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I am generating random vectors $X_1, \dots, X_N$ from a $d$-dimensional multivariate normal $\text N(\mu, \Sigma)$. I would like to know what is the probability that a given point $y \in R^d$ falls within the convex hull of the sample (N > d).

I can't find any result concerning this problem, apart from this answer which covers only a specific point in $R^d$ (the mean). Is anybody aware of any work on this topic?

My final aim is finding the point $y$ at which the $P(y \in \text{ConHull}(X_1, \dots, X_N))$ is maximal. Thanks for any suggestion.

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$ y=\mu $ will maximize the probability of $ y $ being in the convex hull of the sample, since the level sets of the normal distribution are ellipsoids centered at $\mu $.

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  • $\begingroup$ thanks for your answer. I agree that the probability of being in the convex hull is maximal at the $\mu$, but I have no idea about how to prove it. Do think that your statement regarding the level sets can be the base for a formal argument? $\endgroup$ – Jugurtha Mar 9 '14 at 16:44
  • $\begingroup$ A good lemma could be that if a sample leads to $\mu$ not being in the convex hull, translate the sample (maybe add $\mu-c$ where $c$ is the centroid of the sample) and argue that the joint pdf $f_{X_1,\ldots,X_n}$ thereby increases. $\endgroup$ – Bjørn Kjos-Hanssen Mar 9 '14 at 17:04
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This is a partial answer, too long for a comment.

Asymptotically, the convex hull converges (after rescaling) to an ellipsoid and thus the inclusion probability tends to $1$ for any point in $R^d$ (as long as $\Sigma$ is non degenerate). So I assume you do not ask about asymptotics as $N\to \infty$.

Also, by performing a linear transformation you can always put yourself in the situation where $\Sigma=I$, so I will assume in what follows that this is the case.

A general answer for d=2 is given by Jewell and Romano (J. Appl. Prob 19 (1982) pp. 546-561); They show that the probability in question is equal to the coverage problem of the unit circle by random arcs of length $\pi$ whose midpoints are taken from a distribution $G$ that can be computed from your initial data: the midpoint is distributed according to the marginal of $\tan^{-1}(y-y_0)/(x-x_0)$ where $(x_0,y_0)$ is the point that you are trying to cover. In the case of $\Sigma=I$ and $(x_0,y_0)=0$, this gives the uniform distribution which is optimal for the arc covering problem.

I don't know about exact expressions for higher dimension, maybe you can find relevant stuff in http://arxiv.org/pdf/0912.0631.pdf.

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Here are some exact answers for the one-dimensional case $(d=1)$: $$N=2\negthinspace:\ \frac{1}{2}(1-a^2)$$

$$N=3\negthinspace:\ \frac{3}{4}(1-a^2)$$

$$N=4\negthinspace:\ \frac{1}{8}(1-a^2)(7+a^2)$$

$$N=5\negthinspace:\ \frac{5}{16}(1-a^2)(3+a^2)$$

$$N=6\negthinspace:\ \frac{1}{32}(1-a^2)(31+16a^2+a^4)$$

where $$a=\text{erf}\left(\frac{x-\mu}{\sqrt{2}\sigma}\right)$$

I got these using Mathematica, with Expectation[ Boole[Min[a, b] < x < Max[a, b]], {a [Distributed] NormalDistribution[], b [Distributed] NormalDistribution[]}] // FullSimplify and obvious variants.

Perhaps someone else will see a pattern in the results or extend them to higher dimensions.

Update: Exact formulas for higher dimensions do not look promising.

Consider the toy question: what is the probability that $(1/2, 3)$ lies in the convex hull of $(0,1)$, $(1,2)$, and $(a,b)$, where $a$ and $b$ are both normally distributed and independent? The answer is enter image description here

which Mathematica does not simplify further. The answer to the original question with $N=3, d=2$ would require four more integrals beyond that.

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