12_Sim_Hw5_Sol

12_Sim_Hw5_Sol - IEOR 4404 Assignment #5 Solutions...

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Unformatted text preview: IEOR 4404 Assignment #5 Solutions Simulation February 20, 2012 Prof. Mariana Olvera-Cravioto Page 1 of 4 Assignment #5 Solutions 1. First note that an immediate algorithm is to generate an exponential random variable X with rate 1, and return X whenever X < . 05. Doing so is inefficient, however, because the probability of the event { X < . 05 } is small. (Indeed, P ( X < . 05) = 1- e- . 5 .) Instead, we can use the inverse transform method. Let Y [ X | X < . 05]. The cumulative distribution function of Y is given by F ( y ) = P [ Y y ] = P [ X y | X < . 05] = Z y f ( u ) du = 1 1- e- . 05 (1- e- y ) , for 0 < y < . 05. The inverse of this function is given by F- 1 ( y ) =- ln(1- (1- e- . 05 ) y ) . With this, we can implement the inverse transform method to generate Y . The steps of the algorithm are the following: STEP 1. Generate a random number U , uniform on [0 , 1]. STEP 2. Return Y = F- 1 ( U ) =- ln(1- (1- e- . 05 ) U ). The following MATLAB code can be used for estimating E [ X | X < . 05]: >> rand(state,0); >> S=0; >> for i=1:1000 S=S-log(1-rand(1)*( 1-exp(-0.05) ) ); end >> S/1000 ans = 0.0249 The exact value of E [ X | X < . 05] can be computed as follows: E [ X | X < . 05] = Z . 05 xf ( x ) dx = Z . 05 x e- x 1- e- . 05 dx = 1- 1 . 05 e- . 05 1- e- . 05 , where we compute the integral using integration by parts. 2 IEOR 4404, Assignment #5 Solutions 2. We wish to generate X by using the acceptance rejection algorithm with an exponential random variable Y with rate . That is, Y has density...
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This note was uploaded on 03/18/2012 for the course IEOR 4404 taught by Professor C during the Spring '10 term at Columbia.

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12_Sim_Hw5_Sol - IEOR 4404 Assignment #5 Solutions...

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