# Distribution of the sample variance given that $\sigma^2$ is unknown

By Cochran's theorem, if $$y_1,....,y_n\sim\mathcal{N}\left(0,\sigma^2\right)$$ independently with a known variance $$\sigma^2\in\mathbb{R}_{>0}$$, then $$\begin{equation} (n-1)\frac{S^2}{\sigma^2}\sim\chi^2_{n-1}, \end{equation}$$ where $$\begin{equation} S^2=\frac{1}{n-1}\sum_{i=1}^{n}(y_i-\bar{y})^2\sim\frac{1}{n-1}\sum_{i=1}^{n}y_i^2, \end{equation}$$ is the sample variance. However, when the variance $$\sigma^2$$ is unknown, what is the distribution of $$S^2$$? Would this require applying Bayes' theorem such that $$\begin{equation} p\left(S^2\right)=\int_0^{\infty} p\left(S^2\mid \sigma^2\right)p\left(\sigma^2\right)\mathop{d\sigma^2}, \end{equation}$$ and, if so, what would be an appropriate (probably uninformative) prior pdf for $$\sigma^2$$? One option is using Jeffrey's prior, where we have $$p(\sigma^2)\propto\sigma^{-2}$$ and so $$\sigma^2\sim\text{Scale-inv-}\chi^2(\nu,S^2)$$ with $$\nu=n-1$$, since the data is normal. The distribution for $$S^2\mid\sigma^2$$ would be scaled chi-squared which is given as $$\begin{equation} p(S^2\mid\sigma^2)=\frac{2^{-\nu/2}}{S^2\Gamma(\nu/2)}\left(\frac{\nu S^2}{\sigma^2}\right)^{\nu/2}\exp\left(-\frac{\nu S^2}{2\sigma^2}\right), \end{equation}$$ if I haven't made an error with the CDF method. Any help would be much appreciated.

• Whether you know $\sigma^2$ or not, the distribution remains the same: nature doesn't care about your state of mind.
– whuber
Nov 21, 2022 at 13:44
• $$p\left(S^2\right)=\int_0^{\infty} p\left(S^2\mid \sigma^2\right)c\sigma^{-2}\mathop{d\sigma^2} = cS^{-2}$$ With an improper distribution for $\sigma$ you will get that the marginal distribution of $S^2$ will be improper as well. It is the posterior distribution, which is a ratio of two improper distributions, where the constant $c$ disappears. Nov 21, 2022 at 14:01
• @SextusEmpiricus Thank you for that derivation (I would upvote if I could). Nov 22, 2022 at 10:35

One option is using Jeffrey's prior, where we have $$p(\sigma^2)\propto\sigma^{-2}$$

This won't work. With an improper distribution for $$\sigma$$ you will get that the marginal distribution of $$S^2$$ will be improper as well.

More precisely you will get

$$p\left(S^2\right)=\int_0^{\infty} p\left(S^2\mid \sigma^2\right)c\sigma^{-2}\mathop{d\sigma^2} = cS^{-2}$$

It is the posterior distribution, which includes a ratio of two improper distributions $$p(\sigma^2)$$ and $$p(S^2)$$, where the constant $$c$$ disappears.

I derived the above indirectly by using Bayes' formula and the expression for the inverse Gamma distribution.

$$p(\sigma^2|S^2) = \frac{p(\sigma^2) p(S^2|\sigma^2)}{p(S^2)}$$

with $$p(\sigma^2)$$ and $$p(S^2|\sigma^2)$$ known, you can argue that $$p(\sigma^2|S^2) \propto {p(\sigma^2) p(S^2|\sigma^2)}$$ must be an inverse Gamma distribution. Then you use the expression for the inverse gamma distribution to figure out the normalization constant and derive $$p(S^2)$$.