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.


$ 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 $.

  • $\begingroup$ 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

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.


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.


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.