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I want to compute the variance of this estimator $\hat{\sigma}^2 = \frac{n}{N}\sum_{i=1}^{N}\big(R_{i} - \frac{1}{N}\sum_{j=1}^{N}R_{j}\big)^2$, where $R_{1}, \ldots, R_{N}$ are i.i.d such that: $ R_{1}$ ~ $N(\frac{a}{n}, \frac{\sigma^2}{n})$. I already found its expectation which is $E[\hat{\sigma}^2] = \sigma^2 - \frac{\sigma^2}{N}$, but I am get stuck when it comes to calculate the variance, as I have to deal with correlated variables.

Any help highly appreciated! Thank you!

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    $\begingroup$ Use the fact that $\frac{N\hat\sigma^2}{\sigma^2}$ has chi-square distribution with $N-1$ degrees of freedom. The variance of this distribution is $2(N-1)$. $\endgroup$ – NCh Apr 16 at 14:09
  • $\begingroup$ One way: Also: expand the quadratic expressions in a multiple summation. Then write a double multiple summation of variances and covariances. Since your estimate is quadratic in the $R_i$ you will have covariances of form $\text{Cov}(R_iR_j,R_kR_l)$, etc, where some of the subscripts are equal in various interesting ways. NCh's suggestion will help, but doesn't cover terms like $\text{Cov}(R_1R_2,R_1R_2)$. $\endgroup$ – kimchi lover Apr 16 at 15:00
  • $\begingroup$ But, how can we prove or see that $\frac{N\hat{\sigma}^2}{\sigma^2}$ has chi-squared distribution? $\endgroup$ – CLBJ_23 Apr 16 at 15:08
  • $\begingroup$ I checked it out. However, here our random variables, namely $X_i = R_i - \frac{1}{N}\sum_{j=1}^{N}R_j$ are not independent, since $Cov(R_i, \frac{1}{N}\sum_{j=1}^{N}R_j) \neq 0$. So in this occasion, what are we supposed to do? $\endgroup$ – CLBJ_23 Apr 16 at 23:25

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