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!
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