Monte Carlo integration of a smooth function

Question

Estimate \int0^1 e^x\,dx (whose true value is e - 1 \approx 1.71828) by simulation. Sample U1, \dots, Un uniformly on [0,1] and average the function values, since \int0^1 e^x\,dx = E[e^U] for U \sim U(0,1). Write the estimator, then say how many samples you need for roughly 3 correct decimals.

Accepted Answer

The sample-mean estimator is unbiased for the integral because each e^{Ui} has expectation exactly e-1: Convergence. The variance of a single term is \operatorname{Var}(e^U) = E[e^{2U}] - (e-1)^2 = 	frac{e^2-1}{2} - (e-1)^2 \approx 3.1945 - 2.9525 = 0.242. The estimator averages n of them, so its standard error is \sigma{\hat I} = \sqrt{\frac{0.242}{n}} \approx \frac{0.492}{\sqrt n}. Monte Carlo error shrinks like O(1/\sqrt n): for error near 10^{-3} you need n \approx (0.492/0.001)^2 \approx 2.4 	imes 10^{5} samples. Each extra decimal of accuracy costs about 100	imes the samples. Takeawa…

Monte Carlo integration of a smooth function

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