1 Exercise 11.1

Stochastic Di¤erential Equations Exercises

Exercise 11.1. The stochastic process

C_t=C₀e ^Wt :t 0 ; r₀ 0

represents the exchange rate evolution, that isC_tis the timetvalue in the domestic currency of one unit of the foreign currencyfW_t:t 0g is a standard Brownian motion.

a) What is the stochastic di¤erential equation satis…ed byfC_t:t 0g.

b) The price evolution of a risky asset, expressed in the foreign currency, is denoted fX_t:t 0g. It satis…es the SDE

dX_t= X_t dt+ X_t dW_t

wherefW_t :t 0gis a Brownian motion, independent offW_t:t 0g:What is the SDE of the risky asset pricefYt :t 0g, expressed in the domestic currency?

Exercise 11.2.

a) Show that there is a unique solution to

dX_t=a(b X_t)dt+ dW(t): b) Find explicitly this solution.

c) What behavior do you expect if the parameters are:.

a = 1; b= 1; = 1; t= 0;01and X0 = 1 a = 1; b= 1; = 1; t= 0;01 and X₀ = 1 a = 1; b= 1; = 0;1; t= 0;01 and X₀ = 1 a = 1; b= 2; = 0;3; t= 0;01 and X₀ = 1

Exercise 11.3. Describe the behavior of the solution of dX_t= 1

2 X_t dt+p

X_t(1 X_t)dW_t

assuming that X₀ is a random variable taken value in the unit interval [0;1]. Explain intuitively why X_t takes value between 0and 1.

Exercise 11.4. Assume that a particle move randomly in a plan such that its position at time t is given by W_t⁽¹⁾;W_t⁽²⁾ where W⁽¹⁾ and W⁽²⁾ are independent Brownian motions.

The distance to the origin is B_t =

r

W_t^{(1) 2}+ W_t^{(2) 2}; t >0:

Knowing that the quadratic covariation of two independent martingale is nil, use Itô’s lemma to show that

dB_t= 1 2

1

B_tdt+W_t⁽¹⁾

B_t dW_t⁽¹⁾+ W_t⁽²⁾

B_t dW_t⁽²⁾:

Exercise 11.5. Show that the stochastic process fS_t:t 0g where S_t=S₀exp

Z t 0

sds

2

2 t+ W_t and W is a( ;F;fF^t:t 0g; P) Brownian motion satis…es

dS_t= _tS_tdt+ S_tdW_t:

Exercise 11.6. Vasicek model. Show that the stochastic process fr_t:t 0g where r_t= +e ^t r₀ + e ^t

Z t 0

e ^sdW_s

and W is a( ;F;fF^t:t 0g;P) Brownian motion satis…es dr_t = ( r_t)dt+ dW_t: Hint : Let Y_t =Rt

0 e ^sdW_s and show thatY is an Itô process.

Exercise 11.7. Consider the system of stochastic di¤erential equations:

dX_t = ( X_t) dt+ dW_t dY_t = e e Y_t dt+edfW_t where the P Brownian motion W and fW are such that

Corr^P h

W_t;fW_ti

= , 8t >0.

a) Find the unique strong solution of this system.

b) DetermineE^P [X_t]:

c) The parameter is called the long term mean and the parameter is the speed of reversion. Justify these statements.

d) DetermineVar^P[X_t]: e) Determine Corr^P[X_t; Y_t]:

Solutions

1 Exercise 11.1

a) Letf(w) =C0e ^w. Since d

dwf(w) =C₀ e ^w and d²

dw²f(w) =C₀ ²e ^w; f(W_t) =C_t, df

dw(W_t) = C_t and d²f

dw² (W_t) = ²C_t: Using Itô’s lemma, we get

dC_t = C_t dW_t+

2

2 C_t dt:

b) The multiplication rule leads to dY_t = dX_tC_t

= X_t dC_t+C_t dX_t+dhX; Cit

= X_t C_t dW_t+

2

2 C_t dt +C_t( X_t dt+ X_t dW_t)

= Y_t dW_t+

2

2 Y_t dt+ Y_t dt+ Y_t dW_t

= Y_t dW_t+ Y_t dW_t +

2

2 + Y_t dt:

2 Exercise 11.2

a) We need to verify that.

(i) jb(x; t) b(y; t)j+j (x; t) (y; t)j K jx yj; 8t 0 (ii) jb(x; t)j+j (x; t)j K (1 +jxj); 8t 0

(iii) E[X₀²]<1

Veri…cation of (i) :

jb(x; t) b(y; t)j+j (x; t) (y; t)j = ja(b x) a(b y)j+j j

= jaj jx yj:

Veri…cation of (ii) :

jb(x; t)j+j (x; t)j = ja(b x)j+ jabj+jaj jxj+

max (jabj+ ;jaj) (1 +jxj) LetK = max (jabj+ ;jaj).

Veri…cation of(iii) :because the initial condition is not speci…ed, we need to chose a square integrable random variable X₀.

b) Let Z_t=X_t b: Itô’s lemma implies that dZ_t = dX_t

= a(b X_t)dt+ dW (t)

= Z_t+ dW(t):

The stochastic process Z is an Ornstein-Uhlenbeck process and we have shown that the strong solution is

Z_t=Z₀e ^at+ e ^at Z t

0

e^asdW_s: ReplacingZt with Xt b; we get

X_t b = (X₀ b)e ^at+ e ^at Z t

0

e^asdW_s:

That is why the solution Is

X_t=X₀exp ( at) +b(1 exp ( at)) + exp ( at) Z t

0

exp (as)dW_s: Let Y_t =Rt

0 exp (as)dW_s. Y =fY_t:t2[0; T]g is an Itô process¹ with Y₀ = 0, K_s = 0 and H_s=e^as. Moreover,

X_t =X₀exp ( at) +b(1 exp ( at)) + exp ( at)Y_t:

1Rappel. Le processus stochastiqueX =fXt: 0 t Tgest un processus d’Itô si pour toutt2[0; T]; Xt

P p.s.

= X0+ Z t

0

Ksds+ Z t

0

HsdWs

oùX0est une variable aléatoireF⁰ mesurable, les processusK=fKt: 0 t TgetH =fHt: 0 t Tg sontF adaptés,

Z T

0 jKsjds <1et Z T

0

H_s²ds <1P p.s.

Dans le cas qui nous occupe,RT

0 jKsjds=RT

0 0s= 0et RT

0 e^2asds= ¹₂^{e2T a}_a ¹ <1:

Letf(y; t) = X0e ^at+b(1 exp ( at)) + e ^aty and note that

@f

@Y (y; t) = e ^at

@²f

@Y² (y; t) = 0 d

dt X0e ^at+b(1 exp ( at)) + e ^aty = a X0e ^at+be ^at e ^aty

= a b X₀e ^at b+be ^at e ^aty

= a(b f(y; t)):

Since X_t =f(Y_t; t), the fourth version of Itô’s lemma² gives the desired result:

df(Y_t; t) = @f

@Y (Y_t; t) H_t dW_t + @f

@Y (Y_t; t) K_t+@f

@t (Y_t; t) + 1 2

@²f

@Y² (Y_t; t) H_t² dt:

dX_t = df(X_t; t)

= e ^at e^at dW_t+ e ^at 0 +a(b f(Y_t; t)) + 1

20 (e^as)² dt

= dWt+a(b f(Yt; t)) dt

= dW_t+a(b X_t) dt:

3 Exercise 11.3

The drift term ¹₂ X_t induce a reversion behavior toward ¹₂. Indeed, if X_t > ¹₂, then

1

2 X_t < 0; implying that the local behavior of X tends to decrease just after t: On the other hand, if X_t< ¹₂, then ¹₂ X_t >0, implying that Xtends to increase. If X_t = ¹₂, then the drift term is nil, having locally no impact on X.

The di¤usion termp

X_t(1 X_t)reaches a maximum atX_t= ¹₂. It worth zero whenever X_t= 0 or1. If X_t = 0, the drift term is positive and the di¤usion term is nil, implying that

2Rappel :

df(Y_t; t) = @f

@Y (Y_t; t) H_tdW_t + @f

@Y (Yt; t) Kt+@f

@t (Yt; t) +1 2

@²f

@Y²(Yt; t) H_t² dt:

Xt+" is strictly positive. If Xt = 1, then the drift term is negative and the di¤usion term is nil, implying that X_t+" <1:

4 Exercise 11.4

Letf(w1; w2) =p

w²₁ +w²₂:

@f

@w₁ (w₁; w₂) = w₁

pw²₁+w₂² = w₁ f(w₁; w₂);

@f

@w₂ (w1; w2) = w₂

pw²₁+w₂² = w₂ f(w₁; w₂);

@²f

@w²₁ (w₁; w₂) = w²₂ pw₁²+w₂²

3 = w²₂

(f(w₁; w₂))³;

@²f

@w²₂ (w₁; w₂) = w²₁ pw₁²+w₂^{2 3}

= w²₁ (f(w₁; w₂))³; and @²f

@w₁@w₂ (w₁; w₂) = w₁w₂ pw₁²+w₂²

3:

Therefore,

dB_t = df W_t⁽¹⁾; W_t⁽²⁾

= W_t⁽¹⁾ f W_t⁽¹⁾; W_t⁽²⁾

dW_t⁽¹⁾+ W_t⁽²⁾ f W_t⁽¹⁾; W_t⁽²⁾

dW_t⁽²⁾

+1 2

W_t⁽²⁾

2

f W_t⁽¹⁾; W_t^{(2) 3}

d W⁽¹⁾ _t+1 2

W_t⁽¹⁾

2

f W_t⁽¹⁾; W_t^{(2) 3}

d W⁽²⁾ _t

W_t⁽¹⁾W_t⁽²⁾ f W_t⁽¹⁾; W_t⁽²⁾

3dD

W⁽¹⁾;W_t⁽²⁾E

t

= W_t⁽¹⁾ f W_t⁽¹⁾; W_t⁽²⁾

dW_t⁽¹⁾+ W_t⁽²⁾ f W_t⁽¹⁾; W_t⁽²⁾

dW_t⁽²⁾+ 1 2

W_t⁽¹⁾

2

+ W_t⁽²⁾

2

f W_t⁽¹⁾; W_t^{(2) 3} dt

= W_t⁽¹⁾

B_t dW_t⁽¹⁾+ W_t⁽²⁾

B_t dW_t⁽²⁾+ 1 2

B_t² B_t³dt

= W_t⁽¹⁾

B_t dW_t⁽¹⁾+ W_t⁽²⁾

B_t dW_t⁽²⁾+ 1 2

1 B_tdt:

5 Exercise 11.7

a)

X_t = + (x₀ ) exp ( t) + Z t

0

exp ( (t s)) dW_s Y_t = e+ x₀ e exp ( et) +

Z t 0

eexp ( e(t s)) dfW_s b)

E^P [X_t] = E^P + (x₀ ) exp ( t) + Z t

0

exp ( (t s)) dW_s

= + (X₀ ) exp ( t) + E^P Z t

0

exp ( (t s)) dW_s

| {z }

=0

= + (x₀ ) exp ( t):

c) If 0;

tlim!1E^P[X_t] = lim

t!1 + (x₀ ) exp ( t) = ; that is, 8" >0, 9t( ; ")>0such that 8t > t( ; "), E^P[X_t] < ".

Note that if 0 ₁ < ₂, then t( ₁; ") > t( ₂; "), that is, the larger is, more quickly E^P[X_t] revert back to .

d)

Var^P + (x₀ ) exp ( t) + Z t

0

exp ( (t s)) dW_s

= Var^P Z t

0

exp ( (t s)) dW_s

= E^P

" Z t 0

exp ( (t s)) dW_s

2#

= E^P Z t

0

( exp ( (t s)))² ds

= Z t

0

( exp ( (t s)))² ds

=

2

2 1 e ^{2 t} :

e)

Cov^P[Xt; Yt] = Cov^P

"

+ (x₀ ) exp ( t) +Rt

0 exp ( (t s)) dW_s; e+ x₀ e exp ( et) +Rt

0 eexp ( e(t s)) dfW_s

#

= Cov^P Z t

0

exp ( (t s)) dW_s; Z t

0 eexp ( e(t s)) dfW_s

= E^P Z t

0

exp ( (t s)) dW_s

Z t

0 eexp ( e(t s)) dfW_s

= E^P Z t

0

exp ( (t s))eexp ( e(t s)) ds

= e

Z t 0

exp ( ( +e) (t s)) ds

= e

+e 1 e ^{( +e)t} :

Corr^P[X_t; Y_t] = Cov^P[X_t; Y_t] q

Var^P[Xt] q

Var^P[Yt]

=

e

+e 1 e ^{( +e)t} q 2

2 (1 e ^{2 t}) qe²

2e(1 e ^2et)

= 2 p e

+e

1 e ^{( +e)t}

p(1 e ^{2 t}) (1 e ^2et):