Proving Limiting Distributions Markov Chains

Question

Ive got a question:

$P=\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}$

$\pi=(\frac12,\frac12)$

Is $\pi$ a limiting distribution of the Markov chain?

I know that P isnt a limiting distribution but im having diffuculty writing proof for it.

Any help will be appreciated.

Answer 1

The reason why there isn't a limiting distribution for your matrix, is because the eigenvalues are $1$ and $-1$. That means that there is no single dominant eigenvalue and that implies (without further elaboration) that there is no proper equilibrium. In other words, the secrets of a Markov process are to be found in the eigenvalues and eigenvectors of the matrix. So for your proofs, that is the chapter you should go over...

Answer 2

If $ \pi $ was a limiting distribution of this Markov Chain, it would follow that

for any initial distibution $ \pi (0):\ \lim_{n\to \infty} \pi(0)P^n=\pi$.

Now try $ \pi(0)= (0.1\ \ 0.9) $ and it will lead to contradiction.