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 <dannys@hotmail.com> 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