Markov Chain with Normal Transition Matrix Consider a (sub)-stochastic matrix $P$, and the associated Markov chain $X$ with
\begin{align*}
\mathbf P [X_n =y | X_0 = x] = P_{xy}^n.
\end{align*}
Suppose we have the condition $P^T P = P P^T$, i.e. the transition matrix is normal. Is there a probabilistic interpretation of this?
 A: If we assume that $P$ is normal, stochastic and irreducible (in the sense of it being the transition matrix of a Markov chain), then we can show that $P^T$ must also be stochastic (proof  below).
As a consequence we have the following properties:


*

*$X$ is stationary with respect to the uniform distribution, see eg. here

*$P^T$ is also a transition matrix of a Markov chain, and can be seen as the time reversed process $X^{-1}$ with
$$P\left(X^{-1}_{n+1} = y | X^{-1}_{n} = x \right) = p(y,x) = 
 P\left(X_{n+1} = x | X_{n} = y\right) $$
The process $X^{-1}$ is also stationary with respect to the uniform distribution.

*The matrix $P^TP$ is also a transition matrix of a Markov chain, and describes a markov chain $Y$ in which a step of $Y$ is a step of $X$ followed by a step of $X^{-1}$. Normality of $P^T P$ means that $Y$ has a symmetric transition matrix. 



Proof that normality implies $P$ is doubly stochastic
Note that stochasticity of $P$ implies
$$P \textbf{1} = \textbf{1},$$
where $\textbf{1}$ denotes the vector with all entries equal to $1$. Then using the normality condition
$$P^T \textbf{1} = P^T (P\textbf{1}) = P^T P\textbf{1} = P P^T \textbf{1},$$
that is
$$P (P^T\textbf{1}) = P^T \textbf{1},$$
so that $P^T\textbf{1}$ is an eigenvector for $P$ with eigenvalue $1$. However, the Perron-Frobenius theorem asserts that $1$ is a simple eigenvalue, and so the eigenvector $1$ is unique up to scaling. That is
$$P^T \textbf{1} = c \textbf{1}$$
for some constant $c >0$. But then note that
\begin{align*}
cn & = \textbf{1}^T(P^T \textbf{1}) \\
& = \left(\textbf{1}^T(P^T \textbf{1}) \right)^T \\
& = (P^T \textbf{1})^T \textbf{1} \\
& = \textbf{1}^T (P\textbf{1}) \\
& = \textbf{1}^T \textbf{1} \\
& = n,
\end{align*}
so that is $c=1$, and $P^T$ is stochastic.
