Assignment03Sol

Assignment03Sol - Columbia University IEOR 4404 Simulation...

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Columbia University IEOR 4404: Simulation Spring 2009 Solution to Assignment 3 1. Exercise 3 Solution: function SimulateDiscrete(p) % SimulateDiscrete(p) generates a discrete random variable with % probability mass function given by the vector p p=[0.3,0.2,0.35,0.15]; U = rand(); i = 0; total = 0; while( U > total ) i = i + 1; total = total + p(i); end i 2. Exercise 4 Solution: The following MATLAB code prints 95% approximate confidence intervals for the expectation and variance of the total number of hits:
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hits = []; for n=1:10000 list = 1:100; for k=100:-1:1 % generate a random integer between 1 and k i = floor(k*rand()) + 1; % swap values of list(i) and list(k) temp = list(i); list(i) = list(k); list(k) = temp; end % count number of hits hits = [ hits sum( list == 1:100 ) ]; end m = mean( hits ); v = mean( ( hits - mean( hits ) ).^2 ); sigmam = std( hits ); sigmav = std( ( hits - mean( hits ) ).^2 ); CIm = [ m - 1.96 * sigmam/n^.5, m + 1.96 * sigmam/n^.5 ] CIv = [ v - 1.96 * sigmav/n^.5, m + 1.96 * sigmav/n^.5 ] Running the code gives us the following output: CIm = 0.9898 1.0292 CIv = 0.9791 1.0436 We now compute the expectation and variance of the number of hits analytically. Let N denote the number of hits and let n = 100. Then N can be written as N = n X i =1 N i , where N i = ( 1 , card i is the i th card to be turned over 0 , otherwise . (1) Notice that for each i = 1 ,...,n we have P [ N i = 1] = 1 n . This is because the i th card to be turned
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over is equally likely to be any of the cards 1 ,...,n . Therefore, for each i = 1 ,...n , E [ N i ] = 1 · 1 n + 0 · ± 1 - 1 n = 1 n , (2) E [ N 2 i ] = 1 2 · 1 n + 0 · ± 1 - 1 n = 1 n . (3) Combining (2) and (3) gives us for each
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Assignment03Sol - Columbia University IEOR 4404 Simulation...

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