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Initial distribution 0 this is the probability

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Unformatted text preview: e case N = 1". Initial distribution 0 . This is the probability distribution of the Markov chain at time 0. For each state i 2 S, we denote by 0 i the probability PfX0 = ig that the Markov chain starts out in state i. Formally, 0 is a function taking S into the interval 0,1 such that 0 i  0 for all i 2 S and X i2S 0 i = 1: Equivalently, instead of thinking of 0 as a function from S to 0,1 , we could think of 0 as the vector whose ith entry is 0 i = PfX0 = ig. Probability transition rule This is speci ed by giving a matrix P = Pij . If S is the nite set f1; : : : ; N g, say, then P is an N  N matrix. Otherwise, P will have in nitely many rows and columns; sorry. The interpretation of the number Pij is the conditional probability, given that the chain is in state i at time n, say, that the chain jumps to the state j at time n +1. That is, Pij = PfXn+1 = j j Xn = ig: We will also use the notation P i; j  for the same thing. Note that we have written this probability as a function of just i and j , but of course it could depend on n as well. The time homogeneity restriction mentioned in the previous footnote is just the assumption that this probability does not depend on the time n, but rather remains constant over time. Formally, a probability transition matrix is an N  N matrix whose entries are all nonnegative and whose rows sum to 1. Finally, you may be wondering why we bother to arrange these conditional probabilities into a matrix. That is a good question, and will be answered soon. Stochastic Processes J. Chang, March 30, 1999 1.1. SPECIFYING AND SIMULATING A MARKOV CHAIN Page 1-3 1.1 Figure. The Markov frog. We can now get to the question of how to simulate a Markov chain, now that we know how to specify what Markov chain we wish to simulate. Let's do an example: suppose the state space is S = f1; 2; 3g, the initial distribution is 0 = 1=2; 1=4; 1=4, and the probability transition matrix is 01 2 31 1010 1.2 P = 2 @ 1=3 0 2=3 A: 3 1=3 1=3 1=3 Think of a frog hopping among lily pads as in Figure 1.1. How does the Markov frog choose a path? To start, he chooses his initial position X0 according to the speci ed initial distribution 0 . He could do this by going to his computer to generate a uniformly distributed random number U0  Unif0; 1, and then taking 8 1 if 0 U 1=2 0 X0 = : 2 if 1=2 U0 3=4 3 if 3=4 U0 1 We don't have to be fastidious about specifying what to do if U0 comes out be exactly 1 2 or 3 4, since the probability of this happening is 0. For example, suppose that U0 comes out to be 0.8419, so that X0 = 3. Then the frog chooses X1 according to the probability distribution in row 3 of P , namely, 1=3; 1=3; 1=3; to do this, he paws his computer again to generate U1  Unif0; 1 independently of U0 , and takes 8 1 if 0 U 1=3 0 X1 = : 2 if 1=3 U0 2=3 3 if 2=3 U0 1: Suppose he happens to get U1 = 0:1234, so that X1 = 1. Then he chooses X2 according to row 1 of P , so that X2 = 2; there's no choice this time...
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