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Does the increase in importance sampling size guarantee the decrease in importance sampling variance?

Some context here: I'm trying to use importance sampling instead of equal probability sampling to reduce the variance of my estimator. Based on my data, if I have a small sample size, the variance of importance sampling data is smaller than the variance of equal probability sampling. But if I increase to a big sample size, equal probability sampling variance is guaranteed to be smaller, and the variance of importance sampling doesn't feel like so.

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The variance of the estimator in importance sampling still reduces in inverse-proportion to the number of values used in the simulation (assuming the variance is finite). In importance sampling, we estimate a moment quantity $\mu \equiv \mathbb{E}(h(X))$ with $X \sim f$ by generating a set of random variables $Y_1,Y_2,...,Y_m \sim \text{IID }g$ and using the estimator:

$$\hat{\mu}_m \equiv \frac{1}{m} \sum_{i=1}^m \frac{h(Y_i) f(Y_i)}{g(Y_i)}.$$

With a bit of algebra (see below) the variance of the estimator can be found to be:

$$\mathbb{V}(\hat{\mu}_m) = \frac{1}{m} \bigg[ \mathbb{E} \bigg( h(X)^2 \cdot \frac{f(X)}{g(X)} \bigg) - \mathbb{E}(h(X))^2 \bigg].$$

As you can see, this quantity is still proportionate to $1/m$ (assuming this variance is finite) so the large-simulation mechanics of this method are still essentially the same as in the unweighted case.


Deriving the variance of the estimator: Write the estimator as:

$$\hat{\mu}_m = \frac{1}{m} \sum_{i=1}^m \varepsilon(Y_i) \quad \quad \quad \quad \quad \varepsilon (y) \equiv \frac{h(y) f(y)}{g(y)}.$$

Since each of these terms is an unbiased estimator of the moment of interest, they have variance:

$$\begin{align} \mathbb{V}(\varepsilon(Y_i)) = \mathbb{E}(\varepsilon(Y_i)^2) - \mu^2 &= \mathbb{E} \bigg( \bigg( \frac{h(Y_i) f(Y_i)}{g(Y_i)} \bigg)^2 \bigg) - \mu^2 \\[6pt] &= \int \limits_\mathbb{R} \bigg( \frac{h(y) f(y)}{g(y)} \bigg)^2 g(y) \ dy - \mu^2 \\[6pt] &= \int \limits_\mathbb{R} \frac{h(x)^2 f(x)}{g(x)} f(x) \ dx - \mu^2 \\[6pt] &= \mathbb{E} \bigg( h(X)^2 \cdot \frac{f(X)}{g(X)} \bigg) - \mathbb{E}(h(X))^2 \\[6pt] \end{align}$$

We then have:

$$\begin{align} \mathbb{V}(\hat{\mu}_m) &= \mathbb{V} \bigg( \frac{1}{m} \sum_{i=1}^m \varepsilon(Y_i) \bigg) \\[6pt] &= \frac{1}{m^2} \sum_{i=1}^m \mathbb{V} ( \varepsilon(Y_i) ) \\[6pt] &= \frac{1}{m^2} \sum_{i=1}^m \bigg[ \mathbb{E} \bigg( h(X)^2 \cdot \frac{f(X)}{g(X)} \bigg) - \mathbb{E}(h(X))^2 \bigg] \\[6pt] &= \frac{1}{m} \bigg[ \mathbb{E} \bigg( h(X)^2 \cdot \frac{f(X)}{g(X)} \bigg) - \mathbb{E}(h(X))^2 \bigg]. \\[6pt] \end{align}$$

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    $\begingroup$ I think it is important to note that in the derivation we sometimes take the expectation over f and sometimes over g. Would be nice to add that as a subscript. $\endgroup$
    – Jannis
    Commented Dec 16, 2022 at 6:57
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    $\begingroup$ just to note for anyone who is confused by the notation, and would like to have subscripts on the $\mathbb E$ operators as @Jannis is asking for, you can supply those subscripts mentally: when the expression contains variable Y, use subscript g, and when X use f. $\endgroup$
    – postylem
    Commented Feb 12, 2023 at 15:37

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