x is chosen at random from the interval [0, a] (with the uniform distribution). y is chosen similarly from [0, b], and z from [0, c]. The three numbers are chosen independently, and a ≥ b ≥ c. Find the expected value of min(x, y, z).
Solution
Answer: c/2 - c2(1/(6a) + 1/(6b) ) + c3/(12ab).
Let m(k) = prob(x ≥ k and y ≥ k and z ≥ k). We have that prob(x ≥ k) = 1 for k < 0, 1 - k/a for 0 ≤ k ≤ a, 0 for k > a. Similarly for y and z. So m(k) is 1 for k < 0, and (1 - k/a)(1 - k/b)(1 - k/c) for 0 ≤ k ≤ c, and 0 for k ≥ a.
Now the probability that the minimum lies between k and k + δk is m(k) - m(k + δk), so the probability density for min(x, y, z) to equal k is -m'(k) which is: (1/a + 1/b + 1/c) -2k(1/(ab) + 1/(bc) + 1/(ca) ) + 3k2/(abc) for 0 ≤ k ≤ c. Thus the expected value is the integral from 0 to c of k(1/a + 1/b + 1/c) -2k2(1/(ab) + 1/(bc) + 1/(ca) ) + 3k3/(abc) which is c2(1/a + 1/b + 1/c)/2 -2c3(1/(ab) + 1/(bc) + 1/(ca) )/3 + 3c4/(4abc).
© John Scholes
jscholes@kalva.demon.co.uk
25 Aug 2001