Are death-birth stochastic process double stochastic? I'm having an issue to justify a statement.
We have seen in class that a double-stochastic discrete process is a process for which the sum of each column and the sum of each line in the transition matrix is equal to 1.
It has the nice property of giving us directly the stationary distribution (limit probabilities) as $\pi_j = \frac{1}{k+1}$ for $j=0,1,...,k$
Since in a death-birth process, we have the symmetry of source/target states, can i assert that it is double-stochastic?
Thank you in advance!
 A: Suppose the state space is $\mathbb N\cup\{0\}$ with transition probabilities
\begin{align}
\mathbb P(X_{n+1}=i+1\mid X_n = i) =: p_i,\quad i=0,1,\ldots\\
\mathbb P(X_{n+1}=i\mid X_n = i) =: r_i,\quad i=0,1,\ldots\\
\mathbb P(X_{n+1}=i-1\mid X_n = i) =: q_i,\quad i=1,2,\ldots\\
\end{align}
For this to give a valid transition matrix $P$, we must have $r_o+p_0=1$ and $q_i+r_i+p_i=1$ for $i\geqslant 1$. If $\nu(0)=1$ and $\nu(n) = \prod_{i=1}^n\frac{p_{i-1}}{q_i}$ then $\nu$ is an invariant measure for $P$ iff $\sum_{n=0}^\infty \nu(n):=C<\infty$. In this case, there is a unique stationary distribution $\pi$ given by $\pi = \frac1C\nu$.
For $P$ to be a doubly stochastic matrix, we must have $r_o+q_1=1$ and $p_{i-1}+r_i+q_{i+1}=1$ for $i\geqslant 1$. Now, to have $\sum_{n=0}^\infty \nu(n)<\infty$, we must have
$$
\liminf_{n\to\infty} \frac{p_n}{q_{n+1}}<1
$$
(consider the case where the $p_n$ and $q_n$ are constant and so the sum is a geometric series). However, from $r_o+p_0=1$ and $r_0+q_1=1$ we get $p_0=q_1$, and from $q_i+r_i+p_i=1$ and $p_{i-1}+r_i+q_{i+1}=1$ we get $p_{i-1}+q_{i+1} = p_i+q_i$. Now, by induction we see that $p_{i-1}=q_i$, and hence $p_i=q_{i+1}$, for all $i\geqslant 1$. This implies that $p_{i-1}=q_i=\frac12$ for all $i$, and so if $P$ is doubly stochastic, it necessarily cannot have a stationary distribution (for it is null recurrent).
