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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).
 Spring '10
 DURRETT
 The Land

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