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Re: Obtaining densest x% of a Bivariate Distribution

Thank you for the reply.

I meant BinormalDistribution instead of BinomialDistribution. In any event,
after further investigation it looks like what I want is not quickly
possible in Mathematica. It appears the only way to do it is to carry out
an evaluation of the PDF across a fine grid and then identify the densest
portions of evaluated grid. It works, but computationally it is too slow
for my needs (this process needs to be repeated many times across many


On Fri, Jul 27, 2012 at 6:13 PM, Tomas Garza <tgarza10 at> wrote:

>  Well, to start with, BinomialDistribution is not a bivariate
> distribution. Second, a polygon is always a connected region (if that's
> what you mean by continuous region). Now, if you mean a univariate
> distribution, such as the binomial, then I would guess you are looking for
> the shortest interval in the support of the distribution which encloses x%
> of the probability. You won't find an interval that satisfies this
> condition for an arbitrary x. Take, for instance, BinomialDistribution[50,
> 0.5], and tabulate it for a number of integral intervals. There is no
> interval which encloses precisely 95% of the probability.
> Clear[f]; f[k_]:=CDF[BinomialDistribution[50, 0.5], k];
> Then take a close look at the following table:
> Table[{{j, k}, f[k]-f[j]},{j,10,22},{k,30,41}]
> -Tomas
> > From: scottfr at
> > Subject: Obtaining densest x% of a Bivariate Distribution
> > To: mathgroup at
> > Date: Thu, 26 Jul 2012 03:36:25 -0400
> >
> > Hi,
> >
> > If I have a bivariate distribution such as,
> >
> > BinomialDistribution[50, .5]
> >
> > How do I obtain the polygon that encloses that densest x% of the
> > distribution. For example, what Mathematica functions would I use to
> > get the most compact polygon (which may or may not be a continuous
> > region and may or may not contain holes) that encloses 95% of the mass
> > of a bivariate normal distribution or a bivariate mixture function.
> >
> > Thank you,
> > Scott
> >

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