MATH 56A SPRING 2008 STOCHASTIC PROCESSES 31

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1 MATH 56A SPRING 2008 STOCHASTIC PROCESSES 3.3. Invariant probability distribution. Definition.4. A probability distribution is a function π : S [0, ] from the set of states S to the closed unit interval [0, ] so that π(i) =. i S When the set of states is equal to S = {, 2,, s} then the condition is: s π(i) =. i= Definition.5. A probability distribution π is called invariant if πp = P. I.e., π is a left eigenvector for P with eigenvalue..3.. probability distribution of X n. Each X n has a probability distribution. I used the following example to illustrate this. /3 2 /4 The numbers /3 and /4 are transition probabilities. They say nothing about X 0. But we need to start in a random state X 0. This is because we need to understand how the transition from one random state to another works so that we can go from X n to X n+. X 0 will be equal to either 0 or with probability: α = P(X 0 = ), α 2 = P(X 0 = 2). These two numbers are between 0 and (inclusive) and add up to : α + α 2 =. So, α = (α, α 2 ) is a probability distribution. α is the probability distribution of X 0 and is called the initial (probability) distribution. Once the distribution of X 0 is given, the probability distribution of every X n is determined by the transition matrix: Theorem.6. The probability distribution of X n is the vector αp n.

2 32 FINITE MARKOV CHAINS So, in the example, 2 2 P(X 2 = 2) = P(X 0 = i) P(X = j X 0 = i) P(X 2 = 2 X = j) i= j= α i p(i,j) p(j,2) = i,j α i p(i, j)p(j, 2) = (αp 2 ) 2. This is the sum of the probabilities of all possible ways that you can end up at state 2 at time 2. To prove this in general, I used the following probability formula: Lemma.7. Suppose that our sample space is a disjoint union of events B i. Then Ω = B i (.) P(A) = i P(B i )P(A B i ) I drew this picture to illustrate this basic concept that you should already know. Proof of theorem.6. By induction on n. If n = 0 then P n = I is the identity matrix. So, αp n = αp 0 = αi = α. This is the distribution of X n = X 0 by definition. So, the theorem holds for n = 0. Suppose the theorem holds for n. Then, by Equation (.) P(X n = ) P(X n+ = X n = ) B p(,) P(X n+ = ) = +P(X n = 2 ) P(X n+ = X n = 2) B 2 p(2,) = (αp n ) p(, ) + (αp n ) 2 p(2, ) = [(αp n )P ] = ( αp n+).

3 MATH 56A SPRING 2008 STOCHASTIC PROCESSES 33 And similarly, P(X n+ = 2) = (αp n+ ) 2. So, the theorem holds for n +. So, it holds for all n 0. Corollary.8. If the initial distribution α = π is invariant, then X n has probability distribution π for all n. Proof. The distribution of X n is αp n = πp n = πp P P = π =π since every time you multiply P you always have π Perron-Frobenius Theorem. I stated this very important theorem without proof. However, the proof is outlined in Exercise.20 in the book. Theorem.9 (Perron-Frobenius). Suppose that A is a square matrix all of whose entries are positive real numbers. Then, A has a left eigenvector π, all of whose coordinates are positive real numbers. I.e., Furthermore, πp = λπ. (a) π is unique up to a scalar multiple. (If α is another left eigenvector of A with positive real entries then α = Cπ for some scalar C.) (b) λ = λ is a positive real number. (c) The eigenvector λ is larger in absolute value then any other eigenvalue of A: λ 2, λ 3, < λ. (So, λ is called the maximal eigenvalue of A.) (d) lim n π P n = λ n π. π assuming that π is a probability distribution, i.e., π i =. I didn t prove this. However, I tried to explain the last statement. When we raise P to the power n, it tends to look like multiplication by λ n. So, we should divide by λ n. If we know that the rows of the

4 34 FINITE MARKOV CHAINS matrix P n are all the same, what is it? λ n α π ( ) P n = (π λ n, π 2, ) α. = πi α α But π λ P = π. So, π λ n P n = π = ( πi ) α and α = πi π. This theorem applies to Markov chains but with some conditions. First, I stated without proof the fact: Theorem.0. The maximal eigenvalue of P is. More precisely, all eigenvalues of P have λ. Proof. Suppose that P has an eigenvalue λ with absolute value greater than. Then, there is an eigenvector x so that xp = λx. Then xp n = λ n x diverges as n goes to infinity. But this is not possible since the entries of the matrix P n are all between 0 and since P n is the probability transition matrix from X 0 to X n by Theorem.6. I ll explain this proof later. What I did explain is class is that is always an eigenvalue of P. This follows from the fact that the rows of P add up to : p(i, j) =. j This implies that the column vector with all entries is a right eigenvector of P with eigenvalue : P. =. For example, if then P ( ) = P = ( 2/3 ) /3 /4 3/4 ( 2/3 /3 /4 3/4 ) ( ) = ( )

5 MATH 56A SPRING 2008 STOCHASTIC PROCESSES 35 The invariant distribution π is a left eigenvector with eigenvalue. The unique invariant distribution is: ( 3 π = 7, 4 ) 7 This means: πp = ( 3 7, 4 ) ( ) ( 2/3 /3 3 = 7 /4 3/4 7, 4 ) = π. 7 You can find π using linear algebra or a computer. You can also use intuition. In the Markov chain: /3 /4 we can use the Law of large numbers which says that, if there are a large number of people moving randomly, then the proportion who move will be approximately equal to the probability. So, if there are a large number of people in states and 2 then one third of those at will move to 2 and one fourth of those in 2 will move to. If you want the distribution to be stable, the numbers should have a 3:4 ratio. If there are 3 guys at point and /3 of them move, then one guy moves to 2. If there are 4 guys at 2 and /4 of them move then one guy moves from 2 to and the distribution is the same. To make it a probability distribution, the vector (3, 4) needs to be divided by 7. Theorem.. If the Markov chain is aperiodic and irreducible then it satisfies the conclusions of the Perron-Frobenius theorem. Proof. These conditions imply that A = P n has all positive entries for some finite n. Then, the Perron-Frobenius eigenvector for A is the invariant distribution for P. The Perron-Frobenius theorem tells us that the distribution of X n will reach an equilibrium (the invariant distribution) for large n (assuming aperiodic). Then next question is: How long does it take? 2

6 36 FINITE MARKOV CHAINS.4. Transient classes. I asked the question: How long does it take to escape from a transient class? I started with a really simple example: p 2 This is a Markov chain with one transient class {} and one absorbing class {2}. The question is: How long can you stay in the transient class? I was glad to see that students know basic probability: P(X n = X 0 = ) = ( p) n. What happens when n goes to infinity? lim ( n p)n = 0 if p > 0. Proof. And you guys helped me with this proof: So, L = 0. L = lim n ( p) n ln L = lim n n ln ( p) =. < <0 So, the probability of remaining indefinitely in state is zero. In other words, you will eventually escape the transient class with probability one (at least in this example). For future reference I recorded this conclusion as follows. Theorem.2. If the probability of success is p > 0 and if you try infinitely many times, then you will eventually succeed with probability one. But how long does it take? Let Then T := smallest n so that X n = 2. P(T = n X 0 = ) = p( p) n. For example, if n = 3 then, T = 3 which means we have: X 0 X X 2 X 3 p p p 2 P(T = 3 X 0 = ) = p( p) 2.

7 MATH 56A SPRING 2008 STOCHASTIC PROCESSES 37 The numbers p( p) n add up to and give what is called the geometric distribution on the nonnegative integers. From this formula we can calculate the conditional expected value of T : E(T X 0 = ) = np( p) n and, yes, the n = 0 term is zero. This is easy to calculate using a little calculus. First we start with the geometric series g(x) := x n = + x + x 2 + x 3 + = x. Then differentiate: n=0 g (x) = n=0 nx n = n=0 ( x) 2. Applying this formula to the expected value problem, x = p and we get: E(T X 0 = ) = p ( ) n( p) n = p = p ( ( p)) 2 p = 2 p. This was actually intuitively obvious from the beginning. For example: /5 When p = /5 then you expect it to happen in 5 trials. So, E(T ) = 5 = /p.

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