Zvi WienerContTimeFin - 2 slide 1 Financial Engineering Zvi Wiener [email protected] tel:...

Zvi Wiener ContTimeFin - 2 slide 1

Financial Engineering

Zvi [email protected]

tel: 02-588-3049


W - Wiener Process = Brownian Motion

dW ~ N(0, dt)

(dW(t))2 = dt

(dW(t)) dt = 0

dt2 = 0


W - Wiener Process = Brownian MotiondW ~ N(0, dt)

dX = dt + dW Arithmetical BM

dX = Xdt + XdW Geometrical BM

dX = (-X)dt + XdW Mean reverting

dX = (X,t)dt + (X,t)dW diffusion


Arithmetic BM dX = dt + dW

time

X


Geometric BM dX = Xdt + XdW

time

X


Mean Reverting Process

dX = (-X)dt + XdW

time

X


Ito’s lemma

If f = f(X) and dX = dt + dW, then

dWfdtffdf XXXX

2

2


Multivariate Ito’s Lemma

Introduce a second variable Y to the system that follows a diffusion

dX = (X,Y,t)dt + (X,Y,t)dW

dY = (X,Y,t)dt + (X,Y,t)dZ

where Z is another standard Wiener process. We define dZdW = dt as the correlation between the two processes.



It can be shown that

E[dZdW] = dt

(dZdW)2 = 0

Probabilistically dZ can be projected on dW

dZ = dW + (1- 2)1/2de

where de is a standard Wiener process uncorrelated with dW,

de dW = 0



The multivariate Ito’s Lemma:

f = f(X,Y,t)

df = fxdX + fydY + ftdt +

0.5(fxxdX2 + 2fxydXdY + fyydY2)


Multiplication Table

dW dZ dt

dW dt dt 0

dZ dt dt 0

dt 0 0 0

Note: terms of higher order (dt)a with a>1 we set to be zero, since we work with the first order terms only.


Multiplication Table

dX2 = (dt + dW)2 = 2dt

dY2 = (dt + dZ)2 = 2dt

dXdY = (dt + dW) (dt + dZ) = dt

22 5.05.0 dYfdXdYfdXfdYfdXfdf yyxyxxyx

dZfdWf

dtffffff

df

yx

yyxyxxtyx

22 5.05.0


Jump Processes

Diffusion processes are continuous. In order to include jumps we use Poisson processes.

Define q(t) such that q(0) = 0 and is constant until a Poisson event occurs. When there is a Poisson event value of q increases by 1.


Jump Processes

In its simplest form, a Poisson process with a constant intensity parameter is:

dq(t) = 1 with probability dt

= 0 with probability 1-dt

at every moment in time, where dq(t) is the instantaneous change in q in moment t.


Standard Poisson Processq

t


Standard Poisson Processjump[lam_, dt_]:=If[ Random[ ] < lam*dt, 1, 0];

tt = NestList[(# + jump[0.5, 0.1])&, 0, 300];

ListPlot[tt, PlotJoined->True];

50 100 150 200 250 300

5

10

15

20Jump Process


A Random Variable with Compact Support

A random variable has compact support if the domain over which the random variable has positive probability measure is a compact set.

Compact set - closed and bounded.

In any infinity sequence of points there is a convergent subsequence.


Diffusion with Jumps

dX = dt + dW + dq

This means that X has jumps by an amount whenever a Poisson event occurs (with intensity ).


Default Event

A default event can be modeled as a jump to zero value.

If X is the value of the security, then = -X can be interpreted as a default event - price drops to zero and remains there forever.

dX = dt + dW - Xdq


Default Event

dX = dt + dW - Xdq

mean continuous changes dt

continuous variance 2dt

occasional default (probability dt)

A general model of default allows a

spectrum of levels (see D. Duffie).


Jump Diffusion

For a real valued function f(X) the change in function value conditional on the occurrence of an event is f(X+)-f(X).

Therefore the expected change in function is:

dt E[f(X+)-f(X)] + (1- dt) [0] =

E[f(X+)-f(X)]dt


Residual Risk

In financial models, we often assume that residual risk is diversifiable. That is, no investor cares about this risk in the pricing of securities.

We also assume that the timing of the jumps and the level of X are independent of each other. However, we may allow to depend on X or follow its own stochastic process.


Ito’s Lemma with jumps

dqXfXfdtfdXfdXfdf

qtXff

txxx )()(5.0

),,(2


Financial Applications A

Suppose that a security with value V guarantees $1dt every instant of time forever. This is the continuous time equivalent of a risk-free perpetuity of $1. If the risk-free interest rate is constant r, what is the (discounted) value of the security?



1. V = V(t), there are NO stochastic variables.

dV = Vtdt

2. The expected capital gain on V is

ECG = E[dV] = Vtdt

3. The expected cash flows to V is ECF = 1 dt

4. The total return on V is

ECG + ECF = (Vt+1)dt



5. Since there is no risk, the total return must

be equal to the risk-free return on V, or rVdt.

(Vt+1) dt = r V dt

6. Divide both sides by dt:

Vt = rV - 1



Vt = rV - 1

DSolve[ V'[t]==r*V[t]-1, V[t], t ]

V(t) = c Exp[r t] + 1/r

given V(0) one can find c


Financial Applications B

Suppose that X follows a geometric Brownian motion with drift and volatility . A security with value V collects Xdt continuously forever. V represents a perpetuity that grows at an average exponential rate of , but whose risks in cash flow variations are considered diversificable. The economy is risk-neutral, and the risk-free interest rate is constant at r. What is the value of this security?


Financial Applications B1. V = V(X), since V is a perpetual claim, its price does not depend on time.

dV = VxdX + 0.5 VxxdX2,

dX = Xdt + XdW,

dX2= 2X2dt

dV = [XVx+0.5 2X2Vxx]dt +XVxdW



2. The expected capital gain:

ECG = E[dV] = [XVx+0.5 2X2Vxx]dt

since E[dW] = 0

3. The Expected cash flow:

ECF = X dt



4. Total return:

TR = ECG + ECF = [XVx+X+0.52X2Vxx]dt

5. But the return must be equal to the risk free return on the same investment V.

rVdt = [XVx+X+0.52X2Vxx]dt

6. Thus the PDE:

rV = XVx+X+0.52X2Vxx




there are several ways to solve it. One can guess that doubling X will double the price V.

If V is proportional to X, then V = X, Vx= , and Vxx=0, then the equation becomes

r X= X+X

= 1/(r- )

V(X) = X/(r- )


Financial Applications C

Modify example B to provide for a sudden possible drop to zero in the value of V. If a Poisson event occurs, one gives up V in exchange for nothing. The gain is zero and the loss is V, so the change in the value of V is 0-V, or -V. The possibility of this jump in any instant is dt.



1. V = V(X, q), there is no t because of the perpetual nature of V.

dV = VxdX + 0.5 VxxdX2 +[0-V]dq



2. Expected Capital Gain:

ECG = E[dV] = [XVx+0.5 2X2Vxx-V]dt

3. Expected Cash Flow:

ECF = Xdt



4. Total Return:

TR = ECG + ECF =

[XVx + 0.5 2X2Vxx - V + X]dt

5. Return on an alternative investment:

rVdt = [XVx + 0.5 2X2Vxx - V + X]dt



6. PDE:

rV = XVx + 0.5 2X2Vxx - V + X

This is the same equation as in B, but with (r+) instead of r.

Note that one can NOT solve it as a standard Cauchy problem because of a singularity at the origin (X=0).



V = X/(r + - )

We discount the cash flow at a higher rate

(r + ) to compensate for the probability of full default.

Alternatively we can see this as an adjustment of the growth rate to ( - ) and discount at the risk free rate. ( - ) is the certainty-equivalent growth rate.


Financial Applications D

Suppose X follows geometric Brownian

motion, and an independent Poisson process

determines the timing of cash payments equal

to the contemporaneous value of X. Let V

represent the claim to the first cash flow in this

stochastic perpetuity. What is the value of V?



1. V = V(X,q), since V is a perpetual claim, its price does not depend on time.

dV = VxdX + 0.5 VxxdX2 + [X-V]dt

Note: we give up the asset V to receive the payment X.



2. The expected capital gain:

ECG = E[dV] = [XVx+0.5 2X2Vxx+ (X-V)]dt

3. The Expected cash flow:

ECF = 0

There are no continuous cash payments.



4. Total return

TR = ECG + ECF =

[XVx+0.5 2X2Vxx+ X - V]dt

5. Return on an alternative investment:

rVdt = [XVx+0.5 2X2Vxx+ X - V]dt



6. PDE

rV = XVx+0.5 2X2Vxx+ X - V

In Example B we had:




(D) rV = XVx+0.5 2X2Vxx+ X - V

(B) rV = XVx+X+0.52X2Vxx

This is the same equation as in Example B, except for two substitutions:

1. (r + ) instead of r

2. The value of V is multiplied by (the cash flow term is X instead of X).



(D) rV = XVx+0.5 2X2Vxx+ X - V

V = X/(r + - )

Check it!


Conclusions

We have studied Present Value (PV) calculations in continuous time settings.

We have received ODE, since all our models were perpetual (no explicit time dependence).


Conclusions

In a risk-neutral economy:

1. Calculate the expected capital gain on an asset from Ito’s lemma.

2. Add the expected cash flows to get the total return.

3. Set the total return equal to the risk-free return.

4. Solve the appropriate DE.


Exercise 1.1

Assume X follows geometric BM with drift and volatility . Let Y = ln(X).

a. What process does Y follow?

b. What is the distribution of Yu, given Yt (t<u)?

c. What is the expected value of Xu, given Xt?

Hint: if z ~N(, 2), then E[ez] = exp[ + 0.5 2]


Exercise 1.1 - Solution

dX = Xdt + XdW Y = ln(X)

dY = (ln(X))’dX + 0.5 (ln(X))”(dX)2 =

(Xdt + XdW)/X + 0.5 (-1/X2) 2X2dt =

( - 0.5 2)dt + dW - arithmetic BM

Yu ~ N(Yt + ( - 0.5 2)(u-t), 2(u-t))

E[eYu] = Xte(u-t)


Exercise 1.2

Assume X follows arithmetic BM with drift and volatility . A security V pays Xdt forever. If X becomes negative, the holder of the asset must make payments to the security issuer. The economy is risk-neutral, and the risk-free discount rate is r.


Exercise 1.2

a. What is the value of V? (Hint: V is linear in X)

b. Suppose the security holder has the right to

abandon the asset if cash flows become

sufficiently negative, i.e., when X=q (q<0). What

is the value of V?

Hint: V=k1exp[k2(X-q)] + k3X + k4. Note that

k2<0. Also, when X=q, V(X)=0. Check that the

ODE is satisfied.


Exercise 1.2

c. If q can be chosen optimally, what is the value-maximizing choice? Verify the second order conditions.

d. What is the value of the abandonment optimal?

Hint: look at parts a and b.



dX = dt + dW

ECG = E[dV] = Vxdt + 0.52Vxxdt + VxE[dW]

ECF = Xdt, if X > q, otherwise 0.

TR = (Vx + 0.52Vxx + X IX>q)

Vx + 0.52Vxx + X IX>q = rV


Exercise 1.2 - SolutionVx + 0.52Vxx + X IX>q = rV

a. V(X) = x/r + /r2

b. k1= - q/r - /r2, k3= 1/r, k4= /r2

the sign is minus, since k2 must be negative.

2

22

2

2

r

k

V=k1exp[k2(X-q)] + k3X + k4



Vx + 0.52Vxx + X IX>q = rV

2

1*0

rkq

q

V

d. Value of the abandonment option is the difference between values with and without the option, when q is chosen optimally.

*)(

2

21

' qXkerk

valuesoption


Exercise 1.3

Assume that X follows geometric BM, with drift and volatility . The economy is risk-neutral, and the risk-free discount rate is r. A machine prints a certificate worth X(t) at random times t generated by a Poisson arrival process with intensity .

a. What is the value of this machine?

b. What is the value of a contingent claim to the

first certificate printed by the machine?


Exercise 1.3

c. Assume Y follows geometric BM with drift and volatility . The correlation between X and Y is 0. What is the value of a certificate produced by X, if it lets its bearer (only) have X certificates printed by machine Y (worth Y at the time of printing)? Y prints at the same average rate, and the number of certificates is determined by the first arrival time.


Exercise 1.3 - Solution a.

dX = Xdt + XdW

ECG = E[dV] = XVxdt + 0.52X2Vxxdt

ECF = Xdt

TR = (XVx + 0.5X22Vxx + X)dt

XVx + 0.5X22Vxx + X = rV

V(X)= X/(r - )


Exercise 1.3 - Solution b.

dX = Xdt + XdW


ECF = (X-V)dt

TR = (XVx + 0.5X22Vxx + (X-V))dt

XVx + 0.5X22Vxx + (X-V) = rV

V(X)= X/(r + - )


Exercise 1.4

A low-risk health insurance policy holder realizes medical losses at random times according to a Poisson arrival process. The level of the loss is given by X, a process which follows GBM with drift and volatility . The economy is risk-neutral, and the risk-free discount rate is r. Medical expenses occur at a rate dt. There is an additional possibility that the claimant will suddenly become a high-risk claimant.


Exercise 1.4

High-risk claimants experience the same possible losses X, but at a higher frequency dt. The timing of the switch from a low-risk to a high-risk is governed by a Poisson process with intensity parameter .

a. If the policy holder stays low-risk forever, what is the value of the policy today?

b. What is the value of a high-risk policy today?

c. What is the value of a low-risk policy today?



dX = Xdt + XdW


ECF = Xdt



V(X)= X/(r - )



dX = Xdt + XdW


ECF = Xdt



V(X)= X/(r - )


Exercise 1.4 - Solution c.dX = Xdt + XdW


ECF = Xdt + (X/(r - )-V)dt

TR = XVx + 0.5X22Vxx + X + (X/(r - )-V)

XVx + 0.5X22Vxx + X + (X/(r - )-V) = rV

rr

XXV )(


Exercise 1.5

Assume that the value of an index X follows GBM with drift and volatility . An asset V promises that, when X reaches Q, the bearer will be paid R and the asset will be retired. The economy is risk-neutral, and the risk-free rate is r.

a. What is the value of the asset (Hint: V=AX).

b. What are sufficient conditions for > 0?



dX = Xdt + XdW


ECF = 0

TR = (XVx + 0.5X22Vxx)dt

XVx + 0.5X22Vxx = rV

V(Q)=R



V

Q X

R


Exercise 1.5 - SolutionXVx + 0.5X22Vxx = rV

AQ = R

V(X) = AX

AX + 0.52 (-1)AX = rAX

+ 0.52 (-1) = r

0.52 2 + ( - 0.52) - r = 0

2

2222 25.05.0

r

Q

XRXV )(


Exercise 1.6

Assume that the value of an index X follows GBM with drift and volatility . A perpetual call option is written such that when it is exercised (at X=Q), the holder receives Q-E (E - is the exercise price). The economy is risk-neutral, and the risk-free rate is r.

a. What is the value of the option, assuming it is exercised when X = Q? (Hint: V=AX, do not forget the boundary condition V(Q)=Q-E)


Exercise 1.6

b. Assuming that the holder of the call option will act to maximize the current value of the option, what Q will he choose?

c. What are sufficient conditions for > 0?

d. Verify that the value satisfies the DE you derived.

e. What are the comparative static properties of the model?



dX = Xdt + XdW


ECF = 0

TR = (XVx + 0.5X22Vxx)dt

XVx + 0.5X22Vxx = rV

V(Q)=Q-E



V(X) = (Q-E)(X/Q)

Optimal exercise is such that maximizes V(X):

1)1( QEQXQ

V

1*

E

Q


Exercise 1.6 - Solution c, e.

V(X) = (Q-E)(X/Q)

Sufficient condition for > 1, is > 0 and r > .

V increases in X

V decreases in E

V increases in r and

V depends on through


Exercise 1.7

Assume X follows GBM with drift and volatility . Assume Y follows GBM with drift and volatility . The correlation between the Wiener components of the two processes is dZxdZy=dt.

a. Write down the laws of motion of the system.

b. Let V = XY. What process does V follow? Define a new process (v, v and dZv), so that dV/V= vdt + vdZv.


Exercise 1.7

c. What are the correlations of dV with dX, dY?

d. Let W = X/Y. What process does W follow? Organize your results as in b.

e. What are the correlations of dW with dX, dY?

f. Run a theoretical regression of dY/Y on dX/X. What are your coefficients? What is the standard error of the regression? What are the time series properties of the volatility of the projection (i.e., the error term)? What is the theoretical R2?


Exercise 1.8

A security with value V pays ydt continuously until x reaches the point q. y follows ABM with drift and volatility , and x follows ABM with drift and volatility w. The correlation between the Wiener components of the two processes is dZxdZy=dt

a. What DE must V satisfy?

b. What are the boundary conditions?

c. Value the asset.


Exercise 1.8 - SolutionV(X,Y), dWdZ=dt

dY = dt + dW dX = dt + wdZ

ECG = (Vy+0.52Vyy+Vx+0.5w2Vxx+Vxy) dt

ECF = Ydt, if X > q, otherwise 0.

TR = ECG + ECF = alternative return (rV)

Vy+0.52Vyy+Vx+0.5w2Vxx+Vxy + Y= rV

V(q,Y)=0


Exercise 1.9

A firm earns Xdt continuously, where X follows ABM with drift and volatility . This is the only asset of the firm. If X becomes negative, then the firm must decide whether to honor its obligations or abandon its operations. We assume it is optimal to abandon its operations. We assume it is optimal to abandon operations when earnings fall below a constant level q. The firm wishes to sell contingent claims against its earnings.


Exercise 1.9

To value an arbitrary contingent claim, we first value four primitive contingent claims with the following cash flows:

g1(X) = 1

g2(X) = X

g3(X) = I{X>c}

g4(X) = X I{X>c}

Here I is the indicator function. g1 receives $1dt until X=q, then he receives nothing.


Exercise 1.9

g2 receives $Xdt until X=q; if X < 0, the cash is paid instead of received.

g3 receives $1dt if X is above c, and is worthless when X reaches q.

g4 receives $Xdt if X is above c, and is worthless when X reaches q.


Exercise 1.9

Let Yi(X) represent the value of a claim giving rise to cash flows of gi(X). We explicitly allow for the optimal abandonment of cash flows; Yi must satisfy the boundary condition Yi(q) = 0 when X reaches the abandonment point q.

We also require YiX< for all X>q.


Exercise 1.9

a. What is the value of each of the Yi(X)?

Hint 1: V = A1exp(k1X) + A2exp(k2X) + A3X + A4

Hint 2: Assume different forms for X<c and X>c.

Hint 3: The solution must be continuous and differentiable at c.


Exercise 1.9b. The contingent claim holders are stock holders, bond holders, government, third parties. The marginal tax rate is . The distribution of earnings occurs instantaneously and is as follows:

Case Debt Equity Gov. 3rd P.

c<X c (1- )(X-c) (X-c) 0

q<X<c X-k 0 0 k

X<q 0 0 0 0

Value each of the claims using the primitive derivative claims against the earnings of the firm.


Exercise 1.9

c. Find the operating and capital structure policy

(levels of q and c) that maximizes the sum of

debt and equity values. Write down the first

order conditions for an interior maximum only.



V(X) X = dt + dW

ECG = Vxdt + 0.52Vxxdt

ECF = g(X)dt

TR = ECG + ECF = Vx + 0.52Vxx + g(X)

Vx+ 0.52Vxx+ g(X) = rV

V(q) = 0



Debt: D = c Y3 + Y2 - kY1 - Y4 + kY3

Equity: E = (1- )(Y4 - cY3)

Government: G = (Y4 - c Y3)

Others: T = k(Y1 - Y3)

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Zvi WienerContTimeFin - 2 slide 1 Financial Engineering Zvi Wiener [email protected] tel:...

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