f is a continuous real-valued function on the closed interval [0, 1] such that f(1) = 0. A point (a_{1}, a_{2}, ... , a_{n}) is chosen at random from the n-dimensional region 0 < x_{1} < x_{2} < ... < x_{n} < 1. Define a_{0} = 0, a_{n+1} = 1. Show that the expected value of ∑_{0}^{n} (a_{i+1} - a_{i}) f(a_{i+1}) is ∫_{0}^{1} f(x) p(x) dx, where p(x) is a polynomial of degree n which maps the interval [0, 1] into itself (and is independent of f).

**Solution**

The expected value is simply ∫ S dv / ∫ dv, where the integral is taken over the allowed region and S is the sum given.

∫ dv = 1/n! [First integrate 1 from x_{1} = 0 to x_{2}, giving x_{2}/1!. Then integrate that from x_{2} = 0 to x_{3}, giving x_{3}^{2}/2! and so on.]

We now have to attack the main integral. We can take the sum outside the integral, giving a sum of n integrals. Now there at most two variables in each integral. The trick is to use a different order of integration for each integral, so that one integrates last over the variables in the integrand. Starting at the x_{1} end gives us a factor x_{}^{i}/i! and starting at the x_{n} end gives us a factor (1 - x)^{j}/j!. We are then left with the a single final integration from 0 to 1 of a term like f(x) x^{n-i}(1 - x)^{i}/( (n - i)! i! ).

At this point we switch the order of summation and integration again to get:

∫_{0}^{1} x^{n}/n! + x^{n-1}(1 - x)/( n - 1! 1! ) + x^{n-2}(1 - x)^{2}/( n - 2! 2! ) + ... + x (1 - x)^{n-1}/( 1! n - 1! ) dx.

After dividing by the other integral (equivalent to multiplying by n!) we get an integrand which is simply (x + 1 - x)^{n} - (1 - x)^{n} = 1 - (1 - x)^{n}. So we have the required result with p(x) = 1 - (1 - x)^{n}.

© John Scholes

jscholes@kalva.demon.co.uk

1 Jan 2001