# Limiting distribution = Initial Distribution MC

• Feb 16th 2011, 09:05 PM
icobes
Limiting 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, 09:22 PM
CaptainBlack
Quote:

Originally Posted by icobes
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!

It is constant.

CB
• Feb 16th 2011, 09:28 PM
icobes
Thanks! Could you please explain why this is the case.
• Feb 16th 2011, 11:17 PM
CaptainBlack
Quote:

Originally Posted by icobes
Thanks! Could you please explain why this is the case.

Because the limiting distribution is by necessity a stationary state. After a large number of steps the siatribution is $A^n x \approx y$, where $A$ is the transition matrix, $n$ the number of steps, $x$ the initial distribution and $y$ the limiting distribution.

So:

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

(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