# Recall that x has a poisson distribution with mean or

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Unformatted text preview: mplies that U is exponential with parameter µ=( 1 + ··· + n) i so using the result for two random variables P (Ti = min(T1 , . . . , Tn )) = P (S < U ) = i i +µ i = 1 + ··· + n proves the desired result. Let I be the (random) index of the Ti that is smallest. In symbols, i P (I = i) = 1 + ··· + n You might think that the Ti ’s with larger rates might be more likely to win early. However, I and V = min{T1 , . . . Tn } are independent. (2.11) Proof. Let fi,V (t) be the density function for V on the set I = i. In order for i to be ﬁrst at time t, Ti = t and the other Tj > t so fi,V (t) = ie it · Y e jt j 6=i i ·( + ··· + n = P (I = i) · fV (t) = 1 since V has an exponential( 1 + ··· + 1 + ··· + n) n )e ( 1 +·+ n )t distribution. Our ﬁnal fact in this section concerns sums of exponentials. Theorem 2.1. Let ⌧1 , ⌧2 , . . . be independent exponential( ). The sum Tn = ⌧1 + · · · + ⌧n has a gamma(n, ) distribution. That is, the density function of Tn is g...
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## This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).

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