Expected Area of Random Polygon In a Circle 

The area of a unit circle is π, whereas the area of a regular nsided polygon inscribed in the circle is (n/2)sin(2π/n), which of course approaches π as n goes to infinity. Now, suppose that instead of a regular ngon we consider a convex polygon whose vertices are chosen randomly on the perimeter of the circle (assuming a uniform distribution for the probability density over the perimeter). In general the area of a random ngon will be less than the area of the regular ngon, but both of them will approach π as n goes to infinity. Ferry posed the problem of proving that the expected value of the deviation from π for the area of a random ngon is, asymptotically, 6 times the deviation for a regular ngon in the limit as n goes to infinity. 

For sufficiently large values of n the maximum distance between consecutive vertices of the random ngon will almost certainly be small. Of course, for any fixed n, it's possible that the maximum angular gap between neighboring vertices could be anything less than 2π, but the probability of the maximum gap being greater than a certain angular arc, α, can be made arbitrarily small by increasing n. On this basis, we see that the asymptotic behavior of the expected area of a random polygon in the limit of increasing n can be deduced by considering only polygons whose maximum sectors subtend arbitrarily small angles. (To make this rigorous, just consider the volume of the state space with a range of maximum angles.) 

Now, in a unit circle, the area of an arc sector of angle α is (α/2π)π = α/2, whereas the area of the corresponding sector of a polygon inscribed in the circle is sin(α)/2, so the discrepancy is (α−sin(α))/2. Recalling that the power series for sin(x) is x − x^{3}/3! + x^{5}/5! − ... this tells us that the discrepancy is (α^{3}/3! − α^{5}/5! + ...)/2. In the "large n, small angle" limit, the discrepancy for a sector of arc α goes to α^{3}/12. In other words, the discrepancy is proportional to the cube of the subtended angle, and this applies to the sectors of regular and random polygons alike. Thus the ratio of the discrepancies for regular and random ngons approaches the ratio of the sum of cubes of the sector angles in the respective figures. 

Obviously we can focus on partitions of the interval from 0 to 2π, but we'll get the same answer (and it's more convenient) to rescale and consider the partitions of the unit interval from 0 to 1. The sum of the cubes of the regular ngon partition of the unit interval is simply n(1/n)^{3}, which is 1/n^{2}. Our question, then, reduces to finding the expected sum of cubes of the segments of a random partition of the unit interval into n parts. We will call this E_{n}. 

One straightforward way of determining the form of E_{n} would be to simply evaluate the expectation for specific values of n such as 



After evaluating a few of these we can easily discern the pattern, but "discerning the pattern" isn't quite the same as rigorously establishing the formula. To be a bit more rigorous, we could proceed inductively. Obviously for n=1 we have E_{1} = 1. In general, E_{n+1} can be computed from E_{n} by the formula 



This is just the expectation for the (n+1)th part to be of size x, evaluated for all x from 0 to 1. Each of these x values is "weighted" by (1−x)^{(n−1)} because that's the probability of all the other n−1 breakpoints being in the 1−x region. On that condition, the expected sum of the cubes is x^{3} plus the expected sum of cubes for a partition of the region 1x into n parts, which is just E_{n} scaled by the factor (1−x)^{3}. 

If we now define the variable u = 1−x we have du = −dx and the above formula becomes 



Evaluating the elementary integrals we get 



Multiplying through by the right hand denominator gives 



Consequently, each quantity of the form k(k+1)(k+2)E_{k} is 6 greater then the preceding quantity. Therefore, to go up n steps we simply add 6n, so we have 



Thus, decrementing the indices by 1, we've proven that 



This is the expected sum of cubes of the segments comprising a random partition of the unit interval into n parts, assuming the n−1 interior dividing points are randomly selected from a uniform distribution. Recall that E_{n} for a regular polygons is simply 1/n^{2}, so the ratio of the discrepancies for random and regular ngons is 6/(1 + 3/n + 2/n^{2}), which goes to 6 as n increases, completing the proof. 

Incidentally, notice that the solution of this problem makes use of the formula for the expected value of the sum of the cubes of the components of a random partition of the unit interval into n parts. More generally, we can give the expected sum of pth powers of the components of a random partition of the unit interval into n parts. Letting E_{n,p} denote this value, we have 



where the denominator on the right side is the binomial coefficient. (With p=3 this reduces to our expression 6/((n+1)(n+2) for cubes.) This general formula is so simple and useful that it ought to be better known than it is. Obviously the "6" in the answer to Ferry's puzzle arises from the p! with p=3 in this expression. We could also construct other scenarios involving the general expression, such as allowing a unit interval to fall on the floor and break at k random points to give k+1 segments, and then consider the total "volume" given by constructing pdimensional rectangular solids with those edge lengths. 

Another interesting thought suggested by Ferry’s puzzle is to give the exact formula for the expected area of an nsided convex polygon whose vertices are randomly selected points on the unit circle (assuming a uniform distribution on the perimeter). To answer this question we obviously can't make use of the smallangle assumption, so we must use the full sine formulas. We could still use central sectors (recognizing that for sector angles greater than π the area will be negative), but we can also deal with strictly positive areas by taking a center on the perimeter of the circle. 

So, let's take an arbitrary point on the circumference as our "origin" O (without loss of generality), and note that the distance from this point to any other point P on the circle is 2sin(α) where α is the angle between the line OP and the tangent to the circle at O. Now, any convex polygon inscribed in the circle with k+1 vertices can be characterized by a set of k increasing angles α_{1}, α_{2}, .., α_{k} denoting the angles between the tangent at O and the lines P_{1}, P_{2}, .., P_{k} respectively. Also, the area of the polygon is given by the sum of k−1 triangular regions defined by these k rays. Thus for any set of angles the area of the polygon is 



Furthermore, we know that a uniform distribution on the circumference of the circle maps to a uniform distribution of the α angles in the range from 0 to π. Hence we can find the expected area of the polygon by integrating over the region with α_{1} from 0 to π, α_{2} from 0 to α_{1}, and so on. (Here I've reversed the order of the indices, so the angles are strictly decreasing.) Letting E[A_{n}] denote the expected area of a random ngon inscribed in a unit circle, we have 



The integrations are all elementary, and we get the following results for the expected area of a "random ngon" inscribed in a unit circle: 



and so on. Obviously the quantities in parentheses are partial sums of the power series expansions of sin(2π) and 1 − cos(2π). The numerical values of E[A_{n}] for the first several values of n are listed below: 



In general, for odd n we have the explicit "sine" formula for the expected area of a random ngon inscribed in a unit circle: 



where m = (n−1)/2, and for even n we have the "1 − cosine" formula 




where m = (n−2)/2. If we recall that the area of a regular ngon in a unit circle is (n/2)sin(2π/n), we see that the original proposition amounts to the assertion 



Thus, letting sin(x,k) denote the kth partial sum of the power series expansion of sin(x), for odd n we have 


