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Let $X_1, X_2,...,X_n$ be an i.i.d. random sample from $N(0, \sigma^{2})$.

a. Find the variance of $\hat{\sigma}^{2}_{MLE}$

So I found $\hat{\sigma}^{2}_{MLE}$ by taking the derivative of the log of the normal pdf function, but from there I am not sure how to proceed. $\hat{\sigma}^{2}_{MLE}$ comes out to $\frac{\sum_{i=1}^n X_i^{2}}{n}$. From there, would I do $\text{var}\left(\frac{\sum_{i=1}^n X_i^{2}}{n}\right)$ ? How do I compute this? Thanks!

Zhanxiong
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user3325193
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2 Answers2

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Hint: If $Y_1,\ldots,Y_n$ are independent random variables and $a_1,\ldots,a_n$ are real constants, then $$ \mathrm{Var}\left(\sum_{i=1}^n a_iY_i\right)=\sum_{i=1}^n a_i^2\mathrm{Var}(Y_i). $$

Stefan Hansen
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  • I understand that $\mathrm{Var}(\frac{\sum_{i=0}^n X_i^{2}}{n})$ equals to $\sum_{i=1}^n \frac{1}{n^{2}}\mathrm{Var}(X_i^2)$. I am just having a hard time figuring out the $\mathrm{Var}(X_i^2)$ part... – user3325193 Feb 20 '14 at 08:07
  • It's ${\rm E}[X_i^4]-{\rm E}[X_i^2]^2$ which can be computed according to e.g. this. – Stefan Hansen Feb 20 '14 at 08:10
  • So $\mathrm{E}(X_i^4) - \mathrm{E}(X_i^2)^2 = (\mu^4 + 6\mu ^2\sigma^2 +3\sigma^4) - (\mu^2+\sigma^2)^2 = 2\sigma^4$ (because $\mu = 0$) ? $\sum_{i=1}^n \frac{1}{n^{2}}\mathrm{Var}(X_i^2) = \sum_{i=1}^n \frac{1}{n^{2}}2\sigma^4$

    Does it then reduce to $\frac{2}{n}\sigma^4$ ? Thanks

     – user3325193 Feb 20 '14 at 09:50
  • Indeed it does. – Stefan Hansen Feb 20 '14 at 09:59
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Let us use first principles and rederive from scratch while ignoring all prepackaged distributions (textbooks would tell you that a sum of squared standard Gaussian random variables $\sim$ a Chi-square distribution).

Let $X'= \frac{X}{\sqrt{n}}$. Hence $X' \sim N(0,\frac{\sigma^2}{n})$. The pdf of a transformation $Y =X^{'2}$, becomes $f(y)= \frac{\sqrt{\frac{n}{y}} e^{-\frac{n y}{2 \sigma ^2}}}{\sqrt{2 \pi } \sigma }\,, y\in (0,\infty)$. The Characteristic Function of Y, $\mathcal{C}(t)=\frac{1}{\sqrt{1-\frac{2 i \sigma ^2 t}{n}}}$. The distribution of an $n$-summed variable has for characteristic function $\mathcal{C}(t)^n$. Now define the raw moment of the convolution $M(d)=-i^d \frac {\partial^d\mathcal {C(t)^n}} {\partial t^d}\bigg|_{t=0}$. So unless I made a mistake somewhere, $M(1)= \sigma ^2 $ and the variance $M(2)-M(1)^2=\frac{2 \sigma ^4}{n}$.

Nero
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