# Finding stationary distribution on markov chain

We have the state space {0,1,2,3,4,5,...,}

we got the following transition probabilities $$p_{00}=q+r$$

$$p_{01}=p$$

$$p_{i,i-1}=q$$

$$p_{i,i}=r$$

$$p_{i,i+1}=p$$

for $$i>=1$$

where we have $$p+q+r=1$$

FInd the stationary distribuion where $$p=.2,q=.4,r=.4$$ Is the stationary distribution unique

For this one I am kind of stuck I tried using flow balance equations and got in=out and the following

$$t_0=(.2t_1+.8t_0)$$ and I get $$t_0=t_1$$ then I did

$$t_1=.2t_0+.4t_1+.4t_2$$ and using $$t_0=t_1$$ I get $$t_2=t_1$$ and $$t_2=t_0$$

then for state 2 I did

$$t_2=.2t_1+.4*t_2+.4*t_3$$ and I get $$t_2=t_3$$ and so $$t_3=t_0$$

So then I keep getting $$t_i=t_0$$ but I do not think this is right because

$$\sum t_i=\sum t_0$$ is supposed to add up to 1.

I am kind of stuck.

• I think my first flow balance equation mgiht be wrong it $.4t_1+.8t_0=t_0$ I think Commented Mar 8 at 17:35

Your first balance should be $$t_0 = 0.4t_1 +0.4t_0$$ so we get $$t_0 = \frac{2}{3}t_1$$, your other flow balances are correct.