Discrete-time stochastic processes

Letting be the left eigenvector of we have au u and

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: sented by t = j s + ` where 1 ≤ ` < s and j ≥ 0. Iterating (4.15), we get T (k) = T (k + j s), and applying (4.10) to this, T (k + `) = T (k + j s + `) = T (k + t) = T (k). where we have used t = j s + ` followed by (4.14). This is the desired contradiction, since ` < s. Thus s = 1 and T (k) = T (k + 1). Iterating this, T (k) = T (k + n) for all n ≥ 0. (4.16) Since the chain is ergodic, each state j continues to be accessible after k steps. Therefore j must be in T (k + n) for some n ≥ 0, which, from (4.16), implies that j ∈ T (k). Since j is n arbitrary, T (k) must be the entire set of states. Thus Pij > 0 for all n ≥ k and all j . This same argument can be applied to any state i on the given cycle with τ nodes. Any state m not on this cycle has a path to the cycle using at most M − τ steps. Using this path to reach a node i on the cycle, and following this with all the walks from i of length k = (M − 1)τ , we see that M−τ +(M−1)τ Pmj >0 for all j, m. The proof is complete, since M − τ + (M − 1)τ ≤ (M − 1)2 + 1 for all τ , 1 ≤ τ ≤ M − 1, with equality when τ = M − 1. Figure 4.4 illustrates a situation where the bound (M − 1)2 + 1 is met with equality. Note that there is one cycle of length M − 1 and the single node not on this cycle, node 1, is the unique starting node at which the bound is met with equality. 4.3 The Matrix representation The matrix [P ] of transition probabilities of a Markov chain is called a stochastic matrix; that is, a stochastic matrix is a square matrix of non-negative terms in which the elements n in each row sum to 1. We first consider the n step transition probabilities Pij in terms of [P]. The probability of going from state i to state j in two steps is the sum over h of all possible two step walks, from i to h and from h to j . Using the Markov condition in (4.1), 2 Pij = M X Pih Phj . h=1 It can be seen that this is just the ij term of the product of matrix [P ] with itself; denoting 2 n [P ][P ] as [P ]2 , this means that Pij is the (i, j ) element of the matrix [P ]2 . Similarly, Pij is 148 CHAPTER 4. FINITE-STATE MARKOV CHAINS the ij element of the nth power of the matrix [P ]. Since [P ]m+n = [P ]m [P ]n , this means that m Pij +n = M X mn Pih Phj . (4.17) h=1 This is known as the Chapman-Kolmogorov equation. An efficient approach to compute n [P ]n (and thus Pij ) for large n, is to multiply [P ]2 by [P ]2 , then [P ]4 by [P ]4 and so forth and then multiply these binary powers together as needed. The matrix [P ]n (i.e., the matrix of transition probabilities raised to the nth power) is very n important for a number of reasons. The i, j element of this matrix is Pij , which is the probability of being in state j at time n given state i at time 0. If memory of the past dies out with increasing n, then we would expect the dependence on both n and i to disappear n in Pij . This means, first, that [P ]n should converge to a limit as n → 1, and, second, that each row of [P ]n should tend to the same set of probabilities. If this convergence occurs (and we later determine the circumstances under which it occurs), [P ]n and [P ]n+1 will be the same in the limit n → 1 which means lim[P ]n = (lim[P ]n )P . If all the rows of lim[P n ] are the same, equal to some row vector π = (π1 , π2 , . . . , πM ), this simplifies to π = π [P ]. Since π is a probability vector (i.e., its components are the probabilities of being in the various states in the limit n → 1), its components must be non-negative and sum to 1. Definition 4.9. A steady-state probability vector (or a steady-state distribution) for a Markov chain with transition matrix [P ] is a vector π that satisfies X π = π [P ] ; where πi = 1 ; πi ≥ 0 , 1 ≤ i ≤ M. (4.18) i The steady-state probability vector is also often called a stationary distribution. If a probability vector π satisfying (4.18) is taken as the initial probability assignment of the chain at time 0, then that assigment is maintained forever. That is, if Pr {X0 =i} = πi for all i, P then Pr {X1 =j } = i πi Pij = πj for all j , and, by induction, Pr {Xn = j } = πj for all j and all n > 0. If [P ]n converges as above, then, for each starting state, the steady-state distribution is reached asymptotically. There are a number of questions that must be answered for a steady-state distribution as defined above: 1. Does π = π [P ] always have a probability vector solution? 2. Does π = π [P ] have a unique probability vector solution? 3. Do the rows of [P ]n converge to a probability vector solution of π = π [P ]? We first give the answers to these questions for finite-state Markov chains and then derive them. First, (4.18) always has a solution (although this is not necessarily true for infinitestate chains). The answer to the second and third questions is simpler with the following definition: 4.3. THE MATRIX REPRESENTATION 149 Definition 4.10. A unichain is a finite...
View Full Document

This note was uploaded on 09/27/2010 for the course EE 229 taught by Professor R.srikant during the Spring '09 term at University of Illinois, Urbana Champaign.

Ask a homework question - tutors are online