And so on stochastic processes j chang march 30 1999

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Unformatted text preview: . Next, he chooses X3 according to row 2 of P . And so on. . . . Stochastic Processes J. Chang, March 30, 1999 1. MARKOV CHAINS Page 1-4 1.2 The Markov property Clearly, in the previous example, if I told you that we came up with the values X0 = 3, X1 = 1, and X2 = 2, then the conditional probability distribution for X3 is P 8 1=3 for j = 1 for j = 2 2=3 for j = 3, fX3 = j j X0 = 3; X1 = 1; X2 = 2g = : 0 which is also the conditional probability distribution for X3 given only the information that X2 = 2. In other words, given that X0 = 3, X1 = 1, and X2 = 2, the only information relevant to the distribution to X3 is the information that X2 = 2; we may ignore the information that X0 = 3 and X1 = 1. This is clear from the description of how to simulate the chain! Thus, fX3 = j j X2 = 2; X1 = 1; X0 = 3g = PfX3 = j j X2 = 2g for all j . P This is an example of the Markov property. 1.3 Definition. A process X0 ; X1 ; : : : satis es the Markov property if fXn+1 = in+1 j Xn = in ; Xn,1 = in,1; : : : ; X0 = i0 g = PfXn+1 = in+1 j Xn = in g P for all n and all i0 ; : : : ; in+1 2 S. The issue addressed by the Markov property is the dependence structure among random variables. The simplest dependence structure for X0 ; X1 ; : : : is no dependence at all, that is, independence. The Markov property could be said to capture the next simplest sort of dependence: in generating the process X0 ; X1 ; : : : sequentially, each Xn depends only on the preceding random variable Xn,1 , and not on the further past values X0 ; : : : ; Xn,2 . The Markov property allows much more interesting and general processes to be considered than if we restricted ourselves to independent random variables Xi , without allowing so much generality that a mathematical treatment becomes intractable. The Markov property implies a simple expression for the probability of our Markov chain taking any speci ed path, as follows: fX0 = i0 ; X1 = i1 ; X2 = i2; : : : ; Xn = ing = PfX0 = i0 gPfX1 = i1 j X0 = i0 gPfX2 = i2 j X1 = i1 ; X0 = i0 g    PfXn = in j Xn,1 = in,1; : : : ; X1 = i1 ; X0 = i0 g = PfX0 = i0 gPfX1 = i1 j X0 = i0 gPfX2 = i2 j X1 = i1 g    PfXn = in j Xn,1 = in,1g = 0 i0 P i0 ; i1 P i1 ; i2     P in,1 ; in : P Stochastic Processes J. Chang, March 30, 1999 1.2. THE MARKOV PROPERTY Page 1-5 So, to get the probability of a path, we start out with the initial probability of the rst state and successively multiply by the matrix elements corresponding to the transitions along the path. 1.4 Exercise. Let X0 ; X1 ; : : : be a Markov chain, and let A and B be subsets of the state space. 1. Is it true that PfX2 2 B j X1 = x1 ; X0 2 Ag = PfX2 2 B j X1 = x1 g? Give a proof or counterexample. 2. Is it true that PfX2 2 B j X1 2 A; X0 = x0 g = PfX2 2 B j X1 2 Ag? Give a proof or counterexample. The moral: be careful about what the Markov property says! 1.5 Exercise. Let X0 ; X1 ; : : : be a Markov chain on the state space f,1; 0; 1g, and suppose that P i; j  0 for all i; j . What is a necessary and su cient condition for the sequence of ab...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.

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