Recognizing a BrownianMotion

so bothB 1 andB 2are Brownian motions.

Recognizing a Brownian Motion

Theorem 0.62 (Levy) Let B ( t ) ; 0  t  T;be a process on ;F;P), adapted to a filtration

F( t ) ; 0tT, such that:

1 the paths ofB ( t )are continuous,

2. Bis a martingale,

3. hBi( t ) = t; 0tT, (i.e., informallydB ( t ) dB ( t ) = dt).

ThenBis a Brownian motion.

Proof: (Idea) Let0  s < t  T be given We need to show thatB ( t ),B ( s ) is normal, with mean zero and variancet,s, and B ( t ),B ( s )is independent of F( s ) We shall show that the

conditional moment generating function ofB ( t ),B ( s )is

IE

e u(B(t),B(s))

F( s )

= e 1 2u2(t,s) :

Since the moment generating function characterizes the distribution, this shows thatB ( t ),B ( s )

is normal with mean 0 and variance t,s, and conditioning on F( s ) does not affect this, i.e.,

B ( t ),B ( s )is independent ofF( s )

We compute (this uses the continuity condition (1) of the theorem)

de uB(t) = ue uB(t) dB ( t ) + 1 2 u 2 e uB(t) dB ( t ) dB ( t ) ; so

e uB(t) = e uB(s) +Z t

s ue uB(v) dB ( v ) + 1

2 u 2Z t

s e uB(v) |{z}dv:

uses cond 3

233

R0 t ue uB(v) dB ( v )is a martingale (by condition 2), and so

IE t

s ue uB(v) dB ( v )

F( s )

=,

Z s

0 ue uB(v) dB ( v ) + IE t

0 ue uB(v) dB ( v )

F( s )

= 0 :

It follows that

IE

e uB(t)

F( s )

= e uB(s) + 1 2 u 2Z t

s IE

e uB(v)

F( s )

dv:

We define

' ( v ) = IE

e uB(v)

F( s )

;

so that

' ( s ) = e uB(s)

and

' ( t ) = e uB(s) + 1 2 u 2Z t

s ' ( v ) dv;

'0

( t ) = 1 2 u 2 ' ( t ) ; ' ( t ) = ke 1 2u2t : Plugging ins, we get

e uB(s) = ke 1 2u2s =)k = e uB(s),1 2u2s : Therefore,

IE

e uB(t)

F( s )

= ' ( t ) = e uB(s)+12u2(t,s) ;

IE

e u(B(t),B(s))

F( s )

= e 1 2u2(t,s) :

23.1 Identifying volatility and correlation

LetB 1 andB 2be independent Brownian motions and

dS 1

S 1 = r dt +  11 dB 1 +  12 dB 2 ;

dS 2

S 2 = r dt +  21 dB 1 +  22 dB 2 ; Define

 1 =q

 211 +  212 ;

 2 =q

 221 +  222 ;

 =  11  21 +  12  22

 1  2 : Define processesW 1andW 2by

dW 1 =  11 dB 1 +  12 dB 2

 1

dW 2 =  21 dB 1 +  22 dB 2

ThenW 1andW 2have continuous paths, are martingales, and

dW 1 dW 1 = 1  21 (  11 dB 1 +  12 dB 2 ) 2

= 1  21 (  211 dB 1 dB 1 +  212 dB 2 dB 2 )

= dt;

and similarly

dW 2 dW 2 = dt:

Therefore,W 1andW 2are Brownian motions The stock prices have the representation

dS 1

S 1 = r dt +  1 dW 1 ;

dS 2

S 2 = r dt +  2 dW 2 : The Brownian motionsW 1andW 2are correlated Indeed,

dW 1 dW 2 = 1  1  2 (  11 dB 1 +  12 dB 2 )(  21 dB 1 +  22 dB 2 )

= 1  1  2 (  11  21 +  12  22 ) dt

=  dt:

23.2 Reversing the process

Suppose we are given that

dS 1

S 1 = r dt +  1 dW 1 ;

dS 2

S 2 = r dt +  2 dW 2 ; whereW 1 andW 2are Brownian motions with correlation coefficient We want to find

 =

"

 11  12

 21  22

#

so that

0=

"

 11  12

 21  22

# "

 11  21

 12  22

#

=

"

 211 +  212  11  21 +  12  22

 11  21 +  12  22  221 +  222

#

=

"

 21  1  2

 1  2  22

#

A simple (but not unique) solution is (see Chapter 19)

 11 =  1 ;  12 = 0 ;

 21 =  2 ;  22 =q

1, 2  2 : This corresponds to

 1 dW 1 =  1 dB 1 =)dB 1 = dW 1 ;

 2 dW 2 =  2 dB 1 +q

1, 2  2 dB 2

=) dB 2 = dW 2, dW 1

p

1, 2 ; ( 6=1)

If =1, then there is noB 2anddW 2 =  dB 1 =  dW 1 :

Continuing in the case6=1, we have

dB 1 dB 1 = dW 1 dW 1 = dt;

dB 2 dB 2 = 1 1

, 2



dW 2 dW 2,2  dW 1 dW 2 +  2 dW 2 dW 2



= 1 1

, 2



dt,2  2 dt +  2 dt

= dt;

so bothB 1 andB 2are Brownian motions Furthermore,

dB 1 dB 2 = p 1

1, 2 ( dW 1 dW 2,dW 1 dW 1 )

= p 1

1, 2 (  dt, dt ) = 0 :

We can now apply an Extension of Levy’s Theorem that says that Brownian motions with zero

cross-variation are independent, to conclude thatB 1 ;B 2are independent Brownians

233

R0 t ue uB(v) dB ( v )is a martingale... dt

= dt;

so bothB 1 andB 2are Brownian motions... generating function characterizes the distribution, this shows thatB ( t ),B ( s )

is normal with mean and variance t,s, and