2
$\begingroup$

Let $X_1, ... , X_n \stackrel{i.i.d} \sim \operatorname{Normal}(\mu , \sigma^2)$. Show that $$\mathrm{Var}\left(\frac1n \sum_\limits{i=1}^n (X_i - \overline X)^2 \right) = \cfrac{2(n-1)\sigma^4}{n^2}$$

My attempt:

$$\mathrm{Var}\left(\frac1n \sum_\limits{i=1}^n (X_i - \overline X)^2 \right) = \cfrac{1}{n^2} \mathrm{Var}\left( \sum_\limits{i=1}^n (X_i - \overline X)^2 \right) = \cfrac{1}{n^2} \mathrm{Var}\left( \sum_\limits{i=1}^n X_i^2 - 2\overline X\sum\limits_{i=1}^n X_i + \sum\limits_{i=1}^n \overline X^2 \right) = \cfrac{1}{n^2} \mathrm{Var}\left( \sum_\limits{i=1}^n X_i^2 - 2n \overline X^2 + n \overline X^2 \right) = \cfrac{1}{n^2} \mathrm{Var}\left( \sum_\limits{i=1}^n X_i^2 - n \overline X^2 \right) = \mathrm{Var}\left( \frac1n \sum_\limits{i=1}^n X_i^2 - \left(\frac1n \sum\limits_{i=1}^n X_i\right)^2 \right) = \mathrm{Var}(\mathsf{E}(X_i^2) - (\mathsf{E}(X_i))^2) = \mathrm{Var}(\mathrm{Var(X_i)}) = \mathrm{Var(\sigma^2) = 0}$$

Perhaps you can see why I am not satisfied with such an answer. Thanks for the help.

$\endgroup$
3
  • $\begingroup$ Sorry, I screwed up there. .How did you get from $\text{Var}\left( \frac1n \sum_\limits{i=1}^n X_i^2 - \left(\frac1n \sum\limits_{i=1}^n X_i\right)^2 \right)$ to $\text{Var}({E}(X_i^2) - ({E}(X_i))^2)$? $\endgroup$
    – Glen_b
    Apr 22 '20 at 7:04
  • $\begingroup$ @Glen_b-ReinstateMonica yeah I was not sure about that one. I guess arithmetic mean of a random sample is not $E[X]?$ $\endgroup$
    – jeremy909
    Apr 22 '20 at 12:58
  • 1
    $\begingroup$ Indeed not. Consider tossing a fair coin several times. The proportion of heads (the sample mean of $X_i$ if $X_i=1$ for a head and $0$ otherwise) in a sample of n=5 could be $0.6$ or $0.2$, but $E(X_i)$ is $0.5$. $\endgroup$
    – Glen_b
    Apr 23 '20 at 3:46
1
$\begingroup$

Since you are looking at the variance of the sample variance (without Bessel's correction) you are going to get moments up to the fourth order. So your working should ultimately reduce to some linear function of the first four moments of the normal distrbution (i.e., up to its kurtosis). If you would like to learn more about the various moments of the sample mean and sample variance, in a general setting that does not presume normality, you can find full derivations in O'Neill (2014). The result you are trying to prove is a special case of more general results proved in that paper.

$\endgroup$
1
$\begingroup$

$\frac{(n-1)S^2}{\sigma^2}\sim \chi^2_{n-1}$ where $\chi^2_{n-1}$ is the Chi-square distribution with $n-1$ degrees of freedom. See here for details.

$Var(\frac{(n-1)S^2}{\sigma^2})=2(n-1)$

$Var \left( \frac{\sum(X_i-\bar{X})^2}{\sigma^2}\right)=2(n-1)$

$Var (\sum(X_i-\bar{X})^2 )=2(n-1)\sigma^4$

$Var (\frac{1}{n}\sum(X_i-\bar{X})^2 )=\frac{2(n-1)\sigma^4}{n^2}$

$\endgroup$
4
  • $\begingroup$ His statistic is $\bar{S}^2$ which does not incorporate Bessel's correction, so the relevant distributional result is $n \bar{S}^2 / \sigma \sim \chi_{n-1}^2$. $\endgroup$
    – Ben
    Apr 22 '20 at 7:08
  • $\begingroup$ @Ben - Reinstate Monica , It was a hint. $\endgroup$
    – Masoud
    Apr 22 '20 at 16:16
  • $\begingroup$ I think I'm a little behind, because it seems you expect me to know that $\cfrac{(n-1)S^2}{\sigma^2} \sim \chi_{n-1}^2$ $\endgroup$
    – jeremy909
    Apr 22 '20 at 21:50
  • $\begingroup$ @jeremy909 , I edit the answer. I hope it will help you. feel free about ask any details. $\endgroup$
    – Masoud
    Apr 22 '20 at 21:58
1
$\begingroup$

The problem with your deduction lies here:

$$ \mathrm{Var}\left( \frac1n \sum_\limits{i=1}^n X_i^2 - \left(\frac1n \sum\limits_{i=1}^n X_i\right)^2 \right) = \mathrm{Var}(\mathsf{E}(X_i^2) - (\mathsf{E}(X_i))^2) $$

as $\overline{X} \ne E \left( X_i \right)$, just $E \left( \overline{X} \right) = E \left( X_i \right)$. For similar reason $\overline{X^2} \ne E \left( X_i^2 \right)$.

Hint: I would continue from there with \begin{align} \frac{1}{n^2} Var \left( \sum{X_i^2} -\frac{1}{n} \left( \sum X_i \right)^2 \right) = \\ = \frac{1}{n^2} Var \left( \frac{n - 1}{n} \sum{X_i^2} - \frac{2}{n} \sum_{i < j} X_i X_j \right) = \ldots \end{align}

$\endgroup$
2
  • $\begingroup$ I got: $\frac{1}{n^2} Var \left( \sum{X_i^2} -\frac{1}{n} \left( \sum X_i \right)^2 \right) = \frac{1}{n^2} Var \left( \frac{n - 1}{n} \sum{X_i^2} - \cfrac{2}{n} \sum_{i < j} X_i X_j \right)$. Did I do something wrong? Also, I'm not sure where to go from here. No matter how I try and expand it out, I end up getting these horrible expectations like $E[X_i^4]$ and $Cov(\sum X_i^2, \sum\limits_{i<j}X_iX_j)$ $\endgroup$
    – jeremy909
    Apr 22 '20 at 21:52
  • $\begingroup$ No, it was my mistake, corrected. What was your next step from here? Did you expanded $Var \left( \ldots \right)$ to $E \left( \ldots^2 \right) - E^2 \left( \ldots \right)$? $\endgroup$
    – oszkar
    Apr 23 '20 at 3:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.