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A discrete-time Markov chain (M.C.), fX t: t= 0;1;g , is a stochastic process with the Markov property: P(X n+1 = jjX 0 = i 0; ;X n 1 = i n 1;X n= i) = P(X n+1 = jjX n= i) for all time index nand all states i 0; ;i n 1;i;j. State space is the range of possible values for the random variables X t. Assume the state space is nite: f0;1; ;Ngor ...
Let us write \[ V_j(N) := \# \big\{ n < N : X_n = j \} \] for the total number of visits to state \(j\) up to time \(N\). Then we can interpret \(V_j(n)/n\) as the proportion of time up to time \(n\) spent in state \(j\), and its limiting value (if it exists) to be the long-run proportion of time spent in state \(j\).
Definition: The state of a Markov chain at time t is the value ofX t. For example, if X t = 6, we say the process is in state6 at timet. Definition: The state space of a Markov chain, S, is the set of values that each X t can take. For example, S = {1,2,3,4,5,6,7}. Let S have size N (possibly infinite).
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P [Y = k] = pk. Set Y = 0 and Xl = Y + Y1 + · · · + Yl where addition takes place in Z/n. Using. Xl+1 = Yl+1 + Xl, the validity of the Markov property and time stationarity are easily verified and it follows that X , X1, X2 · · · is a Markov chain with state space Z/n = {0, 1, 2, · · · , n − 1}.
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Example (Gambler’s Ruin): Every time a gambler plays a game, he wins $1 w.p. p, and he loses $1 w.p. 1 − p. He stops playing as soon as his fortune is either $0 or $N. The gambler’s fortune is a MC with the following Pij’s: Pi,i+1 = p, i = 1,2,...,N − 1 Pi,i−1 = 1 − p, i = 1,2,...,N − 1 P0,0 = PN,N = 1
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So in the long run, the chain is expected to spend a proportion of π (j) of its time at the state j. Formally, let j be a state, and let I m (j) be the indicator of the event {X m = j}. The proportion of time the chain spends at j, from time 1 through time n, is.
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Jul 18, 2022 · The matrix shows that in the long run, Professor Symons will walk to school 1/3 of the time and bicycle 2/3 of the time. When this happens, we say that the system is in steady-state or state of equilibrium. In this situation, all row vectors are equal. If the original matrix is an n by n matrix, we get n row vectors that are all the same.
