# Confidence interval about a multivariate normal probability

From a given sample $x_i \underset{\mathrm{iid}}{\sim} {\cal N}(\mu, \sigma^2)$ it is possible to get a confidence interval about the probability $\Pr(x_i \geq a)$ for any number $a$, by "inverting" some tolerance intervals.

If ${\boldsymbol x}_i \underset{\mathrm{iid}}{\sim}{\cal N}_p({\boldsymbol \mu}, \Sigma)$ is a sample from a $p$-variate normal distribution, how to get a confidence interval about the probability $\Pr(x_{i1} \geq a_1, \ldots, x_{ip} \geq a_p)$? It is easy to get a credible interval in the Bayesian framework, but what about frequentist methods? Is the bootstrap asymptotically valid? Does there exist another known approach?

EDIT 15/09/2012: The Delta method should be possible. If someone is comfortable with the calculations I would be glad if he posts the solution here ;)

• There's something strange about the notation: given that $(x_i)$ is the sample, then $\Pr(x_i \ge a)$ makes no sense (or is trivially $0$ or $1$). E.g., if the sample is $(3.12, -1.03, 0.17)$, $i=3$, and $a=1$, this expression means nothing other than $\Pr(0.17 \ge 1)$. This makes it difficult to discern what the question is asking. – whuber Sep 11 '12 at 20:45
• @whuber Stephane is using small x to denote a random variable contrary to our custom to use small letters for observaed values and capitals for random variables. – Michael Chernick Sep 11 '12 at 21:37
• Thanks, @Michael, but I still cannot make sense of that interpretation. If the tolerance interval is constructed from the $x_i$, it estimates a property of the distribution, not a probability; if it is actually a prediction interval, then it estimates the probability for an independent result not included within the $x_i$. There's a lot of ambiguity here. – whuber Sep 11 '12 at 21:46
• @whuber Michael is right, I'm using some abusive notations, but I think they are not ambiguous if the reader uses "psychology" to guess my statement (by droping trivial or meaningless possibilities). Such abusive notations are common in the Bayesian literature, I think. – Stéphane Laurent Sep 11 '12 at 21:49
• @Macro In the univariate case I think this method yields very larger confidence intervals than those based on tolerance intervals – Stéphane Laurent Sep 12 '12 at 5:16

## 1 Answer

It is easier to construct confidence ellipsoids for multidimensional normals. The probabilities you want require messy numerical integration, I think, since the regions are open-sided rectangular. I think you could certainly bootstrap the samples and get bootstrap estimates of the joint probabilities. But I think that would not be commonly done since there is an exact answer even though it involves multidimensional numerical integration.

• I don't understand how to derive the confidence interval about the probability of a "rectangle" from a confidence ellipsoid (even with integration) ? – Stéphane Laurent Sep 11 '12 at 21:42
• @StéphaneLaurent I am not saying that. I am just saying that for multivariate normal distributions it is easy to calulate ingegrals for elliptical regions but difficult for rectangular ones. – Michael Chernick Sep 11 '12 at 21:58