# Distribution of the sample variance $S^2$ from a normal population [closed]

Let $$X_1, X_2, X_3, ….., X_n$$ be $$N(\mu, \sigma^2)$$ distributed. Then what is the distribution of $$S^2$$

I have already proven that if $$X_i$$ are $$N(\mu, \sigma^2)$$, then $$\frac{(n-1)S^2}{\sigma^2}$$ is $$\chi^2(n-1)$$. I also know that if it is $$\chi^2(n-1)$$ it is in particular a Gamma$$(\frac{(n-1)}{2}, 2)$$. How should I approach the problem next? I want the more formal distribution, rather than just stating that $$S^2$$ is $$\frac{\sigma^2\chi^2(n-1)}{n-1}$$.

• Simply transform $(n-1)S^2/\sigma^2\to S^2$ (i.e. change variables). – StubbornAtom Sep 10 '19 at 20:04
• How should I do that? I do not understand how. – Pablo Sep 10 '19 at 20:05
• You have already answered your question. To put your statement only slightly differently, $S^2$ is distributed as $\sigma^2/(n-1)$ times a $\chi^2(n-1)$ variate. That's perfectly clear and formal. What would you be looking for as an answer, then? – whuber Sep 10 '19 at 20:07
• @Pablo en.wikipedia.org/wiki/…. – StubbornAtom Sep 10 '19 at 20:08
• When the observations are independent identically distributed with an unknown variance you have (n-1)S$^2$/ $\sigma$$^2$ is a pivotal quantity allowing you to generate confidence intervals or test an hypothesis about the variance. S$^2$ by itself is not pivotal and its distribution depends in the value of the unknown variance. So there is nothing more you can say other than it being proportional to a chi-square distribution. – Michael R. Chernick Sep 10 '19 at 20:29

Maybe this is a useful clue. Let $$n = 5; \sigma=12.$$ Then $$S^2 \sim \mathsf{Gamma}(\text{shape}= \alpha = 2,\, \text{rate} = \lambda = 2/144),$$ which gives $$E(S^2) = \alpha/\lambda = 144 = \sigma^2.$$
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