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# 4-2 - Handbook 6.2 Steady-States Suppose we had X0 = 10 and...

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Handbook 6.2: Steady-States Suppose we had X 0 = bracketleftbig 1 0 bracketrightbig and T = bracketleftbigg 1 / 4 3 / 4 1 / 2 1 / 2 bracketrightbigg . If we were to calculate the next few distribution vectors, here’s what we would find: X 1 = bracketleftbig 0 . 25 0 . 75 bracketrightbig X 2 = bracketleftbig 0 . 4375 0 . 5625 bracketrightbig X 3 bracketleftbig 0 . 3906 0 . 6094 bracketrightbig X 4 bracketleftbig 0 . 4023 0 . 5977 bracketrightbig X 5 bracketleftbig 0 . 3994 0 . 6006 bracketrightbig X 6 bracketleftbig 0 . 4001 0 . 5999 bracketrightbig It seems that as time goes on, the distribution vector gets closer and closer to becoming bracketleftbig 0 . 4 0 . 6 bracketrightbig . The change that the probabilities undergo become smaller and smaller as time goes on. In other words, we can say that the distribution vector is approaching a steady-state . We will now find a way to show that the steady- state distribution vector for T is indeed bracketleftbig 0 . 4 0 . 6 bracketrightbig . If the distribution approaches a steady-state, then somewhere in the future, we will have that X n +1 = X n (the distribution will no longer change). But, we know that X n +1 = X n T . Putting these two ideas together, we have that somewhere down the road that X n = X n T . Let’s suppose that X n = bracketleftbig x 1 x 2 bracketrightbig , where x 1 and x 2 are the steady- state probabilities that we want to find. Substituting this, and T into the last equation we have the following matrix equation : bracketleftbig x 1 x 2 bracketrightbig = bracketleftbig x 1 x 2 bracketrightbig bracketleftbigg 1 / 4 3 / 4 1 / 2 1 / 2 bracketrightbigg If we were to perform the matrix multiplication on the right-hand side, we would have: bracketleftbig x 1 x 2 bracketrightbig = bracketleftbig (1 / 4 x 1 + 1 / 2 x 2 ) (3 / 4 x 1 + 1 / 2 x 2 ) bracketrightbig By comparing the corresponding entries, we get two equations: x 1 = 1 / 4 x 1 + 1 / 2 x 2 x 2 = 3 / 4 x 1 + 1 / 2 x 2 1

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Moving all the variables to one side of the equation, we have: - 3 / 4 x 1 + 1 / 2 x 2 = 0 3 / 4 x 1 - 1 / 2 x 2 = 0 There is one fact that we haven’t used: a condition on the sum of the probabilities . Since x 1
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