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Unformatted text preview: wo jumps by time h
is 1 minus the probability of 0 or 1 jumps
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◆
( h)2
1
e h + he h = 1 (1 + h) 1
h+
+ ...
2!
= ( h)2 /2! + . . . = o(h) That is, when we divide it by h it tends to 0 as h ! 0. Thus, if j 6= i,
ph (i, j )
⇡e
h h u(i, j ) ! u(i, j ) 121 4.1. DEFINITIONS AND EXAMPLES as h ! 0. Comparing the last equation with the deﬁnition of the jump rate in
(4.1) we see that q (i, j ) = u(i, j ). In words we leave i at rate and go to j
with probability u(i, j ).
Example 4.1 is atypical. There we started with the Markov chain and then
computed its rates. In most cases, it is much simpler to describe the system by
writing down its transition rates q (i, j ) for i 6= j , which describe the rates at
which jumps are made from i to j . The simplest possible example is:
Example 4.2. Poisson process. Let X (t) be the number of arrivals up to
time t in a Poisson process with rate . Since arrivals occur at rate in the
Poisson process the number of arrivals, X (t), increases from n to n + 1 at rate
, or in symbols
q (n, n + 1) =
for all n 0
This simplest example is a building block for o...
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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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