# Finding variance of an estimator

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?

• This is an unbiased estimator of what? Feb 12, 2023 at 17:11
• @AaronHendrickson of $\mathbb{E}[\phi(X)]$
– jet
Feb 12, 2023 at 17:12
• Cross-post: stats.stackexchange.com/q/605169/119261. Feb 12, 2023 at 18:13

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.