347952
ρ = −K (T − t)e−r(T−t)Φ(σ√
T − t − w)
≈ −20.2315
Using the linear approximation,
dP = ρdr = (−20.2315)(0.0075) = −0.151737.
Example (2 of 2)
S = 200 σ = 0.30 T = 3/12K = 195 r = 0.0625 t = 0
Using these values
w =ln(S/K ) + (r + σ2/2)(T − t)
σ√
T − t≈ 0.347952
ρ = −K (T − t)e−r(T−t)Φ(σ√
T − t − w)≈ −20.2315
Using the linear approximation,
dP = ρdr = (−20.2315)(0.0075) = −0.151737.
Example (2 of 2)
S = 200 σ = 0.30 T = 3/12K = 195 r = 0.0625 t = 0
Using these values
w =ln(S/K ) + (r + σ2/2)(T − t)
σ√
T − t≈ 0.347952
ρ = −K (T − t)e−r(T−t)Φ(σ√
T − t − w)≈ −20.2315
Using the linear approximation,
dP = ρdr = (−20.2315)(0.0075) = −0.151737.
Homework
I Read Chapter 9.I Exercises: 1–11.
Credits
These slides are adapted from the textbook,An Undergraduate Introduction to Financial Mathematics,3rd edition, (2012).author: J. Robert Buchananpublisher: World Scientific Publishing Co. Pte. Ltd.address: 27 Warren St., Suite 401–402, Hackensack, NJ07601ISBN: 978-9814407441
