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4703-10-Notes-ATV

# 4703-10-Notes-ATV - Copyright c 2010 by Karl Sigman 1 1.1...

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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 μ = 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 μ = E ( X ) , as n → ∞ . But the SLLN does not tell us how good the approximation is; we consider this next. Letting σ 2 = V ar ( X ) denote the variance of the distribution, we conclude that (2) V ar ( X ( n )) = σ 2 n . The central limit theorem asserts that as n → ∞ , the distribution of Z n def = n σ ( X ( n ) - μ ) 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 ) - μ | > z σ n ) P ( | Z | > z ) = 2 P ( Z > z ) . (We can obtain any value of P ( Z > z ) by referring to tables, etc.) For any α > 0 no matter how small (such as α = 0 . 05), letting z α/ 2 be such that P ( Z > z α/ 2 ) = α/ 2, we thus have P ( | X ( n ) - μ | > z α/ 2 σ n ) α, which implies that the unknown mean μ lies within the interval X ( n ) ± z α/ 2 σ n with (approxi- mately) probability 1 - α . This allows us to construct confidence intervals for our estimate: we say that the interval X ( n ) ± z α/ 2 σ n is a 100(1 - α )% confidence interval for the mean μ . Typically, we would use (say) α = 0 . 05 in which case z α/ 2 = z 0 . 025 = 1 . 96, and we thus obtain a 95% confidence interval X ( n ) ± (1 . 96) σ n . The length of the confidence interval is 2(1 . 96) σ n which of course tends to 0 as the sample size n gets larger. In practice we would not actually know the value of σ 2 ; it would be unknown (just as μ 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

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It can be shown that s 2 ( n ) σ 2 , with probability 1, as n → ∞ and that E ( s 2 ( n )) = σ 2 , n 2. So, in practice we would use s ( n ) is place of σ 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 . 1.2 Application to Monte Carlo simulation In Monte Carlo simulation, instead of “collecting” the iid data X 1 , . . . , X n , we simulate it. Moreover, we can choose n as large as we want; n = 10 , 000 for example, so the central limit theorem justification for constructing confidence intervals can safely be used (e.g., in statistics “out in the field” applications, one might only have n = 25 or n = 50 samples!). Thus we can immediately obtain confidence intervals for Monte Carlo estimates.
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4703-10-Notes-ATV - Copyright c 2010 by Karl Sigman 1 1.1...

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