# Summarizing samples of an interval

I have a particular semiparametric model which I'm fitting via MCMC. One of the model parameters I have "semiparametric'ed" away (say $\alpha$) is known to lie between two other parameters, $\theta_1$ and $\theta_2$. Since I have a series of samples $(\theta_1^t, \theta_2^t)$ I also have a series of interval estimates for $\alpha$. What is a reasonable way to summarize these? I can think of doing something like putting down a grid and just recording for each gridpoint whether each sample covers it or not. But I worry about efficiency since I don't have a great sense of what a plausible range is, and I actually have hundreds to thousands of $\alpha$'s.

(I'm also welcoming retags, since I can't seem to find any I like...)

• The approach you describe would compute $F-G$ where $F$ is the empirical cdf of $\theta_1$ and $G$ is the EDF of $\theta_2$. So, if your software reports CDFs (or even histograms), you already have the data you need.
– whuber
May 27 '11 at 14:43
• This would work; the more I think about it the more I think that just taking the difference of the marginal posterior cdf's would be fine. It would give me a point estimate of the interval instead of a distribution over intervals (hopefully that makes sense; it looks weird having written it). I may be overthinking it.
– JMS
May 27 '11 at 14:54
• Also, it's my own software so I can make it report whatever I like - even, unfortunately, incorrect results :)
– JMS
May 27 '11 at 14:55
• There's nothing weird about a distribution over intervals. If you like, think of it as a distribution over the subset of the plane $\{(x,y)|y \ge x\}$. Also, there's a growing literature on interval-based procedures which considers exactly such things. (One's data consist of intervals, but one is agnostic about where in each interval the true value lies. The big application is in the analysis and reduction of errors in numerical computations.)
– whuber
May 27 '11 at 15:02
• @whuber Oh I agree, and I've seen some work on interval estimation of late, but for some reason it just feels weird in my brain to write "point estimate of an interval" even though such a beast is perfectly well defined :) (well maybe it isn't - it's not a point! - but I think you get the idea...)
– JMS
May 27 '11 at 15:07

## 1 Answer

Let's see: $$P(\alpha) = \int P(\alpha, \theta_1, \theta_2) d\theta_1 d\theta_2 = \int P(\alpha | \theta_1, \theta_2) P(\theta_1, \theta_2) d\theta_1 d\theta_2$$ With the good help of Monte Carlo, we can approximate this as $$\frac{1}{n} \sum_t P(\alpha | \theta_1^t, \theta_2^t)$$ With the grid trick, you are doing something like this, but then you are implicitly assuming that $P(\alpha | \theta_1, \theta_2)$ is uniform. Is that an OK assumption for you? If so, it's a good technique if you want to get a view on the distribution of $\alpha$. If you are only out for confidence intervals for $\alpha$, you can probably do better by ordering your intervalsbase on $\theta_1$, and then finding the last value that is only covered by the first $2.5\%$ of you intervals (if you're aiming for $95\%$ confidence), and then similar for the other side.

An OK range of values for your grid would be from the minimum $\theta_1$ up till the maximum $\theta_2$, no? If you mean by 'hundreds of thousands of $\alpha$s that you have that many parameters that you wish to get information about simultaneously, then yes, you are in trouble: you cannot abstract out those parameters, and then hope to automagically and easily get this kind of information back. Univariately though, you're safe.

• Yeah, the way the model is formulated I think it makes sense to take $p(\alpha|\theta_1, \theta_2)$ to be uniform on $(\theta_1, \theta_2)$. The trouble with using the min/max of the $\theta$'s is that they're unbounded :) But, with a little effort we might be able to find a plausible range. In terms of getting the information back, it's actually there; if you look at a traceplot of the $\theta_i$'s they've only got a relatively small gap between them, which doesn't move very much. Part of the reason for the semiparameteric model is exactly that the $\alpha$ parameters are so strongly...
– JMS
May 27 '11 at 14:39
• ... correlated with the $\theta$'s that they're a pain to infer. But if I can get a rough idea of their distribution from a burn-in period I can (maybe) use that to develop a proposal for an independence metropolis scheme to update $\theta$ in a great big block - the order restrictions on $\theta$ induces a lot of autocorrelation in my chain and that should help break it.
– JMS
May 27 '11 at 14:43