Consider the two-state Marcov chain with transition matrix $$P= \ \begin{pmatrix} 1-p & p \\ 1 & 0 \end{pmatrix} $$
Find the limiting Distribution.
Answer:
Let $ \pi=(\pi_1,\pi_2) $ be the distribution vector.
Then , we have
$\pi P=\pi $ which gives us
$ \pi_1 (1-p)+ \pi_2=\pi_1, \\ \pi_1p+0=\pi_2, $
Solving we get
$ p=0 $
Thus,
$ \pi_2=0 , \ \pi_1=constant=c, say$
But since $ \pi_1+\pi_2=1 $ , we see $\pi_1=1, \ \pi_2=0$
Thus the limiting distribution is $ \ \pi=(1,0) $
Am I right?
Best Answer
$$\pi_2 = p\pi_1$$
$$\pi_1 + \pi_2=1$$
$$(p+1)\pi_1=1$$
$$\pi_1=\frac1{p+1}$$
$$\pi_2=\frac{p}{1+p}$$
Note that $p$ is not a variable to be solved, it is fixed and it is given to you.