18.440: Lecture 33 Markov Chains

Transcription

1 18.440: Lecture 33 Markov Chains Scott Sheffield MIT 1

2 Outline Markov chains Examples Ergodicity and stationarity 2

3 Outline Markov chains Examples Ergodicity and stationarity 3

4 Markov chains Consider a sequence of random variables X 0, X 1, X 2,... each taking values in the same state space, which for now we take to be a finite set that we label by {0, 1,..., M}. Interpret X n as state of the system at time n. Sequence is called a Markov chain if we have a fixed collection of numbers P ij (one for each pair i, j {0, 1,..., M}) such that whenever the system is in state i, there is probability P ij that system will next be in state j. Precisely, P{X n+1 = j X n = i, X n 1 = i n 1,..., X 1 = i 1, X 0 = i 0 } = P ij. Kind of an almost memoryless property. Probability distribution for next state depends only on the current state (and not on the rest of the state history). 4

5 Simple example For example, imagine a simple weather model with two states: rainy and sunny. If it s rainy one day, there s a.5 chance it will be rainy the next day, a.5 chance it will be sunny. If it s sunny one day, there s a.8 chance it will be sunny the next day, a.2 chance it will be rainy. In this climate, sun tends to last longer than rain. Given that it is rainy today, how many days to I expect to have to wait to see a sunny day? Given that it is sunny today, how many days to I expect to have to wait to see a rainy day? Over the long haul, what fraction of days are sunny? 5

6 Matrix representation To describe a Markov chain, we need to define P ij for any i, j {0, 1,..., M}. It is convenient to represent the collection of transition probabilities P ij as a matrix: P 00 P P 0M P 10 P P 1M A = P M0 P M1... P MM For this to make sense, we require P ij 0 for all i, j and M j=0 P ij = 1 for each i. That is, the rows sum to one. 6

7 Transitions via matrices Suppose that p i is the probability that system is in state i at time zero. What does the following product represent? P 00 P P 0M 10 P P 1M P ( ) p 0 p 1... p M P M0 P M1... P MM Answer: the probability distribution at time one. How about the following product? ( ) p 0 p 1... p A M n Answer: the probability distribution at time n. 7

8 Powers of transition matrix (n) We write P ij over n steps. From the matrix point of view for the probability to go from state i to state j (n) (n) (n) n P... P 00 P 01 0M P00 P P 0M (n) (n) (n) P 10 P P 1M P 10 P P1M = (n) (n) (n) P P... P P M0 P M1... PMM M0 M1 MM If A is the one-step transition matrix, then A n is the n-step transition matrix. 8

9 Questions What does it mean if all of the rows are identical? Answer: state sequence X i consists of i.i.d. random variables. What if matrix is the identity? Answer: states never change. What if each P ij is either one or zero? Answer: state evolution is deterministic. 9

10 Outline Markov chains Examples Ergodicity and stationarity 10

11 Outline Markov chains Examples Ergodicity and stationarity 11

12 Simple example Consider the simple weather example: If it s rainy one day, there s a.5 chance it will be rainy the next day, a.5 chance it will be sunny. If it s sunny one day, there s a.8 chance it will be sunny the next day, a.2 chance it will be rainy. Let rainy be state zero, sunny state one, and write the transition matrix by ( ).5.5 A =.2.8 Note that ( ) A = ( ) Can compute A =

13 Does relationship status have the Markov property? In a relationship Single It s complicated Married Engaged Can we assign a probability to each arrow? Markov model implies time spent in any state (e.g., a marriage) before leaving is a geometric random variable. Not true... Can we make a better model with more states? 13

14 Outline Markov chains Examples Ergodicity and stationarity 14

15 Outline Markov chains Examples Ergodicity and stationarity 15

16 Ergodic Markov chains Say Markov chain is ergodic if some power of the transition matrix has all non-zero entries. Turns out that if chain has this property, then (n) π j := lim n P ij exists and the π j are the unique non-negative solutions of π M j = k=0 π k P kj that sum to one. This means that the row vector ( ) π = π 0 π 1... π M is a left eigenvector of A with eigenvalue 1, i.e., πa = π. We call π the stationary distribution of the Markov chain. One can solve the system of linear equations M π j = k=0 π k P kj to compute the values π j. Equivalent to considering A fixed and solving πa = π. Or solving (A I )π = 0. This determ ines π up to a multiplicative constant, and fact that π j = 1 determines the constant. 16

17 Simple example ( ).5.5 If A =, then we know.2.8 ( ) ( ).5.5 ( ) πa = π 0 π 1 = π 0 π 1 = π..2.8 This means that.5π 0 +.2π 1 = π 0 and.5π 0 +.8π 1 = π 1 and we also know that π 0 + π 1 = 1. ( Solving these ) equations gives π 0 = 2/7 and π 1 = 5/7, so π = 2/7 5/7. Indeed, ( πa = 2/7 ( ) ).5.5 ( 5/7 = 2/7.2.8 ) 5/7 = π. Recall (that ) ( ) ( ) /7 5/7 π A = = /7 5/7 π 17

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