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Solution to Homework 1

# Solution to Homework 1 - Solution to Homework 1 1 What is...

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Solution to Homework 1 1. What is the distribution of B s + B t , s t ? Solution . E ( B s + B t ) = 0, var ( B s + B t ) = var ( B s )+ var ( B t )+2 E ( B s B t ) = s + t +2 s = 3 s + t . So B s + B t N (0 , 3 s + t ). 2. Compute E [ B t 1 B t 2 B t 3 ] for t 1 < t 2 < t 3 . Solution . Let with 0 = t 0 t 1 t 2 t 3 , Δ t i = t i - t i - 1 and Δ B t i = B t i - B t i - 1 . So Δ B t i ’s are independent N (0 , Δ t i ), and E [ B t 1 ( B t 2 + Δ B t 2 ) B t 3 ] = E B t 1 B t 1 + Δ B t 2 )(Δ B t 1 + Δ B t 2 + Δ B t 3 )] = 0 . 3. Calculate P ( B (2) 0) and P ( B (1) 0 , B (2) 0). Solution . P ( B (2) 0) = 1 / 2 as B (2) N (0 , 2). Next, Let X = B (1) and Y = B (2) - B (1). Then X, Y are i.i.d. N (0 , 1). Hence, P ( B (1) 0 , B (2) 0) = P ( X 0 , X + Y 0) = Z x = -∞ P ( X 0 , X + Y 0 | X = x ) d Φ( x ) = Z 0 x = -∞ P ( Y ≤ - x | X = x ) d Φ( x ) = Z 0 x = -∞ Φ( - x ) d Φ( x ) = Z 0 x = -∞ [1 - Φ( x )] d Φ( x ) = Z 1 / 2 t =0 [1 - t ] dt = ( t - t 2 / 2) | 1 / 2 t =0 = 3 / 8 . 4. Recall that a stochastic process { X t , t 0 } is said to be strictly stationary if ( X t 1 , ..., X t n ) has the same joint distribution as ( X t 1 + a , ..., X t n + a ) for all n , a , t 1 , ..., t n .

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