Homework7Sol.pdf - IEOR-E4707 Spring 2018 Homework 7...

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IEOR-E4707, Spring 2018: Homework 7 Solutions April 17, 2018 Problem 1. Consider the SDE dX s = μ ( s, X s ) ds + σ ( s, X s ) dW s We assume that, just as with a geometric BM, if we begin a process at an arbitrary initial positive value X t = x at an arbitrary initial time t and evolve it forward using this equation, its value at each time T > t could be any positive number but cannot be less than or equal to zero. For 0 t < T , if we solve this equation with initial condition X t = x the random variable X T has density function p ( t, T, x, y ) in the y variable. We assume that p ( t, T, x, y ) = 0 for 0 t < T and y 0. Prove that p ( t, T, x, y ), called the transition density of the SDE, satisfies the Kolmogorov’s backward equation: - p t ( t, T, x, y ) = μ ( t, x ) p x ( t, T, x, y ) + 1 2 σ 2 ( t, x ) p xx ( t, T, x, y )
Problem 2. Continue with Problem 1. Show that p ( t, T, x, y ) satisfies also the Kolmogorov’s forward equation: ∂T p ( t, T, x, y ) = - ∂y ( μ ( T, y ) p ( t, T, x, y )) + 1 2 2 ∂y 2 ( σ 2 ( T, y ) p ( t, T, x, y ))
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Thus, integrate both sides, we will get, h b ( X T ) - h b ( X t ) = Z T t [ h 0 b ( X s ) μ ( s, X s ) + 1 2 σ 2 ( s, X s ) h 00 b ( X s )] ds + Z T t h 0 b ( X s ) σ ( s, X s ) dW s (2) We know that, by the definition of p ( t, T, x, y ), E t,x [ h b ( X T ) - h b ( X t )] = Z -∞ h b ( y ) p ( t, T, x, y ) dy - h ( x ) (3) Also, by using (2), we also can get, E t,x [ h b ( X T ) - h b ( X t )] = Z T t E t,x [ h 0 b ( X s ) μ ( s, X s ) + 1 2 σ 2 ( s, X s ) h

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