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I have a zero mean multivariate normal probability distribution where WLOG each marginal variance is unity and all pairwise correlation coefficient are equal and positive. The number of elements in the random vector is arbitrary. I am interested in the probability distribution of the number of elements which exceed some threshold -- the same threshold for each element. The outcomes for each elements are collectively identically distributed dependent Bernoulli random variables, so by de Finetti's exchangeability theorem this probability distribution has a representation as a mixture of binomial distributions. I'd like to figure out the mixing measure but I haven't the foggiest idea how to start.

(edited to specify zero mean and equality of correlations)

edited Mar 9 2012 at 22:30

asked Mar 9 2012 at 21:54

Cyan
45●4

1

In order that de Finetti's theorem be applicable, "collectively identically distributed" will have to mean exchangeable. That's OK, since the hypotheses given here do imply they're exchangeable, but it doesn't hurt to say things like this clearly. – Michael Hardy Mar 9 2012 at 23:26

mike · Accepted Answer · 2012-04-04 18:40:43Z UTC

The mixing distribution is the same as the long term average . Since your gaussians can be repesented as $Z + X_i$ where $X_i$ are i.i.d., and also independent of $Z$, the long term average for the events I think you are looking at is $\frac 1 n \sum^n 1_{X_i + Z > c}$. By conditioning on $Z$ this is seen to have distribution $\Phi(c - Z)$ and that should be your mixer.

Exchangeable normal distribution mixing measure

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