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If I have an estimator such as the sample variance $$\hat{\sigma}^2_{} = \displaystyle\frac{1}{N}\sum_{k=1}^{N}(x_k-\hat{x})^2$$

How do I calculate the expected value of the estimator if it passes through a non-linear transformation? $$E(\displaystyle\frac{1}{1+\hat{\sigma}^2_{}}) $$

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  • $\begingroup$ Jessica, do you have an answer to your question and could you update it? I have a similar question like you without knowing the information of distribution. thx $\endgroup$ – lzstat Feb 17 '16 at 22:51
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So this is a tough one without knowing the distribution of the $(x_k)$ or whether the $x_k$ are i.i.d or not.

Assuming they are i.i.d. and $\hat{\sigma}^2$ has some distribution $K$, then its not necessarily guaranteed that $\mathbb{E}\big[\tfrac{1}{1+\hat{\sigma}^2}\big]$ is even defined without a little more info on $K$.

Now, if we assume that $x_k$ are i.i.d Normal random variables with mean 0 and variance $\sigma^2$, then each $x_k^2$ will follow a Chi-squared distribution. The random variable $$\frac{1}{\sum\limits_{k-1}^{N}X^2}$$ will follow whats called a scaled-inverse Chi-squared distribution.

$\mathbb{E}\big[\sum\limits_{k-1}^{N}X^2\big]^{-1} = \frac{N \sigma^{-2}}{N-2}$

Now this isn't exactly what you've asked for, but it might be a good first step.

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  • $\begingroup$ Thanks measure. Apparently you can use slutskys theorem and show the sample variance converges to the variance asymptotically. Assuming you have finite mean/variance. $\endgroup$ – jessica Nov 28 '15 at 5:45

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