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Unformatted text preview: Copyright c 2010 by Karl Sigman 1 Introduction to reducing variance in Monte Carlo simulations 1.1 Review of confidence intervals for estimating a mean In statistics, we estimate an unknown mean = E ( X ) of a distribution by collecting n iid samples from the distribution, X 1 ,...,X n and using the sample mean (1) X ( n ) = 1 n n X j =1 X j . This is justified by the strong law of large numbers (SLLN) , which asserts that this estimate converges with probability one (wp1) to the desired = E ( X ) , as n . But the SLLN does not tell us how good the approximation is; we consider this next. Letting 2 = V ar ( X ) denote the variance of the distribution, we conclude that (2) V ar ( X ( n )) = 2 n . The central limit theorem asserts that as n , the distribution of Z n def = n ( X ( n ) ) tends to N (0 , 1), the unit normal distribution. Letting Z denote a N (0 , 1) rv, we conclude that for n sufficiently large, Z n Z in distribution. From here we obtain for any z 0, P (  X ( n )  > z n ) P (  Z  > z ) = 2 P ( Z > z ) . (We can obtain any value of P ( Z > z ) by referring to tables, etc.) For any > 0 no matter how small (such as = 0 . 05), letting z / 2 be such that P ( Z > z / 2 ) = / 2, we thus have P (  X ( n )  > z / 2 n ) , which implies that the unknown mean lies within the interval X ( n ) z / 2 n with (approxi mately) probability 1 . This allows us to construct confidence intervals for our estimate: we say that the interval X ( n ) z / 2 n is a 100(1 )% confidence interval for the mean . Typically, we would use (say) = 0 . 05 in which case z / 2 = z . 025 = 1 . 96, and we thus obtain a 95% confidence interval X ( n ) (1 . 96) n . The length of the confidence interval is 2(1 . 96) n which of course tends to 0 as the sample size n gets larger. In practice we would not actually know the value of 2 ; it would be unknown (just as is). But this is not really a problem: we instead use an estimate for it, the sample variance s 2 ( n ) defined by s 2 ( n ) = 1 n 1 n X j =1 ( X j X n ) 2 . 1 It can be shown that s 2 ( n ) 2 , with probability 1, as n and that E ( s 2 ( n )) = 2 , n 2. So, in practice we would use s ( n ) is place of when constructing our confidence intervals. For example, a 95% confidence interval is given by X ( n ) (1 . 96) s ( n ) n . The following recursions can be derived; they are useful when implementing a simulation requiring a confidence interval: X ( n + 1) = X ( n ) + X n +1 X ( n ) n + 1 , S ( n + 1) 2 = 1 1 n S ( n ) 2 + ( n + 1)( X ( n + 1) X ( n )) 2 ....
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 Fall '07
 sigman

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