A transition probability matrix $\mathbf{P}$ is said to be doubly stochastic if the sum over each column equals one

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A transition probability matrix is defined as doubly stochastic when the sum of each column equals one. For an irreducible and aperiodic chain with M+1 states, the limiting probabilities can be shown to be equal, specifically \(\pi_j = \frac{1}{M+1}\) for each state \(j\). This indicates that in the long run, each state is equally likely to be visited. The problem remains unsolved by participants in the discussion, highlighting a lack of engagement. The original poster has provided their solution for reference.
Chris L T521
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Here's this week's problem.

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Problem: A transition probability matrix $\mathbf{P}$ is said to be doubly stochastic if the sum over each column equals one; that is,\[\sum_i P_{i,j}=1,\qquad\forall j.\]
If such a chain is irreducible and aperiodic and consists of $M+1$ states $0,1,\ldots,M$, show that the limiting probabilities are given by
\[\pi_j=\frac{1}{M+1},\quad j=0,1,\ldots,M.\]

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No one answered this week's question. You can find my solution below.

To show that this is true, we show that $\pi_j=\frac{1}{M+1}$ satisfies the system of equations $\pi_j=\sum\limits_{i=0}^M\pi_iP_{ij}$ and $\sum\limits_{j=0}^M\pi_j=1$. Supposing that $\pi_j=\frac{1}{M+1}$, we see that\[\sum\limits_{j=0}^M\pi_j=\frac{1}{M+1}\sum\limits_{j=0}^M1=\frac{1}{M+1}(M+1)=1\]
and
\[\pi_j=\sum\limits_{i=0}^M\pi_iP_{ij}\implies \sum\limits_{j=0}^M\sum\limits_{i=0}^M\pi_jP_{ij}=(M+1)\pi=1.\]
Thus, $\pi_j$ must be $\frac{1}{M+1}$ for these equations to be satisfied.
 

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