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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 \begin{equation} \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), \end{equation} where \begin{equation} s^{2}=\frac{1}{n-1}\sum_{i=1}^{n}(X_{i}-\bar{X})^{2}, \end{equation} 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?

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    $\begingroup$ It might be helpful for $g(y)$ to be an increasing function of $y$ $\endgroup$
    – Henry
    Commented Oct 11, 2017 at 23:46
  • $\begingroup$ @Henry Why is that so? Does the answer to the question become harder if $g(y)$ is not an increasing function? $\endgroup$
    – Aaron Hendrickson
    Commented Oct 11, 2017 at 23:49
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    $\begingroup$ If you had for example $g(y)=\sin(y)$, I suspect you could have serious problems in some situations $\endgroup$
    – Henry
    Commented Oct 11, 2017 at 23:51
  • $\begingroup$ @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 $s^{2}$?... $\endgroup$
    – Aaron Hendrickson
    Commented Oct 11, 2017 at 23:56
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    $\begingroup$ 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 $n=11, s^2=0.8, \alpha=0.05$ $\endgroup$
    – Henry
    Commented Oct 12, 2017 at 0:01

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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

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