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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.
 Spring '10
 DURRETT
 The Land

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