Consider the markov chain with transition matrix
P=
[0 1]
[1 0]
Show that P^n does not tend to a limit, but that
A_n = (I + P + P^2 +...+ P^n)/(n+1)
does tend to a limit.
thank you!
$\displaystyle P^{2n+1}$ is the identity matrix.
While $\displaystyle P^{2n}$ is the orginal matrix.
So, there is no limit.
You're bouncing back and forth between these two states with probability one.
I+P is the matrix of all ones, and so is $\displaystyle P^{2}+P^3$ and so on .....
Thus you will have a limit of .5 in all four positions of $\displaystyle A_n$
You can show that A_n has a limit by showing that all the entries of A_n have a limit.
The elements of the diagonal of A_n make this sequence:
(1/1, 1/2, 2/3, 2/4, 3/5, 3/6, ...)
As you can see, for n odd, they are always 0.5, for n even they are (1+ n/2) / (n+1).
lim (1 + n/2) / (n+1) = lim 1/(n+1) + lim n / (2*(n+1) = 0 + lim n/2n = 0.5
The elements outside of the diagonal are also 0.5 for n odd (above 0) and for n even they are (n/2) / (n+1), whose limit is 0.5 as shown above.
Therefore lim A_n =
[0.5 0.5]
[0.5 0.5]