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

  • 1
    $\begingroup$ Simply transform $(n-1)S^2/\sigma^2\to S^2$ (i.e. change variables). $\endgroup$ – StubbornAtom Sep 10 '19 at 20:04
  • $\begingroup$ How should I do that? I do not understand how. $\endgroup$ – Pablo Sep 10 '19 at 20:05
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    $\begingroup$ 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? $\endgroup$ – whuber Sep 10 '19 at 20:07
  • $\begingroup$ @Pablo en.wikipedia.org/wiki/…. $\endgroup$ – StubbornAtom Sep 10 '19 at 20:08
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    $\begingroup$ 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. $\endgroup$ – 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 $E(S^2) = \alpha/\lambda = 144 = \sigma^2.$

A simulation in R:

v = replicate(10^5, var(rnorm(5, 100, 12)))
[1] 144.0218   # aprx 144

hist(v, prob=T, col="skyblue2")
  curve(dgamma(x, 2, 2/144), add=T, lwd=2, col="red")

enter image description here


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