Re: Expected value

PARASHAR wrote:

A string of unit length is cut into n random strings.

(1)What is the expected value of the shortest string



I assume the  n-1  cut-points are i.i.d. uniform.  I do it this way:

Let M_n be the length of the shortest amongst n pieces. Let [F_n](u) = P{M_n > u}. Then EM_n = integral(0..1/n, F_n). Note that [F_n](u) = 0 for u >= 1/n.

Let X be the last cut-point, i.e., the largest of the n-1 uniform r.v.'s. Then X has Beta(n-1,1) distribution and M_n = min(X M_(n-1), 1-X), where M_(n-1) and X are independent.

[F_n](u) = EP(M_n > u | X) = EP(min( X M_(n-1), 1-X) > u | X).

If X >= 1-u, P(min( X M_(n-1), 1-X) > u | X) = 0.
If X < 1-u, P(min( X M_(n-1), 1-X) > u | X) = P(X M_(n-1) > u | X) = [F_(n-1)](u/X).

Thus, [F_n](u) = (n-1) integral(x=(n-1) u...1-u, [F_(n-1)](u/x) x^(n-2)) for 0 <= u <= 1/n. F1(u) = 1 for
0 <= u <= 1.

By inspection then induction, [F_n](u) = (1-n u)^(n-1) for 0 <= u <= 1/n. Thus, EM_n = 1/n^2.

--
Stephen J. Herschkorn                        sjherschko@xxxxxxxxxxxx
Math Tutor on the Internet and in Central New Jersey and Manhattan

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