17 give a probabilistic proof of n 1 i n 1 i 2 2 i 2

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17. Give a probabilistic proof of = = = n 1 i n 1 i 2 2 i 2 i nx x ) x (x where = = n 1 i i n x x letting X denote a random variable that is equally likely to take an any of the values x 1 ,…..,x n and then by applying the identify Var (X) = E[X ] – (E[X]) 2 . (10 marks) Sol. X is a random variable, that is equally likely to be taken on any of the value x 1 , …, x n . We have, E[X] = x P{X=x} So, E[X] = = n 1 i i n 1 x = = n 1 i i n x E[X 2 ] = = n 1 i 2 i x n 1 (E[X]) 2 = ( = n 1 i i n x ) 2 = ( ) 2 x ( = = n 1 i i n x x Q ) 2 n 1 i 2 i 2 2 x x n 1 (E[x]) ] E[X = = And then, Var (X) = ] μ) E[(X 2 = )] x E[(x 2 Var (X) = 2 n 1 i i ) x (x n 1 = We have, Var (X) = 2 2 (E[X]) ] E[X
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2 n 1 i i ) x (x n 1 = = 2 n 1 i 2 i x x n 1 = 2 1 ) ( x x n i i = = 2 1 2 x x n i i = 18 . To estimate E[X], X1,……,X16 have been simulated with the following values resulting, 10,11,10.5,11.5,14,8,13,6,15,10,11.5,10.5,12,8,16,5. Based on these data, if we want the standard deviation of the estimator of E[X] to be less than 0.1, roughly how many additional simulation runs will be needed? (10 marks) Solution X1,….,X16 have the resulting values 10, 11, 10.5, 11.5, 14, 8, 13, 6, 15, 10, 11.5, 10.5, 12, 8, 16, 5 respectively. Thus, n = 16 = = n i i n x X 1 = = 16 1 16 i i x = 16 172 = 10.75 We have, S 2 = 1 ) ( 2 1 = n x x n i i Thus, 2 16 1 ) ( x x i i = = 136.4375 095833 . 9 1 16 4375 . 136 2 = = S The standard deviation of the estimator of E[X} to be less than 0.1. 1 . 0 015929873 . 3 = k 15929873 . 30 = k k = 909.5653 k 910 runs 19. Suppose we want to use simulation to estimate 1 0 u x E e e dx θ = = . Using antithetic variables, discuss what kind of improvement is possible. (10 marks) Solution θ = E[e U ] = 1 0 x dx e θ = e-1
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Since the function h(u) = eU is clearly a mom tone function, the arithmetic variable approach leads to a variance reduction. Cov(e U ,e 1-U ) = E[e U e 1-U ] = E[e U ] E [e 1-U ] = e-(e-1) 2 = -0.2342 And, Var(e U ) = E[e² U ] – ( E[e U ])² = 1 0 2x dx e – (e-1) 2 = ( ) 2 2 1 - e - 2 1 e = 0.2420 { } { } ( ) 0.1210 2 e Var 2 U exp U exp Var U 2 1 = = + By the use of arithmetic variable U and 1-U gives in variance of ( ) ( ) 0.0039 2 e e Cov 2 e Var 2 e e Var U - 1 U U U - 1 U = + + = + Therefore, the variance reduction of 9607 percent will occur by using arithmetic variables. 20. Consider a sequence of random numbers and let N be the first one that is greater than its immediate predecessor. That is N= min (n: n 2 , U n >U n-1 ). Explain what kind of improvement is possible on the variance of estimator “e” by using antithetic variable. (20 marks) Solution N= the first one that is greater than its immediate predecessor. N= min (n: n 2 , U n .> U n-1 ) { } { } ! n 1 U ..... U U P n N P n 2 1 = = All possible orderings of U 1 , U 2 ,……., U n are equally likely. P{N=n} = P{U 1 >n-1} – P{N>n}
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