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m507b-hw7soln-s08

# m507b-hw7soln-s08 - MATH 507b ASSIGNMENT 7 SOLUTIONS SPRING...

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MATH 507b ASSIGNMENT 7 SOLUTIONS SPRING 2008 Prof. Alexander Chapter 7: (4.1)(i) Let u < v a and Y s ( ω ) = 1 { ω ( t - s ) ( u,v ) } for s < t , so we have Y T a ( θ T a ω )1 { T a <t } = 1 { ω ( t ) ( u,v ) } 1 { T a <t } . Note that (2 a - v, 2 a - u ) is the “mirror image” of ( u, v ) relative to a , and let Z s ( ω ) = 1 { ω ( t - s ) (2 a - v, 2 a - u ) } . Since B t - s is symmetric, we have E a Y s = E a Z s for all s < t , while Z T a ( θ T a ω )1 { T a <t } = 1 { ω ( t ) (2 a - v, 2 a - u ) } . Therefore letting ϕ ( a, s ) = E a Y s and ψ ( a, s ) = E a Z s we have ϕ = ψ and P 0 ( T a < t, B t ( u, v )) = E 0 (( Y T a θ T a )1 { T a <t } ) = E 0 ( E 0 ( Y T a θ T a | F T a )1 { T a <t } ) = E 0 ( ϕ ( a, T a )1 { T a <t } ) = E 0 ( ψ ( a, T a )1 { T a <t } ) = E 0 ( E 0 ( Z T a θ T a | F T a )1 { T a <t } ) = E 0 (( Z T a θ T a )1 { T a <t } ) = P 0 ( T a < t, B t (2 a - v, 2 a - u )) = P 0 ( B t (2 a - v, 2 a - u )) . (ii) Let f be the joint density of ( M t , B t ). Letting u → -∞ in (i) we get P 0 ( T a < t, B t < x ) = P 0 ( B t > 2 a - x ) for x a , and then f ( a, x ) = - 2 ∂a∂x P 0 ( M t > a, B t < x ) = - 2 ∂a∂x P 0 ( T a < t, B t < x ) = - 2 ∂a∂x P 0 ( B t > 2 a - x ) = - 2 ∂a∂x Z 2 a - x 1 2 πt e - r 2 / 2 t dr = - ∂a 1 2 πt e - (2 a - x ) 2 / 2 t = 2(2 a - x ) 2 πt 3 e - (2 a - x ) 2 / 2 t . 1

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(4.2) Let Y = 1 { T 0 t } so that by the Markov property, P 0 ( R 1 + t ) = E 0 ( Y θ 1 ) = E 0 ( E B 1 Y ) . Using the Reflection Principle we get E a Y = P a ( T 0 t ) = P 0 ( T a t ) = 2 P 0 ( B t ≥ | a | ) = 2 P ( B 1 | a | t ) , which we call ϕ ( a, t ). Thus P 0 ( R 1 + t ) = E 0 ϕ ( B 1 , t ). If we differentiate the left side we
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m507b-hw7soln-s08 - MATH 507b ASSIGNMENT 7 SOLUTIONS SPRING...

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