Estimating Stratified Monte Carlo Variance from Sample Sums

Summary

The document asks how to estimate the variance of a stratified Monte Carlo estimator using each stratum’s sample sum and sum of squared observations. It compares a variance expression involving stratum probabilities and variances with an algorithm that calculates the unbiased sample variance within each stratum. The underlying method is to recover each stratum’s sample variance from its first and second sample moments, then combine those estimates using the appropriate stratum weights and sample allocations.

The question itself gives no worked derivation, numerical example, or answer, so it does not resolve the apparent discrepancy. In particular, the displayed expressions use different-looking probability factors, and the reader must check whether the variance being estimated is for a stratum mean or for the weighted overall estimator. The sample-variance formula also requires at least two observations in a stratum. The material is useful as a pointer to a core variance-estimation issue, but it is an unanswered question rather than a complete tutorial.

Key ideas

• Within-stratum sample variance can be calculated from the sample sum and the sum of squared observations.
• The estimated variance of a stratified estimator depends on how stratum probabilities and sample allocations enter the weighting.
• A stratum sample variance requires more than one observation.
• The document raises a formula interpretation question but does not supply its resolution.

Tags

# Understanding Monte Carlo Simulation with Stratified Sampling


# Understanding Monte Carlo Simulation with Stratified Sampling












I am studying Monte Carlo Simulation with Stratified Sampling following Martin Haugh's lectures notes. As is it explained the variance of the estimator is $$ Var(\hat{\theta}_{st,n}) = \sum_{j=1}^{m} \frac{p_{j}\sigma_{j}^2}{n_{j}} $$ At the footnote of page 12 he specifies that $$Var(\hat{\theta}_{st,n})= \sum_{j=1}^{m} Var ( \frac{\sum_{i=1}^{n_{j}} Y_{i}^{(j)}}{n_{j}})p_{j}^2 $$ and that this quantity can be estimated using both $ sum_{j}=\sum_{i=1}^{n_{j}} Y_{i}^{(j)} $ and $ sumsquares_{j}= \sum_{i=1}^{n_{j}} Y_{i}^{(j)^2} $ . I do not get how this can be done considering the fact that later in the page's 13 algorithm the estimator used is is $\hat{\sigma}_{j}^2 = (sumsquares_{j} - \frac{sum_{j}^2}{n_{j}})/ (n_{j}-1)$

Thanks in advance

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.