Stochastic

9 0 79 21 exponential distribution where on the last

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Unformatted text preview: 1 ET = t fT (t) dt = t · e t dt 0 Z1 t1 = te + e t dt = 1/ 0 (2.3) 0 Integrating by parts with f (t) = t2 and g 0 (t) = e t , we see that Z Z1 ET 2 = t2 fT (t) dt = t2 · e t dt 0 Z1 2 t1 = te + 2te t dt = 2/ 2 0 (2.4) 0 by the formula for ET . So the variance var (T ) = ET 2 (ET )2 = 1/ 2 (2.5) While calculus is required to know the exact values of the mean and variance, it is easy to see how they depend on . Let T = exponential( ), i.e., have an 77 78 CHAPTER 2. POISSON PROCESSES exponential distribution with rate , and let S = exponential(1). To see that S/ has the same distribution as T , we use (2.1) to conclude P (S/ t) = P (S t) = 1 e t = P (T t) Recalling that if c is any number then E (cX ) = cEX and var (cX ) = c2 var (X ), we see that ET = ES/ var (T ) = var (S )/ 2 Lack of memory property. It is traditional to formulate this property in terms of waiting for an unreliable bus driver. In words, “if we’ve been waiting for t units of time then the probability we must wait s m...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.

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