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MSc Financial Engineering 2009 - 2010

Mathematical Methods - Christmas Assignment

Solutions

The solution of exercise 1 was discussed in detail in the last MathMethods Lecture on March 16. Below you will find the solutions of theother exercises (including exercise 2 which was also already discussedon March 16).

Exercise 2. By definition,

g(, t) = E

eiXt

.

We study the time-variation of g: we first observe that by linearity ofexpectations

g

t(, t)dt = g(, t + dt) g(, t)

= E

eiXt+dt E

eiXt

= E

eiXt+dt eiXt

= E

d

eiXt

.

Next, by Itos lemma applied to f(Xt) with f(X) = eix (and therefore

f(x) = ieix, f(x) = (i)2eix =

2eix):

d

eiXt

= ieiXtdXt 1

22eiXt(dXt)

2.

Using the SDE (1) and Itos multiplication table1,

(dXt)2 = ((1 2Xt)dt + 1dW1,t + 2XtdW2,t)2

= (21 + 212Xt + 2X2t )dt.

Plugging this, and the SDE for dXt, into the expression for d

eiXt

,and anticipating that we will be taking expectations, and therefore onlyexplicitly computing the drift term, we find that

d

eiXt

=1

222

2X2t (i2+ 122)Xt + (i11

221

2)

eiXt dt

+ ( )dW1,t + ( )dW2,t.Hence, taking expectations,

g

t(, t) = 1

222

2E

X2t eiXt

(i2+ 122)E

Xte

iXt

+ (i11

221

2)E

eiXt

.(1)

1(dW1,t)2 = (dW2,t)2 = dt, dW1,tdW2,t = dt, and the other products equal to 01

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We now observe, by differentiating both sides ofg(, t) = E

eiXt

withrespect to and differentiating the right hand side under the expecta-

tion sign (which is (morally) allowed since the expectation operator islinear)

g(, t) = E

eiXt

= iE

Xte

iXt

,

and similarly,

2

2g(, t) = E

2

2eiXt

= E

X2t e

iXt

.

Plugging these into the expression (1), we arrive at the following PDEfor g(, t):

g

t=

1

222

2 2g

2+ (i12 2)

g

+ (i1

1

221

2)g.

This PDE has to be solved with initial value condition g(, 0) = eix0 ,where x0 is the value of the process at time 0.

Exercise 3. (a) This simply is the SDE for a Geometric BrownianMotion discussed in Chapter 4 of the Notes, and which is solved byconverting the SDE into one for log X0t : computing d log X

0t using Itos

lemma one finds:

d log X0t = (2 +

1

222)dt + 2dW2,t,

and hence

X0t = Ce(2+

1

222

)t+2W2,t,

with C a constant.

(b) We now interpret the SDE (1) as an inhomogeneous SDE associatedto the homogeneous SDE (9), the inhomogeneous part of (1) being thepart which does not contain Xt, namely1dt + 1dW1,t. We try tosolve this by the method of variation of constants, as would do for anordinary differential equation. We therefore seek a solution to (1) of

the formXt = Ct X

0t ,

with X0t a solution of (9) (with initial value 1, say) and Ct a processto be determined. We assume that Ct is an Ito-process with respect tothe two Brownian motions, that is, that

dCt = atdt + b1,tdW1,t + b2,tdW2,t.

We first derive an SDE for Ct using as input that X0t solves (9), just as

for ODEs. The only twist is that for deriving a product we now haveto use the product rule from Ito calculus:

d(Ct X0t ) = Ct dX0t + X0t dCt + dCt dX0t ;

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cf. exercise 4.25 of Lecture 4, and its solution. Inserting the SDE (9)satisfied by X0t into this expression, we find:

d(Ct X0t ) = Ct(2X0t dt + 2X0t dW2,t) + X0t dCt + dCt dX0t

= 2Xtdt + 2XtdW2,t + X0t dCt + dCt dX0t .Since Xt = CtX

0t as to satisfy (1), this has to equal:

2Xtdt + 2XtdW2,t + 1dt + 1dW1,t.Hence we want that

X0t dCt + dCt dX0t = 1dt + 1dW1,t.

To convert this into an equation of the form dCt = we insertdC

t= dC

t= a

tdt + b

1,tdW

1,t+ b

2,tdW

2,tand derive expressions for

the coefficients. Using Itos multiplication table 2 we find that

dCtdX0t = (( )dt + b1,tdW1,t + b2,tdW2,t) (( )dt + 2dW2,t))

= (2b1,t + 2b2,t) dt,

and therefore

X0t dCt + dCt = (at + 2b1,t + 2b2,t) X0t dt + b1,tX

0t dW1,t + b2,tX

0t dW2,t.

Since this has to be equal to 1dt + 1dW1,t we find, on comparingcoefficients, that

(at + 2(b1,t + b2,t)) X

0t = 1

b1,tX0t = 1b2,tX

0t = 0,

or

b1,t =1X0t

, b2,t = 0,

and

at =1X0t

2b1,t =1 12

X0t.

Hence

dCt = (X0t )

1 ( (1 12)dt + 1dW1,t) ,as was to be shown.

(c) Integrating the last-found expression for dCt with initial value C0 =c R, we find:

Ct = c +

t0

(1 12)(X0s )1 ds + 1(X0s )1dW1,s,

and therefore

Xt = cX0t +

t0

(1 12)(X0s )1X0t ds + 1(X0s )1Xt dW1,s.

2and in particular (dW2,t)2 = dt and dW1,tdW2,t = dt

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Inserting the expression for X0t we found in part (a) (taking C = 1),we finally find

Xt = c e(2+

1

222

)t+2W2,t + (1 12)t

0

e(2+1

222

)(ts)+2(W2,tW2,s)ds

+ 1

t0

e(2+1

222

)(ts)+2(W2,tW2,s) dW1,s .

Exercise 4. (a) By Itos multiplication rules,

(dXt)2 = ( ( )dt + 1dW1,t + 2XtdW2,t)2

= (1dW1,t + 2XtdW2,t)2

= 21(dW1,t)

2

+ 212XtdW1,tdW2,t + 22X

2t (dW2,t)

2

= (21 + 212Xt + 22X

2t )dt.

Since there are no dW-terms, this is also the conditional expectationof (dXt)

2 at time t.

(b) We define Wt as in the exercise or, equivalently, by

dWt =1dW1,t + 2XtdW2,t21 + 212Xt +

22 X

2t

.

Since there is no drift term, this defines a martingale. By the calcula-tions of part (a) above, we find that (dWt)

2 = dt. It then follows from

Levys theorem that Wt is a Brownian motion. Since, clearly,21 + 212Xt +

22X

2t dWt = 1dW1,t + 2XtdW2,t,

the SDE (14) follows.

(c) If = 0 and 1 = 0, then

dXt = 2Xtdt +

21 + 22X

2t dWt.

If Xt = 112 sinh(Zt), with dZt = atdt + btdWt, with coefficients at,

bt to be determined3, then by Itos lemma

dXt = 12cosh(Zt)dZt + 1

2sinh(Zt)(dZt)2

=12

cosh(Zt)(atdt + btdWt) +

1

2sinh(Zt)b

2t dt

=12

at cosh(Zt) +

1

2b2t sinh(Zt)

dt +

12

bt cosh(Zt)dWt.

On the other hand, since

21 + 22X

2t =

21

1 + sinh2(Zt)

= 21 cosh

2(Zt),

3Since Zt = sinh1((2/1)Xt), is the image of the Ito process Xt a differentiable function,

we know from Itos lemma that dZt = atdt + btdWt for suitable at, bt.

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the SDE for Xt can also be written as

dXt = 21

2 sinh(Zt)dt + 1 cosh(Zt)dWt.Comparing the two expressions for dXt, we find that

12

bt = 1 bt = 2,

and12

at cosh(Zt) +

1

2b2t sinh(Zt)

= 2

12

sinh(Zt).

Cancelling out the 1/2 on both sides, and using that bt = 2, we findon solving for at that

at cosh(Zt) = 2 + 1222 sinh(Zt),or

at =

2 +1

222

tanh(Zt).

It therefore follows that

dZt =

2 +1

222

tanh(Zt) + 2dWt,

showing that = 2 +1

222 in formula (17) of the exercise.

(c) We now suppose that

dXt = (1 2Xt)dt +

21 + 212Xt + 22 X

2t dWt,

with not necessarily equal to 0. We complete the square in theexpression under the -sign:

22X2t + 212Xt +

21 =

22

X2t + 2

12

Xt + 2

21

22 2

21

22

+ 21

= 22

Xt +

12

2+ (1 2)21

= 22 Y

2t + (1

2

)21,

where Yt := Xt + 12

. Replacing Xt by Yt 12

in the drift term of the

SDE above, we then find, after a short computation, that Yt satisfies:

dYt = (1 2Yt) dt +21 + 22Y2t dWt,with 1 = 1 + 1

2,

and

21 = (1

2)21.