Finding the stationary distribution for this Discrete Time Markov Chain (DTMC) I have the following discrete-time Markov chain defined on the state space $S:=\{0,1,2,\ldots\}$:
$$p(i,j) = \begin{cases}
1, \quad &\text{if $i=0$ and $j=1$}\\
\frac{1}{2}, \quad &\text{if $j=i-1$ and $i=1,2,\ldots$}\\
\frac{i-1}{2(i+1)}, \quad &\text{if $j=i$ and $i=1,2,\ldots$}\\
\frac{1}{i+1}, \quad &\text{if $j=i+1$ and $i=1,2,\ldots$}\\
0, \quad &\text{otherwise}
\end{cases}$$
In other words, 
$$P = \begin{bmatrix}
0&1\\
1/2 & 0 &1/2\\
 & 1/2& 1/(2*3)&1/3\\
&&1/2&2/(2*4)&1/4\\
&&&1/2&3/(2*5)&1/5\\
&&&\ddots&\ddots&\ddots
\end{bmatrix}$$
I'm asked to find the stationary distribution for this Markov chain. I know that solving $\pi=\pi P$ along with $\sum_{i=0}^\infty \pi_i=1$ for $\pi$ will get me the stationary distribution, but every time I try to compute this, I get stuck. Specifically, I have 


*

*$\pi_0 = 1/2*\pi_1$ 

*$\pi_n = \pi_{n-1}\left(\frac{1}{n}\right) + \pi_n\left(\frac{n-1}{2(n+1)}\right)+ \pi_{n+1}\left(\frac{1}{2}\right)\quad$ for $n \ge 1$
I just don't know where to go from here. Any help would be outstanding. Is there any alternative way to compute stationary distributions other than solving the system $\{\pi = \pi P, \quad \sum_{i=0}^\infty \pi_i = 1\}$?
 A: This looks like a modified Birth-Death Chain which should say "reversible" to you.  Reversible chains only require solving the detailed balance equations, which in general is a lot easier than solving the global balance equations.    
so you need to solve
$\pi_i P_{i,j} = \pi_j P_{j,i}$ 
for the top left corner (i.e. for states 0 and 1) you have
$\pi_0 1 = \pi_1 \frac{1}{2}$ 
now consider natural number $n \geq 2$, the detailed balance equations give
$\pi_{n-1}\frac{1}{n} =\pi_n\frac{1}{2}$
or
$\pi_{n-1}\frac{2}{n} =\pi_n$ 
at this point you can try applying this for small values of $n$ and form a guess:
$\pi_{1}\frac{1}{2}\cdot\frac{2^{n}}{n!} = \pi_{1}\frac{2^{n-1}}{n!} =\pi_n$
(note this technically actually nolds for the case of n=1 and n=0 as well)  
for n = 2 this reads
$\pi_{1}\frac{2}{2!}= \pi_1 =\pi_2$ which is equivalent to previously stated $\pi_{n-1}\frac{2}{n} =\pi_n$ when n = 2.  This is our base case.  
now for $n\geq 3$
$\pi_{1}\frac{2^{n-1}}{n!} = \big(\pi_{1}\frac{2^{n-2}}{(n-1)!}\big)\frac{2}{n} =\big(\pi_{n-1}\big)\frac{2}{n} =\pi_n$ 
where the middle inequality follows by induction hypothesis.  
The final thing is, for a positive recurrent chain with one communicating class, the $\pi_i$'s must all sum to one, so  
$1 =  \sum_{n=0}^\infty \pi_n  =  \sum_{n=0}^\infty \pi_1\frac{1}{2}\frac{2^{n}}{n!}= \pi_1\frac{1}{2}\big(\sum_{n=0}^\infty \frac{2^{n}}{n!}\big)= \pi_1 \frac{1}{2} e^2 $ 
so $\pi_1 = \frac{2}{e^2}$
and  $\pi_n =\frac{2}{e^2}\frac{2^{n-1}}{n!}=\frac{2^{n}}{n!e^2}$ 
A: This is a birth-death process and so has an invariant measure given by $\nu(1)=1$ and  $$\nu(n) = \prod_{j=0}^{n-1}\frac{p_j}{q_{j+1}},$$
where $p_j=\mathbb P(X_{n+1}=j+1\mid X_n=j)$ and $q_j = \mathbb P(X_{n+1}=j-1\mid X_n=j)$. (I leave it to the reader to check that this is an invariant measure.) So the process has a stationary distribution if and only if $\nu$ is summable, that is,
$$
\sum_{n=0}^\infty\prod_{j=0}^{n-1}\frac{p_j}{q_{j+1}}<\infty.
$$
If this sum is finite with value $C$, define $\pi = \frac1C\nu$. Then $\pi$ is a stationary distribution for the Markov chain.
