Consider the transition matrix
$ P = \begin{bmatrix} 1-p&p\\ q&1-q \end{bmatrix} $
for general $2$-state Markov Chain $(0 \le p, q\le 1)$.
(a) Find the limiting distribution (if it exists) if $p + q = 1$.
(b) Find the limiting distribution (if it exists) if $p + q \ne 1$.
I myself calculated (a) with the answer $(\frac{q}{p+q}, \frac{p}{p+q}) = (q, p )$.
But, I couldn't understand the problem (b).
If $p+q \ne 1$, then the answer should be $(\frac{q}{p+q}, \frac{p}{p+q})$. Isn't that so?
What else could have been derived?
Since:
$$\begin{bmatrix} q&p \end{bmatrix} \begin{bmatrix} 1-p&p\\ q&1-q \end{bmatrix} =\begin{bmatrix} q - pq + pq & qp + p - pq \end{bmatrix} = \begin{bmatrix} q&p \end{bmatrix} $$
We have $v = [q\ \ p]$ is a right eigenvector with eigenvalue $1$, and the limiting distribution is a rescaled version of $v$ whose entries sum to $1$, i.e. $\pi = {v \over \sum_i v_i} = [{q \over p+q}\ \ {p \over p+q}]$.
None of the above depends on the value of the sum $p+q$; i.e. it is valid whether $p+q = 1$ or $= \sqrt{2}$ or $= 10^{-4}$ or any other value.
Nice trick question though!
Ooh, as @Phicar pointed out, the case of $p=q=1 \implies P^n$ does not converge. If you define the "limiting distribution" as $\lim P^n \pi_0$ then it does not exist. But if you define the "limiting distribution" as the unique probabilistic vector s.t. $\pi P = \pi$ then it does exist (and $\pi = [{q \over p+q}\ \ {p \over p+q}]$ as always). Not sure the terminology in this corner case...
Come to think of it, the $p=q=0$ case is even more problematic as any $\pi P = \pi$ (since $P=$ identity matrix).