I am trying to prove that $s^2=\frac{1}{n-1}\sum^{n}_{i=1}(X_i-\bar{X})^2$ is a consistent estimator of $\sigma^2$ (variance), meaning that as the sample size $n$ approaches $\infty$ , $\text{var}(s^2)$ approaches 0 and it is unbiased.
I understand how to prove that it is unbiased, but I cannot think of a way to prove that $\text{var}(s^2)$ has a denominator of n. Does anyone have any ways to prove this?
The question asks:
A random sample of size n is taken from a normal population with variance $\sigma^2$. Show that the statistic $s^2$ is a consistent estimator of $\sigma^2$
So far I have gotten:
$\text{var}(s^2) = \text{var}(\frac{1}{n-1}\Sigma X^2-n\bar X^2)$
$= \frac{1}{(n-1)^2}(\text{var}(\Sigma X^2) + \text{var}(n\bar X^2))$
$= \frac{n^2}{(n-1)^2}(\text{var}(X^2) + \text{var}(\bar X^2))$
But as I do not know how to find $Var(X^2) $and$ Var(\bar X^2)$, I am stuck here (I have already proved that $S^2$ is an unbiased estimator of $Var(\sigma^2)$)
Source : Edexcel AS and A Level Modular Mathematics S4 (from 2008 syllabus) Examination Style Paper Question 1. This is for my own studies and not school work.
[self-study]
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