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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?

owen88
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1 Answers1

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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.

owen88
  • 4,670