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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?

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$$\pi_2 = p\pi_1$$

$$\pi_1 + \pi_2=1$$

$$(p+1)\pi_1=1$$

$$\pi_1=\frac1{p+1}$$

$$\pi_2=\frac{p}{1+p}$$

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

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