A telephone wire stretched tight between two poles placed ten meters apart is a favorite resting spot for a flock of crows.

Suppose one morning two crows land on the wire, each at a random spot (the probability is uniformly distributed). With a bucket of paint and a brush you mark the stretch of wire between them. A certain length of wire will have been painted.

On average, what length of wire would you expect to have painted? Assume that each bird is a single point along the line, and so has no width.

Suppose instead that a dozen crows landed on the wire, each at an independent, random location, and you painted the stretch of wire between each bird and its nearest neighbor. On average, what total length of wire would you expect to have painted now?

And if a thousand crows landed?

A computer-generated solution could be found, but bonus points will be awarded for a formal proof!

A standard reference on statistics is Harald Cramer's Mathematical Methods of Statistics, a Princeton "Landmarks in Mathematics" book first published in the mid-40's. Section 28.6, pp. 370-378, covers extreme values and range of statistical samples. (28.6.9), p. 372, gives the expected value of the difference of the vth extremes x and y of n independent samples  drawn from a population uniformly distributed over [a,b] as

E(x-y)=(1-2*v/(n+1))*(b-a).

The most extreme samples (max and min) correspond to v=1 so that for these the expected value is (1-2/(n+1))*(b-a)=[(n-1)/(n+1)]*(b-a).

Posted by Richard on 2004-06-09 00:13:58