Stochastic

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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).

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