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Unformatted text preview: Columbia University IEOR 4404: Simulation Fall 2009 Solution to Assignment 7 1. Exercise 8.20 Solution: (a) E [ I ] = P [ Y < g ( X )] = P [ bU 2 < g ( U 1 )] = 1 P [ bU 2 < g ( U 1 )  U 1 = x ] dx = 1 P [ bU 2 < g ( x )] dx = 1 P [ U 2 < g ( x ) b ] dx = 1 g ( x ) b dx = 1 b 1 g ( x ) dx (b) First, we compute Var( bI ): Var( bI ) = b 2 Var( I ) = b 2 ( E [ I 2 ] E [ I ] 2 ) = b 2 ( E [ I ] E [ I ] 2 ) = b 2 [ 1 b 1 g ( x ) dx 1 b 2 ( 1 g ( x ) dx ) 2 ] = b 1 g ( x ) dx ( 1 g ( x ) dx ) 2 Next, we compute Var( g ( U )): Var( g ( U )) = E [ g ( U ) 2 ] E [ g ( U )] 2 = 1 g ( x ) 2 dx ( 1 g ( x ) dx ) 2 Thus, since g ( x ) b for 0 x 1, Var( bI ) = b 1 g ( x ) dx ( 1 g ( x ) dx ) 2 1 g ( x ) 2 dx ( 1 g ( x ) dx ) 2 = Var( g ( U )) . (1) 2. Exercise 8.22 (a) Repeat n simulation runs. In run i , for 1 i n , use independent random numbers U 1 ,i and U 2 ,i to generate independent binomial ( n,p ) random variables X i and Y i using the inverse transform algorithm (see page 57 of the textbook). Estimate by 1 n n i =1 e X i Y i (b) XY can be used as a control variate. In order to show that using XY as a control variate gives us an estimator with smaller variance, it suces to show that XY is correlated to e XY . But this follows from the fact that e XY is a strictly increasing function of XY (see page 140). Since E [ XY ] = n 2 p 2 , our estimator for is e XY + c ( XY n 2 p 2 ) . for c < 0. (c) Recall the Taylor series expansion of f ( x ) = e x : e x = 1 + x + x 2 2! + x 3 3! + . Thus, e XY can be approximated as e XY 1 + XY + ( XY ) 2 2! . Intuitively, since 1+ XY +( XY ) 2 / 2 provides a better approximation to e XY than 1+ XY , XY + ( XY ) 2 / 2 is even more correlated to e XY than XY is. Thus, the control variate XY + ( XY ) 2 / 2 should perform better than the control variate XY . (d) Recall that the moment generating function of a binomial ( n,p ) random variable X is given by M X ( t ) = [ pe t + (1 p )] n ....
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 Spring '10
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