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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 $P$ is doubly stochastic, show that $P^n$ is doubly stochastic $\forall n\in\Bbb{N}$

I was thinking to prove this with induction but I got stuck in the base case showing that every column's sum equals one.

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    You can show that every row's sum equals one? Then you can use the fact that $P^T$ is also doubly stochastic. – 2017-02-28
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    @JyChen Thanks, did I use your suggestion correctly? – 2017-03-02

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Base Case: $n=2$

I want to show that the sum of every column of $P^2$ equals $1$

let $J$ be the number of rows of P

Expanding the elements of the product of $P^2$

$\sum_{j\in J}P^2=\sum_{j\in J}(\sum_{s\in J}P_{js}P_{sk})=\sum_{s\in J}P_{sk}\sum_{j\in J}P_{js}$ $\ \ \ \ ,(k\in J$)

Here we use the fact that P is double stochastic, therefore:

$\sum_{s\in J}P_{sk}\sum_{j\in J}P_{js}=\sum_{s\in J}P_{sk}=1$

We suppose it is true for m = n:

$\sum_{j\in J}P^{m+1}=\sum_{j\in J}(\sum_{s\in J}P_{js}P_{sk}^m)=\sum_{s\in J}P_{sk}^m\sum_{j\in J}P_{js}$ $\ \ \ \ ,(k\in J$)

Here we use our induction hypothesis:

$\sum_{s\in J}P_{sk}^m\sum_{j\in J}P_{js}=\sum_{s\in J}P_{sk}^m=1$

$q.e.d.$

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    Your proof is right, but it is a direct way to prove the assertion. In fact, my suggestion is that $(P^n)^T=(P^T)^n$, then the sum of each column of $(P^n)^T$ is equal to the sum of each row of $(P^T)^n$. Since $P^T$ is also a stochastic matrix, we can know the sum of each row of $(P^T)^n$ is 1. – 2017-03-03