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Suppose I'm trying to estimate $\mathbb{E}[\phi(X)]$ where $X$ is an $N(0,\sigma^2)$ r.v. by using the estimator

$$\theta = \frac{1}{n \sigma} \sum_{i=1}^n \exp(-Y_i^2(1/2\sigma^2 - 1/2))\phi(Y_i)$$

where $Y_i$ are i.i.d $N(0,1)$ r.v.'s and $\phi$ is s.t. $\phi(X)$ has finite mean and variance.

How do I show that this is unbiased? I've currently tried manipulating the thing inside the sum to try and get something that resembles a normal pdf which integrates to 1, but there's a couple of things missing that means this doesn't work.

Hints much appreciated

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  • $\begingroup$ What is $\phi$? Just an arbitrary measurable function? $\endgroup$
    – Math1000
    Feb 16 '15 at 23:54
  • $\begingroup$ @Math1000 yup, with $\phi(X)$ having finite mean and variance $\endgroup$
    – Taimur
    Feb 16 '15 at 23:55
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$$ \begin{eqnarray*} \mathbb{E}\left[\theta\right] &=& \mathbb{E}\left[ \frac{1}{n \sigma} \sum_{i=1}^n \exp(-Y_i^2(1/2\sigma^2 - 1/2))\phi(Y_i)\right] \\ && \\ &=&\frac{1}{n \sigma} \mathbb{E}\left[ \sum_{i=1}^n \exp(-Y_i^2(1/2\sigma^2 - 1/2))\phi(Y_i)\right] \\ && \\ &=&\frac{1}{n \sigma} \sum_{i=1}^n \mathbb{E}\left[ \exp(-Y_i^2(1/2\sigma^2 - 1/2))\phi(Y_i)\right] \\ && \\ &=&\frac{1}{\sigma} \mathbb{E}\left[ \exp(-Y^2(1/2\sigma^2 - 1/2))\phi(Y)\right] \\ && \\ &=&\frac{1}{\sigma\sqrt{2\pi}}\int\limits_{-\infty}^{\infty}e^{-y^2(\frac{1}{2\sigma^2} - \frac{1}{2})}\phi(y)e^{-\frac{1}{2}y^2}\mathrm dy \\ && \\ &=&\frac{1}{\sigma\sqrt{2\pi}}\int\limits_{-\infty}^{\infty}e^{-\frac{1}{2}\frac{y^2}{\sigma^2}}\phi(y)\mathrm dy && \\ &&\\ &=&\mathbb{E}\left[\phi(X)\right]\,, \end{eqnarray*} $$

where $Y\sim\mathcal{N}(0,1)$.

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