10SimHw10Solution - IEOR 4404 Simulation Prof Mariana...

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IEOR 4404 Assignment #10 Solutions Simulation April 16, 2010 Prof. Mariana Olvera-Cravioto Page 1 of 5 Assignment #10 Solutions 1. (a) Let X be a uniform random variable, over [0 , 1]. Then, it is well known that E [ X ] = 1 / 2 and V ar [ X ] = 1 / 12. Instead of the raw estimator described in the problem, we would like to reduce the variance of the estimator by using X as a control variable. That is, we estimate the mean E [ Y ] where Y = I + c * ( X - E [ X ]) ; it is better to do so since E [ Y ] = E [ I ], but V ar [ Y ] < V ar [ I ]. The value of c * is given by Equation (8.1) on p.147 of the book: c * = - cov ( I, X ) V ar [ X ] . Let us compute the resulting variance reduction. The variance reduction is given at the bottom of p.147: V ar [ Y ] V ar [ I ] = 1 - cov 2 ( I, X ) V ar [ X ] V ar [ I ] First, note that V ar [ I ] is given by: V ar [ I ] = E [ I 2 ] - E [ I ] 2 = P ( X a ) - ( P ( X a )) 2 = a - a 2 , since P ( X a ) = a when X U [0 , 1]. There remains to compute cov ( I, X ), which is given by: cov ( I, X ) = E [ IX ] - E [ I ] E [ X ] = E [ IX | X a ] P ( X a ) - E [ I ] E [ X ] ; but, E [ IX | X a ] P ( X a ) = E [ X | X a ] P ( X a ) = R a 0 xdx = a 2 / 2, we get: cov ( I, X ) = a 2 / 2 - a/ 2 , which is negative for a < 1, as expected. The variance reduction is given by: V ar [ Y ] V ar [ I ] = 1 - 3( a - a 2 ) . (b) Let X exp (1). We repeat the same steps as above. First, note that E [ X ] = V ar [ X ] = 1.
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