From: israel@math.ubc.ca (Robert Israel) Subject: Re: Expected index Date: 23 Nov 2000 20:54:32 GMT Newsgroups: sci.math Summary: [missing] In article <0r8l0ozpdaye@forum.mathforum.com>, Dan berger wrote: >Let X1,X2,... be independent, identically distributed random variables >which are uniformly distributed on [0,1]. What is the >expected index of the first partial sum >which is greater than 1? More generally, let X_i have a distribution given by a density f(t) on [0,infinity). Let y(t) be the expected index of the first partial sum > t. Then by a "first step analysis", y(t) = 1 + int_0^t f(x) y(t-x) dx = 1 + (f*y)(t), where * denotes convolution. The Laplace transform of y(t) is Y(s) where Y(s) = 1/s + F(s) Y(s) and F(s) is the Laplace transform of f(t), so Y(s) = 1/(s (1 - F(s))). In this case, F(s) = (1 - exp(-s))/s so Y(s) = 1/(exp(-s) + s - 1) = sum_{n=0}^infinity (-1)^(n) exp(-ns)/(s-1)^(n+1), and y(t) = sum_{n=0}^infinity (n-t)^n Heaviside(t-n) exp(t-n)/n! In particular, for 0 <= t <= 1 we have only the n=0 term: y(t) = exp(t). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2