Finding the variance of this estimator

Question

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?

Answer 1

$\DeclareMathOperator{\Var}{Var}$

You can perform the similar calculation for variance by noting $\Var(Y) = E[Y^2] - (E[Y])^2$ for any random variable $Y$: \begin{align} & E\left[\exp \left[-2Y_i^2\left(\frac{1}{2\sigma^2}-\frac{1}{2}\right)\right]\phi(Y_i)^2\right] \\ =& \int_{-\infty}^\infty \exp\left[ -y^2\left(\frac{1}{\sigma^2}-1\right)\right]\phi(y)^2 \cdot \frac{1}{\sqrt{2\pi}} \exp(-y^2/2)dy \\ =& \int_{-\infty}^\infty \phi(y)^2 \cdot \frac{1}{\sqrt{2\pi}}\exp\left(-y^2\left(\frac{1}{\sigma^2} - \frac{1}{2}\right)\right)dy \end{align}

Now here comes the problem, this integral diverges if $\sigma^2 > 2$. If $\sigma^2 \leq 2$, then you can further simplify by making a change of variable of $y$ to match the standard normal density $\varphi(x) = \frac{1}{\sqrt{2\pi}}e^{-x^2/2}$.

Answer 2

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

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

Now, consider the expectation above. We have

$$\mathbb{E}\left[\frac{1}{\sigma}\exp \left[-Y_i^2\left(\frac{1}{2\sigma^2}-\frac{1}{2}\right)\right]\phi(Y_i)\right] = \int_\infty^\infty \frac{1}{\sigma}\exp\left[ -y^2\left(\frac{1}{2\sigma^2}-\frac{1}{2}\right)\right]\phi(y) \cdot\frac{1}{\sqrt{2\pi}} \exp(-y^2/2) \, dy$$

$$= \int_\infty^\infty \phi(y) \cdot \frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{1}{2}\left(\frac{y}{\sigma}\right)^2\right) \, dy$$

$$ = \mathbb{E}[\phi(X)]$$ QED