# 15 Markov Chains: Limiting Probabilities

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1 MARKOV CHAINS: LIMITING PROBABILITIES 67 Markov Chains: Limiting Probabilities Example Assume that the transition matrix is given by 7 2 P = 6 Recall that the n-step transition probabilities are given by powers of P So let s look at some large powers of P, beginning with 6 3 P = Then, to four decimal places P 8 and subsequent powers are the same to this precision The matrix elements appear to converge and the rows become almost identical Why? What determines the limit? These questions will be answered in this chapter We say that a state i S has period d if () Pii n > implies that d n, and (2) d is the largest positive integer that satisfies () Example 2 Simple random walk on Z, with p (,) The period of any state is 2 because the walker can return to her original position in any even number of steps, but in no odd number of steps Example 3 Random walk on vertices of a square Again, the period of any state is 2, for the same reason Example Random walk on vertices of triangle The period of any state is because the walker can return in two steps (one step out and then back) or three steps (around the triangle) Example Deterministic cycle In a chain with n states,,,n, which moves from i ti (i + ) mod n with probability, has period n So, any period is possible However, if the following two transition probabilities are changed: P = 9 and P =, then the chain has period In fact, the period of any state i with P ii > is trivially It can be shown that a period is the same for all states in the same class If a state, and therefore its class, has period, it is called aperiodic If the chain is irreducible, we call the entire chain aperiodic if all states have period

2 MARKOV CHAINS: LIMITING PROBABILITIES 68 For a state i S, let R i = min{n : X n = i} be the first time, after time, that the chain is in i S Also, let f (n) i = P(R i = n X = i) be the p m f of R i when the starting state is i itself (in which case we may call R i the return time) We can connect these to a familiar quantity, f i = P(ever reenter i X = i) = n= f (n) i, so that the state i is recurrent exactly when n= f(n) i = Then we define m i = E[R i X = i] = n= nf (n) i If the above series converges, ie, m i <, then we say that i is positive recurrent It can be shown that positive recurrence is also a class property: a state shares it with all members of its class Thus an irreducible chain is positive recurrent if each of its states is It is not hard to show that a finite irreducible chain is positive recurrent In this case there must exist a m and an ǫ > so that i can be reached from any j in m steps with probability at least ǫ Then P(R i n) ( ǫ) n/m, which goes to geometrically fast We now state the key theorems Some of these have rather involved proofs (although none are all that difficult), which we will merely sketch or omit altogether Theorem Proportion of the time spent at i Assume that the chain is irreducible and positive recurrent Let N n (i) be the number of visits to i in the time interval from through n Then, in probability N n (i) lim =, n n m i Proof The idea is quite simple: once the chain visits i, it returns on the average once per m i time steps, hence the proportion of time spent there is /m i We skip a more detailed proof A vector of probabilities, π i, i S, such that i S π i = is called an invariant, or stationary, distribution for a Markov chain with transition matrix P if π i P ij = π j for all j S i S

3 MARKOV CHAINS: LIMITING PROBABILITIES 69 In matrix form, if we put π into a row vector [π,π 2,], then [π,π 2,] P = [π,π 2,], thus [π,π 2,] is the left eigenvector of P, for eigenvalue More important for us is the probabilistic interpretation If π i is the p m f for X, that is, P(X = i) = π i, for all i S, it is also p m f for X, and then for all other X n, that is, P(X n = i) = π i, for all n Theorem 2 Existence and uniqueness of invariant distributions An irreducible positive recurrent Markov chain has a unique invariant distribution, which is given by π i = m i In fact, an irreducible chain is positive recurrent if and only if a stationary distribution exists The formula for π should not be a surprise: if the probability that the chain is in i is always π i, then one should expect the proportion of time spent at i, which we already know to be /m i, to be equal to π i We will not, however, go deeper into the proof Theorem 3 Convergence to invariant distribution If a Markov chain is irreducible, aperiodic, and positive recurrent, then, for every i,j S, lim P n n ij = π j Recall that P n ij = P(X n = j X = i), and note that the limit is independent of the initial state Thus the rows of P n are more and more similar to the row vector π as n becomes large The most elegant proof of this theorem uses coupling, an important idea first developed by a young French probabilist Wolfgang Doeblin in the late 93s (Doeblin s life is a very sad story An immigrant from Germany, he died as a soldier in the French army in 9, at the age of 2 He made significant mathematical contributions during his army service) Start with two independent copies of the chain two particles moving from state to state according to transition probabilities one started from i, the other using initial distribution π Under the stated assumptions, they will eventually meet Afterwards, the two particles move together in unison, that is, they are coupled Thus the difference between the two probabilities at time n is bounded above by twice the probability that coupling does not happen by time n, which goes to We will not go into greater details, but, as we will see in the next example, periodicity is necessary Example 6 Deterministic cycle with a = 3 has transition matrix P =

4 MARKOV CHAINS: LIMITING PROBABILITIES 7 This is an irreducible chain, with invariant distribution π = π = π 2 = 3 (as it is very easy to check) Moreover P 2 =, P 3 = I, P = P, etc Although the chain does spend /3 of the time at each state, the transition probabilities are a periodic sequence of s and s and do not converge Our final theorem is mostly a summary of results for the special, and for us the most common, case Theorem Convergence theorem for finite state space S Assume the Markov chain with a finite state space is irreducible There exists a unique invariant distribution given by π i = m i 2 For every i, and irrespective of the initial state, in probability n N n(i) π i, 3 If the chain is also aperiodic, then for all i and j, P n ij π j If the chain is periodic with period d, then for every pair i,j S, there exists an integer r, r d, so that lim P md+r m ij = dπ j and so that Pij n = for all n such that n r mod d Example 7 We begin by our first example, Example That was clearly an irreducible, and also aperiodic (note that P > ) chain The invariant distribution [π,π 2,π 3 ] is given by 7π + π 2 = π 2π + 6π 2 +π 3 = π 2 π = π 3 This system has infinitely many solutions, and we need to use π + π 2 + π 3 =

5 MARKOV CHAINS: LIMITING PROBABILITIES 7 to get the unique solution π = 2 37, π 2 = 37, π 3 = 2 37 Example 8 General two-state Markov chain Here S = {, 2} and and we assume that < α,β < and after a little algebra, P = [ α α β β ], απ + βπ 2 = π ( α)π + ( β)π 2 = π 2 π + π 2 = π = π 2 = β + β α α β + α Here are a few common follow-up questions: Start the chain at In the long run, what proportion of time does the chain spend at 2? Answer: π 2 (and does not depend on the starting state) Start the chain at 2 What is the expected return time to 2? Answer: π 2 In the long run, what proportion of time is the chain at 2, while at the previous time it was at? Answer: π P 2, as it needs to be at at the previous time, and then make a transition to 2 (again, the answer does not depend on the starting state) Example 9 Assume that a machine can be in states labeled, 2, 3, and In states and 2, the machine is up, working properly In states 3 and, the machine is down, out of order Suppose that the transition matrix is P = (a) Compute the average length of time the machine remains up after it goes up (b) Compute the proportion of times that the system is up, but down at the next time step (this is called the breakdown rate)

6 MARKOV CHAINS: LIMITING PROBABILITIES 72 We begin with computing the invariant distribution, which works out to be π = 9 8, π 2 = 2 π 3 = 8, π = 3 8 The breakdown rate is then the answer to (b) π (P 3 + P ) + π 2 (P 23 + P 2 ) = 9 32, Now let u be the average stretch of time machine remains up and d be the average stretch of time it is down We need to compute u to answer (a) We will achieve this by figuring out two equations that u and d must satisfy For the first equation, we use that the proportion of time the system is up is u u + d = π + π 2 = 2 8 For the second equation, we use that there is a single breakdown in every interval of time consisting of the stretch of up time followed by the stretch of down time, ie, from a repair to the next repair This means u + d = 9 32, the breakdown rate from (b) The system of two equations gives d = 2 and, to answer (a), u = 9 Computing the invariant distribution amounts to solving a system of linear equations Nowadays this is not a big deal, even for enormous systems; still, it is worthwhile to observe that there are cases when the invariant distribution is very easy to identify We call a square matrix with nonnegative entries doubly stochastic if the sum of the the entries in each row and in each column is Proposition Invariant distribution in a doubly stochastic case 8, If a transition matrix for an irreducible Markov chain with a finite state space S is doubly stochastic, its (unique) invariant measure is uniform over S Proof Assume that S = {,,m}, as usual If [,,] is the row vector with m s, then [,,]P is exactly the vextor of column sums, thus [,,] This vector is preserved by right multiplication by P and then so is m [,,] This vector specifies the uniform p m f on S Example Simple random walk on a circle Pick a probability p (,) Assume that a points labeled,,,a are arranged on a circle clockwise From i, the walker moves to i + (with a identified with ) with probability p and to i (with identified with a )

7 MARKOV CHAINS: LIMITING PROBABILITIES 73 with probability p The transition matrix is p p p p P = p p p p p p and is doubly stochastic Moreover, the chain is aperiodic if a is odd and otherwise periodic with period 2 Therefore, the proportion of time the walker spends at any state is a, which is also the limit of Pij n for all i and j if a is odd If a is even, then P n ij = if (i j) and n have a different parity, while if they are of the same parity, Pij n 2 a Assume that we change the transition probabilities a little: assume that, only when she is at, the walker stays at with probability r (,), moves to with probability ( r)p and to a with probability ( r)( p) The other transition probabilities are unchanged Clearly now the chain is aperiodic for any a, but the transition matrix is no longer doubly stochastic What happens is the invariant distribution? The walker spends a longer time at ; if we stop the time while she transitions from to the chain is the same as before, and spends equal proportion of time in all states It follows that our perturbed chain spends the same proportion of time in all states except, where it spends a Geometric( r) time at every visit Therefore π is larger by the factor r than other π i The row vector with invariant distributions thus is [ r + a r ] [ = +( r)(a ) ] r r +( r)(a ) +( r)(a ) Thus we can still determine invariant distribution if only self-transitions P ii are changed Problems Consider the chain in Problem 2 of Chapter 2 (a) Determine the invariant distribution (b) Determine lim n P n Why does it exist? 2 Consider the chain in Problem of Chapter 2, with the same initial state Determine the proportion of time the walker spends at a 3 Roll a fair die n times and let S n be the sum of the numbers you roll Determine, with proof, lim n P(S n mod 3 = ) Peter owns two pairs of running shoes Each morning he goes running He is equally likely to leave from his front or back door Upon leaving the house, he chooses a pair of running shoes

8 MARKOV CHAINS: LIMITING PROBABILITIES 7 at the door from which he leaves, or goes running barefoot if there are no shoes there On his return, he is equally likely to enter at each door, and leaves his shoes (if any) there (a) What proportion of days he runs barefoot? (b) What proportion of days is there at least one pair of shoes at the front door (before he goes running)? (c) Now also assume that the pairs of shoes are green and red and that he chooses a pair at random if he has a choice What proportion of mornings does he run in green shoes? Prof Messi does one of three service tasks every year, coded as, 2, and 3 The assignment changes randomly from year to year as a Markov chain with transition matrix Determine the proportion of years that Messi has the same assignment as the previous two years 6 Consider the Markov chain with states,,2,3,, which transition from state i > to one of the states,,i with equal probability, and transition from to with probability Show that all Pij n converge as n and determine the limits Solutions to problems The chain is irreducible and aperiodic Moreover, (a) π = [ 2, 2, 2] 6 and (b) the limit is π = 2 2 The chain is irreducible and aperiodic Moreover, π = [, 8, 8, ] 2 and the answer is π + π 3 = 3 3 Consider S n mod 3 This is a Markov chain with states,,2 and transition matrix (To get next row, shift cyclically to the right) This is a doubly stochastic matrix, with π i = 3, for all i So the answer is 3 Consider the Markov chain with states given by the number of shoes at the front door Then P = 3 2 3

9 MARKOV CHAINS: LIMITING PROBABILITIES 7 This is a doubly stochastic matrix, with π = π = π 2 = 3 Answers: (a) π 2 + π 2 2 = 3 ; (b) π + π 2 = 2 3 ; (c) π + π 2 + π 2 2 = 3 Solve for the invariant distribution: π = [ 6 9 7, 7 7, 7] The answer is π P 2 +π 2 P π 3 P 2 33 = 6 As the chain is irreducible and aperiodic, Pij n converges to π j, j =,,2,3,, where π is given by π = [ 2 37, 6 37, 37, 3 37, 2 37]

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