5
$\begingroup$

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 ;)

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – whuber Sep 11 '12 at 20:45
  • 1
    $\begingroup$ @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. $\endgroup$ – Michael Chernick Sep 11 '12 at 21:37
  • $\begingroup$ 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. $\endgroup$ – whuber Sep 11 '12 at 21:46
  • $\begingroup$ @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. $\endgroup$ – Stéphane Laurent Sep 11 '12 at 21:49
  • 1
    $\begingroup$ @Macro In the univariate case I think this method yields very larger confidence intervals than those based on tolerance intervals $\endgroup$ – Stéphane Laurent Sep 12 '12 at 5:16
-1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I don't understand how to derive the confidence interval about the probability of a "rectangle" from a confidence ellipsoid (even with integration) ? $\endgroup$ – Stéphane Laurent Sep 11 '12 at 21:42
  • 1
    $\begingroup$ @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. $\endgroup$ – Michael Chernick Sep 11 '12 at 21:58

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.