In a Markov chain, if the limiting distribution is identical to the initial Distribution, then what does this say about the distribution of Xn? (the n is a subscript).

Thanks!

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- Feb 16th 2011, 08:05 PMicobesLimiting distribution = Initial Distribution MC
In a Markov chain, if the limiting distribution is identical to the initial Distribution, then what does this say about the distribution of Xn? (the n is a subscript).

Thanks! - Feb 16th 2011, 08:22 PMCaptainBlack
- Feb 16th 2011, 08:28 PMicobes
Thanks! Could you please explain why this is the case.

- Feb 16th 2011, 10:17 PMCaptainBlack
Because the limiting distribution is by necessity a stationary state. After a large number of steps the siatribution is $\displaystyle A^n x \approx y$, where $\displaystyle $$ A$ is the transition matrix, $\displaystyle $$n$ the number of steps, $\displaystyle $$x$ the initial distribution and $\displaystyle $$y$ the limiting distribution.

So:

$\displaystyle Ay \approx A(A^n x) = A^{n+1} x \approx y$

(In fact because of the limits involved the left most term is identical to the right most term and not just approximatly equal, this could be made explicit by putting the appropriate limits in the above)

CB