m507b-hw8soln-s08

m507b-hw8soln-s08 - MATH 507b ASSIGNMENT 8 SOLUTIONS SPRING...

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MATH 507b ASSIGNMENT 8 SOLUTIONS SPRING 2008 Prof. Alexander (1) The mean of I ( f,t ) = R t 0 f ( s ) dW s is 0 since it’s a martingale and I ( f, 0) = 0, so the variance is EI ( f,t ) 2 = R t 0 Ef ( s ) 2 ds . In particular: (a) For f ( s ) = | W s | 1 / 2 , we have Ef ( s ) 2 = E | W s | = s 1 / 2 EW 1 = ( 2 s π ) 1 / 2 , and the variance is ( 2 π ) 1 / 2 R t 0 s 1 / 2 ds = 2 3 ( 2 π ) 1 / 2 t 3 / 2 . (b) For f ( s ) = ( W s + s ) 2 we have Ef ( s ) 2 = E ( W 4 s + 4 sW 3 s + 6 s 2 W 2 s + 4 s 3 W s + s 4 ) = s 2 EW 4 1 + 6 s 3 EW 2 1 + s 4 = 3 s 2 + 6 s 3 + s 4 , and the variance is R t 0 (3 s 2 + 6 s 3 + s 4 ) ds = t 3 + 3 2 t 4 + 1 5 t 5 . (2)(a) Let H ( x ) = e x so Z t = H ( Y t ). Then dY t = f ( t ) dW t - 1 2 f ( t ) 2 dt so by Ito’s Lemma, dZ t = H 0 ( Y t ) dY t + 1 2 H 00 ( Y t ) f ( t ) 2 dt = e Y t ± f ( t ) dW t - 1 2 f ( t ) 2 dt ² + 1 2 e Y t f ( t ) 2 dt = e Y t f ( t ) dW t . (b) Yes, because by (a),
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m507b-hw8soln-s08 - MATH 507b ASSIGNMENT 8 SOLUTIONS SPRING...

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