# Expected number of intersections within a circle

Consider a circle. We perform random walks on the boundary of this circle. Without loss of generality, consider the circle to be centered at the origin, and our random walk starts $$0^{\circ}$$. You return to the starting point after $$n$$ steps.

We will make $$n$$ total steps. Each step you randomly traverse to some point on the boundary. There is a straight line that connects the points for each step. So for example, if you traverse to $$90^{\circ}$$ on the first step, there will be a straight line going from $$0^{\circ}$$ to $$90^{\circ}$$, and then if you go to $$155^{\circ}$$ afterwards, there will then be another line going from $$90^{\circ}$$ to $$155^{\circ}$$.

What is the expected number of intersections created by these line segments?

I think this question may be somewhat similar to Expected number of intersection points when $n$ random chords are drawn in a circle, but here all the $$n$$ line segments are connected. Which means two consecutive line segments cannot intersect one another. So this reduces the maximum number of intersections from $$\frac{n(n-1)}{2}$$ to $$\frac{(n-1)(n-2)}{2}$$, I believe. The probability of an intersection between these pairs is still $$1/3$$.

So does the solution simply just become $$\frac{1}{6}(n-1)(n-2)$$? I saw some discussions offline that it could be $$\frac{1}{6}n(n-3)$$.

• Is the random walk a uniform random step between 0 and 360 degrees? Mar 3, 2021 at 15:55
• I'm not sure, but let's assume that. Also when I think about it, does it actually matter how it's distributed as long as each position has a non-zero probability? I was thinking that, by symmetry, it does not matter. Mar 3, 2021 at 15:57
• I doubt it. Imagine the step is log-normally distributed. You will most likely have a minimum number of steps before an intersection likely. Because with small steps, you don’t have possibility of an intersection. Also, with log-normal you will have non zero probability of stepping to any point on the circle. Mar 3, 2021 at 16:03
• @Ralff When you say you "doubt" it, you mean you think the distribution DOES make a difference right? Mar 3, 2021 at 16:19
• Yes. I should have been more clear and not have said “doubt”.... Mar 3, 2021 at 16:22

Because you end at the starting point, there are $$\binom{n}{2} - n = \frac{n(n-3)}{2}$$ pairs of non-adjacent chords, and each pair of these chords intersects with probability $$\frac13$$ (here you use that the points are sampled uniformly, as in Expected number of intersection points when $n$ random chords are drawn in a circle). By linearity of expectation, the total expected number of intersections is $$\frac{n(n-3)}{6}$$ (coinciding intersections, i.e., three or more chords passing through the same point, happen with probability $$0$$).
Your formula $$\frac{(n-1)(n-2)}{6}$$ would be correct if the walk would not necessarily end at the starting point. Note that for $$n=3$$ your formula gives $$\frac13$$, whereas the answer should be $$0$$ because the walk ends at the starting point.
• What does the $\binom{n}{2}$ represent physically in this case? Mar 3, 2021 at 16:38
• There are $n$ chords, so there are $\binom{n}{2}$ pairs of chords. Mar 3, 2021 at 16:42
• Sorry, that was a typo. I meant the $n$. It seems you're subtracting the number of pairs of adjacent chords? Mar 3, 2021 at 16:46
• Indeed -- there are $n$ such pairs Mar 3, 2021 at 19:33