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

I have no idea how to prove it but

  • If a chain is irreducible then all states communicate i.e $$P_{ij}>0\space \text{and}\space P_{ji}>0\space\forall i,j$$

  • If $d$ denotes the period of any state, if a chain is irreducible, then $d(i)=1\forall i$

If $P_{(M+1)\times (M+1)}$ matrix and $\pi$ is the stationary distribution $$\pi_j=\sum_iP_{ij}\pi_i\space j=0,1,\dots,M+1$$

but how I can get this expression?



We first must note that $\pi_j$ is the unique solution to $\pi_j=\sum \limits_{i=0} \pi_i P_{ij}$ and $\sum \limits_{i=0}\pi_i=1$.

Let's use $\pi_i=1$. From the double stochastic nature of the matrix, we have $$\pi_j=\sum_{i=0}^M \pi_iP_{ij}=\sum_{i=0}^M P_{ij}=1$$ Hence, $\pi_i=1$ is a valid solution to the first set of equations, and to make it a solution to the second we must normalize it by dividing by $M+1$.

Then by uniqueness as mentioned above, $\pi_j=\dfrac{1}{M+1}$. $$ \blacksquare$$

Note : To understand this proof, one must recall the definition of a stationary distribution.

A vector $\mathbf{\pi}$ is called a stationary distribution vector of a Markov process if the elements of $\mathbf{\pi}$ satisfy: $$ \mathbf{\pi} = \mathbf{\pi} \cdot \mathbf{P}, \sum_{i \in S} \pi_{i} = 1 \text{ , and } \pi_{i} > 0\text{ }\forall \text{ } i \in S $$ Note that a stationary distribution may not exist, and may not be unique.

  • $\begingroup$ what they mean by "limiting probabilities" $\endgroup$ – Roland Jul 14 '15 at 23:41
  • $\begingroup$ @JohnSheridan A limiting probability is essentially the value of a stationary distribution for a given state i.e. $\pi_{i}$. Do you know what a stationary distribution is? $\endgroup$ – miradulo Jul 14 '15 at 23:44
  • $\begingroup$ I'm not entirely sure, but it's kind of convergence, like $$p^n(i,j)\rightarrow \pi$$ when $n\rightarrow\infty$ That is it? $\endgroup$ – Roland Jul 14 '15 at 23:47
  • 3
    $\begingroup$ @JohnSheridan Added what I believe to be a good definition to my answer for you. $\endgroup$ – miradulo Jul 14 '15 at 23:57

We first must note that $\pi_j$ is the unique solution to $\pi_j=\sum \limits_{i=0}^M \pi_i P_{ij}$ and $\sum \limits_{i=0}\pi_i=1$ ...$(\star)$

Again, since doubly stochastic, $\sum \limits_{i=0}^M P_{ij}=1$.

From, $\pi_j=\sum \limits_{i=0}^M \pi_i P_{ij}$ and $\sum \limits_{i=0}^M P_{ij}=1$, we can get, $1-\pi_j=\sum \limits_{i=0}^M (1-\pi_i) P_{ij}$ or, $\dfrac{1-\pi_j}{M}=\sum \limits_{i=0}^M \dfrac{(1-\pi_i)}{M} P_{ij}$. This implies $\Big( \dfrac{1-\pi_0^*}{M},\dfrac{1-\pi_1^*}{M},\dots,\dfrac{1-\pi_M^*}{M}\Big)$ is also a solution if $(\pi_0^*,\pi_1^*,\dots, \pi_M^*)$ is a solution of $(\star)$. Because of uniqueness, we get $$\dfrac{1-\pi_j}{M}=\pi_j\Rightarrow \pi_j=\dfrac{1}{M+1},\forall j$$


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