MFE6516 Stochastic Calculus for Finance · StochasticCalculus MultivariateStochasticCalculus Black-Scholes-MertonModel MFE6516StochasticCalculusforFinance WilliamC.H.Leon NanyangBusinessSchool

Stochastic CalculusMultivariate Stochastic Calculus

Black-Scholes-Merton Model

MFE6516 Stochastic Calculus for Finance

William C. H. Leon

Nanyang Business School

December 11, 2017

1 / 77 William C. H. Leon MFE6516 Stochastic Calculus for Finance



1 Stochastic CalculusIto IntegralIto-Doeblin FormulaSome Applications

2 Multivariate Stochastic CalculusMultivariate Brownian MotionMultivariate Ito-Doeblin Formula

3 Black-Scholes-Merton ModelBlack-Scholes-Merton EquationParabolic Partial Differential Equation




Ito IntegralIto-Doeblin FormulaSome Applications

Ito Integral for Simple Integrands

Let(Ω,F ,

(Ft

),P

)be a filtered probability space, W be a Brownian

motion, h be an adapted process and T be a positive constant.

Let Π = {t0, t1, t2, . . . , tm} be a partition of [0,T ] and consider a simpleprocess

h(t) =

m∑i=1

h(ti−1) 1[ti−1,ti )(t).





For 0 ≤ t ≤ T , define

I (t) =

∫ t

0

h(u) dW (u)

as follow:

I (t) =

k−1∑i=1

h(ti−1)(W (ti )−W (ti−1)

)+ h(tk−1)

(W (t)−W (tk−1)

),

if tk−1 ≤ t < tk , and

I (T ) =

m∑i=1

h(ti−1)(W (ti )−W (ti−1)

).





Some Properties of Ito Integral

I (t) =

∫ t

0

h(u) dW (u)

1 Ito integral is a martingale.

2 E (I (t)) = 0 for all t ≥ 0.

3 (Ito isometry) E(I 2(t)

)= E

(∫ t

0h2(u) du

)for all t ≥ 0.

4 (Quadratic variation) [I , I ](t) =∫ t

0h2(u) du for all t ≥ 0.





Proof

Let 0 ≤ s < t ≤ T where s ∈ [tl−1, tl ), t ∈ [tk−1, tk) and l ≤ k . Then

I (t) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

I (s) + h(tl−1)(W (t)−W (s)

)if l = k ;

I (s)+h(tl−1)(W (tl )−W (s)

)+

k−1∑i=l+1

h(ti−1)(W (ti )−W (ti−1)

)+ h(tk−1)

(W (t)−W (tk−1)

)if l < k .

Thus,

E( I (t) | F(s)) = I (s) and E (I (t)) = 0.





Let Di = W (ti )−W (ti−1), for i = 1, 2, . . . , k − 1, andDk = W (t)−W (ti−1). Then

I 2(t) =

(k∑

i=1

h(ti−1)Di

)2

=

k∑i=1

h2(ti−1)D2i + 2

∑∑1≤i<j≤k

h(ti−1)h(tj−1)DiDj .

Note that

1 h2(ti−1) ∈ F(ti−1) and Di is independent of F(ti−1), for 1 ≤ i ≤ k .

2 h(ti−1)h(tj−1)Di ∈ F(tj−1) and Dj is independent of F(tj−1), for1 ≤ i < j ≤ k .





E(I 2(t)

)=

k∑i=1

E(h2(ti−1)D

2i

)+ 2

∑∑1≤i<j≤k

E (h(ti−1)h(tj−1)DiDj)︸︷︷︸=0

=k−1∑i=1

E(h2(ti−1)

)(ti − ti−1) + E

(h2(tk−1)

)(t − tk−1)

= E

(k−1∑i=1

h2(ti−1)(ti − ti−1) + h2(tk−1)(t − tk−1)

)

= E

(∫ t

0

h2(u) du

).





Let ΠS = {s0, s1, s2, . . . , sm} be a partition of [tj−1, tj ]. Then

lim‖ΠS‖→0

m∑i=1

(I (si )− I (si−1)

)2

= lim‖ΠS‖→0

h2(tj−1)

m∑i=1

(W (si )−W (si−1)

)2

= h2(tj−1)(tj − tj−1

)=

∫ tj

tj−1

h2(u) du.

Thus,

[I , I ](t) =

∫ t

0

h2(u) du.





Ito Integral for General Integrands

Suppose the adapted process h is square integrable, i.e.

E

(∫ T

0

h2(u) du

)< ∞.

Choose a sequence of simple process hn such that

limn→∞

E

(∫ T

0

∣∣hn(u)− h(u)∣∣2 du

)= 0.

For 0 ≤ t ≤ T , define∫ t

0

h(u) dW (u) = limn→∞

∫ t

0

hn(u) dW (u).





Properties of Ito Integral

Fix T > 0. Let h be an adapted and square integrable process. Then theIto integral I (t) =

∫ t

0h(u) dW (u) as defined earlier has the following

properties:

1 (Continuity) Paths of I (t), as a function of t, are continuous.

2 (Adaptivity) I (t) is F(t)-measurable for all t.

3 (Linearity) Ito integral of a linear combination is the linearcombination of Ito integrals.

4 (Martingale) Ito integral is a martingale.

5 (Ito isometry) E(I 2(t)

)= E

(∫ t

0h2(u) du

)for all t ≥ 0.

6 (Quadratic variation) [I , I ](t) =∫ t

0h2(u) du for all t ≥ 0.





Exercise

Compute ∫ T

0

W (u) dW (u).

Hint: Let n ∈ Z+ and ti =

Tni , for i = 0, 1, 2, . . . , n. Consider the simple

process

hn(t) =

n∑i=1

W (ti−1) 1[ti−1,ti )(t).





Solution

∫ T

0

W (u) dW (u) = limn→∞

∫ T

0

hn(u) dW (u)

= limn→∞

n∑i=1

W (ti−1)(W (ti )−W (ti−1)

)

= limn→∞

1

2

(W 2(T )−

n∑i=1

(W (ti )−W (ti−1)

)2)

=1

2

(W 2(T )− [W ,W ](T )

)=

1

2

(W 2(T )− T

).





Remark

Recall

Var (W (t)) = t = [W ,W ](t).

Let

I (t) =

∫ t

0

W (u) dW (u) =1

2

(W 2(t)− t

).

Then

Var (I (t)) =1

4Var

(W 2(t)

)=

1

2t2

[I , I ](t) =

∫ t

0

W 2(u) du.





Ito-Doeblin Formula for Brownian Motion

Let f (t, x) be a C 1,2 function and W be a Brownian motion. Then, forevery T ≥ 0,

f(T ,W (T )

)=f

(0,W (0)

)+

∫ T

0

ft(t,W (t)

)dt

+

∫ T

0

fx(t,W (t)

)dW (t)︸︷︷︸

Ito integral

+1

2

∫ T

0

fxx(t,W (t)

)dt;

and in differential form

df(T ,W (T )

)=ft(T ,W (T )) dT

+ fx(T ,W (T )) dW (T ) +1

2fxx(T ,W (T )) dT .





Sketch of Proof

Let Π = {t0, t1, t2, . . . , tn} be a partition of [0,T ]. Then

f(T ,W (T )

)− f(0,W (0)

)=

n∑i=1

(f(ti ,W (ti )

)− f(ti−1,W (ti−1)

)).

By Taylor’s Theorem,

f(ti ,W (ti )

)− f(ti−1,W (ti−1)

)=ft

(ti−1,W (ti−1)

)(ti − ti−1

)+ fx

(ti−1,W (ti−1)

)(W (ti )−W (ti−1)

)+

1

2ftt

(ti−1,W (ti−1)

)(ti − ti−1

)2+ ftx

(ti−1,W (ti−1)

)(ti − ti−1

)(W (ti )−W (ti−1)

)+

1

2fxx

(ti−1,W (ti−1)

)(W (ti )−W (ti−1)

)2+ ◦(higher order).





Since

lim‖Π‖→0

n∑i=1

ft(ti−1,W (ti−1)

)(ti − ti−1

)=

∫ T

0

ft(t,W (t)

)dt,

lim‖Π‖→0

n∑i=1

fx(ti−1,W (ti−1)

)(W (ti )−W (ti−1)

)=

∫ T

0

fx(t,W (t)

)dW (t),

lim‖Π‖→0

n∑i=1

ftt(ti−1,W (ti−1)

)(ti − ti−1

)2= 0,

lim‖Π‖→0

n∑i=1

ftx(ti−1,W (ti−1)

)(ti − ti−1

)(W (ti )−W (ti−1)

)= 0,

lim‖Π‖→0

n∑i=1

fxx(ti−1,W (ti−1)

)(W (ti )−W (ti−1)

)2=

∫ T

0

fxx(t,W (t)

)dt,

we have the Ito-Doeblin formula.





Example

Let f (x) = 12x

2. Then

1

2W 2(T ) = f

(W (T )

)− f(W (0)

)=

∫ T

0

fx(W (t)

)︸︷︷︸W (t)

dW (t) +1

2

∫ T

0

fxx(W (t)

)︸︷︷︸1

dt

=

∫ T

0

W (t) dW (t) +1

2T .





Example

Let s, r and σ be positive constants, and(W (t)

)t≥0

be a Brownian

motion. Define

f(t,W (t)

)= s e

(r− 12σ

2)t+σW (t), for t ≥ 0.

Show that

f(t,W (t)

)= s +

∫ t

0

r f(u,W (u)

)du +

∫ t

0

σ f(u,W (u)

)dW (u).





Solution

Consider f(t, x

)= s e

(r− 12σ

2)t+σx ∈ C 1,2. Since

ft(t, x

)=

(r − 1

2σ2)f(t, x

),

fx(t, x

)= σ f

(t, x

), and fxx

(t, x

)= σ2 f

(t, x

),

the Ito-Doeblin formula says

df(u,W (u)

)=

(r − 1

2σ2)f(u,W (u)

)︸︷︷︸=ft

(u,W (u)

) du + σ f(u,W (u)

)︸︷︷︸=fx

(u,W (u)

) dW (u) +1

2σ2 f

(u,W (u)

)︸︷︷︸=fxx

(u,W (u)

) du

= r f(u,W (u)

)du + σ f

(u,W (u)

)dW (u).





Ito Process

Let W be a Brownian motion. An Ito process is a stochastic process ofthe form

X (t) = X (0) +

∫ t

0

g(u) du +

∫ t

0

h(u) dW (u)

where X (0) is non-random, and g and h are adapted stochasticprocesses.

1 Ito process are continuous.

2 Almost all stochastic processes, except those with jumps, are Itoprocess.

3 [X ,X ](t) =∫ t

0h2(u) du.





Proof

Let R(t) =∫ t

0g(u) du and I (t) =

∫ t

0h(u) dW (u). Then

[X ,X ](t) = [R ,R](t)︸︷︷︸=0

+ 2 [R , I ](t)︸︷︷︸=0

+ [I , I ](t)

=

∫ t

0

h2(u) du.





Integration w.r.t. Ito Process

Let X be an Ito process where

X (t) = X (0) +

∫ t

0

g(u) du +

∫ t

0

h(u) dW (u)

and Γ be an adapted process. Define the integral with respect to an Itoprocess ∫ t

0

Γ(u) dX (u) =

∫ t

0

Γ(u)g(u) du +

∫ t

0

Γ(u)h(u) dW (u)︸︷︷︸Ito process

.





Ito-Doeblin Formula for Ito Process

Let f (t, x) be a C 1,2 function, W be a Brownian motion and X be an Itoprocess where

X (t) = X (0) +

∫ t

0

g(u) du +

∫ t

0

h(u) dW (u).

Then for every T ≥ 0

f(T ,X (T )

)= f

(0,X (0)

)+

∫ T

0

ft(t,X (t)) dt

+

∫ T

0

fx(t,X (t)) dX (t) +1

2

∫ T

0

fxx(t,X (t)) d [X ,X ](t)

= f(0,X (0)

)+

∫ T

0

ft(t,X (t)) dt +

∫ T

0

fx(t,X (t))g(t) dt

+

∫ T

0

fx(t,X (t))h(t) dW (t) +1

2

∫ T

0

fxx(t,X (t))h2(t) dt.





In differential form,

dX (T ) = g(T ) dT + h(T ) dW (T )

and

df(T ,X (T )

)= ft(T ,X (T )) dT

+ fx(T ,X (T )) dX (T ) +1

2fxx(T ,X (T )) d [X ,X ](T )

= ft(T ,X (T )) dT + fx(T ,X (T ))g(T ) dT

+ fx(T ,X (T ))h(T ) dW (T ) +1

2fxx(T ,X (T ))h2(T ) dT .





Integration by Parts Formula

Let X and Y be Ito processes. Then for every T ≥ 0

X (T )Y (T ) = X (0)Y (0)

+

∫ T

0

X (t) dY (t) +

∫ T

0

Y (t) dX (t) + [X ,Y ](T ).

In particular,

X 2(T ) = X 2(0) + 2

∫ T

0

X (t) dX (t) + [X ,X ](T ).






d(X (T )Y (T )

)= X (T ) dY (T ) + Y (T ) dX (t) + dX (T ) dY (T )

and

d(X 2(T )

)= 2X (T ) dX (T ) + dX (T ) dX (T )





Proof

Using the Ito-Doeblin formula for f (x) = x2, we have

X 2(T ) = X 2(0) + 2

∫ T

0

X (t) dX (t) + [X ,X ](T ).

The integration by parts formula then follows from

X (T )Y (T ) =1

2

((X (T ) + Y (T )

)2 − X 2(T )− Y 2(T )).





Ito Integral of Deterministic Integrand

Let h be a non-random function of time t and

I (t) =

∫ t

0

h(u) dW (u).

For each t ≥ 0, the random variable

I (t) ∼ N(0,

∫ t

0

h2(u) du

).





Proof

Since I is a martingale,

E (I (t)) = I (0) = 0,

and by Ito isometry,

Var (I (t)) = E(I 2(t)

)=

∫ t

0

h2(u) du.





Let u ∈ R, and define

X (t) = −1

2

∫ t

0

(u h(s)

)2

ds︸︷︷︸=u2 Var(I (t))

+

∫ t

0

u h(s) dW (s)︸︷︷︸=u I (t)

.

From the Ito-Doeblin formula,

d(eX (t)

)= u h(t) eX (t) dW (t),

eX (t) = 1 +

∫ t

0

u h(s) eX (s) dW (s),

therefore,

E

(eX (t)

)= 1 ⇒ E

(eu I (t)

)= e

12 u

2Var(I (t)).





Generalized Geometric Brownian Motion

Let W be a Brownian motion, and μ and σ be adapted stochasticprocesses. The stochastic differential equation

dS(t) = S(t)(μ(t) dt + σ(t) dW (t)

)has the solution

S(t) = S(0) e∫

t

0

(μ(u)− 1

2σ2(u)

)du+

∫t

0σ(u) dW (u).

If μ and σ are deterministic functions, then

ln

(S(t)

S(0)

)∼ N

(∫ t

0

(μ(u)− 1

2σ2(u)

)du,

∫ t

0

σ2(u) du

).





Verification

Let

X (t) =

∫ t

0

(μ(u)− 1

2σ2(u)

)du +

∫ t

0

σ(u) dW (u).

Then

dX (t) =

(μ(t)− 1

2σ2(t)

)dt + σ(t) dW (t)

and

d [X ,X ](t) = σ2(t) dt.





Let f (x) = S(0) ex . Then

fx(x) = f (x) and fxx(x) = f (x).


df(X (t)

)= fx

(X (t)

)dX (t) +

1

2fxx

(X (t)

)d [X ,X ](t)

dS(t) = S(t)(μ(t) dt + σ(t) dW (t)

).





Ornstein-Uhlenbeck or Mean Reverting Process

Let W be a Brownian motion, and α, β and σ be positive constants.The stochastic differential equation

dR(t) = β(α− R(t)

)dt + σ dW (t)

has the solution

R(t) = e−β tR(0) + α

(1− e

−β t)+ σ e−β t

∫ t

0

eβ u dW (u)

∼ N(e−β tR(0) + α

(1− e

−β t),σ2

2β

(1− e

−2β t))

.





Verification

Let

X (t) =

∫ t

0

eβ u dW (u) ∼ N

(0, e

2βt−12β

).

Then

dX (t) = eβ t dW (t) and d [X ,X ](t) = e

2β t dt.

Let

f (t, x) = e−β tR(0) + α

(1− e

−β t)+ σ e−β tx

Then

ft(t, x) = β(α− f (t, x)

),

fx(t, x) = σ e−β t and fxx(t, x) = 0.






df(t,X (t)

)= ft(t,W (t)) dt + fx(t,X (t)) dX (t)


)dt + σ dW (t).





Solving Linear Stochastic Differential Equation

Let W be a Brownian motion, and α, β, γ and σ be deterministicfunctions. The linear stochastic differential equation

dX (t) =(α(t) + β(t)X (t)

)dt +

(γ(t) + σ(t)X (t)

)dW (t)

has the solution

X (t) = e

∫t

0

(β(u)− 1

2σ2(u)

)du+

∫t

0σ(u) dW (u) ×

[X (0)

+

∫ t

0

e−

∫u

0

(β(v)− 1

2σ2(v)

)dv−

∫u

0σ(v) dW (v)

(α(u)− γ(u)σ(u)

)du

+

∫ t

0

e−

∫u

0

(β(v)− 1

2σ2(v)

)dv−

∫u

0σ(v) dW (v)γ(u) dW (u)

].





Proof

Consider solutions of the form X (t) = X1(t)X2(t) where

dX1(t) = β(t)X1(t) dt + σ(t)X1(t) dW (t),

dX2(t) = a(t) dt + b(t) dW (t)

for some functions a(t) and b(t). Then

dX (t) = X1(t) dX2(t) + X2(t) dX1(t) + d [X1,X2](t)

=( =α(t)︷︸︸︷(

a(t) + σ(t)b(t))X1(t) + β(t)X (t)

)dt

+(b(t)X1(t)︸︷︷︸

=γ(t)

+ σ(t)X (t))dW (t)





Since

a(t) =α(t)− γ(t)σ(t)

X1(t)and b(t) =

γ(t)

X1(t),

therefore,

X1(t) = e

∫t

0

(β(u)− 1

2σ2(u)

)du+

∫t

0σ(u) dW (u)

and

X2(t) = X (0) +

∫ t

0

α(u)− γ(u)σ(u)

X1(u)du +

∫ t

0

γ(u)

X1(u)dW (u).





Mean Reverting Process

Let W be a Brownian motion, and α, β and σ be positive constants.The stochastic differential equation


)dt + σ

√R(t) dW (t)

has the “solution”

R(t) = e−β tR(0) + α

(1− e


∫ t

0

eβ u

√R(u) dW (u),

and

E (R(t)) → α and Var (R(t)) → ασ2

2β

as t → ∞.





Verification

Let f (t, x) = eβ tx . Then

ft(t, x) = β f (t, x), fx(t, x) = eβ t and fxx(t, x) = 0.


df(t,R(t)

)= ft(t,R(t)) dt + fx(t,R(t)) dR(t),

d(eβ tR(t)

)= αβ e

β t dt + σ eβ t√

R(t) dW (t).

Hence,

eβ tR(t) = R(0) + α

(eβ t − 1

)+ σ

∫ t

0

eβ u

√R(u) dW (u).





Since

R(t) = e−β tR(0) + α

(1− e


∫ t

0

eβ u

√R(u) dW (u)︸︷︷︸

Ito integral

,

thus

E (R(t)) = e−β tR(0) + α

(1− e

−β t)

and

limt→∞

E (R(t)) = α.





Var (R(t)) = E

((σ e−β t

∫ t

0

eβ u

√R(u) dW (u)

)2)

= σ2e−2β t

E

(∫ t

0

e2β uR(u) du

)

= σ2e−2β t

∫ t

0

e2β u

E (R(u)) du

=ασ2

2β

(1− 2 e−β t + e

−2β t)+ R(0)

σ2

β

(e−β t − e

−2β t)

and

limt→∞

Var (R(t)) =ασ2

2β.




Multivariate Brownian MotionMultivariate Ito-Doeblin Formula

Definition of Multivariate Brownian Motion

A d-dimensional Brownian motion is a process

W (t) =(W1(t),W2(t), . . . ,Wd (t)

)�,

for t ≥ 0, such that:

1 Each Wi (t) is a 1-dimensional Brownian motion.

2 Wi (t) and Wj(t) are independent for i = j .

Associated with{W (t)

}t≥0

is a filtration{F(t)

}t≥0

such that:

3 F(s) ⊂ F(t) for 0 ≤ s < t.

4 W (t) is F(t)-measurable for each t ≥ 0.

5 W (t)−W (s) is independent of F(s) for 0 ≤ s < t.





Cross Variation

For a multivariate Brownian motion W ,

dWi (t) dWj (t) =

{dt if i = j ;

0 if i = j .





2-Dimensional Ito-Doeblin Formula

Let f (t, x , y) be a C 1,2,2 function, W = (W1,W2)� be a two dimensional

Brownian motion, and X and Y be Ito processes where

X (t) = X (0) +

∫ t

0

α1(u) du +

∫ t

0

σ11(u) dW1(t) +

∫ t

0

σ12(u) dW2(t)

Y (t) = Y (0) +

∫ t

0

α2(u) du +

∫ t

0

σ21(u) dW1(t) +

∫ t

0

σ22(u) dW2(t).





Then

f(t,X (t),Y (t)

)= f

(0,X (0),Y (0)

)+

∫ t

0

ft(u,X (u),Y (u)

)du

+

∫ t

0

fx(u,X (u),Y (u)

)dX (u) +

∫ t

0

fy(u,X (u),Y (u)

)dY (u)

+1

2

∫ t

0

fxx(u,X (u),Y (u)

)d [X ,X ](u)

+1

2

∫ t

0

fyy(u,X (u),Y (u)

)d [Y ,Y ](u)

+

∫ t

0

fxy(u,X (u),Y (u)

)d [X ,Y ](u).






dX (t) = α1(t) dt + σ11(t) dW1(t) + σ12(t) dW2(t),

dY (t) = α2(t) dt + σ21(t) dW1(t) + σ22(t) dW2(t)

and

df(t,X (t),Y (t)

)= ft

(t,X (t),Y (t)

)dt + fx

(t,X (t),Y (t)

)dX (t)

+ fy(t,X (t),Y (t)

)dY (t) +

1

2fxx

(t,X (t),Y (t)

)d [X ,X ](t)

+1

2fyy

(t,X (t),Y (t)

)d [Y ,Y ](t) + fxy

(t,X (t),Y (t)

)d [X ,Y ](t).





df (·) =[ft(·) + α1(t)fx(·) + α2(t)fy (·)

+1

2

(σ211(t) + σ2

12(t))fxx(·) + 1

2

(σ221(t) + σ2

22(t))fyy (·)

+(σ11(t)σ21(t) + σ12(t)σ22(t)

)fxy (·)

]dt

+(σ11(t)fx(·) + σ21(t)fy (·)

)dW1(t)

+(σ12(t)fx(·) + σ22(t)fy (·)

)dW2(t).





Multi-Dimensional Ito-Doeblin Formula

Let f (t, x) : R+ × Rm �→ R be a function with continuous partial

derivatives ft , fxi and fxi xj , for 1 ≤ i , j ≤ m, and X be a m-dimensional Itoprocess with

dX(t) = α

(t,X(t)

)dt + σ

(t,X(t)

)dW(t)

where α = (αi )m×1, σ = (σij)m×d and W = (Wi )d×1 is a d-dimensionalBrownian motion. Then

df =

⎛⎝ft +

m∑i=1

αi fxi +1

2

m∑i=1

m∑j=1

d∑k=1

σikσjk fxi xj

⎞⎠ dt +

m∑i=1

d∑k=1

σik fxidWk

=

(ft + (∇Xf )

�α+

1

2Tr

(σ

�HX f σ

))dt + (∇Xf )

�σ dW(t).





Levy Characterization of Brownian Motion

Let the stochastic processes Mi be martingales, for i = 1, 2, . . . , d .Suppose Mi (0) = 0, Mi (t) has continuous paths, and

[Mi ,Mj ](t) =

{t if i = j ,

0 otherwise,

for 1 ≤ i , j ≤ d and t ≥ 0. Then M =(Mi

)is a d-dimensional Brownian

motion.





Proof

Suppose d = 1. Let f (t, x) = eux− 1

2 u2t . Then

df(t,M(t)

)=

=− 12 u

2f (t,x)︷︸︸︷ft(t,M(t)

)dt +

=u f (t,x)︷︸︸︷fx(t,M(t)

)dM(t) + 1

2

=u2f (t,x)︷︸︸︷fxx

(t,M(t)

) =dt︷︸︸︷d [M,M](t),

= u f(t,M(t)

)dM(t),

f(t,M(t)

)= 1 +

∫ t

0

u f(s,M(s)

)dM(s)︸︷︷︸

Martingale

.

Hence,

E

(eu M(t)

)= e

12 u

2tE(f(t,M(t)

))= e

12 u

2t .





Let 0 ≤ τ < t. Then

d(euM(t)

)= u e

u M(t) dM(t) +1

2u2 eu M(t) dt

eu M(t) = e

u M(τ) + u

∫ t

τ

eu M(s) dM(s)︸︷︷︸Martingale

+1

2u2

∫ t

τ

eu M(s) ds.

Therefore,

E

(eu

(M(t)−M(τ)

) ∣∣∣∣ F(τ)

)= 1 +

1

2u2

∫ t

τ

E

(eu

(M(s)−M(τ)

) ∣∣∣∣ F(τ)

)ds

d

dt

(E

(eu

(M(t)−M(τ)

) ∣∣∣∣ F(τ)

))=

1

2u2 E

(eu

(M(t)−M(τ)

) ∣∣∣∣ F(τ)

).

Hence,

E

(eu

(M(t)−M(τ)

) ∣∣∣∣ F(τ)

)= e

12 u

2(t−τ).





Let 0 ≤ τ1 < t1 ≤ τ2 < t2. Then

E

(eu2

(M(t2)−M(τ2)

)+u1

(M(t1)−M(τ1)

))

= E

(E

(eu2

(M(t2)−M(τ2)

) ∣∣∣∣ F(τ2)

)eu1

(M(t1)−M(τ1)

))

= e12 u

22(t2−τ2) E

(E

(eu1

(M(t1)−M(τ1)

) ∣∣∣∣ F(τ1)

))

= e12 (u

22(t2−τ2)+u2

1(t1−τ1))

Hence increments over non-overlapping intervals are independent.





Suppose d = 2. From the one-dimensional result, both M1 and M2 areBrownian motions. It remains to show that they are independent. Letf (t, x , y) = e

u1x+u2y−12 (u

21+u2

2)t . Then

df(t,M1(t),M2(t)

)=

=− 12 (u

21+u2

2)f (t,x,y)︷︸︸︷ft(t,M1(t),M2(t)

)dt

+

=u1 f (t,x,y)︷︸︸︷fx(t,M1(t),M2(t)

)dM1(t) +

=u2 f (t,x,y)︷︸︸︷fy(t,M1(t),M2(t)

)dM2(t)

+ 12

=u21 f (t,x,y)︷︸︸︷

fxx(t,M1(t),M2(t)

) =dt︷︸︸︷d [M1,M1](t) +

=u1u2f (t,x,y)︷︸︸︷fxy

(t,M1(t),M2(t)

) =0︷︸︸︷d [M1,M2](t)

+ 12

=u22 f (t,x,y)︷︸︸︷

fyy(t,M1(t),M2(t)

) =dt︷︸︸︷d [M2,M2](t),

= f(t,M(t)

)(u1 dM1(t) + u2 dM2(t)

).





Since

f(t,M1(t),M2(t)

)= 1 +

Martingale︷︸︸︷∫ t

0

u1 f(s,M1(s),M2(s)

)dM1(s)

+

∫ t

0

u2 f(s,M1(s),M2(s)

)dM2(s)︸︷︷︸

Martingale

,

thus,

E

(eu1 M1(t)+u2 M2(t)

)= e

12 (u

21+u2

2)t E(f(t,M1(t),M2(t)

))= e

12 (u

21+u2

2)t .




Black-Scholes-Merton EquationParabolic Partial Differential Equation

Risk-Free Asset Price Process

Let M(t), t ≥ 0, denotes the risk-free asset price process that is modeledby

dM(t) = r M(t) dt

where r is a constant. The ordinary differential equation has the solution

M(t) = er t

assuming M(0) = 1.





Risky Asset Price Process

Let S(t), t ≥ 0, denotes the risky asset price process that is modeled bya geometric Brownian motion

dS(t) = S(t)(μ dt + σ dW (t)

)where W is a Brownian motion, and μ and σ > 0 are constants. Thestochastic differential equation has the solution

S(t) = S(0) e

(μ− 1

2σ2)t+σW (t)

and

ln

(S(t)

S(0)

)∼ N

((μ− 1

2σ2)t, σ2t

).





Portfolio Value Process

Consider a portfolio strategy(h(t), v(t)

), t ≥ 0, where h(t) and v(t)

represent the unit holdings at time t in the risky and risk-free assetsrespectively.

Let X (t), t ≥ 0, denotes the portfolio value process of the abovestrategy. Then

X (t) = h(t) S(t) + v(t)M(t)

and

dX (t) = h(t) dS(t) + v(t) dM(t)

= X (t) r dt︸︷︷︸Riskless Return

+ h(t) S(t)(μ− r

)dt︸︷︷︸

Risk Compensation

+ h(t) S(t)σ dW (t)︸︷︷︸Risk / Volatility

.





Verification

Since

X (t +Δt)− X (t)

= h(t)(S(t +Δt)− S(t)

)︸︷︷︸=S(t)

(μΔt+σΔW (t)

) + v(t)(M(t +Δt)−M(t)

)︸︷︷︸

=r M(t)Δt

,

letting Δt → 0, we have


= h(t) S(t)(μ dt + σ dW (t)

)+ v(t)M(t)︸︷︷︸

=X (t)−h(t)S(t)

r dt

= X (t) r dt + h(t) S(t)(μ− r

)dt + h(t) S(t)σ dW (t).





Self-Financing

The continuous-time self-financing condition


can equivalently be stated as

S(t) dh(t) + dS(t) dh(t) +M(t) dv(t) + dM(t) dv(t)︸︷︷︸=0

= 0.





Proof

Using Ito-Doeblin formula,

dX (t)︸︷︷︸=h(t) dS(t)+v(t) dM(t)

= h(t) dS(t) + S(t) dh(t) + d [S , h](t)

+ v(t) dM(t) +M(t) dv(t) + d [M, v ](t),

hence

S(t) dh(t) + dS(t) dh(t) +M(t) dv(t) + dM(t) dv(t) = 0.





Verification

Since(h(t +Δt)− h(t)

)S(t +Δt) +

(v(t +Δt)− v(t)

)M(t +Δt) = 0,

S(t)(h(t +Δt)− h(t)

)+

(S(t +Δt)− S(t)

)(h(t +Δt)− h(t)

)+M(t)

(v(t +Δt)− v(t)

)+

(M(t +Δt)−M(t)

)(v(t +Δt)− v(t)

)= 0,

letting Δt → 0, we have

S(t) dh(t) + dS(t) dh(t) +M(t) dv(t) + dM(t) dv(t) = 0.





Self-Financing Condition

For t ≥ 0, the value of a portfolio at time t is

X (t) =

m∑i=0

hi (t) Si (t),

where Si (t) is the price of asset i and hi (t) is the unit of asset i in theportfolio at time t, for i = 0, 1, . . . ,m.

The trading strategy(h0(t), h1(t), . . . , hm(t)

)t≥0

is self-financing if

dX (t) =m∑i=0

hi (t) dSi (t),

for t ≥ 0.





Option Value Process

Consider a simple option that pays g(S(T )

)at time T .

Suppose the value of the option at time t is given by C(t, S(t)

), for

0 ≤ t ≤ T . Then

dC(t, S(t)

)=

(Ct

(t, S(t)

)+ μS(t)CS

(t, S(t)

)+

1

2σ2S2(t)CSS

(t, S(t)

))dt

+ σS(t)CS

(t, S(t)

)dW (t).





Verification

Since

dS(t) = S(t)(μ dt + σ dW (t)

)and d [S , S ](t) = σ2S2(t) dt,

from the Ito-Doeblin formula,

dC(t, S(t)

)= Ct

(t, S(t)

)dt + CS

(t, S(t)

)dS(t) +

1

2CSS

(t, S(t)

)d [S , S ](t)

=

(Ct

(t, S(t)

)+ μ S(t)CS

(t, S(t)

)+

1

2σ2S2(t)CSS

(t, S(t)

))dt

+ σ S(t)CS

(t, S(t)

)dW (t).





Equating Evolutions

Consider a hedging portfolio that consists of a short one unit of optionand h(t) units of the underlying at time t, 0 ≤ t ≤ T .

The portfolio value at time t is

X (t) = h(t) S(t)− C(t, S(t)

)and

dX (t) = h(t) dS(t) +

=0 Self-Financing︷︸︸︷S(t) dh(t) + dS(t) dh(t)− dC

(t, S(t)

)=

(μ h(t) S(t)− Ct

(t, S(t)

)− μ S(t)CS

(t, S(t)

)− 1

2σ2S2(t)CSS

(t, S(t)

))dt

+ σ S(t)(h(t)− CS

(t, S(t)

))dW (t).





Choose h(t) = CS

(t, S(t)

). Then

dX (t) =

(−Ct

(t, S(t)

)− 1

2σ2S2(t)CSS

(t, S(t)

))dt

= r X (t) dt.

Therefore

−Ct

(t, S(t)

)− 1

2σ2S2(t)CSS

(t, S(t)

)= r

(S(t)CS

(t, S(t)

)− C(t, S(t)

)).





Black-Scholes-Merton Partial Differential Equation

The value of the option C (t, s) satisfies the Black-Scholes-Merton partialdifferential equation

Ct(t, s) + r s Cs(t, s) +1

2σ2s2Css(t, s) = r C (t, s),

for 0 ≤ t < T and 0 ≤ s < ∞, with terminal condition

C (T , s) = g(s).





Second-Order Partial Differential Equation

General second-order partial differential equation in two variables, τ andx , has the form

A fττ + B fτx + C fxx + · · · = 0.

The partial differential equation is

1 Elliptic if B2 − 4AC < 0;

2 Parabolic if B2 − 4AC = 0;

3 Hyperbolic if B2 − 4AC > 0.

The Black-Scholes-Merton partial differential equation is an example of abackward parabolic partial differential equation that can be transformedinto a form analogous to the heat equation.





Homogeneous Heat Equation

The following initial value problem on (−∞,∞):

fτ (τ, x) = a fxx(τ, x), 0 < τ < ∞, −∞ < x < ∞,

f (0, x) = g(x);

has the solution

f (τ, x) =1√4πaτ

∫ ∞

−∞

e−

(x−y)2

4aτ g(y) dy .





The following initial value problem on [0,∞) with homogenous Dirichletboundary conditions:

fτ (τ, x) = a fxx(τ, x), 0 < τ < ∞, 0 ≤ x < ∞,

f (0, x) = g(x),

f (τ, 0) = 0;

has the solution

f (τ, x) =1√4πaτ

∫ ∞

0

(e−

(x−y)2

4aτ − e−

(x+y)2

4aτ

)g(y) dy .





European Call Option

The value of a European call option C (t, s), with strike price K andmaturity time T , satisfies the Black-Scholes-Merton partial differentialequation

Ct(t, s) + r s Cs(t, s) +1

2σ2s2Css(t, s)− r C (t, s) = 0

for 0 ≤ t < T and 0 ≤ s < ∞, with terminal condition

C (T , s) = (s − K )+.





Transformation

Under the following transformation:

x = ln s and τ = T − t;

the Black-Scholes-Merton partial differential equation becomes

Cτ (τ, x)−(r − 1

2σ2

)Cx(τ, x)− 1

2σ2Cxx(τ, x) + r C (τ, x) = 0

for 0 < τ ≤ T and −∞ < x < ∞, with initial condition

C (0, x) = ( ex − K )+.





Let

C (τ, x) = eατ+βx f (τ, x)

where

α = −r − 1

2σ2

(r − 1

2σ2

)2

and β =1

2− r

σ2.

Then the Black-Scholes-Merton partial differential equation reduces tothe heat equation

fτ (τ, x) =1

2σ2fxx(τ, x)

for 0 < τ ≤ T and −∞ < x < ∞, with initial condition

f (0, x) = e−βx( ex − K )+.





Solution

f (τ, x) =1√

2πσ2τ

∫ ∞

lnK

e−

(x−y)2

2σ2τ

(e(1−β)y − K e

−βy)dy

and

C (τ, x) = eατ+βx f (τ, x)

= exN(d+)− e

−rτK N(d−)

where

d± =x − lnK + (r ± 1

2σ2)τ

σ√τ

.


Documents

MFE6516 Stochastic Calculus for Finance · StochasticCalculus MultivariateStochasticCalculus Black-Scholes-MertonModel MFE6516StochasticCalculusforFinance WilliamC.H.Leon NanyangBusinessSchool