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Unformatted text preview: IEOR 4703: Solutions to Homework 7 1. Consider the problem of estimating ( x ) = E [ e Z I { Z x } ] for x 1 and where Z N(0 , 1). Let X := e Z I { Z x } . (a) Show that ( x ) (1 ( x )) e x where ( . ) is the CDF of a standard normal random variable. SOLUTION: Observe that e z is an increasing function and e Z 1 { Z x } e x 1 { Z x } Hence, ( x ) = E [ e Z 1 { Z x } ] e x E [ 1 { Z x } ] = e x P ( Z x ) = (1 ( x )) e x (b) Show that E [ X 2 ] (1 ( x )) e 2 x SOLUTION: E [ X 2 ] = E [ e 2 Z 1 { Z x } ] e 2 x E [ 1 { Z x } ] = e 2 x P ( Z x ) = (1 ( x )) e 2 x (c) Consider doing importance sampling with alternative density g ( y ) having the density of W N( x, 1). (That is, we shift the mean by x .) Show that ( x ) = E [ Y ] where Y = e x 2 / 2 e W e Wx I { W>x } . Further show that E [ Y ] e x 2 / 2 E [ e W I { W>x } ] . SOLUTION: Let h ( y ) = e y I { y > x } . Letting f ( y ) denote the N (0 , 1) density and g ( y ) the N ( x, 1) density, it is easily verified that the likelihood ratio L ( y ) = f ( y ) /g ( y ) is given by L ( y ) = e y 2 / 2 e ( y x ) 2 / 2 = e x 2 / 2 yx , and thus h ( W ) L ( W ) = e x 2 / 2 e W e Wx I { W>x } = Y. From the general theory of importance sampling, we know that E ( X ) = E ( h ( W ) L ( W )) = E ( Y ) . Since e Wx < 0 when W > x , we have Y e x 2 / 2 e W I { W>x } yielding E [ Y ] e x 2 / 2 E [ e W I { W>x } ] . 1 (d) Using the fact that W may be expressed as Z + x , show that ( x ) e 1 ( x 1) 2 / 2 . (In particular ( x ) 0 as x , which can also be proved directly using the monotone convergence theorem on E [ e Z I { Z x } ].) SOLUTION: As e z + x e z + x for z 0 and x 1, E [ e W...
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This note was uploaded on 03/14/2012 for the course IEOR 4703 taught by Professor Sigman during the Spring '07 term at Columbia.
 Spring '07
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