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09Stoch6

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

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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 (det Σ) 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 0 ) where Σ 0 = A Σ A t . Let { B t } t 0 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)

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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 (det Σ) 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 0 · · · 0 0 t 2 - t 1 · · · 0 . . . . . . . . . . . . 0 0 · · · 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 (det Σ 0 ) 1 / 2 exp ( - 1 2 ( x t Σ 0 - 1 x ) ) (2.2.3) where Σ 0 = 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 . . . B t n = 1 0 0 · · · 0 1 1 0 · · · 0 . . . . . . . . . 1 1 1 · · · 1 B t 1 B t 2 - B t 1 . . . B t n - B t n - 1 and notice that cov( B t i , B t j ) = min { t i , t j } . By using this and the construction of the probability space in the last section, we can conclude the existence of a probability space for the Brownian motion.

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

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