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!

Suppose the length of the wire is L, and let X[1], ... X[n] be the location of the i'th bird, where X[i]~U[0, L].  Then the total length being painted is X(n)-X(1), where X(i) denote the ordered statistic of X[i].  That is, X(1) is the shortest position from the left, X(2) is the 2nd, ..., X(n) is the furthest position.  Then the expected value of the total length is given by E[X(n)-X(1)] = E[X(n)] - E[X(n)].

It's easy to find the distribution for X(n) and X(1)

P(X(n)<x) = P(X[1]<x, ..., X[n]<x) = product(P(X[i]<x)) = (x/L)^n.

P(X(1)<x) = 1-P(X(1)>x) = 1-[(L-x)/L]^n

So by taking the derivative of the cdf w.r.t x,
pdf for X(n) = (n/L)(x/L)^(n-1) = n*x^(n-1)/L^n
pdf for X(1) = (n/L)[(L-x)/L]^(n-1) = n*(L-x)^(n-1)/L^n

So E[X(n)] = ¡ìx*f(X(n), x)dx, x=0..L
= L*n/(n+1)
and E[X(1)] =¡ìx*f(X(1), x)dx, x=0..L
= L/(n+1)

Therefore, the expected length = L*n/(n+1)-L/(n+1) = L*(n-1)/(n+1)

Posted by Bon on 2004-07-12 17:52:56