This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Handbook 6.2: SteadyStates • Suppose we had X = 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 . 25 0 . 75 bracketrightbig X 2 = bracketleftbig . 4375 0 . 5625 bracketrightbig X 3 ≈ bracketleftbig . 3906 0 . 6094 bracketrightbig X 4 ≈ bracketleftbig . 4023 0 . 5977 bracketrightbig X 5 ≈ bracketleftbig . 3994 0 . 6006 bracketrightbig X 6 ≈ bracketleftbig . 4001 0 . 5999 bracketrightbig • It seems that as time goes on, the distribution vector gets closer and closer to becoming bracketleftbig . 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 steadystate . We will now find a way to show that the steady state distribution vector for T is indeed bracketleftbig . 4 0 . 6 bracketrightbig . • If the distribution approaches a steadystate, 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 righthand 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 • Moving all the variables to one side of the equation, we have:...
View
Full
Document
This note was uploaded on 02/05/2011 for the course MATH 377 taught by Professor Stephenlang during the Spring '11 term at University of Victoria.
 Spring '11
 stephenlang
 Calculus, Vectors

Click to edit the document details