# Are confidence intervals for the sample variance transformation respecting?

Given the random sample $\{X_{1},X_{2},\dots,X_{n}\}$ where $X_{i}\sim\mathcal{N}(\mu,\sigma^{2})$, the confidence intervals for $\sigma^{2}$ is $$\left(\dfrac{(n-1)s^2}{\chi^2_{\alpha/2,n-1}} \leq \sigma^2 \leq \dfrac{(n-1)s^2}{\chi^2_{1-\alpha/2,n-1}}\right),\\ \left(a \leq \sigma^2 \leq b\right),$$ where $$s^{2}=\frac{1}{n-1}\sum_{i=1}^{n}(X_{i}-\bar{X})^{2},$$ and $\bar{X}$ is the sample mean.

If I have some arbitrary transformation $g(s^{2})$, can I simply compute the confidence interval of this transformation to be $g(a)\leq g(\sigma^{2})\leq g(b)$? If not, why?

• It might be helpful for $g(y)$ to be an increasing function of $y$ – Henry Oct 11 '17 at 23:46
• @Henry Why is that so? Does the answer to the question become harder if $g(y)$ is not an increasing function? – Aaron Hendrickson Oct 11 '17 at 23:49
• If you had for example $g(y)=\sin(y)$, I suspect you could have serious problems in some situations – Henry Oct 11 '17 at 23:51
• @Henry I see your point. So maybe a sufficient condition for $g(y)$ for this to work is that it is strictly increasing or decreasing over the domain of $s^{2}$?... – Aaron Hendrickson Oct 11 '17 at 23:56
• If $g(y)$ is decreasing then you will need to reverse the direction of the inequalities and the ends of the confidence interval. Incidentally, with a similar thought, I would be interested in what you thought the values of $a$ and $b$ were in your first statement if for example $n=11, s^2=0.8, \alpha=0.05$ – Henry Oct 12 '17 at 0:01

See this link for what the distribution of $g(s_{N}^{2})$ will be when $N$ is large. It should be fairly easy to prove from here that the answer is no