Unformatted text preview: NUOUS TIME MARKOV CHAINS Since pt (i, 2) = 1 pt (i, 1) it is su ceient to compute pt (i, 1). Doing the ﬁrst
column of matrix multiplication on the right, we have
p0 (1, 1) =
t
p0 (2, 1)
t pt (1, 1) + pt (2, 1) = = µpt (1, 1) (pt (1, 1) pt (2, 1)) µpt (2, 1) = µ(pt (1, 1) pt (2, 1)) (4.12) Taking the di↵erence of the two equations gives
[pt (1, 1) pt (2, 1)]0 = ( + µ)[pt (1, 1) pt (2, 1)] Since p0 (1, 1) = 1 and p0 (2, 1) = 0 we have
pt (1, 1) pt (2, 1) = e ( +µ)t Using this in (4.12) and integrating
t pt (1, 1) = p0 (1, 1) +
pt (2, 1) = p0 (2, 1) + µ+ e µ+ e µ
+
e
+µ µ+ = (µ+ )s
0
t = (µ+ )s
0 µ
µ+ µ
e
µ+ (µ+ )t (µ+ )t As a check on the constants note that p0 (1, 1) = 1 and p0 (2, 1) = 0. To prepare
for the developments in the next section note that the probability of being in
state 1 converges exponentially fast to the equilibrium value µ/(µ + ).
There are not many examples in which one can write down solutions of the
Kolmogorov’s di↵erential equation. A remarkable exception is:
Example 4.9. Yule process. In this model each particle spli...
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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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