Let {$Xt;t = 0, 1, . . .$} be a Markov chain with state space $S_X=${$0, 1, 2$}, initial distribution $p$, and transition matrix $P$, where $P$ is
\begin{bmatrix}0&2/3&1/3\\1/4&0&3/4\\1/2&1/2&0\end{bmatrix}
Assume that the initial distribution of the chain is $\pi=\frac{1}{11}(3,4,4)$.
Let $Y_t=X_{10−t}$ for $t = 0, 1, . . . , 10.$ Then, {$Y_t;t = 0, 1, . . . , 10$} is also a Markov chain.
I am tasked to compute the transition matrix of {$Y_t;t = 0, 1, . . . , 10$}.
I understand that this is done using
$P(Y_1=j|Y_0=i)=\frac{P(Y_0=i|Y_1=j)P(Y_1=j)}{P(Y_0=i)}=\frac{P(X_{10}=i|X_9=j)P(X_9=j)}{P(X_{10}=i)}=\frac{P_{j,i}\pi_j}{\pi_i}$
which should yield the transition matrix
\begin{bmatrix}0&1/3&2/3\\1/2&0&1/2\\1/4&3/4&0\end{bmatrix}
However, I don't understand how to interpret $\frac{P_{j,i}\pi_j}{\pi_i}$ and how to translate this into the transition matrix. Does anyone know how to help? How do I find $\pi_j$?
First note that the initial distribution $\pi$ is the unique stationary distribution for $P$, that is, $\pi=\pi P$. In general, a Markov chain with stationary distribution $\pi$ is reversible if and only if $\pi_i P_{i,j} = \pi_j P_{j,i}$ for all states $i$,$j$ - which we can verify here by a straightforward computation.
The transition probabilities $\hat P_{i,j}=\frac{\pi_j}{\pi_i}P_{j,i}$ are that of the reversed chain.