Say I pick $n$ i.i.d. random standard normal points in $\mathbb{R}^d$. Roughly, as long as $n$ is much smaller than exponential in $d$, with high probability all points will be on the convex hull. This is because with high probability they will all be near the radius $\sqrt{d}$ sphere and all almost orthogonal, and thus each point is the furthest in its own direction from the origin. Let $p(n,d)$ be the failure probability that at least one point is in the interior of the convex hull.

Question: What's the best upper bound on $p(n,d)$ as a function of $n$ and $d$? I care most about the regime $d \gg 1$, $n \in O(\operatorname{poly}(d))$.

  • $\begingroup$ I think that by "all points will be on the convex hull" you mean, that no $X_i \in co(X_1,\ldots,X_n)^0$, when $X_1,\ldots,X_n$ are the generated random $X_i$, It seems that the condition $X_i \not\in ext(co(X_1,\ldots,X_n))$ gives the same probabilities. $\endgroup$ Jul 9 '20 at 11:01
  • $\begingroup$ Yes, I was glossing over the boundary because it is measure 0. $\endgroup$ Jul 9 '20 at 11:31

It's not too bad to see that the probability is at most $2n^2 e^{-d/2e}$. Let $x_1,\ldots,x_n$ be the points. We will use a union bound, so it is sufficient to examine the probability that $x_1$ is in the convex hull of $x_2,\ldots,x_n$. This happens if and only if there are $\lambda_j \in [0,1]$ with $\sum \lambda_j = 1$ and $$x_1 = \sum_{j = 2}^n \lambda_j x_j\,.$$

Take an inner product with $x_1$ to see that this implies $$\| x_1 \|_2^2 = \sum_{j = 2}^n \lambda_j \langle x_1, x_j \rangle.$$

Thus $$P(x_1 \in \mathrm{conv}(x_2,\ldots,x_n)) \leq P( \|x_1\|^2 \leq \max_{j \geq 2} |\langle x_1, x_j \rangle|).$$

If we divide by $\|x_1\|$, the RHS probability bound becomes

$$P\left(\|x_1\| \le \max_{j \ge 2} \left|\left<x_1/\|x_1\|, x_j \right>\right|\right).$$

$\|x_1\|^2 \sim \chi^2_d$ and $\left<x_1/\|x_1\|, x_j\right> \sim N(0,1)$, so from $\chi^2_d$ and $N(0,1)$ tail bounds we have

\begin{align*} P(\|x_1\| \le t\sqrt{d}) &\le \left(t e^{(1-t^2)/2}\right)^d \\ P\left(\left<x_1/\|x_1\|, x_j\right> \ge t\sqrt{d}\right) &\le \frac{\left(e^{-t^2/2}\right)^d}{t\sqrt{2\pi d}} \end{align*}

for any $t \in (0,1)$. Matching the base of the exponents gives \begin{align*} t e^{(1-t^2)/2} &= e^{-t^2/2} \\ t &= e^{-1/2} \approx 0.606531 \end{align*}

whence union bounding shows \begin{align*}P(x_1 \in \mathrm{conv}(x_2,\ldots,x_n)) &\le (n-1) \left(1 + \frac{1}{\sqrt{2\pi d/e}}\right) e^{-d/2e} \\ &< 2ne^{-d/2e} \end{align*} and so $$P(\exists~j \text{ s.t. }x_j \in \mathrm{conv}(x_1,\ldots,x_{j-1},x_{j+1},\ldots,x_n)) < 2n^2 e^{-d/2e}.$$

I do not know if this is optimal, but it's worth noting that it's basically the strategy you suggested. When $n$ is exponentially large in $d$ the probability does not tend to $0$ provided the exponent is big enough, which is where this bound breaks.

  • 1
    $\begingroup$ Thanks, yes this the argument I had in mind, but nicely expressed. I’m interesting in knowing $c$; if you don’t feel like writing that out okay if I edit to fill in later? $\endgroup$ Jul 9 '20 at 18:46
  • 1
    $\begingroup$ Yeah absolutely feel free to edit it in. $\endgroup$
    – Marcus M
    Jul 9 '20 at 21:26
  • $\begingroup$ I think the exponent can be tightened from $e^{-d/2e}$ to $2^{-d/2}$, but don't have time to complete the proof. $\endgroup$ Jul 14 '20 at 20:23

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