09Stoch6

09Stoch6 - 2.2. BROWNIAN MOTION AND SAMPLE PATHS 53 2.2...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 2.2. BROWNIAN MOTION AND SAMPLE PATHS 53 2.2 Brownian motion and sample paths For a normal r.v. X ∼ N (0 ,σ 2 ), the symmetry implies E ( X 2 k +1 ) = 0, and the integration by parts yields E ( X 2 k ) = 1 · 3 · 5 ··· (2 k- 1) · σ 2 k . (2.2.1) We also need the following elementary properties of the normal r.v.’s :- Suppose X 1 ∼ N ( μ 1 ,σ 2 1 ), X 2 ∼ N ( μ 2 ,σ 2 2 ), and they are independent, then X 1 + X 2 ∼ N ( μ 1 + μ 2 ,σ 2 1 + σ 2 2 ).- Suppose X = ( X 1 , ··· ,X n ) is a n-variate normal r.v. with distribution N ( μ , Σ) where Σ is a symmetric, positive definite n × n-matrix. The density function is given by f ( x ) = 1 (2 π ) n 2 (detΣ) 1 2 exp (- 1 2 ( x- μ ) t Σ- 1 ( x- μ ) ) . By a direct calculation, we have Σ = [cov( X i ,X j )] where cov( X i ,X j ) = E (( X i- μ i )( X j- μ j )).- Suppose X ∼ N ( μ , Σ). Let Y = A X + c where A is non-singular, then by a change of variable, Y has density g ( y ) = | det A |- 1 f ( A- 1 y- c ). A direct substitution yields Y ∼ N ( A μ + c , Σ ) where Σ = A Σ A t . Let { B t } t ≥ be the Brownian motion defined as in Section 2.1, then the process is stationary in the sense that the distribution B t- B s depends only on the difference t- s . Since B t has distribution N (0 ,t ), it follows that E ( B t ) = 0 , E ( B 2 t ) = t. Moreover, by using independence, we have E ( B s B t ) = min( s,t ) (2.2.2) 54 CHAPTER 2. BROWNIAN MOTION This can be checked directly: assume s < t , then E ( B s B t ) = E ( B s ( B s + ( B t- B s )) ) = E ( B 2 s ) + E ( ( B s ( B t- B s ) ) = E ( B 2 s ) + E ( B s ) E ( B t- B s ) = s. For 0 < t 1 < t 2 ··· < t n , the joint distribution of ( B t 1 ,B t 2- B t 1 , ··· ,B t n- B t n- 1 ) is given by f t 1 ··· t n ( x 1 , ··· ,x n ) = n Y i =0 1 p 2 π ( t i- t i- 1 ) exp ‡- ( x i- x i- 1 ) 2 2( t i- t i- 1 ) · = 1 √ 2 π (detΣ) 1 / 2 exp (- 1 2 ( z t Σ- 1 z ) ) where z = ( x 1 , x 2- x 1 , ··· , x n- x n- 1 ) and Σ = t 1 ··· t 2- t 1 ··· . . . . . . . . . . . . ··· t n- t 1 . On the other hand the distribution of the n- variate random vector ( B t 1 , B t 2 , ··· B t n ) is given by g t 1 , ··· ,t n ( x ) = 1 √ 2 π (detΣ ) 1 / 2 exp (- 1 2 ( x t Σ- 1 x ) ) (2.2.3) where Σ = t 1 t 1 t 1 ··· t 1 t 1 t 2 t 2 ··· t 2 t 1 t 2 t 3 ··· t 3 . . . . . . . . . . . . . . . t 1 t 2 t 3 ··· t n . This follows from the transformation of the above multivariate normal random 2.2. BROWNIAN MOTION AND SAMPLE PATHS 55 vector : B t 1 B t 2 ....
View Full Document

This note was uploaded on 05/23/2010 for the course MATH 987 taught by Professor Cheung,cecilia during the Spring '08 term at CUHK.

Page1 / 12

09Stoch6 - 2.2. BROWNIAN MOTION AND SAMPLE PATHS 53 2.2...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online