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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.