# Thread: 2nd Markov chain question

1. ## 2nd Markov chain question

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!

2. $P^{2n+1}$ is the identity matrix.

While $P^{2n}$ 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 $P^{2}+P^3$ and so on .....

Thus you will have a limit of .5 in all four positions of $A_n$

3. 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]