Function of a confidence interval

Question

If we have a confidence interval for a given parameter $\theta$ given as $[\theta_l, \theta_u]$ ($l$ is for lower and $u$ is for upper) at confidence level $\gamma$, and we have a monotone (Borel) measurable function $f(\cdot)$, can we claim that $[f(\theta_l), f(\theta_u)]$ is a confidence interval for the transformed parameter $f(\theta)$ at the same level $\gamma$? Can we say anything about the confidence interval if $f$ is not monotone?

Answer 1

If we use the definition of confidence interval from wikipedia we have that $$ \mathbb{P}(\theta_l \leq \theta \leq \theta_u) = \gamma $$ If we apply $f(\cdot)$ to all inequalities in the probability operator we immediately get $$ \mathbb{P}(f(\theta_l) \leq f(\theta) \leq f(\theta_u)) = \gamma $$ I found this so quick and easy that I am amazed I hadn't seen it before. Could anyone please confirm if this is correct?

As for the case when $f$ is not monotone, I do not know what can be said in general.

Answer 2

Here is one case where a monotone transformation has been widely used in practice.

Suppose data are normal with $\mu$ and $\sigma$ both unknown. Then $(n-1)S^2/\sigma^2 \sim \mathsf{Chis}(\mathrm{df} = n-1.$

If $L$ and $U$ are chosen so that $P(L \le (n-1)S^2/\sigma^2 \le U) = 0.95,$ where $S^2$ is the sample variance, one has a 95% CI for $\sigma^2$ of the form $\left(\frac{(n-1)S^2}{U}, \frac{(n-1)S^2}{L}\right).$

For $\sigma>0$ the square-root transformation is monotone, so it follows that a 95% CI for $\sigma$ is $\left(\sqrt{\frac{(n-1)S^2}{U}}, \sqrt{\frac{(n-1)S^2}{L}}\right).$