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Suppose $X$ is distributed with a unimodal pdf $f(x)$ and let $Y = g(X)$ for some strictly monotone function $g$. Hence $g$ is invertible. Is there an analytically tractable relationship between the highest posterior interval of $X$ and the highest posterior interval of $Y$? For example, let $X$ be $\chi^2$ with $d$ degrees of freedom, and let $Y = \sqrt(X)$.

Comparison to transformations on equal-tailed intervals

Clearly, the answer would be yes if we were considering equal-tailed intervals, since if (as above) $P(X<q) \leq 1−α/2$, say, then $P(Y^2<q) = P(Y < \sqrt q) \leq 1−α/2$.

Therefore, if both $X$ and $Y$ are symmetric and unimodal, then the HPI case is equivalent to the equal-tailed case, and inverting $g$ provides a mapping between intervals without ever needing to evaluate anything about the distribution of $Y$. I would be interested in any less trivial relationship between the HPI intervals.

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  • $\begingroup$ Could you clarify what "analytically tractable" might mean, given that you have framed this question completely abstractly with minimal conditions on $g$? We aren't even assuming it's continuous. Please note, too, that $Y$ need not be unimodal--even when $g$ is smooth. $\endgroup$ – whuber Sep 21 '18 at 18:54
  • $\begingroup$ I would be interested in any expression for the HPD of Y that is some function of the HPD of X, as opposed to one that requires solving the HPD equations anew after applying change-of-variables on $Y = g(X)$. If some additional conditions on $g$ would be helpful, like one-to-one and differentiable, I am not opposed to hypothesizing them. For my part, I suspect that the answer to this question is that there is not any such relationship between the two HPD regions, unless $f$ is symmetric, in which case the HPD correspond to equal-tailed intervals. $\endgroup$ – Andrew M Sep 25 '18 at 2:33
  • $\begingroup$ You are providing less information than you might think. Even when $f$ is symmetric, $g(X)$ can be asymmetric and have arbitrarily many modes while still being a smooth one-to-one function. Much depends on exactly what you mean by "highest posterior interval." I have understood it to be a region $\mathcal R$ for which $\Pr(Y\in\mathcal R)=1-\alpha$ and $y_0\in\mathcal{R},\ y_1\notin\mathcal{R}$ imply $f_Y(y_0)\ge f_Y(y_1).$ But in this sense your "clearly ... yes" conclusion is not true. So, what is your meaning? $\endgroup$ – whuber Sep 25 '18 at 17:03
  • $\begingroup$ As I said, if you think that imposing more structure on $g$ could substantially sharpen this question, I am not opposed. For my part, I would find it helpful if you could link to, or cite reference to some of your examples that you saying show weakness of the conditions on $g$, as I am unfamiliar with some of them. I am also confused by your assertion about the equal-tailed intervals: $P(Y < q) \leq 1-\alpha$, say, then $P(X < g^{-1}(q)) \leq 1 -\alpha$. This expression did not require (formally) any changes-of-variables, instead I was able to leverage properties of $g$. $\endgroup$ – Andrew M Sep 25 '18 at 18:10
  • $\begingroup$ The concept of "equal-tailed" really makes sense only when the distribution is unimodal. Your post and comments sound more and more like you're still trying to formulate a question, which currently sounds vague and broad. Is there any way you could focus it, perhaps by appealing to a specific related problem you might be facing? $\endgroup$ – whuber Sep 25 '18 at 18:20

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