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Unformatted text preview: distribution with rate equal to the sum of the rates 1 + · · · n .
In the last paragraph we have computed the duration of a race between exponentially distributed random variables. We will now consider: “Who ﬁnishes
ﬁrst?” Going back to the case of two random variables, we break things down
according to the value of S and then using independence with our formulas
(2.1) and (2.2) for the distribution and density functions, to conclude
Z1
P (S < T ) =
fS (s)P (T > s) ds
Z0 1
=
e s e µs ds
0
Z1
( + µ)e ( +µ)s ds =
(2.9)
=
+µ 0
+µ 79 2.1. EXPONENTIAL DISTRIBUTION where on the last line we have used the fact that ( + µ)e ( +µ)s is a density
function and hence must integrate to 1.
From the calculation for two random variables, you should be able to guess
that if T1 , . . . , Tn are independent exponentials, then
P (Ti = min(T1 , . . . , Tn )) = i
1 + ··· + (2.10)
n That is, the probability of i ﬁnishing ﬁrst is proportional to its rate.
Proof. Let S = Ti and U be the minimum of Tj , j 6= i. (2.8) i...
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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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