Solutions to Examples on

Stochastic Differential Equations

December 4, 2012

2

Q 1. LetW1(t) andW2(t) be two Wiener processes with correlated increments ∆W1 and ∆W2 such

that E [∆W1∆W2] = ρ∆t. Prove that E [W1(t)W2(t)] = ρ t. What is the value of E [W1(t)W2(s)]?

Answer

Let 0 = t0 < t1 < t2 < · · · < tn = t be a dissection of the interval [0, t], then

W(1)t =

n∑k=1

∆W(1)k , W

(2)t =

n∑k=1

∆W(2)k .

where W(1)k and W

(2)k are zero mean Gaussian deviates satisfying E[W (1)

k W(2)k ] = ρ(tk − tk−1). Thus

E[W (1)t W

(2)t ] = E

[( n∑k=1

∆W(1)k

)( n∑j=1

∆W(2)j

)]=

n∑j,k=1

E[∆W

(1)k ∆W

(2)j

]

However, ∆W(1)k and ∆W

(2)j are independent Gaussian deviates if k = j and so E

[∆W

(1)k ∆W

(2)j

]=

ρ(tk − tk−1)δ(k − j). Thus

E[W (1)t W

(2)t ] =

n∑k=1

ρ(tk − tk−1) = ρ(tn − t0) = ρ t .

Q 2. Let W (t) be a Wiener process and let λ be a positive constant. Show that λ−1W (λ2t) and

tW (1/t) are each Wiener processes.

Answer

This question concerns the properties of a random variable under changes of variable. We need to

show that each random variable is Gaussian distributed with mean value zero and variance t.

(a) Here t is a parameter and W (λ2t) is a Gaussian random variable with mean value zero and

variance λ2t. Let Y = λ−1W (λ2t) then

fY = fWdW

dY= λ fW = λ

1√2πλ2t

exp(− W 2

2λ2t

)=

1√2πt

exp(− λ2Y 2

2λ2t

)=

1√2πt

exp(− Y 2

2t

).

Thus Y is a Gaussian deviate with mean value zero and variance t.

(b) Here t is again a parameter and W (1/t) is a Gaussian random variable with mean value zero

and variance 1/t. Let Y = tW (1/t) then

fY = fWdW

dY=

1

tfW =

1

t

1√2π(1/t)

exp(− W 2

2(1/t)

)=

1√2πt

exp(− (Y/t)2

2λ2(1/t)

)=

1√2πt

exp(− Y 2

2t

).

Thus Y is a Gaussian deviate with mean value zero and variance t.

3

Q 3. Suppose that (ε1 , ε2) is a pair of uncorrelated N (0, 1) deviates.

(a) By recognising that ξx = σx ε1 has mean value zero and variance σ2x, construct a second deviate

ξy with mean value zero such that the Gaussian deviate X = [ ξx , ξy ]T has mean value zero

and correlation tensor

Ω =

σ2x ρ σxσy

ρ σxσy σ2y

where σx > 0, σy > 0 and | ρ | < 1.

(b) Another possible way to approach this problem is to recognise that every correlation tensor is

similar to a diagonal matrix with positive entries. Let

α =σ2x − σ2y

2, β =

1

2

√(σ2x + σ2y)

2 − 4(1− ρ2)σ2xσ2y =

√α2 − ρ2σ2xσ

2y .

Show that

Q =1√2β

√β + α −

√β − α√

β − α√β + α

is an orthogonal matrix which diagonalises Ω, and hence show how this idea may be used to

find X = [ ξx , ξy ]T with the correlation tensor Ω.

(c) Suppose that (ε1, . . . , εn) is a vector of n uncorrelated Gaussian deviates drawn from the dis-

tribution N (0, 1). Use the previous idea to construct an n-dimensional random column vector

X with correlation structure Ω where Ω is a positive definite n× n array.

Answer

(a) Let X = [ ξx , ξy ]T then ξx has variance σ2x and so we may write ξx = σx ε1 without any loss in

generality. The task is now to find ξy. The idea is to write ξy = α ε1+β ε2. It now follows that

E [ ξx ξy ] = E [σx ε1 (α ε1 + β ε2) ] = ασx

E [ ξy ξy ] = α2 + β2

Therefore, choose ασx = ρ σxσy and α2 + β2 = σ2y . Thus α = ρ σy and β2 = σ2y(1 − ρ2). One

possible vector deviate is

X =[σx ε1 , ρ σy ε1 +

√1− ρ2 σy ε2

]T.

4

(b) To check that Q is an orthogonal matrix, it is enough to observe that the two columns of Q are

orthogonal to each other, and that each column of Q is a unit vector. Thus

QTΩQ =1

2β

√β + α

√β − α

−√β − α

√β + α

σ2x ρ σxσy

ρ σxσy σ2y

√β + α −

√β − α√

β − α√β + α

Without substituting for α and β, the computation of QTΩQ simplifies to

1

2β

(β + α)σ2x + (β − α)σ2y + 2ρσxσy√β2 − α2 (σ2y − σ2x)

√β2 − α2 + 2αρσxσy

(σ2y − σ2x)√β2 − α2 + 2αρσxσy (β + α)σ2y + (β − α)σ2x − 2ρσxσy

√β2 − α2

.

We need to demonstrate that this is a diagonal matrix. Consider therefore

(σ2y − σ2x)√β2 − α2 + 2αρσxσy = −2α

√β2 − α2 + 2αρσxσy

= 2α(ρ σxσy −√β2 − α2) .

It follows directly from the definition of β that this entry is zero. Consequently, QTΩQ is a

diagonal matrix. By noting that, first that σ2x−σ2y = 2α and, second that ρ σxσy =√β2 − α2,

the array QTΩQ now becomes

1

2β

β(σ2x + σ2y) + 2α2 + 2ρσxσy√β2 − α2 0

0 β(σ2y + σ2x)− 2α2 − 2ρσxσy√β2 − α2

=

1

2β

β(σ2x + σ2y) + 2β2 0

0 β(σ2y + σ2x)− 2β2

=

1

2

σ2x + σ2y + 2β 0

0 σ2y + σ2x − 2β

It is again obvious from the definition of β that σ2y+σ

2x ≥ 2β and so the entries of this diagonal

matrix are non-negative. Let

Y =1√2

[ √σ2x + σ2y + 2β ε1 ,

√σ2x + σ2y − 2β ε2

]Twhere ε1 ∼ N(0, 1) and ε2 ∼ N(0, 1), then the random variable X = QY has mean value zero

and covariance tensor QTΩQ.

(c) Since Ω is a symmetric positive definite n × n array, it is possible to find an n × n orthog-

onal matrix Q such that Ω = QDQT in which D is a diagonal matrix whose entries are the

eigenvalues of Ω in some order. Since Ω is positive definite, then each entry of D is positive.

Let Y = [√λ1 ε1, · · · ,

√λn εn ]

T where λk > 0 is the k-th diagonal entry of D. Consider the

properties of X = QY . Clearly

E [X ] = E [QY ] = QE [Y ] = 0 ,

E [XXT ] = E [QY Y TQT ] = QE [Y Y T ]QT = QDQT = Ω .

5

Thus X has the required properties.

Q 4. Let X be normally distributed with mean zero and unit standard deviation, then Y = X2 is

said to be χ2 distributed with one degree of freedom.

(a) Show that Y is Gamma distribution with λ = ρ = 1/2.

(b) What is now the distribution of Z = aX2 if a > 0 and X ∼ N(0, σ2).

(b) If X1, · · · , Xn are n independent Gaussian distributed random variables with mean zero and

unit standard deviation, what is the distribution of Y = XTX where X is the n dimensional

column vector with k-th entry Xk.

Answer

(a) Since Y = X2 then clearly Y ≥ 0 and so every non-zero value of Y arises from either X or −X.

Therefore, the density of Y is

fY (y) = 2fX(x)dX

dY=

2√2π

e−x2/2 1

2y−1/2 =

1√2π

y−1/2 e−y/2 .

Evidently the distribution function for Y is the special case of the Gamma distribution in which

λ = ρ = 1/2.

(b) Since Z = aX2 with a > 0 then clearly Z ≥ 0 and so once again every non-zero value of Z

arises from either X or −X. The density of Z is now

fZ(z) = 2fX(x)dX

dZ=

2√2π σ

e−x2/2σ2 1

2√az−1/2 =

1√2π

√a σ

z−1/2 e−z/2aσ2.

The distribution function for Z is now the Gamma distribution with ρ = 1/2 and λ = (2aσ2)−1.

(c) If X1, · · · , Xn are each normally distributed with mean zero and unit standard deviation, then

Y = X21 +X2

2 + · · ·+X2n is likewise Gamma distributed with parameters λ = 1/2 and ρ = n/2.

This result follows from the previous example. In this case we say that Y is χ2 distributed with

n degrees of freedom.

The chi-squared distribution with n degrees of freedom plays an important role in statistical

hypothesis testing in which deviates are assigned to “bins”.

Q 5. Calculate the bounded variation (when it exists) for the functions

(a) f(x) = |x| x ∈ [−1, 2] (b) g(x) = log x x ∈ (0, 1]

(c) h(x) = 2x3 + 3x2 − 12x+ 5 x ∈ [−3, 2] (d) k(x) = 2 sin 2x x ∈ [π/12, 5π/4]

(e) n(x) = H(x)−H(x− 1) x ∈ R (f) m(x) = (sin 3x)/x x ∈ [−π/3, π/3] .

6

Answer

The simplest approach here is to draw each function.

(a) The bounded variation of f(x) = |x| for x ∈ [−1, 2] is 1 + 2 = 3.

(b) The function g(x) = log x→ −∞ as x→ 0+.

(c) Here we need to find the stationary values of h(x) = 2x3+3x2−12x+5 for x ∈ [−3, 2]. Clearly

dh/dx = 6x2 + 6x − 12 = 6(x2 + x − 2) = 6(x + 2)(x − 1) and so h(x) has turning values at

x = −2 and x = 1. Thus h(−3) = 4, h(−2) = 25, h(1) = −2 and h(2) = 9 and so the bounded

variation is

|4− 25|+ |25− (−2)|+ |(−2)− 9| = 59 .

(d) We need to draw the function in our mind. The points of interest are h(π/12) = 2 sin(π/6) = 1,

h(π/4) = 2 sin(π/2) = 2, h(3π/4) = 2 sin(3π/2) = −2 and h(5π/4) = 2 sin(5π/2) = 2. Thus

the variation is 1 + 4 + 4 = 9.

(e) The Heaviside function or step function H(x) jumps from zero to 1 as x passes through x = 0

while the step function H(x − 1) jumps from zero to 1 as x passes through x = 1. Thus n(x)

is the top hat function and its variation is 2.

(f) The function m(x) = (sin 3x)/x is zero at the end points of the interval [−π/3, π/3]. It takes

the value 3 as x→ 0 and therefore the variation is 6.

Q 6. Use the definition of the Riemann integral to demonstrate that

(a)

∫ 1

0x dx =

1

2, (b)

∫ 1

0x2 dx =

1

3.

You will find useful the formulae∑n

k=1 k = n(n+ 1)/2 and∑n

k=1 k2 = n(n+ 1)(2n+ 1)/6.

Answer

Choose a uniform dissection of [0, 1] in which the interval is divided into n sub-intervals of length

1/n by the points xk = k/n where 0 ≤ k ≤ n.

(a) In the first example we shall pedantic and take a different value of the function in each subin-

terval of the dissection in order to illustrate in this simple case that the value of the integral is

independent of the choice of dissection point. In the following partial sum λk ∈ [0, 1] so that

the partial sum is

Sn =1

n

n∑k=1

f(ξk) =1

n

n∑k=1

xk−1λk + (1− λk)xk =1

n

n∑k=1

(k − 1)

nλk +

k

n(1− λk) .

7

By simple algebra Sn becomes

Sn = − 1

n2

n∑k=1

λk +1

n2n(n+ 1)

2= − 1

n2

n∑k=1

λk +1

2

(1 +

1

n

).

Clearly 0 ≤∑n

k=1 λk ≤ n and therefore

1

2

(1− 1

n

)≤ Sn ≤ 1

2

(1 +

3

n

)→ lim

n→∞Sn =

1

2.

(b) Here we take the left hand endpoint of each interval to deduce that∫ 1

0x2 dx = lim

n→∞

1

n

n∑k=1

k2

n2= lim

n→∞

1

n3

n∑k=1

k2 = limn→∞

1

n3n(n+ 1)(2n+ 1)

6=

1

3.

Q 7. The functions f , g and h are defined on [0, 1] by the formulae

(a) f(x) =

x sin(π/x) x > 0

0 x = 0

(b) g(x) =

x2 sin(π/x) x > 0

0 x = 0

(c) h(x) =

(x/ log x) sin(π/x) x > 0

0 x = 0, 1

Determine which functions have bounded variation on the interval [0, 1].

Answer

Either provide a proof that f is a function of bounded variation over [0, 1] or provide a dissection Dn

over which it is clear that f is not a function of bounded variation as n→ ∞.

(a) Let Dn be the dissection of [0, 1] with nodes x0 = 0, xk = 2/(2n− 2k + 1) when 1 ≤ k ≤ n− 1

and xn = 1. Clearly f(x0) = f(xn) = 0 while

f(xk) =2

2n− 2k + 1sin((n− k)π + π/2) =

2(−1)n−k

2n− 2k + 1.

The variation of f over Dn is therefore

V(f) =k=n∑k=0

∣∣∣ 2(−1)n−k

2n− 2k + 1

∣∣∣ = k=n∑k=0

2

2n− 2k + 1=

k=n∑k=0

2

2k + 1.

This series diverges, and so f is not of bounded variation over [0, 1].

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(b) The stationary values of g occur where

dg

dx= 2x sin(π/x)− π cos(π/x) = 0 → tan(π/x) = π/(2x) .

The solution to this equation is x1 = π/θ1, x2 = π/θ2, · · · where tan(θk) = θk/2. Let D be the

dissection based on the points 1 = x0 > x1 > x2 > · · · > 0, then g either decreases or increases

between the nodes of D. Consequently, the variation of g over D is

V(g) =∞∑k=1

π2

θ2k| sin θk | ≤

∞∑k=1

π2

θ2k(1)

It is easy to demonstrate that θ1 > π/4, θ2 > 5π/4 and in general that θk > kπ − 3π/4.

Therefore π/θk < 4/(4k − 3). Finally, it follows from (1) that

V(g) ≤∞∑k=1

π2

θ2k≤

∞∑k=1

16

(4k − 3)2<∞ .

This series converges, and so g is of bounded variation over [0, 1].

(c) Let Dn be the dissection of [0, 1] with nodes x0 = 0, xk = 2/(2n− 2k + 1) when 1 ≤ k ≤ n− 1

and xn = 1. Clearly h(x0) = h(xn) = 0 while

h(xk) =2

(2n− 2k + 1) log(2/(2n− 2k + 1))sin((n− k)π + π/2)

=2(−1)n−k

(2n− 2k + 1) log(2/(2n− 2k + 1))

The variation of h over Dn therefore satisfies

V(h) ≥k=n−1∑k=1

∣∣∣ 2(−1)n−k

(2n− 2k + 1) log(2/(2n− 2k + 1))

∣∣∣=

k=n−1∑k=1

2

(2n− 2k + 1) log(n− k + 1/2)

=

k=n−1∑k=1

2

(2k + 1) log(k + 1/2)

This series can be condensed. The condensed series will diverge, and so h is not a function of

bounded variation over [0, 1].

Q 8. Prove that ∫ b

a(dWt)

2G(t) =

∫ b

aG(t) dt

9

Answer

Let Dn denote the dissection a = t0 < t1 < · · · < tn = b of the finite interval [a, b] and let

Φ =

∫ b

a(dWt)

2G(t) =

∫ b

aG(t) dt .

The value of Φ is based on the mean square limiting value of the partial sum

Sn =

n∑k=1

G(tk−1) (Wk −Wk−1)2

Since (Wk −Wk−1) =√tk − tk−1 εk where εk ∼ N(0, 1) then

limn→∞

E [Sn ] = limn→∞

E[ n∑k=1

G(tk−1) (tk − tk−1) ε2k

]= lim

n→∞

n∑k=1

G(tk−1) (tk − tk−1) E [ ε2k ]

= limn→∞

n∑k=1

G(tk−1) (tk − tk−1)

=

∫ b

aG(t) dt .

To demonstrate mean square convergence one needs to show that

limn→∞

E[ ( n∑

k=1

G(tk−1) , (tk − tk−1) (ε2k − 1)

)2 ]= 0 .

The calculation proceeds as follows.

limn→∞

E[ ( n∑

k=1

G(tk−1) , (tk − tk−1) (ε2k − 1)

)2 ]= lim

n→∞E[ n∑k=1

n∑j=1

G(tk−1) (tk − tk−1) (ε2k − 1)G(tj−1) (tj − tj−1) (ε

2j − 1)

]= lim

n→∞

n∑k=1

n∑j=1

G(tk−1)G(tj−1) (tk − tk−1)(tj − tj−1)E [ (ε2k − 1) (ε2j − 1) ] .

(2)

The value of this expected value is zero whenever j = k. Thus expression (2) simplifies to

limn→∞

E[ ( n∑

k=1

G(tk−1) , (tk − tk−1) (ε2k − 1)

)2 ]= lim

n→∞

n∑k=1

G2(tk−1) (tk − tk−1)2 E [ ε4k − 2ε2k + 1 ]

= 2 limn→∞

n∑k=1

G2(tk−1) (tk − tk−1)2 = 0 .

(3)

10

Q 9. Prove that ∫ b

aWn dWt =

1

n+ 1

[W (b)n+1 −W (a)n+1

]− n

2

∫ b

aWn−1 dt .

Answer

First apply the identity∫ b

af(t,Wt) o dWt =

∫ b

af(t,Wt) dWt +

1

2

∫ b

a

∂f(t ,Wt)

∂Wdt .

with f(t,Wt) =Wnt to obtain∫ b

aWnt o dWt =

∫ b

aWnt dWt +

n

2

∫ b

aWn−1t dt .

Now apply the identity∫ b

a

∂g(t,Wt)

∂Wto dWt = g(b,W (b) )− g(a,W (a) )−

∫ b

a

∂g

∂tdt .

with g(t,Wt) =Wnt to obtain∫ b

aWnt o dWt =

1

n+ 1

[Wn+1(b)−Wn+1(a)

].

The final result is therefore∫ b

aWnt dWt =

1

n+ 1

[Wn+1(b)−Wn+1(a)

]− n

2

∫ b

aWn−1t dt .

Q 10. The connection between the Stratonovich and Ito integrals relied on the claim that

limn→∞

E[ n∑k=1

∂f(tk−1Wk−1)

∂W

((Wk−1/2 −Wk−1)(Wk −Wk−1)−

tk − tk−1

2

) ]2= 0

provided E [ |∂f(t,W )/∂W |2 ] is integrable over the interval [a, b]. Verify this unsubstantiated claim.

Answer

The calculation begins with the consideration of

ψn =[ n∑k=1

∂f(tk−1Wk−1)

∂W

((Wk−1/2 −Wk−1)(Wk −Wk−1)−

tk − tk−1

2

)]2.

This function is expanded out to give the double summation

ψ =

n∑j,k=1

[∂f(tk−1Wk−1)

∂W

((Wk−1/2 −Wk−1)(Wk −Wk−1)−

tk − tk−1

2

) ][∂f(tj−1Wj−1)

∂W

((Wj−1/2 −Wj−1)(Wj −Wj−1)−

tj − tj−1

2

) ]

11

which is now divided into the case j = k and the case j = k to get

ψ =n∑k=1

[∂f(tk−1Wk−1)

∂W

]2[(Wk−1/2 −Wk−1)(Wk −Wk−1)−

tk − tk−1

2

]2+ 2

n∑1≤j<k≤n

∂f(tj−1Wj−1)

∂W

((Wj−1/2 −Wj−1)(Wj −Wj−1)−

tj − tj−1

2

)∂f(tk−1Wk−1)

∂W

(Wk−1/2 −Wk−1)(Wk −Wk−1)−

tk − tk−1

2

)When j = k then (Wj−1/2 −Wj−1)(Wj −Wj−1)− (tj − tj−1)/2 and Wk−1/2 −Wk−1)(Wk −Wk−1)−(tk − tk−1)/2 are independent random deviates with mean value zero. Therefore the expectation of

the double sum is zero. To appreciate why this is the case consider the expectation being taken

term-by-term starting at k = n and working downwards. Define εk−1 =Wk−1/2 −Wk−1 then[(Wk−1/2 −Wk−1)(Wk −Wk−1)−

tk − tk−1

2

]2=

[− εk−1εk−1/2 + ε2k−1/2 −

tk − tk−1

2

]2= ε2k−1ε

2k−1/2 + ε4k−1/2 +

(tk − tk−1)2

4− (tk − tk−1)(−εk−1εk−1/2 + ε2k−1/2)− 2ε3k−1/2εk−1

Now E[(Wk−1/2 −Wk−1)(Wk −Wk−1)− (tk − tk−1)/2 ]2 = 3(tk − tk−1)

2/4 and therefore

E[ψn] ≤3max |tk − tk−1|

4

n∑k=1

[∂f(tk−1Wk−1)

∂W

]2(tk − tk−1) → 0

as n→ ∞.

Q 11. Compute the value of

Φ =

∫ b

aWt dWt

where the definition of the integral is based on the choice ξk = (1− λ)tk−1 + λtk and λ ∈ [0, 1].

Answer

Let Dn denote the dissection a = t0 < t1 < · · · < tn = b of the finite interval [a, b] and let

Φ =

∫ b

aWt dWt

where the limiting procedure is based on the choice ξk = (1− λ)tk−1 + λtk with λ ∈ [0, 1]. The task

is to find Φ such that

limn→∞

E[ (

Φ−n∑k=1

W (ξk) (Wk −Wk−1))2 ]

(4)

The algebraic manipulations

2Wk+λ−1Wk = W 2k+λ−1 +W 2

k − (Wk+λ−1 −Wk)2

2Wk+λ−1Wk−1 = W 2k+λ−1 +W 2

k−1 − (Wk+λ−1 −Wk−1)2

12

are required to proceed further. Start by subtracting the first expression from the second, and then

sum from k = 1 to k = n to obtain

2

n∑k=1

Wk+λ−1 (Wk −Wk−1 ) =

n∑k=1

W 2k − (Wk+λ−1 −Wk)

2 −W 2k−1 + (Wk+λ−1 −Wk−1)

2

= W 2(b)−W 2(a) +n∑k=1

(Wk+λ−1 −Wk−1)2 − (Wk+λ−1 −Wk)

2

= W 2(b)−W 2(a) +n∑k=1

(ξk − tk−1) ε2k − (tk − ξk) η

2k

where εk and ηk are uncorrelated N (0, 1) deviates. The expected value of the summation on the right

hand side of this expression is (2λ− 1)(b− a) and therefore the contention is that

Φ =1

2

[W 2(b)−W 2(a) + (2λ− 1)(b− a)

].

To prove this assertion, we substitute this expression for Φ in equation (4). We need to show that

1

2limn→∞

E[ (

(2λ− 1)(b− a)− λ

n∑k=1

ε2k(tk − tk−1) + (1− λ)

n∑k=1

η2k (tk − tk−1))2 ]

= 0

The calculation proceeds as follows:-

limn→∞

E[ (

(2λ− 1)(b− a)− λ

n∑k=1

ε2k(tk − tk−1) + (1− λ)

n∑k=1

η2k (tk − tk−1))2 ]

= limn→∞

E[ (

λn∑k=1

(ε2k − 1)(tk − tk−1)− (1− λ)n∑k=1

(η2k − 1) (tk − tk−1))2 ]

= limn→∞

E[λ2

n∑k=1

n∑j=1

(ε2k − 1)(ε2j − 1)(tk − tk−1)(tj − tj−1)

− 2λ(1− λ)

n∑k=1

n∑j=1

(η2k − 1)(ε2j − 1)(tk − tk−1)(tj − tj−1)

+ (1− λ)2n∑k=1

n∑j=1

(η2k − 1)(η2j − 1)(tk − tk−1)(tj − tj−1)]

In the usual way, if k = j then (ξ2j − 1) and (ξ2k − 1) are uncorrelated, otherwise they are correlated.

Thus the previous argument continues with

limn→∞

E[ (

(2λ− 1)(b− a)− λ

n∑k=1

ε2k(tk − tk−1) + (1− λ)

n∑k=1

η2k (tk − tk−1))2 ]

= limn→∞

E[λ2

n∑k=1

(ε2k − 1)2(tk − tk−1)2 + (1− λ)2

n∑k=1

(η2k − 1)2(tk − tk−1)2]

As has been seen previously, E [ (ε2k − 1)2 ] = E [ ε4k − 2ε2k + 1 ] = 2 with an identical argument for

13

E [ (η2k − 1)2 ]. Consequently,

limn→∞

E[ (

(2λ− 1)(b− a)− λn∑k=1

ε2k(tk − tk−1) + (1− λ)n∑k=1

η2k (tk − tk−1))2 ]

= 2[λ2 + (1− λ)2] limn→∞

n∑k=1

(tk − tk−1)2 = 0 .

Hence it has been proved that∫ b

aWt dWt =

1

2

[W 2(b)−W 2(a) + (2λ− 1)(b− a)

]with the choice ξk = (1− λ)tk−1 + λtk and λ ∈ [0, 1].

Q 12. Solve the stochastic differential equation

dxt = a(t) dt+ b(t) dWt , x0 = X0

where X0 is constant. Show that the solution xt is a Gaussian deviate and find its mean and variance.

Answer

The formal solution to this problem is

x(t) = X0 +

∫ t

0a(s) ds+

∫ t

0b(s) dWs

where the second integral is to be interpreted as an Ito integral. The fact that x(t) is a Gaussian

deviate stems from the fact that the stochastic part of x is simply a weighted combination of Gaussian

deviates, and is therefore a Gaussian deviate. The mean is

µ(t) = X0 +

∫ t

0a(s) ds

and the variance is

E [ (x− µ)2 ] = E[ ∫ t

0b(s) dWs

∫ t

0b(s) dWs

]=

∫ t

0b2(s) ds .

Q 13. The stochastic integrals I01 and I10 defined by

I10 =

∫ b

a

∫ s

0dWu ds , I01 =

∫ b

a

∫ s

0du dWs .

occur frequently in the numerical solution of stochastic differential equations. Show that

I10 + I01 = (b− a)(Wb −Wa) .

Answer

14

The double integrals for I10 and I01 may be converted to the single integrals

I10 =

∫ b

a

∫ s

adWu ds =

∫ b

a(Ws −Wa) ds ,

I01 =

∫ b

a

∫ s

adu dWs =

∫ b

a(s− a) dWs .

It is clear that I10 is a Riemann integral and I01 is a Riemann-Stieltjes integral. Both integrals may

be computed using the standard rules of integral calculus. Thus

I01 =

∫ b

a(s− a) dWs

=[(s− a)(Ws −Wa)

] ba−

∫ b

a(Ws −Wa) ds

= (b− a)(Wb −Wa)− I10 .

Thus I01 + I10 = (b− a)(Wb −Wa).

Q 14. The stochastic integrals I111 defined by

I111 =

∫ t

a

∫ s

a

∫ u

adWw dWu dWs

occurs frequently in the numerical solution of stochastic differential equations. Show that

I111 =(Wb −Wa)

6

[(Wb −Wa)

2 − 3(b− a)].

Answer

The evaluation of I111 proceeds in a number of steps, the first of which is the conversion of the triple

integral to the double integral

I111 =

∫ b

a

∫ s

a

∫ u

adWw dWu dWs =

∫ b

a

∫ s

a(Wu −Wa) dWu dWs .

where Ws =W (s) etc. The double integral may now be further integrated to yield

I111 =1

2

∫ b

a[ (Ws −Wa)

2 − (s− a) ] dWs

=[ 16(Ws −Wa)

3] ba− 1

2

∫ b

a(Ws −Wa) ds−

1

2

∫ b

a(s− a) dWs

=1

6(Wb −Wa)

3 − 1

2( I10 + I01 )

where I10 and I01 are defined in the previous problem which asserts that

I10 + I01 = (b− a)(Wb −Wa) .

15

The final conclusion is therefore that

I111 =(Wb −Wa)

6

[(Wb −Wa)

2 − 3(b− a)].

Q 15. Show that the value of the Riemann integral

I10 =

∫ b

a

∫ s

0dWu ds

may be simulated as a Gaussian deviate with mean value zero, variance (b− a)3/3 and such that its

correlation with (Wb −Wa) is (b− a)2/2.

Answer

It has already been shown that

ξ = I10 =

∫ b

a

∫ s

adWu ds =

∫ b

a(Ws −Wa) ds .

First, it is clear that ξ is a Gaussian process since it is simply the sum of Gaussian processes. The

mean value of ξ is zero and its variance is

E [ ξ2 ] = E[ ∫ b

a(Ws −Wa) ds

∫ b

a(Wu −Wa) du

]= E

[ ∫ b

a

∫ b

a(Ws −Wa) (Wu −Wa) ds du

]=

∫ b

a

∫ b

aE[(Ws −Wa) (Wu −Wa)

]ds du

=

∫ b

a

∫ b

amin (s− a, u− a) ds du

= 2

∫ b

a

∫ u

a(s− a) ds du

=

∫ b

a(u− a)2 du =

(b− a)3

3.

The correlation between (Wb −Wa) and ξ is given by

E[(Wb −Wa)

∫ b

a(Ws −Wa) ds

]=

∫ b

aE[(Wb −Wa) (Ws −Wa)

]ds

=

∫ b

a(s− a) ds =

(b− a)2

2.

In conclusion, ξ = I10 is simulated as a Gaussian deviate with mean value zero, variance (b − a)3/3

and such that it has correlation (b− a)2/2 with (Wb −Wa).

Q 16. Solve the degenerate Ornstein-Uhlenbeck stochastic initial value problem

dx = −αx+ σ dW , x(0) = x0

16

in which α and σ are positive constants and x0 is a random variable. Deduce that

E [X ] = E [x0 ] e−α t , V [X ] = V [x0 ] e

−2αt +σ2

2α[ 1− e−2αt) ] .

Answer

The solution may be obtained by applying Ito’s lemma to the function F = x eαt. Ito’s lemma for

this SDE is

dF =[Ft(t, x)− αxFx(t, x)

]dt+ σ Fx(t, x) dWt

and in this case yields

dF = σ eαt dWt .

Thus the solution is

x(t) = x0 e−αt + σ

∫ t

0e−α(s−t) dWs

The mean solution is

E [x(t) ] = E[x0 e

−αt + σ

∫ t

0e−α(s−t) dWs

]= E [x0 ] e

−αt .

Consequently,

x(t)− E [x(t) ] = (x0 − E [x0 ]) e−αt + σ

∫ t

0e−α(s−t) dWs

Assuming that x0 is uncorrelated with the Wiener process, it follows that

V [x(t) ] = V [x0 ] e−2αt + σ2 E

[ ∫ t

0e−α(s−t) dWs

∫ t

0e−α(u−t) dWu

]= V [x0 ] e

−2αt + σ2∫ t

0

∫ t

0e−α(s−t) e−α(u−t)E [ dWs dWu ]

= V [x0 ] e−2αt + σ2

∫ t

0e−2α(s−t) ds

= V [x0 ] e−2αt +

σ2

2α[ 1− e−2αt) ] .

Q 17. It is given that the instantaneous rate of interest satisfies the equation

dr = µ(t, r) dt+√g(t, r) dW . (5)

Let B(t, R) be the value of a zero-coupon bond at time t paying one dollar at maturity T . Show that

B(t, R) is the solution of the stochastic differential equation

∂B

∂t+ µ(t, r)

∂B

∂r+g(t, r)

2

∂2B

∂r2= rB (6)

with terminal condition B(T, r) = 1.

17

(a) The CIR equation takes the form of equation (8) with µ(t, r) = α(θ− r) and g(t, r) = σ2r. It is

given that the anzatz B(t, r) = exp[β0(T − r) + β1(T − t)r] satisfies equation (6) provided the

coefficient functions β0(T − t) and β1(T − t) satisfy a pair of ordinary differential equations.

Determine these equations with their boundary conditions.

(b) Solve these ordinary differential equations and deduce the value of B(t, R) where R is the spot

rate of interest.

Answer

Ito’s lemma for B(t, r) gives

dB =∂B

∂tdt+

∂B

∂rdr +

g(t, r)

2

∂2B

∂r2dt → E[dB] =

(∂B∂t

+ µ(t, r)∂B

∂r+g(t, r)

2

∂2B

∂r2

)dt

But E[dB] = rB dt and therefore B(t, r) satisfies the partial differential equation

∂B

∂t+ µ(t, r)

∂B

∂r+g(t, r)

2

∂2B

∂r2= rB .

Q 18. It is given that the solution of the initial value problem

dx = −αx+ σ x dW , x(0) = x0

in which α and σ are positive constants is

x(t) = x(0) exp[− (α+ σ2/2)t+ σW (t)

]Show that

E [X ] = x0 e−α t , V [X ] = e−2α t(eσ

2 t − 1)x20 .

Answer

The expected value of X is

E [X ] = E[x(0) exp[−(α+ σ2/2)t+ σW (t) ]

]= x(0) e−(α+σ2/2)t E [ e−σW (t) ]

The task is now to compute

E [ e−σW (t) ] =1√2π t

∫ ∞

−∞e−σW e−W

2/2t dW

=1√2π t

eσ2t/2

∫ ∞

−∞e−(W+σ t)2/2t dW = eσ

2t/2 .

Consequently it follows that

E [X ] = x(0) e−(α+σ2/2)t eσ2t/2 = x(0) e−α t .

18

The variance of X is

V [X ] = E[ (

x(0) exp[−(α+ σ2/2)t+ σW (t) ]− x(0) e−(α t)2 ]

= x2(0) e−2α t E[ (

exp[−σ2t/2 + σW (t) ]− 1)2 ]

= x2(0) e−2α t−σ2t E[ (

eσW (t) − eσ2t/2

)2 ]= x2(0) e−2α t−σ2t E [ e2σW (t) − 2eσW (t) eσ

2t/2 + eσ2t ]

= x2(0) e−2α t−σ2t[ e2σ2 t − 2eσ

2t + eσ2t ]

= x2(0) e−2α t [ eσ2 t − 1 ] .

Q 19. Benjamin Gompertz (1840) proposed a well-known law of mortality that had the important

property that financial products based on male and female mortality could be priced from a single

mortality table with an age decrement in the case of females. Cell populations are also well-know to

obey Gompertzian kinetics in which N(t), the population of cells at time t, evolves according to the

ordinary differential equationdN

dt= αN log

(MN

),

where M and α are constants in which M represents the maximum resource-limited population of

cells. Write down the stochastic form of this equation and deduce that ψ = logN satisfies an OU

process. Further deduce that mean reversion takes place about a cell population that is smaller than

M , and find this population.

Answer

The stochastic form of this SDE is

dN = αN log(MN

)dt+ σN dW ,

Ito’s lemma applied to ψ = logN gives

dψ =dN

N− σ2N2

2N2dt = α log

(MN

)dt+ σ dW − σ2

2dt =

(α logM − αψ − σ2

2

)dt+ σ dW .

Thus ψ satisfies an OU equation with mean state ψ = logM − σ2/2α which in turn translates into

the population N = Me−σ2/2α. The standard solution of the OU equation may be used to write

down the general solution for ψ and consequently the general solution for N(t). The result of this

calculation is that

N(t) = exp[(1− e−αt)ψ + e−αtψ0 + σ

∫ t

0e−α(t−s) dWs

]= exp

[(1− e−αt) log (Me−σ

2/2α) + e−αt logN0 + σ

∫ t

0e−α(t−s) dWs

]= (N0)

e−αt

(Me−σ2/2α) (1−e

−αt) exp[σ

∫ t

0e−α(t−s) dWs

]= Me−σ

2/2α(N0

M

) e−αt

exp[σ2 e−αt

2α+ σ

∫ t

0e−α(t−s) dWs

].

19

Q 20. It is given that the instantaneous rate of interest, r(t), is driven by a two-factor model in

which r(t) = r1(t) + r2(t) where r1(t) and r2(t) satisfy the equations

dr1 = α1(θ1 − r1) dt+ σ1√r1 dW1 ,

dr2 = α2(θ2 − r2) dt+ σ2√r2 dW2 ,

(7)

where dW1 and dW2 are independent increments in the Wiener processesW1 andW2. Let B(t, R1, R2)

be the value of a zero-coupon bond at time t paying one dollar at maturity T . Construct the partial

differential equation satisfied by B(t, R1, R2) and state the terminal condition.

Answer

Ito’s lemma for B(t, r1, r2) gives

dB =∂B

∂tdt+

∂B

∂r1dr1 +

∂B

∂r2dr2 +

1

2

(σ21r1

∂2B

∂r21dt+ σ22r2

∂2B

∂r22dt)

→ E[dB] =(∂B∂t

+ α1(θ1 − r1)∂B

∂r1+ α2(θ2 − r2)

∂B

∂r2+σ21r12

∂2B

∂r21+σ22r22

∂2B

∂r22

)dt

But E[dB] = (r1 + r2)B dt and therefore B(t, r1, r2) satisfies the partial differential equation

∂B

∂t+ α1(θ1 − r1)

∂B

∂r1+ α2(θ2 − r2)

∂B

∂r2+σ21r12

∂2B

∂r21+σ22r22

∂2B

∂r22= (r1 + r2)B .

Q 21. Solve the stochastic differential equation

dxt = a(t) dt+ b(t) dWt , x0 = X0

where X0 is constant. Show that the solution xt is a Gaussian deviate and find its mean and variance.

Answer

The formal solution to this problem is

x(t) = X0 +

∫ t

0a(s) ds+

∫ t

0b(s) dWs

where the second integral is to be interpreted as an Ito integral. The fact that x(t) is a Gaussian

deviate stems from the fact that the stochastic part of x is simply a weighted combination of Gaussian

deviates, and is therefore a Gaussian deviate. The mean is

µ(t) = X0 +

∫ t

0a(s) ds

and the variance is

E [ (x− µ)2 ] = E[ ∫ t

0b(s) dWs

∫ t

0b(s) dWs

]=

∫ t

0b2(s) ds .

20

Q 22. Solve the stochastic differential equation

dxt = − xt dt

1 + t+dWt

1 + t, x0 = 0 .

Answer

The Ito lemma for the SDE

dxt = − xt dt

1 + t+dWt

1 + t

is

dF =[Ft(t, x)−

xt Fx(t, x)

1 + t+

Fxx2(1 + t)2

]dt+

Fx(t, x)

1 + tdWt

The idea is to choose F (t, x) so that the coefficients of dt and dWt are functions of t alone. Clearly

Fx(t, x) = p(t)x + q(t) is the most general expression for F for which the coefficient of dWt is a

function of t alone. with this choice, the coefficient of dt becomes

Ft(t, x)−xt Fx(t, x)

1 + t+

Fxx2(1 + t)2

=dp

dtx+

dq

dt− p(t)xt

1 + t.

Choose p(t) = 1 + t and q(t) = 0 to obtain the result

d( (1 + t)x ) = dWt → (1 + t)x =Wt + x0 .

The final solution is therefore x(t) =W (t)/(1 + t).

Q 23. Solve the stochastic differential equation

dxt = −xt dt2

+√1− x2t dWt , x0 = a ∈ [−1, 1] .

Answer

The Ito lemma for the SDE

dxt = −xt dt2

+√

1− x2t dWt

is

dF =[Ft(t, x)−

xt Fx(t, x)

2+

(1− x2t )Fxx2

]dt+

√1− x2t Fx(t, x)

1 + tdWt

Choose F (t, x) so that the coefficients of dt and dWt are functions of t alone. This requires that

Fx =p(t)√1− x2

→ F = p(t) sin−1 x+ q(t) .

With this choice, the coefficient of dt becomes

Ft(t, x)−xt Fx(t, x)

2+

(1− x2t )Fxx2

=dp

dt

1√1− x2

+dq

dt− x p(t)

2√1− x2

+p(t)x

2√

1− x2t

=dp

dt

1√1− x2

+dq

dt.

21

Choose p(t) = 1 and q(t) = 0 to obtain the result

d( sin−1 x ) = dWt → sin−1 x =Wt + sin−1 a .

The final solution is therefore

x(t) = sin(W (t) + sin−1 a) =√

1− a2 sinW (t) + a cosW (t) .

Q 24. Solve the stochastic differential equation

dxt = dt+ 2√xt dWt , x(0) = x0 .

Answer

The Ito lemma for this SDE is

dF = [Ft(t, x) + Fx(t, x) + 2xFxx ] dt+ 2√xFx(t, x) dWt

The idea is to choose F (t, x) so that the coefficients of dt and dWt are functions of t. If 2√xFx(t, x) =

p(t) then F (t, x) = p(t)√x+ q(t). The coefficient of dt is

Ft(t, x) + Fx(t, x) + 2xFxx =√xdp

dt+dq

dt+p(t)

2√x− 2x

p(t)

4x√x=

√xdp

dt+dq

dt.

Choose p(t) = 1 and q(t) = 0. Thus F =√x and the Ito’s lemma gives

dF = dWt , →√x =

√x0 +Wt

The solution to the SDE is therefore xt = (Wt +√x0 )

2.

Q 25. Solve the stochastic differential equation

dx = [ a(t) + b(t)x ] dt+ [ c(t) + d(t)x ] dWt

by making the change of variable y(t) = x(t)ϕ(t) where ϕ(t) is to be chosen appropriately.

Answer

dy =[x dϕ+ (a+ bx)ϕ

]dt+ [ c(t) + d(t)x ]ϕdWt

Choose ϕ to be the solution of the equation

dϕ = −ϕ [ b(t) dt+ d(t) dWt ]

so that y satisfies

dy = ϕ [ a(t) dt+ c(t) dWt ] → y(t) = y(0) +

∫ t

0ϕ(s) a(s) ds+

∫ t

0ϕ(s) c(s) dWs .

22

The determine ϕ, we make the substitution z = log ϕ to obtain

dz =[ 1

ϕ(−ϕ b(t))− 1

2ϕ2(d2ϕ2)

]dt+

1

ϕ(−d(t)ϕ) dWt

= −[b(t) +

d(t)2

2

]dt− d(t) dWt .

Q 26. In an Ornstein-Uhlenbeck process, x(t), the state of a system at time t, satisfies the stochastic

differential equation

dx = −α(x−X) dt+ σ dWt

where α and σ are positive constants and X is the equilibrium state of the system in the absence of

system noise. Solve this SDE. Use the solution to explain why x(t) is a Gaussian process, and deduce

its mean and variance.

Answer

For any function y = y(x, t), Ito’s lemma gives

dy =[ ∂y∂t

− α(x−X)∂y

∂x+σ2

2

∂2y

∂x2

]dt+ σ

∂y

∂xdWt .

Take y = (x−X)eαt then

dy =[α(x−X) eαt − α(x−X) eαt

]dt+ σ eαt dWt = σ eαt dWt .

Integration of this Ito equation yields

(x−X)eαt = (x(0)−X) + σ

∫ t

0eαs dWs → x = X + (x(0)−X)e−αt + σ

∫ t

0e−α(t−s) dWs .

We observe that if x(0) is constant or is itself a Gaussian deviate then x(t) is simply a sum of Gaussian

deviates and so is a Gaussian deviate. The mean value of x(t) is

µ(t) = E[X + (x(0)−X)e−αt + σ

∫ t

0e−α(t−s) dWs

]= X + (x(0)−X)e−αt .

The variance of x(t) is now computed from

E[(x− µ)2

]= E

[ ((x(0)− x(0))e−αt + σ

∫ t

0e−α(t−s) dWs

)2 ]= E

[(x(0)− x(0))2 e−2αt + 2(x(0)− x(0)) e−αt σ

∫ t

0e−α(t−s) dWs

+ σ2∫ t

0e−α(t−s) dWs

∫ t

0e−α(t−u) dWu

]= σ20 e

−2αt + σ2∫ t

0e−2α(t−s) ds .

23

Q 27. Let x = (x1, . . . , xn) be the solution of the system of Ito stochastic differential equations

dxk = ak dt+ bk α dWα

where the repeated greek index indicates summation from α = 1 to α = m. Show that x =

(x1, . . . , xn) is the solution of the Stratonovich system

dxk =[ak −

1

2bj αbk α,j

]dt+ bk α dWα .

Let ϕ = ϕ(t,x) be a suitably differentiable function of t and x. Show that ϕ is the solution of the

stochastic differential equation

dϕ =[ ∂ϕ∂t

+ ak∂ϕ

∂xk

]dt+

∂ϕ

∂xkbk α dWα

where

ak = ak −1

2bk α,j bj α .

Answer

The SDE dxi = ai dt+ bi α dWα has formal solution

xi(t) = xi(0) +

∫ t

t0

ai(s,xs) ds+

∫ t

t0

b i α (s,xs) dWαs .

The task is to relate the Ito integral in this solution to the corresponding Stratonovich integral. Each

Wiener process behaves separately.∫ t

t0

b i α (s,xs) dWαs = lim

n→∞

n∑k=1

bi α(tk−1/2,xk−1/2) (Wαk −Wα

k−1 )

= limn→∞

n∑k=1

bi α(tk−1,xk−1) (Wαk −Wα

k−1 )

+ limn→∞

n∑k=1

∂bi α(tk−1,xk−1)

∂t(tk−1/2 − tk−1) (W

αk −Wα

k−1 )

+ limn→∞

n∑k=1

∂bi α(tk−1,xk−1)

∂x(j)(x

(j)k−1/2 − x

(j)k−1) (W

αk −Wα

k−1 ) + · · ·

=

∫ t

t0

b i α (s,xs) dWαs

+ limn→∞

n∑k=1

∂bi α(tk−1,xk−1)

∂x(j)(x

(j)k−1/2 − x

(j)k−1) (W

αk −Wα

k−1 )

It follows directly from the stochastic differential equation that

x(j)k−1/2 − x

(j)k−1 = aj(tk−1,xk−1) (tk−1/2 − tk−1) + bj β(tk−1,xk−1)(W

βk−1/2 −W β

k−1 ) + · · ·

24

and therefore the value of the second contribution to the Stratonovich integral is

limn→∞

n∑k=1

∂bi α(tk−1,xk−1)

∂x(j)bj β(tk−1,xk−1)(W

βk−1/2 −W β

k−1 ) (Wαk −Wα

k−1 )

= limn→∞

1

2

n∑k=1

b i α ,j (tk−1,xk−1) b j β(tk−1,xk−1) δα ,β ds

=1

2

∫ t

t0

b i α ,j (t,x) b j α(t,x) ds .

In conclusion,∫ t

t0

b i α (s,xs) dWαs =

∫ t

t0

b i α (s,xs) dWαs +

1

2

∫ t

t0

b i α ,j (t,x) b j α(t,x) ds

where repetition of α implies summation over the independent Wiener processes. The formal solution

of the SDE with the Ito integral replaced by the Stratonovich integral is therefore

xi(t) = xi(0) +

∫ t

t0

(ai(s,xs)−

1

2b i α,j (s,xs) bj α(s,xs)

)ds+

∫ t

t0

b i α (s,xs) dWαs .

Thus the Stratonovich form of the SDE is obtained from the Ito form by replacing the drift component

ai by the modified drift ai = ai − b i α,j bj α/2.

Let ϕ = ϕ(t,x) then

dϕ =∂ϕ

∂tdt+ ϕ,j dx

(j) +1

2ϕij dx

(i) dx(j) + · · ·

=∂ϕ

∂tdt+ ϕ,j (aj dt+ bj α dWα ) +

1

2ϕij bi α dWα bj β dWβ + · · ·

=∂ϕ

∂tdt+ ϕ,j (aj dt+ bj α dWα ) +

1

2ϕij bi α bj α dt .

Now write

bj α dWα = bj α dWα −+1

2bj α ,k bk α dt

to obtain

dϕ =(∂ϕ∂t

+ ϕ,j aj −1

2ϕ,j bj α ,k bk α +

1

2ϕij bi α bj α

)dt+ ϕ,j bj α dWα

=(∂ϕ∂t

+ ϕ,j aj +1

2ϕij bi α bj α

)dt+ ϕ,j bj α dWα .

Q 28. The displacement x(t) of the harmonic oscillator of angular frequency ω satisfies x = −ω2x.

Let z = x+ iωx. Show that the equation for the oscillator may be rewritten

dz

dt= iωz .

The frequency of the oscillator randomised by the addition of white noise of standard deviation σ

to give random frequency ω + σξ(t) where ξ ∼ N(0, 1). Determine now the stochastic differential

equation satisfied by z.

25

Solve this SDE under the assumption that it should be interpreted as a Stratonovich equation, and

use the solution to construct expressions for

(a) E [ z(t) ] (b) E [ z(t) z(s) ] (c) E [ z(t) z(s) ] .

Answer

Let z = x+ iωx then

dz

dt=d2x

dt2+ i ω

dx

dt= −ω2 x(t) + i ω

dx

dt= i ω

( dxdt

+ i ω x(t))= i ω z(t) .

Let ω be replaced by ω ++σξ(t) in the equation z = iωz to obtain

dz = i(ω + σξ(t))z dt → dz = iωz dt+ iσz dWt .

We interpret this as a Stratonovich SDE and not an ITO SDE. With this interpretation, the SDE

has solution

z(t) = z(0) exp[ iωt+ iσW (t) ] .

(a)

z(t) = E[z(0) exp[ iωt+ iσW (t) ]

]= E [ z(0) ] eiωt E [ eiωW (t) ]

However, E [ eαW (t) ] = eα2t/2 and so apply this result yields the final answer

z(t) = E [ z(0) ] eiωt e−σ2t/2 = E [ z(0) ] e(iω−σ

2/2)t .

(b)

E [ z(t) z(s) ] = E[z(0) e iωt+iσW (t) z(0) e iωs+iσW (s)

]= E [ z2(0) ] e iω(t+s) E [ eiσ(W (s)+W (t)) ] .

The difficulty here lies in the fact that W (t) and W (s) are correlated. Suppose t < s, then

W (t) and W (s)−W (t) are independent deviates. Thus

E [ eiσ(W (s)+W (t)) ] = E [ e2iσW (t)+iσ(W (s)−W (t)) ]

= E [ e2iσW (t) ] E [ eiσ(W (s)−W (t)) ]

= e−2σ2t e−σ2(s−t)/2

= e−σ2(s+t)/2−σ2t .

More generally, E [ eiσ(W (s)+W (t)) ] = e−σ2(s+t)/2−σ2min (t,s). The final result is therefore

E [ z(t) z(s) ] = E [ z2(0) ] e (iω−σ2/2)(t+s)−σ2min (t,s) .

26

(c)

E [ z(t) z(s) ] = E[z(0) e iωt+iσW (t) z(0) e−iωs−iσW (s)

]= E [ z(0) z(0) ] e iω(t−s) E [ eiσ(W (t)−W (s)) ]

SinceW (t)−W (s) is a Gaussian deviate with variance |t−s|, then it follows from the observation

E [ eαW (t) ] = eα2t/2 that

E [ z(t) z(s) ] = E [ z(0) z(0) ] e iω(t−s)−σ2|t−s|/2 .

Q 29. It has been established in a previous example that the price B(t, R) of a zero-coupon bond

at time t and spot rate R paying one dollar at maturity T satisfies the Bond Equation

∂B

∂t+ µ(t, r)

∂B

∂r+g(t, r)

2

∂2B

∂r2= rB

with terminal condition B(T, r) = 1 when the instantaneous rate of interest evolves in accordance

with the stochastic differential equation dr = µ(t, r) dt+√g(t, r) dW .

The popular square-root process proposed by Cox, Ingersol and Ross, commonly called the CIR

process, corresponds to the choices µ(t, r) = α(θ − r) and g(t, r) = σ2r. It is given that the anzatz

B(t, r) = exp[β0(T − r) + β1(T − t)r] is a solution of the Bond Equation in this case provided the

coefficient functions β0(T−t) and β1(T−t) satisfy a pair of ordinary differential equations. Determine

these equations with their boundary conditions.

Solve these ordinary differential equations and deduce the value of B(t, R) where R is the spot rate

of interest.

Answer

Let B(t, r) = exp[β0(T − r) + β1(T − t)r] and define τ = T − t then

∂B

∂t= −

(dβ0dτ

+ rdβ1dτ

)B(t, r) ,

∂B

∂r= β1B(t, r) ,

∂2B

∂r2= β21 B(t, r) .

Substituting these calculations into the bond equation gives

−r dβ1dτ

− αrβ1 +σ2r

2β21 = r

which in turn leads to the equations

dβ0dτ

= αθβ1

dβ1dτ

=σ2

2β21 − αβ1 − 1

with initial conditions β0(0) = β1(0) = 0.

The solution of these equations is a mechanical exercise in which β1 is computed first followed by β0.

27

Q 30. In a previous exercise it has been shown that the price B(t, R1, R2) of a zero-coupon bond

at time t paying one dollar at maturity T satisfies the partial differential equation

∂B

∂t+ α1(θ1 − r1)

∂B

∂r1+ α2(θ2 − r2)

∂B

∂r2+σ21r12

∂2B

∂r21+σ22r22

∂2B

∂r22= (r1 + r2)B ,

when the instantaneous rate of interest, r(t), is driven by a two-factor model in which r(t) = r1(t) +

r2(t) in which r1(t) and r2(t) evolve stochastically in accordance with the equations

dr1 = α1(θ1 − r1) dt+ σ1√r1 dW1 ,

dr2 = α2(θ2 − r2) dt+ σ2√r2 dW2 ,

(8)

where dW1 and dW2 are independent increments in the Wiener processes W1 and W2.

Given that the required solution has generic solution B(t, r1, r2) = exp[β0(T − r) + β1(T − t)r1 +

β2(T − t)r2], construct the ordinary differential equations satisfied by the coefficient functions β0, β1

and β2. What are the appropriate initial conditions for these equations?

Answer

Let B(t, r) = exp[β0(T − r) + β1(T − t)r1 + β2(T − t)r2] and define τ = T − t then

∂B

∂t= −

(dβ0dτ

+ r1dβ1dτ

+ r2dβ2dτ

)B(t, r1, r2)

∂B

∂r1= β1B(t, r1, r2) ,

∂B

∂r2= β2B(t, r1, r2) ,

∂2B

∂r21= β21 B(t, r1, r2) ,

∂2B

∂r22= β22 B(t, r1, r2) .

Substituting these calculations into the bond equation gives

−(dβ0dτ

+ r1dβ1dτ

+ r2dβ2dτ

)+ α1(θ1 − r1)β1 + α2(θ2 − r2)β2 +

σ21r12

β21 +σ22r22

β22 = (r1 + r2) .

which in turn leads to the equations

dβ0dτ

= α1θ1β1 + α2θ2β2 ,

dβ1dτ

=σ212β21 − α1β1 − 1 ,

dβ2dτ

=σ222β22 − α2β2 − 1 ,

with initial conditions β0(0) = β1(0) = β2(0) = 0.

Q 31. The position x(t) of a particle executing a uniform random walk is the solution of the

stochastic differential equation

dxt = µdt+ σ dWt , x(0) = X ,

28

where µ and σ are constants. Find the density of x at time t > 0.

Answer - Intuitive approach

The SDE can be integrated immediately to get

xt = X +

∫ t

0µ(s) ds+ σWt , x(0) = X .

Consequently xt−X−µt is a Gaussian random deviate with mean value zero and variance σ2t. Thus

f(x, t) =1

σ√2πt

exp[− (x−X − µt)2

2σ2t

].

Answer - PDE approach

If x satisfies the SDE dxt = µdt + σ dWt then the density f(x, t) of x at time t satisfies the partial

differential equation∂f

∂t=σ2

2

∂2f

∂x2− µ

∂f

∂x.

The task is to solve this equation with initial condition f(x, 0) = δ(x−X). In absence of intuition,

take either the Fourier transform of f with respect to x or the Laplace transform of f with respect

to t. For example, let

f(t;ω) =

∫ ∞

−∞f(x, t) eiω x dx

then

df

dt=

σ2

2

∫ ∞

−∞

∂2f

∂x2eiω x dx− µ

∫ ∞

−∞

∂f

∂xeiω x dx

=(−ω2σ2

2+ µ i ω

)f

with initial condition f(0;ω) = eiω X . Clearly the solution of this first order ODE is

f(t;ω) = exp(−ω2σ2 t

2+ (X + µ t) i ω

).

However, this is the characteristic function of the Gaussian distribution with meanX+µ t and variance

σ2 t. Thus

f(x, t) =1

σ√2π t

e−(x−X−µ t)2/2σ2 t .

Q 32. The position x(t) of a particle executing a uniform random walk is the solution of the

stochastic differential equation

dxt = µ(t) dt+ σ(t) dWt , x(0) = X ,

where µ and σ are now prescribed functions of time. Find the density of x at time t > 0.

Answer - Intuitive approach

29

This is a repeat of the previous example. The SDE can be still be integrated immediately to get

xt = X +

∫ t

0µ(s) ds+

∫ t

0σ(s) dWs , x(0) = X .

In this case xt −X −∫ t

0µ(s) ds is an N

(0,

∫ t

0σ2(s) ds

)Gaussian random deviate. Thus

f(x, t) =1√

2π

∫ t

0σ2(s) ds

exp

[−

(x−X −

∫ t

0µ(s) ds

)2

2

∫ t

0σ2(s) ds

].

Answer - PDE approach

If x satisfies the SDE dxt = µ(t) dt + σ(t) dWt then the density f(x, t) of x at time t satisfies the

partial differential equation∂f

∂t=σ2(t)

2

∂2f

∂x2− µ(t)

∂f

∂x.

The task is to solve this equation with initial condition f(x, 0) = δ(x − X). The procedure is

precisely the same as the previous question, except that the Fourier transform of f is now the

preferred approach. Let

f(t;ω) =

∫ ∞

−∞f(x, t) eiω x dx

then

df

dt=

σ2(t)

2

∫ ∞

−∞

∂2f

∂x2eiω x dx− µ(t)

∫ ∞

−∞

∂f

∂xeiω x dx

=(−ω2σ2(t)

2+ µ(t) i ω

)f

with initial condition f(0;ω) = eiω X . Clearly the solution of this first order ODE is

f(t;ω) = exp[−ω2

2

∫ t

0σ2(u) du+

(X +

∫ t

0µ(u) du

)i ω

].

This function is now the characteristic function of the Gaussian distribution with mean and variance

X +

∫ t

0µ(u) du ,

∫ t

0σ2(u) du

Thus

f(x, t) =1√2π

1√∫ t

0σ2(u) du

exp

−(x−X −

∫ t

0µ(u) du

)2

2

∫ t

0σ2(u) du

.

30

Q 33. The state x(t) of a particle satisfies the stochastic differential equation

dxt = a dt+ b dWt , x(0) = X ,

where a is a constant vector of dimension n, b is a constant n ×m matrix and dW is a vector of

Wiener increments with m×m covariance matrix Q. Find the density of x at time t > 0.

Answer - Intuitive approach

This is again a repeat of the previous example. In matrix notation this SDE can be integrated

immediately to get

Xt = X0 +At+BW .

where Xt, X0 and A are N dimensional column vectors, B is a constant N ×M matrix and Wt is

an M dimensional column vector of correlated Wiener processes. Clearly Xt is an N dimensional

Gaussian deviate with expected value X0 +At. The covariance of Xt is therefore

E[(Xt −X0 −At)(Xt −X0 −At)T] = E[B(WtWTt )B

T] = BQBTt = Gt

where Qdt = E[dWtdWTt ]. Thus

f(X, t) =1

(2π t)N/21

|G |1/2exp

[− (X −X0 −At)G−1 (X −X0 −At)T

2t

].

Answer - PDE approach

If x satisfies the SDE dxt = a dt+ b dWt then the density f(x, t) of x at time t satisfies the partial

differential equation∂f

∂t=g jk2

∂2f

∂xj ∂xk− ak

∂f

∂xk.

where

g jk =m∑

r,s=1

b jrQ rs b ks .

The task is to solve this equation with initial condition f(x, 0) = δ(x−X). The n-dimensional Fourier

transform of f with respect to x is defined by the formula

f(t;ω) =

∫Rn

f(x, t) eiω ,x dx

in which ω is an n-dimensional vector. By taking the Fourier transform of the Kolmogorov equation

satisfied by f(x, t), it follows that

df

dt=

gjk2

∫Rn

∂2f

∂xj ∂xkeiω ,x dx− ak

∫Rn

∂f

∂xkeiω ,x dx

=(−g jkωjωk

2+ akωk i

)f

with initial condition f(0;ω) = eiω .X . Clearly

f(t;ω) = exp(−g jkωjωk t

2+ (Xk + ak t) i ωk

).

31

By way of variety, the probability density function f(x, t) is computed by direct inversion of the

Fourier transform using the identity

f(x, t) =1

(2π)n

∫Rn

f(t;ω) e−iω .x dω

=1

(2π)n

∫Rn

exp(− 1

2

[g jkωjωk t+ 2(xk −Xk − ak t) i ωk

) ]dω .

Let v = x−X− a t, then by observing that G = [gjk t] is a symmetric positive definite matrix, it is

straight forward algebra to demonstrate that

g jk ωjωk t+ (xk −Xk − ak t) i ωk = (ω + ivG−1)G (ω + ivG−1)T + vG−1 vT .

Therefore,

f(x, t) = exp[− vG−1 vT

2

] 1

(2π)n

∫Rn

exp[− 1

2(ω + ivG−1)G (ω + ivG−1)T

]dω

= exp[− vG−1 vT

2

] 1

(2π)n

∫Rn

exp[− 1

2ξG ξT

]dξ .

In order to evaluate this integral, observe that since G is symmetric and positive definite then there

exists a non-singular matrix F such that G = FFT. Thus

ξG ξT = ξ FFT ξT = (ξ F ) (ξ F )T = η ηT , η = ξ F .

Changing variables from ξ to η = ξ F gives

f(x, t) = exp[− vG−1 vT

2

] 1

(2π)n

∫Rn

exp[− 1

2ξG ξT

]dξ

= exp[− vG−1 vT

2

] 1

(2π)n

∫Rn

1

detFexp

[− η ηT

2

]dη

= exp[− vG−1 vT

2

] 1

(2π)n

∫Rn

1

detFexp

[− η21 + η22 + · · ·+ η2n

2

]dη .

Now |F |2 = |G | = tn | g | where g = [g jk]. Thus |F | = tn/2 | g |1/2. Furthermore,∫Rn

exp[− η21 + η22 + · · ·+ η2n

2

]dη =

[ ∫Rn

exp(−η2/2) dη]n

= (2π)n/2 .

In conclusion, the probability density function of x is

f(x, t) =1

(2π t)n/21

| g |exp

[− (x−X− a t) g−1 (x−X− a t)T

2t

], g = [g jk] .

Q 34. The state x(t) of a system evolves in accordance with the stochastic differential equation

dxt = µx dt+ σ x dWt , x(0) = X ,

32

where µ and σ are constants. Find the density of x at time t > 0.

Answer - Intuitive approach

The SDE is the geometric random walk. Take Y = logX and use Ito’s lemma to construct the SDE

satisfied by Yt. Ito’s lemma yields

dY =dY

dXdX +

σ2X2

2

d2Y

dX2dt → dY =

1

X(µX dt+ σX dWt) +

σ2X2

2

(−1

X2dt)

which simplifies to give

dY = (µ− σ2/2) dt+ σ dWt .

This equation has solution Y = Y0+(µ−σ2/2) t+σWt. Consequently Y is a Gaussian deviate with

mean value Y0 + (µ− σ2/2) t and variance σ2t. The density of Y is thus

fY (Y, t) =1

σ√2π t

exp[− (Y − Y0 − (µ− σ2/2)t )2

2σ2t

].

The transformation Y = logX is now used to map f(Y, t) into

fX(X, t) =1

σ√2π t

1

Xexp

[− ( log(X/X0)− (µ− σ2/2)t )2

2σ2t

].

Answer - PDE approach

If the state x(t) of a system satisfies the stochastic differential equation dx = µx dt+ σ x dWt with µ

and σ constants, then f(x, t), the density of x at time t > 0, satisfies the partial differential equation

∂ f(x, t)

∂t=

1

2

∂2

∂x2

(σ2x2f(x, t)

)− ∂

∂x

(µxf(x, t)

).

Let z = log x then∂f

∂x=∂f

∂z

1

x,

∂2f

∂x2=∂2f

∂z21

x2− ∂f

∂z

1

x2.

Thus it is seen that f satisfies the modified PDE

∂ f

∂t=

1

2

∂2

∂x2

(σ2x2f(x, t)

)− ∂

∂x

(µxf(x, t)

)=

σ2

2

∂

∂x

(2xf(x, t) + x2

∂f

∂x

)− ∂

∂x

(µxf(x, t)

)=

σ2

2

(2f(x, t) + 4x

∂f

∂x+ x2

∂2f

∂x2

)− µ

(x∂f

∂x+ f

)=

σ2x2

2

∂2f

∂x2+ (2σ2 − µ)x

∂f

∂x+ (σ2 − µ)f

=σ2

2

(∂2f∂z2

= −∂f∂z

)+ (2σ2 − µ)

∂f

∂z+ (σ2 − µ)f

=σ2

2

∂2f

∂z2+ (

3

2σ2 − µ)

∂f

∂z+ (σ2 − µ)f .

Now take the Fourier transform of this equation with respect to z to deduce that

f(t;ω) =

∫Rf(z, t) ei ω z dz

33

satisfies the ordinary differential equation

df

dt=

(− σ2ω2

2− i (

3

2σ2 − µ)ω + (σ2 − µ)

)f

with initial condition f(0, ω) = X−1 ei ω logX . Clearly,

f(t, ω) = X−1 exp(− σ2ω2 t

2+ i (logX − 3t

2σ2 + µ t)ω + (σ2 − µ) t

)with inverse transform

f = e(σ2−µ) t 1

σX√2π t

exp[−

(z − logX +3t

2σ2 − µ t)2

2σ2t

]

= e(σ2−µ) t 1

σX√2π t

exp[−

(log x− logX +3t

2σ2 − µ t)2

2σ2t

]

= e(σ2−µ) t 1

σX√2π t

exp[−

(log x/X +(σ22

− µ)t+ σ2 t)2

2σ2t

]

= e(σ2−µ) t 1

σX√2π t

exp[−

(log x/X +(σ22

− µ)t)2

2σ2t− 2

(log x/X +(σ22

− µ)t)

2σ2tσ2 t− σ2 t

2

]

= e(σ2−µ) t 1

σX√2π t

exp[−

(log x/X +(σ22

− µ)t)2

2σ2t− log x/X −

(σ22

− µ)t− σ2 t

2

]

=1

σX√2π t

exp[−

(log x/X +(σ22

− µ)t)2

2σ2t− log x/X

]

=1

σX√2π t

exp[−

(log x/X +(σ22

− µ)t)2

2σ2t

]exp(− log x/X)

=1

σ x√2π t

exp[−

(log x/X +(σ22

− µ)t)2

2σ2t

].

Q 35. The state x(t) of a system evolves in accordance with the Ornstein-Uhlenbeck process

dx = −α (x− β) dt+ σ dWt , x(0) = X ,

where α, β and σ are constants. Find the density of x at time t > 0.

Answer - Intuitive approach

The SDE is reorganised into the form dx + α (x − β) dt = σ dWt. Ito’s lemma is now applied to

(x− β)eαt to obtain

d[(x− β)eαt] = σeαt dWt → (x− β)eαt = (X0 − β) + σ

∫ t

0eαsdWs .

34

which simplifies to give

x(t) = β + (X0 − β)e−αt + σ

∫ t

0e−α(t−s)dWs .

Thus x(t) is a gaussian deviate with mean value β + (X0 − β)e−αt and variance

σ2∫ t

0e−2α(t−s)ds =

σ2(1− e−αt)

2α.

The final conclusion is that x has probability density function

f(x, t) =1

σ

√α

π(1− e−αt)exp

[− α(x− β − (X0 − β)e−αt )2

σ2(1− e−αt)

].

Answer - PDE approach

If x(t) evolves in accordance with the stochastic differential equation dx = −α (x − β) dt + σ dWt

then the density of x at time t > 0 satisfies the partial differential equation

∂f

∂t=σ2

2

∂2f

∂x2+ α

∂

∂x

((x− β)f

)and the initial condition f(x, 0) = δ(x−X). Let

f(t;ω) =

∫Rf(x, t) ei ω x dx

then by taking the Fourier transform of the partial differential equation, it follows that f(t;ω) satisfies

the ordinary differential equation

∂f

∂t= −σ

2ω2

2f + α

∫R

∂

∂x

((x− β) f(x, t)

)ei ω x dx

= −σ2ω2

2f − i α ω

∫R(x− β) f(x, t) ei ω x dx

= −σ2ω2

2f + i α β ω f − αω

∂f

∂ω

with initial condition f(0, ω) = ei ω X . Clearly ϕ(t, ω) = log f satisfies the first order partial differential

equation∂ϕ

∂t+ αω

∂ϕ

∂ω= −σ

2ω2

2+ i α β ω

with initial condition ϕ(0, ω) = i ω X. We may solve this equation by characteristic methods. Take

η = logω − α t and ξ = ω then

∂ϕ

∂t+ αω

∂ϕ

∂ω= −α∂ϕ

∂η+ αω

( ∂ϕ∂ξ

+1

ω

∂ϕ

∂η

)= α ξ

∂ϕ

∂ξ= −σ

2ξ2

2+ i α β ξ .

This equation is now integrated to obtain

ϕ = −σ2ξ2

4α+ i β ξ + ψ(η)

35

where the initial condition yields

i ω X = −σ2ω2

4α+ i β ω + ψ(logω) → ψ(η) = i eη (X − β) +

σ2 e2η

4α.

It therefore follows that

ϕ = −σ2ω2

4α+ i β ω + i (X − β) elogω−α t +

σ2

4αe2(logω−α t)

= −σ2ω2

4α+ i β ω + i (X − β)ω e−α t +

σ2ω2

4αe−2α t

= −σ2ω2

4α( 1− e−2α t ) + i ω (β + (X − β) e−α t ) .

Thus f(t;ω) = expϕ corresponds to a Gaussian distribution with mean β+(X−β) e−α t and variance

σ2(1− e−2α t)/2α, that is

f(x, t) =1

σ

√α

π(1− e−2α t)exp

[− α

(x− β − (X − β) e−α t)2

σ2(1− e−2α t)

].

Q 36. If the state of a system satisfies the stochastic differential equation

dx = a(x, t) dt+ b(x, t) dW , x(0) = X ,

write down the initial value problem satisfied by f(x, t), the probability density function of x at time

t > 0. Determine the initial value problem satisfied by the cumulative density function of x.

Answer

Let f(x, t) be the probability density function corresponding to the distribution of the states of the

stochastic differential equation

dx = a(x, t) dt+ b(x, t) dW , x(0) = X ,

then f(x, t) satisfies the partial differential equation

∂f

∂t=

1

2

∂2(b2f)

∂x2− ∂(af)

∂x, f(x, 0) = δ(x−X) .

Let F (x, t) be the cumulative distribution function of f(x, t) then

F (x, t) =

∫ x

−∞f(u, t) du → f(x, t) =

∂F

∂x.

The equation satisfied by F (x, t) is therefore

∂2F

∂x∂t=

1

2

∂2

∂x2

(b2∂F

∂x

)− ∂

∂x

(a∂F

∂x

)→ ∂

∂x

[∂F

∂t− 1

2

∂

∂x

(b2∂F

∂x

)+ a

∂F

∂x

]= 0 .

Thus it follows that F satisfies

∂F

∂t=

1

2

∂

∂x

(b2∂F

∂x

)− a

∂F

∂x+ ψ(t) .

36

However,

F (x, t) → 1 ,∂F

∂x= f → 0 ,

∂2F

∂x2=∂f

∂x→ 0 , x→ ∞

and therefore ψ(t) = 0. In conclusion, F (x, t) satisfies the partial differential equation

∂F

∂t=

1

2

∂

∂x

(b2∂F

∂x

)− a

∂F

∂x

with initial condition

F (x, 0) =

∫ x

−∞f(u, 0) du =

0 x < X ,

1/2 x = X ,

1 x > X .

Q 37. Cox, Ingersoll and Ross proposed that the instantaneous interest rate r(t) should follow the

stochastic differential equation

dr = α(θ − r)dt+ σ√r dW , r(0) = r0 ,

where dW is the increment of a Wiener process and α, θ and σ are constant parameters. Show that

this equation has associated transitional probability density function

f(t, r) = c(vu

)q/2e−(

√u−

√v)2 e−2

√uv Iq(2

√uv) ,

where Iq(x) is the modified Bessel function of the first kind of order q and the functions c, u, v and

the parameter q are defined by

c =2α

σ2(1− e−α(t−t0)), u = cr0 e

−α(t−t0) , v = cr , q =2αθ

σ2− 1 .

Answer

The transitional probability density function for the stochastic differential equation

dr = α(θ − r)dt+ σ√r dw ,

satisfies the partial differential equation

∂f

∂t=σ2

2

∂2 [rf ]

∂r2− α

∂[(θ − r)f ]

∂r, f(r, 0) = δ(r −R) .

Since the variance is proportional to r (and cannot be negative), the sample space of r is (0,∞). Let

f(t;ω) =

∫ ∞

0f(r, t) e−ω r dr

37

then the Laplace transform of the Kolmogorov equation gives

∂f(t;ω)

∂t=

∫ ∞

0

∂

∂r

[σ22

∂ [rf ]

∂r− α (θ − r)f

]e−ω r dr

=[ ( σ2

2

∂ [rf ]

∂r− α (θ − r)f

)e−ω r

]∞0

+ ω

∫ ∞

0

[σ22

∂ [rf ]

∂r− α (θ − r)f

]e−ω r dr

= ω

∫ ∞

0

[σ22

∂ [rf ]

∂r− α (θ − r)f

]e−ω r dr

=σ2ω

2

[rf e−ω r

]∞0

+σ2ω2

2

∫ ∞

0rf e−ω r dr − ωαθ f(t;ω) + ωα

∫ ∞

0rf e−ω r dr

= −σ2ω2

2

∂f(t;ω)

∂ω− ωαθ f(t;ω)− ωα

∂f(t;ω)

∂ω

= −[ σ2ω2

2+ ωα

] ∂f(t;ω)∂ω

− ωαθ f(t;ω) .

Let ϕ(t, ω) = log f(t;ω) then clearly ϕ is the solution of the partial differential equation

∂ϕ

∂t+

( σ2ω2

2+ ωα

) ∂ϕ∂ω

= −ωαθ

with initial condition ϕ(0, ω) = log f(0;ω) = −ωR. Take ξ = ω and η = logω − log(2α+ σ2ω)− α t

then

∂ϕ

∂t+

( σ2ω2

2+ ωα

) ∂ϕ∂ω

= −α ∂ϕ∂η

+( σ2ω2

2+ ωα

)( ∂ϕ∂ξ

+∂ϕ

∂η

( 1

ω− σ2

2α+ σ2ω

))= −α ∂ϕ

∂η+σ2ω2 + 2ωα

2

( ∂ϕ∂ξ

+∂ϕ

∂η

2α

ω(2α+ σ2ω)

)=

σ2ω2 + 2ωα

2

∂ϕ

∂ξ.

Therefore ϕ satisfies the partial differential equation

σ2ω2 + 2ωα

2

∂ϕ

∂ξ= −ωαθ → ∂ϕ

∂ξ= − 2αθ

σ2ξ + 2α.

Thus the general solution for ϕ is

ϕ = −2αθ

σ2log(σ2ξ + 2α) + ψ(η)

where the initial condition gives

−ωR = −2αθ

σ2log(σ2ω + 2α) + ψ(logω − log(2α+ σ2ω) ) .

Let λ = log(ω/(2α+ σ2ω)) then

ω

2α+ σ2ω= eλ → ω =

2α

e−λ − σ2.

Thus

ψ(λ) =−2αR

e−λ − σ2+

2αθ

σ2log

( 2α e−λ

e−λ − σ2

).

38

Bearing in mind that the task is to compute f(t;ω) = eϕ, it follows that

f(t;ω) = (σ2ω + 2α)−(1+q) eψ(η)

= (σ2ω + 2α)−(1+q)( 2α

1− σ2 eη

)q+1exp

[ −2αR

e−λ − σ2

]=

( 2α

(σ2ω + 2α)(1− σ2 eη)

)q+1exp

[ −2αReη

1− σ2 eη

]where the parameter q = 2αθ/σ2 − 1. Now

eη = e−α tω

2α+ σ2ω

which further simplifies f(t;ω) to obtain

f(t;ω) =( 2α

2α+ σ2ω(1− e−αt)

)q+1exp

[ −2αRω e−αt

2α+ σ2 ω(1− e−αt)

].

Let functions c, u, v be defined by

c =2α

σ2(1− e−αt), u = cR e−αt) , v = cr

then

f(t;ω) =( c

c+ ω

)q+1exp

[ −cRω e−αtc+ ω

]=

( c

c+ ω

)q+1exp

[ −uωc+ ω

]=

( c

c+ ω

)q+1e−u exp

[ cu

c+ ω

].

Since f(t;ω) is a function of (c+ ω), it follows that

f(r, t) = e−u e−cr L−1[(c/ω)q+1 exp(cu/ω) ;ω → r

].

To complete this calculation, we compute the Laplace transform of rq/2 Iq(2√ucr) where Iq(x) is the

modified Bessel function of argument x. The result is

L[rq/2 Iq(2

√ucr) ; ω] =

∫ ∞

0rq/2 Iq( 2

√ucr ) e−ω r dr

=

∫ ∞

0rq/2

( (2√ucr )q

2qΓ(q + 1)

∞∑k=0

(ucr)k

k! (q + 1)k

)e−ω r dr

=(uc)q/2

Γ(q + 1)

( ∞∑k=0

(uc)k

k! (q + 1)k

∫ ∞

0rq+k e−ω r dr

)=

(uc)q/2

Γ(q + 1)

( ∞∑k=0

(uc)k

k! (q + 1)k

Γ(k + q + 1)

ωq+k+1

)=

1

c

( uc

)q/2 (c/ω)q+1

Γ(q + 1)

( ∞∑k=0

(uc/ω)k

k!Γ(q + 1)

)=

1

c

( uc

)q/2(c/ω)q+1 exp(uc/ω) .

39

Thus

f(r, t) = c e−u e−cr( cu

)q/2rq/2 Iq(2

√ucr)

= c e−u e−cr(cru

)q/2Iq(2

√ucr)

= c(vu

)q/2e−(

√u−

√v)2 e−2

√uv Iq(2

√uv)

when cr is replaced by v.

Q 38. Consider the problem of numerically integrating the stochastic differential equation

dx = a(t, x) dt+ b(t, x) dW , x(0) = X0 .

Develop an iterative scheme to integrate this equation over the interval [0, T ] using the Euler-

Maruyama algorithm.

It is well-know that the Euler-Maruyama algorithm has strong order of convergence one half and

weak order of convergence one. Explain what programming strategy one would use to demonstrate

these claims.

Answer

Let N be the number of steps to be taken in advancing the solution from t = 0 to t = T , then

∆t = T/N . The standard deviation of each Wiener step is therefore σ =√∆ and the iterative

scheme is captured by the pseudo-code

1. Initialize x at X0;

2. Iterate N times x − > a(k∆t, x)∆t+ σb(k∆t, x) ξ where ξ ∼ N(0, 1).

3. Final value of x is x(T )

Note that there is no requirement to store intermediate values of x.

Strategy First it is necessary to simulate the integration process a large number of times, say M

times, by which is meant that x(T ) will be simulated M times with each simulation based on an

underlying realisation of the Wiener process. Let x∆tk (T ) be the value of x(T ) returned by the kth

simulation when using the step size ∆t. In computing x∆tk (T ) we choose a very fine resolution of the

interval [0, T ], say involving N small time steps, and construct the corresponding series of N Wiener

increments. These increments, appropriately packaged to build the Wiener increments ∆W∆t over

intervals of duration ∆t, define the realisation of the underlying Wiener process needed to compute

x∆tk (T ) and, of course, by integrating at the finest resolution, i.e. with ∆t = T?N , one computes

numerically ones best estimate of the true value of xk(T ) against which numerical error will be

measured. Needless to say, if the SDE has an exact solution expressed in terms of W (T ), then this

solution would be taken as the exact solution against which error is to be estimated.

40

For each realisation the error ek = x∆tk (T )− xk(T ) is computed. To test for strong convergence one

plots log(∑M

k=1 | ek |)/M against log∆t and to test for weak convergence one plots log

∣∣∣∑Mk=1 ek

∣∣∣/Magainst log∆t. The former plot will be an approximate straight line with gradient one half and the

latter plot will be an approximate straight line of gradient one.

Q 39. Compute the stationary densities for the following stochastic differential equations.

(a) dX = (β − αX) dt+ σ√X dW

(b) dX = −α tanX dt+ σ dW

(c) dX = [(θ1 − θ2) cosh(X/2)− (θ1 + θ2) sinh(X/2)] cosh(X/2) dt+ 2 cosh(X/2) dW

(d) dX =α

Xdt+ dW

(e) dX =( αX

−X)dt+ dW

Answer

The stationary density f(x) satisfies

1

2

d(gf)

dx−µf = 0 → d(gf)

dx=

2µ

g(gf) → d log(gf)

dx=

2µ

g→ f(x) =

A

gexp

(∫2µ

gdx

),

where A is a constant which takes the value which ensures that f integrates to one.

(a) Here µ = (β − αX) and g(x) = σ2X. Thus

f(x) =A

σ2xexp

(∫2(β − αx)

σ2 xdx

)=

A

σ2xexp

(∫2β

σ2x− 2α

σ2dx

)=

A

σ2x2β/σ

2−1 e−2αx/σ2.

(b) Here µ = −α tanX and g(x) = σ2. Thus

f(x) =A

σ2exp

(− 2

∫tanx

σ2dx

)=

A

σ2exp

(2log | cosx|

σ2

)=

A

σ2( cos2 x)1/σ

2

.

(c) Here µ = [(θ1 − θ2) cosh(X/2)− (θ1 + θ2) sinh(X/2)] cosh(X/2) and and g(x) = 4 cosh2(X/2).

Thus

f(x) =A

4 cosh2(x/2)exp

(∫[(θ1 − θ2) cosh(x/2)− (θ1 + θ2) sinh(x/2)] cosh(x/2)

4 cosh2(x/2)dx

)=

A

4 cosh2(x/2)exp

(∫θ1 − θ2

4− θ1 + θ2

4

sinh(x/2)

cosh(x/2)dx

)=

A

4 cosh2(x/2)exp

(∫(θ1 − θ2)x

4− θ1 + θ2

2log cosh(x/2)

)=

A

4 cosh2+(θ1+θ2)/2(x/2)exp

((θ1 − θ2)x

4

).

41

(d) Here µ = α/X and g(x) = σ2. Thus

f(x) =A

σ2exp

(∫2α

σ2xdx

)=

A

σ2exp

(2ασ2

log x)=

A

σ2x2α/σ

2.

(e) Here µ = (α/X −X) and g(x) = σ2. Thus

f(x) =A

σ2exp

(2

∫(α/x− x)

σ2dx

)=

A

σ2exp

(2ασ2

log x− x2

σ2

)=

A

σ2x2α/σ

2e−x

2/σ2.