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Unformatted text preview: In continuous time, as in discrete time, the transition probability satisﬁes
Theorem 4.1. Chapman–Kolmogorov equation.
ps (i, k )pt (k, j ) = ps+t (i, j )
k Why is this true? In order for the chain to go from i to j in time s + t, it must
be in some state k at time s, and the Markov property implies that the two
parts of the journey are independent.
Proof. Breaking things down according to the state at time s, we have
P (Xs+t = j |X0 = i) =
P (Xs+t = j, Xs = k |X0 = i)
k Using the deﬁnition of conditional probability and the Markov property, the
P (Xs+t = j |Xs = k, X0 = i)P (Xs = k |X0 = i) =
pt (k, j )ps (i, k )
k k (4.1) shows that if we know the transition probability for t < t0 for any
t0 > 0, we know it for all t. This observation and a large leap of faith (which we
will justify later) suggests that the transition probabilities pt can be determined
from their derivatives at 0:
q (i, j ) = lim h!0 ph (i, j )
h for j 6= i (4.1) If this limit exists (and it will in all the cases we consider) we will call q (i, j )
the jump rate from i to j . To explain this name we will compute the:
Jump rates for Example 4.1. The probability of at least t...
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