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Comparing Sampling Methods for Estimating Probabilities in Exponential Distribution

This study evaluates various sampling techniques for estimating the probability of a random variable ( X ) from an exponential distribution, specifically for ( P(X > 2) ). The exact probability is deduced to be ( e^{-2} approx 0.1353 ). Different sampling methods, including direct sampling and importance sampling, yielded varying results, highlighting stability in accepted samples compared to probability estimates. Additionally, bootstrapping is applied to estimate the accuracy of ( P(X > 2) ), demonstrating fluctuations in sample estimates and standard deviations.

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Comparing Sampling Methods for Estimating Probabilities in Exponential Distribution

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  1. Basic problem • A random variable X follows the exponential distribution, p(x)=exp(-x) for x=>0. Check how different ways of sampling will compare in terms of accuracy for estimating the probability of x>2 with 1,000 samples. • Exact value of probability is exp(-2)=0.1353.

  2. From actual distribution x=exprnd(1,1,1000); y=sum((sign(x-2)+1)/2)=142 x=exprnd(1,1,1000); y=sum((sign(x-2)+1)/2)=126 x=exprnd(1,1,1000); y=sum((sign(x-2)+1)/2)=138 x=exprnd(1,1,1000); y=sum((sign(x-2)+1)/2)=115 x=exprnd(1,1,1000); y=sum((sign(x-2)+1)/2)=154 x=exprnd(1,1,1000); y=sum((sign(x-2)+1)/2)=134 • Estimatedrelativeaccuracybasedon 142 is • Exactrelativeaccuracyis

  3. Rejection sampling from a gamma distribution • Pick a gamma distribution with a=1, b=2. • Need M=2 for bounding. x=gamrnd(1,2,1,1000); p=exppdf(x); q=gampdf(x,1,2); ratio=p./(2*q); ratio(1:10) accepttest=rand(1,1000); accept=(sign(ratio-accepttest)+1)/2; acceptsample=x.*accept; exceed=sum(sign(acceptsample-2)+1)/2=72; 64; 61 nsamples=sum(sign(acceptsample))=490;519; 516 prob=exceed/nsamples=0.1469;0.1233; 0.1182 Repeated 3 times

  4. Question • Why the number of accepted samples is more stable than the estimate of the probability?

  5. Importance sampling from same distribution ratio=p./q; exceedsamples=(sign(x-2)+1)/2; exceed=sum(exceedsamples.*ratio)=145.4; 131.7; 132.4 %To get estimate of probability divide by 1,000 ratio=ratio/sum(ratio); exceed=sum(exceedsamples.*ratio)=0.1499; 0.1311; 0.1279 %With normalized weight get probability directly.

  6. Bootstrapping • Illustrate bootstrapping for estimating accuracy of probability of x>2 from actual distribution. x=exprnd(1,1,1000); y=sum((sign(x-2)+1)/2)=143 for i=1:100 xs=datasample(x,1000); y(i)=sum((sign(xs-2)+1)/2); end mean(y)=143.0500 std(y)=10.4566 y(1:10)=159 140 134 121 149 147 126 163 138 141

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