08SimHw4sol - IEOR 4404 Assignment #4 Solutions Simulation...

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Unformatted text preview: IEOR 4404 Assignment #4 Solutions Simulation 20th October 2008 Prof. Mariana Olvera-Cravioto Page 1 of 4 Assignment #4 Solutions 1. (a) Inverse transform: F ( x ) = , if x <- 1 , 1 2 ( x 3 + 1) , if- 1 x < 1 , 1 , if x 1 Therefore, we see that the inverse function is given by F- 1 ( u ) = (2 u- 1) 1 / 3 for 0 u 1. The algorithm is then: step 1: Generate U Uniform(0,1). step 2: Return X = (2 U- 1) 1 / 3 . Acceptance-Rejection: choose g ( x ) = max- 1 t 1 f ( t ) = 3 / 2 and h ( x ) the uniform density on [- 1 , 1]. Then, an algorithm is given by: step 1: Generate Y Uniform(-1,1). step 2: Generate U 2 Uniform(0,1) independent of Y . step 2: If U 2 f ( Y ) /g ( Y ) then return X = Y , otherwise, go to step 1. Note that in order to generate Y in STEP 1 we generate U 1 Uniform (0,1), independent of U 2 , and return Y =- 1 + 2 U 1 . Composition: Write f ( x ) = 1 f 1 ( x ) + 2 f 2 ( x ), where 1 = 2 = 1 / 2, f 1 ( x ) = 3 x 2 1 (0 x 1) and f 2 ( x ) = 3 x 2 1 (- 1 x< 0) , where 1 denotes the indicator function (equal to 1 over the specified set, and 0 otherwise). Note that F 1 ( x ) = x 3 +1 for- 1 x 0, and F 2 ( x ) = x 3 for 0 x 1. We can generate X 1 from distribution F 1 via the inversion transform method with F- 1 1 ( u ) = ( u- 1) 1 / 3 , and X 2 from distribution F 2 via the inversion transform method with F- 1 2 ( u ) = u 1 / 3 . An algorithm is then: step 1: Generate independent U 1 ,U 2 Uniform(0,1). step 2: If U 1 1 / 2, return X = ( U 2- 1) 1 / 3 , otherwise return X = U 1 / 3 2 . (b) Inverse Transform: F ( x ) = , if x < , x 2 2 a (1- a ) , if 0 x < a, 2 x- a 2(1- a ) , if a x < 1- a, 2 a (1- a )- ( x- 1) 2 2 a (1- a ) , if 1- a x < 1 , 1 , if x 1 and its inverse is given by: F- 1 ( u ) = p 2 a (1- a ) u, if 0 u <...
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08SimHw4sol - IEOR 4404 Assignment #4 Solutions Simulation...

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