# hwk9 - E3106 Solutions to Homework 9 Columbia University...

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E3106, Solutions to Homework 9 Columbia University Exercise 10.23 . Since standard Brownian motion B ( t ) is a Martingale and T is a stopping time for B ( t ), it follows from the martingale stopping theorem (Exercise 19) that E ( B ( T )) = E ( B (0)) = 0 . Since B ( t )= X ( t ) μt σ , it follows that E ( X ( T ) μT )=0 or E ( T )= 1 μ E ( X ( T )) . (1) Let p denote the probability that { X ( t ) ,t 0 } hits A before it hits B .B y the result of part (b) of Exercise 22, we have 1= E (exp { 2 μX ( T ) / σ 2 } ) = E (exp { 2 μX ( T ) / σ 2 }| X ( t )h its A before B ) p + E (exp { 2 μX ( T ) / σ 2 }| X ( t )h its B before A )(1 p ) =exp { 2 μA/ σ 2 } p +exp { 2 μB/ σ 2 } (1 p ) , where the last equality follows from the de f nition of T . The above equation yields p = 1 e 2 μB/ σ 2 e 2 μA/ σ 2 e 2 μB/ σ 2 . (2) Now, from equation (1) and (2), we obtain E ( T )= 1 μ E ( X ( T )) = 1 μ [ E ( X ( T

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hwk9 - E3106 Solutions to Homework 9 Columbia University...

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