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DISCRETE-TIME MARKOV CHAINS
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The Markov chain {Xn , n = 0, 1, } is completely determined by the probability distribution of the initial state X0 and the one-step transition probabilities pij In applications of Markov chains the art is: (a) to choose the state variable(s) such that the Markovian property (311) holds, (b) to determine the one-step transition probabilities pij Once this (dif cult) modelling step is done, the rest is simply a matter of applying the theory that will be developed in the next sections The student cannot be urged strongly enough to try the problems at the end of this chapter to acquire skills to model new situations Let us return to the drunkard s walk Example 311 (continued) The drunkard s random walk In this example we have already de ned the state variable as the position of the drunkard The process {Xn } with Xn denoting the state just after the nth step of the drunkard is indeed a discrete-time Markov chain The one-step transition probabilities are as follows For any interior state (x, y) with N < x, y < N , we have p(x,y)(v,w) =
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for (v, w) = (x + 1, y), (x 1, y), (x, y + 1), (x, y 1), otherwise
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For any boundary state (x, N ) with N < x < N , we have p(x,y)(v,w) =
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For the boundary state (x, N ) with N < x < N , (N, y) and (N, y) with N < y < N , the one-step transition probabilities follow similarly For the corner point (x, y) = (N, N ), we have p(x,y)(v,w) =
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for (v, w) = (N 1, N ), (N, N 1), otherwise
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Similarly, for the corner points (x, y) = ( N, N ), ( N, N ) and (N, N ) A variant of the drunkard s random walk problem is the problem in which the drunkard never chooses the same direction as was chosen in the previous step Then we have to augment the state with an extra state variable in order to satisfy the Markovian property The state of the drunkard after each step is now de ned as (x, y, z), where (x, y) denotes the position of the drunkard and z {N, S, W, L} denotes the direction of the last step Letting Xn be the state of the drunkard s process just after the nth step (with the convention X0 = (0, 0)), the stochastic process {Xn } is a discrete-time Markov chain It is left to the reader to write down the one-step transition probabilities of this process
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THE MODEL
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Example 312 A stock-control problem The Johnson hardware shop carries adjustable-joint pliers as a regular stock item The demand for this tool is stable over time The total demand during a week has a Poisson distribution with mean The demands in the successive weeks are independent of each other Each demand that occurs when the shop is out of stock is lost The owner of the shop uses a so-called periodic review (s, S) control rule for stock replenishment of the item The inventory position is only reviewed at the beginning of each week If the stock on hand is less than the reorder point s, the inventory is replenished to the order-up point S; otherwise, no ordering is done Here s and S are given integers with 0 s S The replenishment time is negligible What is the average ordering frequency and what is the average amount of demand that is lost per week These questions can be answered by the theory of Markov chains In this example we take as state variable the stock on hand just prior to review Let Xn = the stock on hand at the beginning of the nth week just prior to review, then the stochastic process {Xn } is a discrete-time Markov chain with the nite state space I = {0, 1, , S} It will be immediately clear that the Markovian property (311) is satis ed: the stock on hand at the beginning of the current week and the demand in the coming week determine the stock on hand at the beginning of the next week It is not relevant how the stock level uctuated in the past To nd the one-step transition probabilities pij = P {Xn+1 = j | Xn = i} we have to distinguish the cases i s and i < s In the rst case the stock on hand just after review equals i, while in the second case the stock on hand just after review equals S For state i s, we have pij = P {the demand in the coming week is i j } = e i j , (i j )! j = 1, , i
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Note that this formula does not hold for j = 0 Then we have for i s, pi0 = P {the demand in the coming week is i or more}
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