The General Case

David Freedman
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David Freedman: University of California, Department of Statistics

Chapter 9 in Markov Chains, 1983, pp 297-328 from Springer

Abstract: Abstract My object in this section is to present Blackwell’s (1958) example of a standard stochastic semigroup, all of whose states are instantaneous. For other examples of this phenomenon, see Sections 3.3 of ACM and Section 2.12 of B & D. To begin with, consider the matrix $$ Q = \left( {\begin{array}{*{20}{c}} { - \lambda } & \lambda \\ \mu & { - \mu } \\ \end{array} } \right) $$ on {0, 1}, with λ and μ nonnegative, λ + μ positive. There is exactly one standard stochastic semigroup P on {0, 1} with P′(0) = Q, namely: 1 $$ P\left( {t,0,0} \right) = \frac{\mu }{{\mu + \lambda }} + \frac{\lambda }{{\mu + \lambda }}{e^{{ - \left( {\mu + \lambda } \right)t}}}P\left( {t,0,1} \right) = 1 - P\left( {t,0,0} \right)P\left( {t,1,1} \right) = \frac{\lambda }{{\mu + \lambda }} + \frac{\mu }{{\mu + \lambda }}{e^{{ - 1\left( {\mu + \lambda } \right)t}}}P\left( {t,1,0} \right) = 1 - P\left( {t,1,1} \right) $$ One way to see this is to use (5.29): define P by (1); check P is continuous, P(0) is the identity, P′(0) = Q, and P(t + s) = P(t) · P(s). Dull computations in the last step can be avoided by thinking: it is enough to do μ + λ = 1 by rescaling time; since P(u) is 2 × 2 and stochastic when u is t or s or t + s, it is enough to check that P(t + s) = P(t) · P(s) on the diagonal; by interchanging μ and λ, so 0 and 1, it is enough to check the (0, 0) position. This is easy.

Keywords: Stationary Transition; Markov Chain; Markov Property; Measurable Subset; Nonnegative Real Number (search for similar items in EconPapers)
Date: 1983
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DOI: 10.1007/978-1-4612-5500-0_9

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