example of an inhomogenenous markov chain I have found a very interesting exercice about inhomogeneous markov chains, which I think you are also interested in and hope someone can solve it.
Let $X_{1},X_{2},...$ be a inhomogeneous markov chain with state space $\{0,1\}$ with transition probabilities
$$
\mathbb{P}(X_{n+1} = 1 | X_{n} = 0) = \frac{1}{n^{2}}\\
\mathbb{P}(X_{n+1} = 0 | X_{n} = 1) = \frac{1}{n^{2}}.
$$


*

*$X := \lim_\limits{n \to \infty} X_{n}$ almost surely exists.

*If $\mathbb{P}(X_{1} = 1) = \mathbb{P}(X_{1} = 0) = \frac{1}{2}$, then $\mathbb{P}(X = 1) = \frac{1}{2}$.

*If $X_{1} \equiv 1$ then $\mathbb{P}(X = 1) > \frac{1}{2}$.


Im very interested how you would solve it. The solution was discussed in the lecture but I hope that I see a different approach. Maybe with matrix calculation

So the markov matrices is,
$$
P_n = \begin{pmatrix}
1-\frac{1}{n^2} & \frac{1}{n^2} \\
\frac{1}{n^2} & 1-\frac{1}{n^2}
\end{pmatrix} = 
\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}+\begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}\frac{1}{n^2}
$$ Hence
$$
P_n = I + A\frac{1}{n^2}
$$ Now this means that all matrices $P_n$ can be simultaniously diagonalize. Note that for $v_1 = (1,1)^t$ $Av_1 = 0$ and for $v_2 = (1,-1)^t$ we have
$$
Av_2 = -2 v_2
$$ Hence to find the final probability matrix we need to calculate
$$
\prod_{i=1}^nP_i \to c = \prod_{i=1}^n\left( 1 - \frac{2}{n^2} \right )
$$ Note https://www.wolframalpha.com/input/?i=prod+(1-2%2Fi%5E2)
Then we get probability matrix as
$$
 \prod_{i=1}^nP_i = \frac{1}{2}\begin{pmatrix}1&1\\1&-1\end{pmatrix}\begin{pmatrix}1&0\\0&c\end{pmatrix}\begin{pmatrix}1&1\\1&-1\end{pmatrix} = \begin{pmatrix}\frac{1+c}{2}&\frac{1-c}{2}\\\frac{1-c}{2}&\frac{1+c}{2}\end{pmatrix},
$$ with $c=-0.22$. Now the convergence of the product proves the limit and the evaluation of the matric putting numbers of c in it lead to the answers of 2,3, 3 looks wrong, note that I included the case $n=1$ this case just swaps the states and from this we note that 3 is proven.
