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ofer zeitouni
ofer zeitouni
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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 the ellipsoid$R^d$ (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.

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 the ellipsoid. 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.

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.

Source Link
ofer zeitouni
ofer zeitouni
  • 7.6k
  • 1
  • 22
  • 38

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 the ellipsoid. 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.