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I'm not sure how to express the variance of this estimator. Here's the setup.

We have $X\sim N(0,\sigma^2)$ and want to estimate $\mathbb{E}[\phi(X)]$ where $\phi : \mathbb{R}\to\mathbb{R}$ is some function such that $\mathbb{E}[\phi(X)]$ has finite mean and variance. We have iid samples $Y_1,\dots, Y_n \sim N(0,1)$.

This is the estimator proposed:

$$\hat{\theta} = \frac{1}{n\sigma}\sum_{i=1}^{n} \exp\left[-Y_i^2\left(\frac{1}{2\sigma^2}-\frac{1}{2}\right)\right]\phi(Y_i)$$.

I have shown this estimator is unbiased. But I'm not sure how to express its variance. The most I can say is that since the $Y_i$ are iid, we have

$$Var(\hat{\theta})=\frac{1}{n^2\sigma^2}\sum_{i=1}^{n} Var\left(\exp \left[-Y_i^2\left(\frac{1}{2\sigma^2}-\frac{1}{2}\right)\right]\phi(Y_i)\right)$$.

How can I express this further?

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I presume you mean $\phi(X)$ has finite mean and variance, as $\mathbb{E}\phi(x)$ is a constant.

You know the estimator is unbiased, so you know the mean is finite, so $$Var(\exp[-Y_i^2(\frac{1}{2\sigma^2}-\frac{1}{2})]\phi(Y_i)) < \infty$$ if and only if $$\mathbb{E}(\exp[-Y_i^2(\frac{1}{2\sigma^2}-\frac{1}{2})]\phi(Y_i))^2 < \infty$$

We can rewrite the inside in terms of the PDF for $X$ and $Y_i$: $$\exp[-Y_i^2(\frac{1}{2\sigma^2}-\frac{1}{2})] = \frac{f_X(Y_i)}{f_Y(Y_i)}$$ So rewriting it like this, and expressing the expectation as an integral, we get: $$\int f_Y(y)(\frac{f_X(y)}{f_Y(y)}\phi(y))^2 dy$$ $$= \int f_X(y)\frac{f_X(y)}{f_Y(y)}\phi(y)^2 dy$$ And again, $$\frac{f_X(y)}{f_Y(y)} = \exp[-y^2(\frac{1}{2\sigma^2}-\frac{1}{2})]$$ So if $\sigma^2 \le 1$ you can use this to show the integral is finite, as you already know $Var\phi(X)$ is finite. However, if $\sigma^2>1$ it seems like it would be possible to find $\phi$ such that the integral is infinite.

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