Stationary distribution of MC

A markov chain on states 0,1,... has transition probabilities $P_{ij}=\frac{1}{i+2}$ for $j=0,1,...,i,i+1$. Find the stationary distribution.

My attempt:

First I calculate the equations

$\Pi_0$=$\Pi_0/2+\Pi_1/3+\Pi_2/4+\Pi_3/5+...$ and so on

But then I don't know how to continue, I mean how to solve for $\Pi_k$.

Well, instead of saying "and so on" write them down. You will see that they are all very similar to one another. For instance we have $$\Pi_0 = \frac{\Pi_0}{2}+\frac{\Pi_1}{3}+ \frac{\Pi_2}{4}+\ldots\\\Pi_1 = \frac{\Pi_0}{2}+\frac{\Pi_1}{3}+ \frac{\Pi_2}{4}+\ldots\\\Pi_2 = \frac{\Pi_1}{3}+\frac{\Pi_2}{4}+ \frac{\Pi_3}{5}+\ldots$$

So we have $\Pi_0 = \Pi_1$ and $\Pi_1 = \frac{\Pi_0}{2}+\Pi_2.$ So immediately we have $\Pi_2= \Pi_0/2.$ Now we can write $\Pi_2 = \Pi_1/3+\Pi_3$ so that $\Pi_3 = \Pi_0/6.$ Then $\Pi_3 = \Pi_2/4 + \Pi_4$ so that $\Pi_4 = \Pi_0/24.$ See the pattern?

To see the factorial come out more clearly, try adding the equations together. If you add the equations from $\Pi_{i+1}$ on you get $$\Pi_{i+1}+\Pi_{i+2}+\ldots = \frac{\Pi_i}{i+1} + 2\frac{\Pi_{i+1}}{i+2} +3\frac{\Pi_{i+2}}{i+3}+\ldots$$ which rearranges to $$\frac{\Pi_{i}}{i+1} = i\left(\frac{\Pi_{i+1}}{i+2} +\frac{\Pi_{i+2}}{i+3}+\ldots\right)$$ and can be further rewritten to $$\Pi_i = i\left(\frac{\Pi_i}{i+1}+\frac{\Pi_{i+1}}{i+2} +\frac{\Pi_{i+2}}{i+3}+\ldots\right)$$ But we can plug in $i-1$ for $i$ in the first equation to give $$\frac{\Pi_{i-1}}{i} = (i-1)\left(\frac{\Pi_i}{i+1}+\frac{\Pi_{i+1}}{i+2} +\frac{\Pi_{i+2}}{i+3}+\ldots\right) = (i-1)\frac{\Pi_i}{i}$$ where we plugged in the second equation. This simplifies to $$\Pi_i = \frac{\Pi_{i-1}}{i-1}.$$

(There might be an easier way to get this 1st order recursion equation that I'm missing...)

• but the answer from the book is $\Pi_i = \frac{e^{-1}}{i!} ,i \ge 0$ or is it wrong? Commented May 15, 2017 at 2:55
• @AaronMartinez That's exactly what I get for the answer. I just haven't done all the work for you. Commented May 15, 2017 at 2:57
• and why not? why didn't you do the whole work for me? hahaha just kidding man.. Commented May 15, 2017 at 3:03
• @AaronMartinez Sorry being grumpy. I've shown you how to get a recursion equation for the $\Pi$'s. If you solve it and normalize it so that $\sum_i \Pi_i = 1$ that will give the book's answer. Commented May 15, 2017 at 3:11
• yep thanks. Also if you don't mind, can you take a look to my unanswered question here? math.stackexchange.com/questions/2281275/… Commented May 15, 2017 at 3:17