Consider the two-state Marcov chain with transition matrix $$P= \ \begin{pmatrix} 1-p & p \\ 1 & 0 \end{pmatrix} $$

Find the limiting Distribution.


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 $


$ \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?


$$\pi_2 = p\pi_1$$

$$\pi_1 + \pi_2=1$$




Note that $p$ is not a variable to be solved, it is fixed and it is given to you.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.