7
$\begingroup$

When summarizing a one dimensional continuous distribution (e.g. a posterior distribution) it is common to use either an equal tailed interval (aka quantile-based) or a highest density interval. The 95% equal tailed interval conceptually corresponds to the median in that when the coverage of the interval $\rightarrow 0\%$ the interval converges to the median. In the same way the highest density interval corresponds to the mode as the mode is the point of the highest density. But another popular point summary of a continuous distribution is the mean and my question is:

What is the interval that corresponds to the mean?

That is:

  • What is that interval called?
  • How is it defined/calculated?

If someone has a comment on the conundrum why it is the case that the mean is a really popular way of summarizing a distribution while the corresponding interval is not as popular (as is my impression).

$\endgroup$
  • $\begingroup$ I was interested in the more general question of any continuous distribution, not just symmetrical distributions. $\endgroup$ – Rasmus Bååth Mar 12 '14 at 23:30
  • $\begingroup$ @gung the median is the x-value corresponding to 50% of the way through the CDF on the y-axis. The midrange is 50% of the way through the CDF on the x-axis. Either way, they don't usually coincide with the mean. But perhaps I misunderstood your intent. $\endgroup$ – Glen_b Mar 12 '14 at 23:33
5
$\begingroup$

I don't know if the interval centered on the mean has a special name, but I can think of more than one way that one could define an interval centered on the mean in the sense that the interval will converge to the mean as the width of the inverval goes to zero. The most simple would perhaps be the interval $$C = (\mu-k, \mu+k),$$ where $\mu$ is the mean and $k$ is the smallest value for which $P(x\in C)\ge 1-\alpha$ holds.

More general, one could define the mean centered interval as $$C = (\mu-a, \mu+b),$$ where $a$ and $b$ are positive and chosen by some procedure such that again the interval has the desired credibility.

However, I think there are good reasons for mainly considering the highest density region and the equal tailed interval. To quote Christian Robert (The Bayesian Choice):

To consider only HPD [highest posterior density] regions is motivated by the fact that they minimize the volume among $\alpha$-credible regions and, therefore, can be envisioned as optimal solutions in a decision setting.

Highest density regions are not necessarily connected intervals, which can happen for example when the distribution is multimodal. On this, Robert writes:

When (...) the confidence region is not connected (...), the usual solution is to replace the HPD $\alpha$-credible region by an interval with equal tails.

$\endgroup$
  • $\begingroup$ The highest density interval does not only correspond to the mode in that the HDI converges to the mode when the coverage area gets smaller. Isn't also the HDI the best interval under a L0 (aka 1-0) loss function (as the mode is the best point estimate under a L0 loss function)? Therefore I would expect that the interval that "corresponds" to the mean would be related to the L2 (quadratic) loss function somehow. $\endgroup$ – Rasmus Bååth Mar 13 '14 at 12:29
  • $\begingroup$ The symmetric interval around the mean will minimize the maximum expected quadratic loss within the credible set (because it is symmetric). It will not minimize the average expected quadratic loss in the set, but you could also construct a credible set with that property. $\endgroup$ – Mikkel N. Schmidt Mar 13 '14 at 16:15
  • $\begingroup$ Ok, but an equal tailed interval and HDI interval does corresponds to minimized expected L1 and L0 loss in the whole set? So what I'm looking for is the interval that mimimizes the expected L2 loss? Still I'm wondering if it has a name, how it is calculated, and why it isn't more widely used... $\endgroup$ – Rasmus Bååth Mar 13 '14 at 21:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.