Importance Sampling Variance vs Importance sampling Size 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.

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