I can't understand why the variance of Chi-square distribution comes from sample variance should be this way

Question

I have some problems with calculating variance of Chi-square distribution. As I know, the variance of it ought to be the 2*m, where m refers to the Degrees of freedom. So, if there are two sample variances, say $s1$ and $s2$, I think each of them has

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

and I believe each of them belongs to $\chi^{2}(n_{i}-1)$. And there is my question, I can't understand why the Var($s_{i}^2$) should be $\frac{2\sigma^{4}}{(n_{i}-1)}$. I mean I think it maybe it should be ${2\sigma^{2}}{(n_{i}-1)}$?

Answer 1

Suppose you have a random sample of size $n$ from a normal distribution with mean $\mu$ and variance $\sigma^2$. Let $s^2$ be the sample variance. If we write $$X = \frac{(n-1)s^2}{\sigma^2}$$ then $X \sim \chi^2_{n-1}$ as you have stated in your question. Now $${\rm var}(X) = 2(n-1)$$ and $$s^2 = \frac{\sigma^2}{n-1}X.$$ Hence $${\rm var}(s^2)=\left(\frac{\sigma^2}{n-1}\right)^2 {\rm var}(X)=\frac{\sigma^4}{(n-1)^2}2(n-1)=\frac{2\sigma^4}{n-1}.$$

Naturally we would expect that the sample variance should become more precise as the sample size increases, so it makes perfect sense that the $n-1$ factor should be in the denominator of the variance of $s^2$ rather than in the numerator.