= P{X 0. = i} (1) If the MC has stationary transition probabilities then, = i} = P{X n+1

Transcription

1 Properties of Markov Chains and Evaluation of Steady State Transition Matrix P ss V. Krishnan - 3/9/2 Property 1 Let X be a Markov Chain (MC) where X {X n : n, 1, }. The state space is E {i, j, k, }. The one step transition probability is the probability of X 1 j given that X i. Or, P{X 1 j X i} (1) If the MC has stationary transition probabilities then, P{X 1 j X i} P{X n+1 j X n i} (2) If P is the matrix of transition probabilities then is the i,j th element in P as shown below; p 11 p 12 p 1j p 1n p 21 p 22 p 2j p 2n P p i1 p i2 p in (3) p n1 p n2 p nj p nn Property 2 Let {Xn : n, 1, }be a Markov Chain with Markov Matrix P. Then the probability of going from state i to state j in n steps is the n-step transition probability given by, P n (i, j) P{X n j X i} (4) where P n (i, j) is the i,j th element in the n-step probability transition matrix P n. Property 3 Let {Xn : n, 1, }be a Markov Chain with Markov Matrix P. Let the initial probability row vector be p i P{X i} p 1 p 2 p j p n. Then, P{X n j} p i P j (5) Property 4 The first passage time T ij is a random variable defining the time it takes to reach a fixed state j for the first time from the initial state i and is defined by, T ij min{n 1 : X i, X n j} (6) (n) If X n i, then T ii represents the first return to state i. If f ij represents the first passage probability to state j from state i in n steps the, the probability of eventually reaching state j is given by, f ij n 1 f (n) ij n P{T ij < X i} (7) If f ij 1, then the passage to state j is certain and is called a recurrent state and if it is < 1 then the first passage to j is uncertain and the state is called transient. Thus, we can define a first passage probability matrix F whose elements can be filled with either 1 or or values in between. 1 and can be filled by inspecting the state transition diagram but the in between values have to be calculated. Similarly, the number of crossings of state j from the state i is another random variable N ij with N ii being at least 1. We can define the expected number of visits (returns) to state j from state i by r ij defined by,

2 r ij E{N ij X i} (8) Thus, we can define the return matrix R whose elements will be, or any value in between. Again the values and can be filled by inspection of the state transition diagram and the in between values require calculation. As in the first passage probabilities, the state j is called recurrent if r ij and transient if r ij <, r ij and f ij are connected by the relationship, r ij 1 1 f jj ; i j, Expected # of returns to state j f ij 1 f jj ; i j, Expected # of passages to state j i, j 1, 2, n (9) Property 5 Let X be a Markov chain with matrix P and let C be a subset of the state space E, i.e, C E. Then C is closed if j C P ij 1, for all i C (1) Property 6 A closed set of states C E containing no proper subsets that are closed is called irreducible. If the number of states within an irreducible set are finite then each state in C is recurrent. For an irreducible set every state communicates with every other state. Property 7 Let X be a Markov chain with matrix P. If all the states are irreducible then a steady state row-vector defined by π can be determined from πp π, Or, π I P (11) Property 8 Let X be a Markov chain with finite state space E and k distinct irreducible sets. Let C a be the a-th irreducible set and P a the corresponding Markov matrix restricted to C a. The matrix F will define the first passage probabilities. a. If j is a transient state, then n P{X n j X i} (12) b. If i and j both belong to the a-th irreducible set, n P{X n j X i} π(j) (13) if is a row vector of steady state probabilities then, 1 (14) P a and π(i) i C a c. If state j is recurrent and i is not in its irreducible set, n P{X n j X i} f ij π(j) (15) where π(j) is given by eq.(14) d. If state j is recurrent and X is in the same irreducible set as j, then n 1 n 1 n I(X m X j) π(j) (16) m

3 where I(X m X j) 1 if Xm j and for any other state. e. If state j is recurrent, then. E T jj 1 π(j) (17) Property 9 Let X be a Markov chain and let B be the finite set of all transient states. Let Q be the transition matrix restricted to set B. Then for i, j A r ij E{N ij X i} (I Q) 1 (i, j) (18) Calculation of F matrix The matrix P is rewritten so that each irreducible recurrent set is treated as a single state with probability 1. To determine the probability of reaching a recurrent state from a transient state we have to find the probability of reaching the appropriate irreducible set. Thus, the rewritten matix P is written as, 1 P 1 1 b 1 b 2 b 3 Q where b j is the vector is the one-step probability of going from transient state i to the irreducible set a given by, b a (i) (2) j C a Example: The P matrix is given by;.6.4 P The corresponding P matrix is given by, 1 Property 1 P P a P b B 1 B 2 Q Let X be a Markov chain with the Markov matrix P given in the reduced form of eq. (19). Then for a transient state i and a recurrent state j we have, for each j in the irreducible set. (19) (21) (22) (I Q) 1 b a (i) (23)

4 In the above example Q and b 1.2 b 2.3.9, (I Q) (I Q) 1 b 1.25, (I Q) 1 b (24) Property 11 From property 9 we can write for any Markov chain X 1 1 ; i j rjj f ij i, j 1, 2, n (25) r ij ; i j r jj For the above example the R matrix can be written as, R (26) 1.11 The last block is obtained from property 9, viz., (I Q) From property (11) we can form the block matrix for the F matrix as F (27) From eqs. (24) and (27) we can formulate the F matrix as follows bearing in mind that for recurrent states the entries are all F Calculation of Steady State Transition Matrix P We are now in a position to calculate the steady state transition matrix using properties (8b) and (8c). The steady state probabilities for the irreducible states P a and P b are obtained from property (8b). In eq. (21) the matrices P a, P b and Q are: P a : P b : Q (29) (28)

5 We solve for the steady state probability vectors from a P a a and bp b b the normalization equations π a1 + π a2 1 and π b1 + π b2 + π b3 1. Thus we have, π a I P a π a1 π a2 : π b I P b π b1 π b2 π b (3) After discarding one equation from each one of eqs.(3) and adding the normalization equations we can write the following linear equations..3.7 π.4.7 π b1 a1 and 1 1 π π b2 (31) a π b3 1 and solving for a and b we have, π π a b and π π b (32) a π b3.826 Now we have to determine the steady state probabilities for the transient states from property (8c), the F matrix of eq. (28) and eqs. (32) as follows: P ss Or, P ss (33) This matrix is the same as calculated by using recursive techniques.