Probability of a random cyclic quadrilateral enclosing a fixed point in its circle

Question

I finally found a single integral solving the natural generalisation of the problem discussed here:

For $n\ge1$ pick $n+2$ points uniformly at random on the unit circle. What is the probability $P_n(x)$ of the convex hull of the points containing $X=(x,0)$ where $0\le x\le1$?

If the convex hull does not contain $X$ exactly one of the following cases must hold:

• All $n+2$ points are on the same side of the $x$-axis ($OX$). This happens with probability $2^{-(n+1)}$.
• There is a point $V$ with Weierstrass parameter $v>0$ and another point $T$ with parameter $t<k/v$ where $k=\frac{x-1}{x+1}$, and none of the remaining points' parameters are between $t$ and $v$. All points thus lie "left" of the segment $TV$ which lies "left" of $X$. The probability of a point falling in the arc $TV$ is $\frac{\arctan v-\arctan t}\pi$, while $t,v$ are Cauchy-distributed and can be selected from the pool of points in $(n+2)(n+1)$ ways. This case's contribution to the overall probability is thus $$\frac{(n+2)(n+1)}{\pi^2}\int_0^\infty\int_{-\infty}^{k/v}\frac1{1+v^2}\frac1{1+t^2}\left(1-\frac{\arctan v-\arctan t}\pi\right)^n\,dt\,dv$$
• There is a point $V$ with Weierstrass parameter $v>0$ and another point $T$ with parameter $k/v<t<0$ and all of the remaining points' parameters are between $t$ and $v$, i.e. they are "right" of $TV$ which in turn is "right" of $X$. By similar reasoning to the previous case the present case's contribution is $$\frac{(n+2)(n+1)}{\pi^2}\int_0^\infty\int_{k/v}^0\frac1{1+v^2}\frac1{1+t^2}\left(\frac{\arctan v-\arctan t}\pi\right)^n\,dt\,dv$$

Putting everything together we have $$P_n(x)=1-2^{-(n+1)}-\frac{(n+2)(n+1)}{\pi^2}\left(\int_0^\infty\int_{-\infty}^{k/v}\frac1{1+v^2}\frac1{1+t^2}\left(1-\frac{\arctan v-\arctan t}\pi\right)^n\,dt\,dv + \int_0^\infty\int_{k/v}^0\frac1{1+v^2}\frac1{1+t^2}\left(\frac{\arctan v-\arctan t}\pi\right)^n\,dt\,dv\right)$$ This can be turned into a single integral by expanding and solving the $t$ integral. Define the polynomial $$Q(V,K)=\sum_{i=0}^n\sum_{j=0}^i\binom ni\binom ij\frac{(-V)^{i-j}}{\pi^i}\frac{K^{j+1}-(-\pi/2)^{j+1}}{j+1}+\sum_{i=0}^n\binom ni\frac{V^{n-i}}{\pi^n}\frac{(-K)^{i+1}}{i+1}$$ Then $$P_n(x)=1-2^{-(n+1)}-\frac{(n+2)(n+1)}{\pi^2}\int_0^\infty\frac{Q(\arctan v,\arctan k/v)}{1+v^2}\,dv$$

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I already worked out that $$P_1(x)=\frac14-\frac3{2\pi^2}\operatorname{Li}_2(x^2)$$ The new surprising thing, though, is that numerical calculations strongly suggest $$\boxed{P_2(x)=2P_1(x)}$$ i.e. it is exactly twice as likely that four random points on the unit circle will enclose $X$ than will three random points, no matter where $X$ is. This relation does not extend to $P_3(x)$ and beyond.

How can the boxed relation be proved?

Answer 1

Let $S_1$ be the unit circle with its normed Lebesgue measure invariated by rotations, so that $\int_{S^1} 1=1$. Integrals will be always written w.r.t. some normed measure below. The measure is omitted, as it is sometimes the case in measure theory, when $\int f$ is seen as a functional in $f$ w.r.t. to some fixed measure $\mu$, it stays then explicitly for $\int f\; d\mu$.

Let $C(n)=C(n,S^1)$ the space of configurations of $n$ points in $S^1$. The symmetric group $S(n)$ acts on $C(n)$ by permutation of the components. The quotient space $C(n)/S(n)$ obtained by identifying points in the same orbit can be seen as the subspace of $C(n)$ with points $(P_1,P_2,\dots,P_n)$ in cyclic order, further taken modulo cyclic permutations. Consider now the space $C_0(n)$ of configurations of $n$ points in $S^1$ which are in cyclic order. (Call such a configuration cyclic.) It is a subspace of $C(n)$, but when the measure is also part of the game, take it normed to total mass one (probability mass).

A function $f$ on $C(n)$ which is $S(n)$-invariant can be seen then, and its expectation can be measured "as it is" in any of the spaces $C(n)$, $C(n)/S(n)$, $C_0(n)$.

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We will be concerned only with $C(3)$, and $C(4)$, the spaces of configurations of three, respectively four points in $S^1$, their quotients w.r.t. the permutation groups $S(3)$, $S(4)$, and the cousins $C_0(3)$, $C_0(4)$ with a restriction to cyclic configurations.

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The wanted relations between the two probabilities can be traced back to the following geometric property:

Lemma: Let $X$ be a point in the unit disk. Let $A,B,C,D$ be four points in this cyclic order on $S^1$, the boundary of the unit disk.

• If $X$ is not in the convex hull $(ABCD)$ of the points $A,B,C,D$, then $X$ is in exactly two among the four solid, convex hull triangles $$ (ABC)\ ,\ (BCD)\ ,\ (CDA)\ ,\ (DAB)\ . $$ (Up to conventions regarding boundaries, and the definition of "in" that includes or excludes them. This amounts to ignoring a set of configurations $(A,B,C,D)\in C_0(4)$ with zero probability.)
• If $X$ is not in the convex hull $(ABCD)$ of the points $A,B,C,D$, then it is also in none of the triangles. In particular, $$ 2\cdot 1_{X\in(ABCD)} = 1_{X\in(ABC)} + 1_{X\in(BCD)} + 1_{X\in(CDA)} + 1_{X\in(DAB)} \ . $$ Above, $1_E$ is the characteristic function of an event $E$, and the event $X\in (ABCD)$ for instance is the set of configurations in $C_0(4)$ containing $X$ in their convex hull.

This is transparent from the picture:

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We can now conclude using the geometric symmetry: $$ \begin{aligned} 2P_2(X) &=2\int_{C(4)}1_{X\in(ABCD)}\\ &=2\int_{C_0(4)}1_{X\in(ABCD)}\\ &=\int_{C_0(4)}( 1_{X\in(ABC)} + 1_{X\in(BCD)} + 1_{X\in(CDA)} + 1_{X\in(DAB)} )\\ &=4\int_{C_0(4)}1_{X\in(ABC)}\\ &=4\int_{C_0(3)}1_{X\in(ABC)}\\ &=4\int_{C(3)}1_{X\in(ABC)}\\ &=4P_1(X)\ . \end{aligned} $$ When passing from $C_0(4)$ to $C_0(3)$ we use the transport formula which connects the measure $\mu$ on $C_0(4)$ with the transported measure $\mu\circ\pi^{-1}$ on $C_0(3)$, where $\pi$ is the natural projection $(A,B,C,D)\to (A,B,C)$, and for an event $E$ in $C_0(3)$ its probability is the same one as for the preimage $\pi^{-1}(E)$ when taking the probability from $C_0(4)$.

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When five points are taken, the same argument fails. For instance, if we consider triangles of the shape $(A_kA_{k+1}A_{k+2})$ then two cyclicly consecutive such triangles cover only the marked regions. For the other regions we need "an other kind of triangles", in any case, the above scheme breaks.