How to prove $s^2$ is a consistent estimator of $\sigma^2$?

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?

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.

• Please add the [self-study] tag & read its wiki. Also, you can use $\LaTeX$ markup on this site, see math.meta.stackexchange.com/questions/5020/…. Nov 14 '20 at 9:51
• The decomposition of the variance is incorrect in several aspects. Nov 14 '20 at 10:55
• @Xi'an On the third line of working, I realised I did not put a ^2 on the n on the numerator of the fraction Nov 14 '20 at 11:02
• @MrDerpinati, please have a look at my answer, and let me know if it's understandable to you or not. Nov 14 '20 at 11:47
• @Xi'an My textbook did not cover the variation of random variables that are not independent, so I am guessing that if $X_i$ and $\bar X_n$ are dependent, $Var(X_i +\bar X_n) = Var(X_i) + Var(\bar X_n)$ ? Nov 14 '20 at 14:18

It's a very well known result that :

If $$X_1, X_2, \cdots, X_n \stackrel{\text{iid}}{\sim} N(\mu,\sigma^2)$$ , then $$Z_n = \dfrac{\displaystyle\sum(X_i - \bar{X})^2}{\sigma^2} \sim \chi^2_{n-1}$$ Thus, $$\mathbb{E}(Z_n) = n-1$$ and $$\text{var}(Z_n) = 2(n-1)$$ .

If you wish to see a proof of the above result, please refer to this link.

Now, since you already know that $$s^2$$ is an unbiased estimator of $$\sigma^2$$ , so for any $$\varepsilon>0$$ , we have :

\begin{align*} &\mathbb{P}(\mid s^2 - \sigma^2 \mid > \varepsilon )\\ &= \mathbb{P}(\mid s^2 - \mathbb{E}(s^2) \mid > \varepsilon )\\ &\leqslant \dfrac{\text{var}(s^2)}{\varepsilon^2}\\ &=\dfrac{1}{(n-1)^2}\cdot \text{var}\left[\sum (X_i - \overline{X})^2)\right]\\ &=\dfrac{\sigma^4}{(n-1)^2}\cdot \text{var}\left[\frac{\sum (X_i - \overline{X})^2}{\sigma^2}\right]\\ &=\dfrac{\sigma^4}{(n-1)^2}\cdot\text{var}(Z_n)\\ &=\dfrac{\sigma^4}{(n-1)^2}\cdot 2(n-1) = \dfrac{2\sigma^4}{n-1} \stackrel{n\to\infty}{\longrightarrow} 0 \end{align*}

Thus, $$\displaystyle\lim_{n\to\infty} \mathbb{P}(\mid s^2 - \sigma^2 \mid > \varepsilon ) = 0$$ , i.e. $$s^2 \stackrel{\mathbb{P}}{\longrightarrow} \sigma^2$$ as $$n\to\infty$$ , which tells us that $$s^2$$ is a consistent estimator of $$\sigma^2$$ .

Note : I have used Chebyshev's inequality in the first inequality step used above. Hope my answer serves your purpose. Thank you.

• Since the OP is unable to compute the variance of $Z_n$, it is neither well-know nor straightforward for them. I thus suggest you also provide the derivation of this variance. Nov 14 '20 at 12:15
• I feel like I have seen a similar answer somewhere before in my textbook (I couldn't find where!) but the method is very different. As I am doing 11th/12th grade (A Level in the UK) maths, to me, this seems like a university level answer, and thus I do not really understand this. Thank you for your input, but I am sorry to say I do not understand. Nov 14 '20 at 14:07
• I guess there isn't any easier explanation to your query other than what I wrote. Also, what @Xi'an is talking about surely needs a proof which isn't very elementary (I've mentioned a link). In fact, the definition of Consistent estimators is based on Convergence in Probability. Do you know what that means ? Nov 14 '20 at 19:59