I have a fairly simple discrete Markov Chain with a transition probability matrix that looks like this:
\begin{bmatrix}0&{n_1}&0&0&0&...\\ {n_2}&0&{n_3}&0&0&...\\ 0&{n_4}&0&{n_5}&0&...\\ ...&...&...&...&...&...\\ ...&0&0&0&n_k&0\end{bmatrix}
Basically, just a long line of states, all of which can go into a neighboring one (First one into second, second one into the first and the third and so on). I know all of transitional probabilities - for simplicity's sake, here I have replaced them with just ${n_i}$.
The problem here, is that I have N states - their amount may vary. I need to find a stationary distribution for this chain, and while making a system of N equations is trivial, I'm not sure how exactly I can solve it.
Is there, perhaps, an easier way to find a stationary distribution of such a chain? Or I should just bruteforce it with an equation system?
Hint
Assuming your transition matrix is row-stochastic, you must have $\ n_1=n_k=1\ $ and $\ n_{2i+1}=1-n_{2i}\ .$ If $\ \pi\ $ is the stationary distribution, then the first and last equations of $$ \pi^T\begin{bmatrix}0&{n_1}&0&0&0&...\\ {n_2}&0&{n_3}&0&0&...\\ 0&{n_4}&0&{n_5}&0&...\\ ...&...&...&...&...&...\\ ...&0&0&0&n_k&0\end{bmatrix}=\pi^T $$ give you $\ \pi_2=\pi_ 1\ $ and $\ \pi_{k-1}=\pi_k\ .$ The second of the equations gives \begin{align} n_2\pi_1+n_3\pi_3&=\pi_2\\ &=n_2\pi_2+(1-n_2)\pi_3\ , \end{align} from which you get $\ \pi_3=\pi_2\ .$ Does that suggest a guess for the remaining entries of $\ \pi\ ?$