Transcript No Slide Title

Financial Engineering
Zvi Wiener
[email protected]
tel: 02-588-3049
Zvi Wiener
ContTimeFin - 2
slide 1
W - Wiener Process = Brownian Motion
dW ~ N(0, dt)
(dW(t))2 = dt
(dW(t)) dt = 0
dt2 = 0
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W - Wiener Process = Brownian Motion
dW ~ N(0, dt)
dX = dt + dW
Arithmetical BM
dX = Xdt + XdW
Geometrical BM

dX = (-X)dt + X dW Mean reverting
dX = (X,t)dt + (X,t)dW
diffusion
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slide 3
Arithmetic BM
dX =  dt +  dW
X


time
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Geometric BM
dX = Xdt + XdW
X
time
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slide 5
Mean Reverting Process

dX = (-X)dt + X dW
X

time
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slide 6
Ito’s lemma
If f = f(X) and dX = dt + dW, then



df   f X 
f XX dt   f X dW
2


2
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slide 7
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.
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Multivariate Ito’s Lemma
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
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Multivariate Ito’s Lemma
The multivariate Ito’s Lemma:
f = f(X,Y,t)
df = fxdX + fydY + ftdt +
0.5(fxxdX2 + 2fxydXdY + fyydY2)
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slide 10
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.
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slide 11
Multiplication Table
dX2 = (dt + dW)2 = 2dt
dY2 = (dt + dZ)2 = 2dt
dXdY = (dt + dW) (dt + dZ) =    dt
df  f x dX  f y dY  0.5 f xxdX 2  f xydXdY  0.5 f yydY 2
df 
f
x

 f y  f t  0.5 2 f xx   f xy  0.5 2 f yy dt
 f x dW  f y dZ
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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.
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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.
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q
Standard Poisson Process
t
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Standard Poisson Process
jump[lam_, dt_]:=If[ Random[ ] < lam*dt, 1, 0];
tt = NestList[(# + jump[0.5, 0.1])&, 0, 300];
ListPlot[tt, PlotJoined->True];
Jump
Process
20
15
10
5
50
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150
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200
250
300
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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.
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Diffusion with Jumps
dX = dt + dW + dq
This means that X has jumps by an amount 
whenever a Poisson event occurs (with
intensity ).
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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
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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).
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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
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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.
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Ito’s Lemma with jumps
f  f ( X , t , q)
df  f x dX  0.5 f xxdX  f t dt   f ( X   )  f ( X )dq
2
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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?
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Financial Applications A
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
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Financial Applications A
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
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Financial Applications A
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
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slide 27
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?
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Financial Applications B
1. 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
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Financial Applications B
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
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Financial Applications B
4. Total return:
TR = ECG + ECF =
[XVx+X+0.52X2Vxx]dt
5. But the return must be equal to the risk free
return on the same investment V.
rVdt = [XVx+X+0.52X2Vxx]dt
6. Thus the PDE:
rV = XVx+X+0.52X2Vxx
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Financial Applications B
rV = XVx+X+0.52X2Vxx
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- )
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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.
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Financial Applications C
1. V = V(X, q), there is no t because of the
perpetual nature of V.
dV = VxdX + 0.5 VxxdX2 +[0-V]dq
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Financial Applications C
2. Expected Capital Gain:
ECG = E[dV] = [XVx+0.5 2X2Vxx-V]dt
3. Expected Cash Flow:
ECF = Xdt
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slide 35
Financial Applications C
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
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Financial Applications C
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).
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slide 37
Financial Applications C
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.
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slide 38
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?
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Financial Applications D
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.
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slide 40
Financial Applications D
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.
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slide 41
Financial Applications D
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
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slide 42
Financial Applications D
6. PDE
rV = XVx+0.5 2X2Vxx+ X - V
In Example B we had:
rV = XVx+X+0.52X2Vxx
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slide 43
Financial Applications D
(D)
rV = XVx+0.5 2X2Vxx+ X - V
(B)
rV = XVx+X+0.52X2Vxx
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).
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slide 44
Financial Applications D
(D)
rV = XVx+0.5 2X2Vxx+ X - V
V = X/(r +  - )
Check it!
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slide 45
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).
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slide 46
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.
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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]
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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)
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slide 49
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.
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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.
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Exercise 1.2
c. If q can be chosen optimally, what is the valuemaximizing choice? Verify the second order
conditions.
d. What is the value of the abandonment optimal?
Hint: look at parts a and b.
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Exercise 1.2 - Solution
dX = dt + dW
ECG = E[dV] = Vxdt + 0.52Vxxdt + VxE[dW]
ECF = Xdt,
if X > q,
otherwise 0.
TR = (Vx + 0.52Vxx + X IX>q)
Vx + 0.52Vxx + X IX>q = rV
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Exercise 1.2 - Solution
Vx + 0.52Vxx + X IX>q = rV
a. V(X) = x/r + /r2
b. k1= - q/r - /r2,
k2 
k4= /r2
k3= 1/r,
     2r
2

2
2
the sign is minus, since k2 must be negative.
V=k1exp[k2(X-q)] + k3X + k4
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Exercise 1.2 - Solution
Vx + 0.52Vxx + X IX>q = rV
V
1
 0  q*  
q
rk 2
d. Value of the abandonment option is the
difference between values with and without the
option, when q is chosen optimally.
1 k2 ( X q*)
option' s value  
e
rk 2
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slide 55
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?
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slide 56
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.
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slide 57
Exercise 1.3 - Solution a.
dX = Xdt + XdW
ECG = E[dV] = XVxdt + 0.52X2Vxxdt
ECF = Xdt
TR = (XVx + 0.5X22Vxx + X)dt
XVx + 0.5X22Vxx + X = rV
V(X)= X/(r - )
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Exercise 1.3 - Solution b.
dX = Xdt + XdW
ECG = E[dV] = XVxdt + 0.52X2Vxxdt
ECF = (X-V)dt
TR = (XVx + 0.5X22Vxx + (X-V))dt
XVx + 0.5X22Vxx + (X-V) = rV
V(X)= X/(r +  - )
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slide 59
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 riskneutral, 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.
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slide 60
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?
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slide 61
Exercise 1.4 - Solution a.
dX = Xdt + XdW
ECG = E[dV] = XVxdt + 0.52X2Vxxdt
ECF = Xdt
TR = (XVx + 0.5X22Vxx + X)dt
XVx + 0.5X22Vxx + X = rV
V(X)= X/(r - )
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Exercise 1.4 - Solution b.
dX = Xdt + XdW
ECG = E[dV] = XVxdt + 0.52X2Vxxdt
ECF = Xdt
TR = (XVx + 0.5X22Vxx +  X)dt
XVx + 0.5X22Vxx + X = rV
V(X)= X/(r - )
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slide 63
Exercise 1.4 - Solution c.
dX = Xdt + XdW
ECG = E[dV] = XVxdt + 0.52X2Vxxdt
ECF = Xdt + (X/(r - )-V)dt
TR = XVx + 0.5X22Vxx + X + (X/(r - )-V)
XVx + 0.5X22Vxx + X + (X/(r - )-V) = rV
 

V (X ) 


r    r 

X
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slide 64
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?
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slide 65
Exercise 1.5 - Solution
dX = Xdt + XdW
ECG = E[dV] = XVxdt + 0.52X2Vxxdt
ECF = 0
TR = (XVx + 0.5X22Vxx)dt
XVx + 0.5X22Vxx = rV
V(Q)=R
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slide 66
Exercise 1.5 - Solution
V
R
Q
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X
slide 67
Exercise 1.5 - Solution
XVx + 0.5X22Vxx = rV
AQ = R
V(X) = AX
AX + 0.52 (-1)AX = rAX
+ 0.52 (-1) = r
0.52 2 + ( - 0.52) - r = 0

0.5   
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2
  0.5   2r
2 2

2
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2
X
V ( X )  R 
Q
slide 68

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 riskneutral, 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)
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slide 69
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?
Zvi Wiener
ContTimeFin - 2
slide 70
Exercise 1.6 - Solution a.
dX = Xdt + XdW
ECG = E[dV] = XVxdt + 0.52X2Vxxdt
ECF = 0
TR = (XVx + 0.5X22Vxx)dt
XVx + 0.5X22Vxx = rV
V(Q)=Q-E
Zvi Wiener
ContTimeFin - 2
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Exercise 1.6 - Solution b.
V(X) = (Q-E)(X/Q)
Optimal exercise is such that maximizes V(X):
V


 1
 X (1   )Q  EQ
Q


E
Q* 
 1
Zvi Wiener
ContTimeFin - 2
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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 
Zvi Wiener
ContTimeFin - 2
slide 73
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.
Zvi Wiener
ContTimeFin - 2
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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?
Zvi Wiener
ContTimeFin - 2
slide 75
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.
Zvi Wiener
ContTimeFin - 2
slide 76
Exercise 1.8 - Solution
V(X,Y),
dWdZ=dt
dY = dt + dW
dX = dt + wdZ
ECG = (Vy+0.52Vyy+Vx+0.5w2Vxx+Vxy) dt
ECF = Ydt,
if X > q,
otherwise 0.
TR = ECG + ECF = alternative return (rV)
Vy+0.52Vyy+Vx+0.5w2Vxx+Vxy + Y= rV
V(q,Y)=0
Zvi Wiener
ContTimeFin - 2
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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.
Zvi Wiener
ContTimeFin - 2
slide 78
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.
Zvi Wiener
ContTimeFin - 2
slide 79
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.
Zvi Wiener
ContTimeFin - 2
slide 80
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.
Zvi Wiener
ContTimeFin - 2
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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.
Zvi Wiener
ContTimeFin - 2
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Exercise 1.9
b. 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.
Zvi Wiener
ContTimeFin - 2
slide 83
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.
Zvi Wiener
ContTimeFin - 2
slide 84
Exercise 1.9 - Solution a.
X = dt + dW
V(X)
ECG = Vxdt + 0.52Vxxdt
ECF = g(X)dt
TR = ECG + ECF = Vx + 0.52Vxx + g(X)
Vx+ 0.52Vxx+ g(X) = rV
V(q) = 0
Zvi Wiener
ContTimeFin - 2
slide 85
Exercise 1.9 - Solution b.
Debt:
D = c Y3 + Y2 - kY1 - Y4 + kY3
Equity:
E = (1- )(Y4 - cY3)
Government:
G =  (Y4 - c Y3)
Others:
T = k(Y1 - Y3)
Zvi Wiener
ContTimeFin - 2
slide 86