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Are not all distinct prove that the bootstrap

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are not all distinct). Prove that the bootstrap approximation to the mean square error MSE(F) is identical to the usual estimate of the mean square error. (10 marks)
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Solution The mean of F e is ( ) = = = n 1 i i e n x X F θ Therefore, the bootstrap estimate of the mean square error is ( ) = = 2 n 1 i i Fe e x - n x E F MSE X1, X 2 , …X n are independent random variables each distributed according to Fe. Since [ ] x X E n x E Fe n 1 i i Fe = = = It follow that ( ) ( ) ( ) [ ] ( ) [ ] ( ) [ ] ( ) ( ) ( ) 2 2 n 1 i i e 2 n 1 i i 2 Fe 2 Fe Fe Fe Fe n 1 i i Fe e n x F MSE so, and x n 1 x - X E X E - X E X Var now n X Var n x Var F MSE = = = = = = = = = x x We have the usual estimate of the mean square error, ( ) 1) - (n n / x /n S 2 n 1 i i 2 = = x . So, the bootstrap approximation is almost identical. 16. A service system in which no new customers are allowed to enter after 5 Pm. Suppose that each day follow the same probability law and estimate the long-sum average amount of time a customer spends in the system. And then prove that the MSE of estimate of average time a customer spends in the system is identical to the bootstrap approximation. (20 marks) Solution Let W i = the amount of time the ith entering customer spends in the system, I 1. n W ...... W W limit θ n 2 1 n + + + Let N i = the number of customers that arrive on day I,
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n 1 1 - i 1 2 1 1 1 1 N ... N 1 N N i N N 2 N 1 N 1 N 2 1 1 W ...... W W D 1, i form general In W ...... W W D W ...... W W D + + + + + + + + + = + + + = + + + = D i is the sum of the times in the system of all arrivals on day i. M 2 1 M 2 1 n N ...... N N D ...... D D limit θ + + + + + + The average time in the system of all customers arriving in the first m days. ( ) ( ) m N ...... N m D ...... D limit θ M 1 M 1 n + + + + The random variables ( ) M 1 D ...... D + + are all independent and identically distributed, as are the random variables ( ) M 1 N ...... N + + . By the strong law of large numbers, [ ] [ ] N E D E θ To estimate we can that simulate the system over k days Esitmate value of [ ] ( ) k D ...... D D D D E k 2 1 + + + = = . Esitmate value of [ ] ( ) k N ...... N N N N E k 2 1 + + + = = Esitmate of ( ) ( ) k 1 k 1 N ...... N D ...... D N D θ + + + + = This is the average time in the system of all arrivals during the first k days. = = = 2 1 i 1 i e θ - N D E ) MSE(F k i k i The empirical distribution function, { } [ ] [ ] = = = = = = = = = = k 1 i i Fe k 1 i i Fe i i Fe k n n N E , k d d D E hence, k 1, .... , i , k 1 n N , d D P ( ) n d Fe θ =
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