I need to prove that the limiting distribution of $(n-1)S^2/\sigma^2$ is a Normal, where $S^2$ is the sample variance.
However, I have no clue on how to do it.
We know that $(n-1)S^2/\sigma^2 \sim \chi^2(n-1)$, however, I'm not even sure this fact can help me...
Any help would be appreciated
EDIT:I needed to add some info. Sorry. $X_1,...,X_n$ are iid Normal random variables, with mean $\mu$, and variance $\sigma^2$
If $X_1,X_2,\ldots$ are i.i.d. standard normal, then $$ Y_n=\sum_{i=1}^n X_i^2\sim \chi^2(n) $$ for all $n\geq 1$. By the CLT $$ \frac{1}{\sqrt{n\sigma^2}}\sum_{i=1}^n (X_i^2-{\rm E}[X_i^2])\to \mathcal{N}(0,1) $$ in distribution as $n\to\infty$, where $\sigma^2=\mathrm{Var}(X_i^2)$. Since $X_i\sim\mathcal{N}(0,1)$ this is equivalent to $$ \frac{1}{\sqrt{2n}}(Y_n-n)\to\mathcal{N}(0,1) $$ in distribution as $n\to\infty$ showing that the sequence $(Y_n)$ properly normalized converges in distribution to a normal limit. That is, for large $n$, $$ Y_n\sim \mathcal{N}(n,2n) $$ approximately.
We have that (since we use the biased-corrected expression for the sample variance),
$$\frac {(n-1)S^2}{\sigma^2} = \sum_{i=1}^n\left(\frac {X_i-\bar X}{\sigma}\right)^2 \equiv W_n \sim \chi^2_{n-1}$$
with
$$ E(W_n) = n-1,\;\; SD(W_n) = \sqrt {2(n-1)}$$
Since this is a finite-sample result we can manipulate the exrpession without reservations.
$$\frac {(n-1)S^2}{\sigma^2} = W_n \Rightarrow nS^2 = \sigma^2W_n + S^2 \Rightarrow nS^2 -n\sigma^2 = \sigma^2W_n + S^2 - (n-1)\sigma^2 -\sigma^2 $$
$$\Rightarrow n\left(S^2 -\sigma^2\right) = \sigma^2\left(W_n -(n-1)\right) + (S^2 - \sigma^2)$$
$$\Rightarrow \sqrt n\left(S^2 -\sigma^2\right) = \sqrt {\frac {n-1}{n}}\cdot \sqrt2\sigma^2\left(\frac {W_n -(n-1)}{\sqrt {2(n-1)}}\right)+ \frac 1{\sqrt n}(S^2 - \sigma^2)$$
The second term on the right goes to zero asymptotically, while the term $\sqrt {\frac {n-1}{n}}$ tends to unity. The term in the big parenthesis is a standardized sum of random variables so it tends to a standard normal. So
$$\sqrt n\left(S^2 -\sigma^2\right) \xrightarrow{d} N(0, 2\sigma^4)$$
which is the known "asymptotic normality" result for $S^2$ if the underlying sample is normal.
So approximately and for large but finite $n$ (not going to infinity, i.e. not at the limit) we are pardoned if we write
$$Q_n \equiv \frac {(n-1)S^2}{\sigma^2} \sim_{approx} N\left(n-1, \frac {2(n-1)^2}{n}\right)$$
The approximating normal has the same expected value as the finite-sample distribution and slightly less variance, (the difference becoming smaller and smaller as $n$ increases).
But wait, the left-hand side is always a positive quantity... in what sense can its distribution be approximated by a normal? The answer lies in the fact that the mean of the approximating normal is $n-1$. As $n$ increases it shifts further to the right (positive orthant), while its standard deviation is of order $\sqrt n$ only. Meaning that less and less probability mass is left to live in the negative orthant. For example already for $n=50$, based on the normal approximation we have that $\hat P(Q_n< 0) < 0.0000004$ so we are "losing" a negligible probabiliy mass (since of course $Q_n$ cannot be negative).
How good an approximation is it? Indicatively, again for $n=50$ we have $P(Q_n \leq 65.28)=0.95$, while using the approximation we have $\hat P(Q_n \leq 66.34) = 0.95$. In an Hypothesis testing framework, this distortion means that if you use the approximation, you will tend to reject the null less often that you should (under the approximation, the significance level for a one-tailed test here would in reality be $0.04$ instead of the nominal $0.05$).
