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After reading this question, I read in the reference provided (Hyndman, 1996, The American Statistician) the following:

It follows inmediately from the definition that the boundary of an HDR consists of those values of the sample space with equal density.

How does it follow from the definition ?

Let $f(x)$ be the density function of a random variable $X$. Then the $100(1-\alpha)\%$ HDR is the subset $R(f_\alpha)$ of the sample space of $X$ such that $$R(f_\alpha) = \{x\colon f(x)\geq f_\alpha\},$$ where $f_\alpha$ is the largest constant such that $$P\big(X\in R(f_\alpha)\big)\geq 1-\alpha.$$

Stephan Kolassa
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cladelpino
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1 Answers1

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It doesn't, at least not in general. This only holds for continuous densities.

If the density $f$ is continuous, then the boundary of an HDR $$R(f_\alpha) = \{x\colon f(x)\geq f_\alpha\}$$ (which is the pre-image of the half-open interval $[f_\alpha,\infty[$ under $f$) is given by $\{x\colon f(x)=f_\alpha\}$.

For a counterexample with a non-continuous density, define

$$ f(x) := \begin{cases} 0, & x<0 \\ \tau, & x=0 \\ \frac{1}{(x+1)^2}, & x>0 \end{cases}$$

This integrates to 1, so it is a density. Every HDR will be of the form $[0,x_\alpha]$, where $x_\alpha>0$ depends on $\alpha$, and obviously, we will typically not have $\tau=f(0)=f(x_\alpha)$.

Alternatively, use any discrete distribution, where this statement will usually also not hold.

Stephan Kolassa
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  • Is it sensible to assign/define a density value to the point of discontinuity in a density function? – Sextus Empiricus Jan 15 '19 at 11:03
  • @MartijnWeterings: I don't think it's very useful. If we don't assign it a value, then the density may not even be defined at the boundary of the HDR. (Which already holds at $\infty$.) – Stephan Kolassa Jan 15 '19 at 11:05
  • I am a bit puzzled about the *meaning* of such assigned value. It may have a meaning to the mathematicians that only care about functions, but I wonder whether it is correct from the applied point of view. How can you speak about the density in a point of discontinuity? Just using some value because otherwise it would be undefined doesn't sound like a good argument to me. – Sextus Empiricus Jan 15 '19 at 11:08
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    Well, we can remove/undefine/redefine a countable number of points from a continuous density, anyway, so one could say that talking about the density's value *at specific points* makes countably infinitely little sense, anyway. – Stephan Kolassa Jan 15 '19 at 11:11
  • Another reference, (Meeker and Escobar 1995) , claim there is a "... close relationship between likelihood based confidence intervals and Bayesian highest posterior density (HPD) intervals (e.g., Casella and Berger 1990, p. 424; Severini 1991)" Does this mean that there is a procedure to establish $f_\alpha$ in relation to the maximum density $f_{MAP}$, based on some ditributional assumptions (like it is done for the likelihood)? – cladelpino Jan 15 '19 at 17:55
  • That doesn't ring a bell. Perhaps you want to post a separate question on that, it's a different topic. – Stephan Kolassa Jan 15 '19 at 17:57