miniliv4

2022-09-30

Expected number of uniformly random points in unit square is in convex position.

If n points are uniformly generated in a unit square, following famous Erdős–Szekeres theorem. The probability of them be in convex position is $({\textstyle (}\genfrac{}{}{0ex}{}{2n-2}{n-1}{\textstyle )}/n!)}^{2$

My interest is to find the expected number of points be in convex position, and I am enumerating i points in the convex position and add all the possible is.

Here is what I did, $P(i)={({\textstyle (}\genfrac{}{}{0ex}{}{2i-2}{i-1}{\textstyle )}/i!)}^{2}$, the probability of having i points in convex position.

And the expected number of points be in convex position can be achieved via $\sum _{i=1}^{n}{\textstyle (}\genfrac{}{}{0ex}{}{n}{i}{\textstyle )}\ast P(i)$

And I was using WolframAlpha to get an idea of this number, for some reason the expected number is bigger than n, which is impossible

Can someone help me where did I do wrong?

If n points are uniformly generated in a unit square, following famous Erdős–Szekeres theorem. The probability of them be in convex position is $({\textstyle (}\genfrac{}{}{0ex}{}{2n-2}{n-1}{\textstyle )}/n!)}^{2$

My interest is to find the expected number of points be in convex position, and I am enumerating i points in the convex position and add all the possible is.

Here is what I did, $P(i)={({\textstyle (}\genfrac{}{}{0ex}{}{2i-2}{i-1}{\textstyle )}/i!)}^{2}$, the probability of having i points in convex position.

And the expected number of points be in convex position can be achieved via $\sum _{i=1}^{n}{\textstyle (}\genfrac{}{}{0ex}{}{n}{i}{\textstyle )}\ast P(i)$

And I was using WolframAlpha to get an idea of this number, for some reason the expected number is bigger than n, which is impossible

Can someone help me where did I do wrong?

Samantha Braun

Beginner2022-10-01Added 9 answers

Step 1

You've computed the expected number of subsets of points in convex position. For example, when $n\le 3$ all ${2}^{n}-1$ nonempty subsets are in convex position, so the sum equals ${2}^{n}-1$

Let X be the maximum cardinality of all sets in convex position. By union bound on all sets of size k we have:

$P(X\ge k)\le {\textstyle (}\genfrac{}{}{0ex}{}{n}{k}{\textstyle )}{\left(\frac{1}{k!}{\textstyle (}\genfrac{}{}{0ex}{}{2k-2}{k-1}{\textstyle )}\right)}^{2}$

Step 2

So we get an upper bound on the expectation of X:

$\begin{array}{rl}E(X)& =\sum _{k=1}^{n}P(X\ge k)\\ & \le \sum _{k=1}^{n}min(1,{\textstyle (}\genfrac{}{}{0ex}{}{n}{k}{\textstyle )}{\left(\frac{1}{k!}{\textstyle (}\genfrac{}{}{0ex}{}{2k-2}{k-1}{\textstyle )}\right)}^{2})\end{array}$

Getting a lower bound is more complicated, I think this upper bound should be reasonably tight. Why? Most of the sum on the right hand side comes from the terms which equal 1, the terms less than 1 decay rapidly. The expected number of sets of size k in convex position equals $(}\genfrac{}{}{0ex}{}{n}{k}{\textstyle )}{\left(\frac{1}{k!}{\textstyle (}\genfrac{}{}{0ex}{}{2k-2}{k-1}{\textstyle )}\right)}^{2$, so when this quantity is much bigger than 1, it's a reasonable to guess that X will be bigger than k with high probability.

You've computed the expected number of subsets of points in convex position. For example, when $n\le 3$ all ${2}^{n}-1$ nonempty subsets are in convex position, so the sum equals ${2}^{n}-1$

Let X be the maximum cardinality of all sets in convex position. By union bound on all sets of size k we have:

$P(X\ge k)\le {\textstyle (}\genfrac{}{}{0ex}{}{n}{k}{\textstyle )}{\left(\frac{1}{k!}{\textstyle (}\genfrac{}{}{0ex}{}{2k-2}{k-1}{\textstyle )}\right)}^{2}$

Step 2

So we get an upper bound on the expectation of X:

$\begin{array}{rl}E(X)& =\sum _{k=1}^{n}P(X\ge k)\\ & \le \sum _{k=1}^{n}min(1,{\textstyle (}\genfrac{}{}{0ex}{}{n}{k}{\textstyle )}{\left(\frac{1}{k!}{\textstyle (}\genfrac{}{}{0ex}{}{2k-2}{k-1}{\textstyle )}\right)}^{2})\end{array}$

Getting a lower bound is more complicated, I think this upper bound should be reasonably tight. Why? Most of the sum on the right hand side comes from the terms which equal 1, the terms less than 1 decay rapidly. The expected number of sets of size k in convex position equals $(}\genfrac{}{}{0ex}{}{n}{k}{\textstyle )}{\left(\frac{1}{k!}{\textstyle (}\genfrac{}{}{0ex}{}{2k-2}{k-1}{\textstyle )}\right)}^{2$, so when this quantity is much bigger than 1, it's a reasonable to guess that X will be bigger than k with high probability.

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