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I have a lot of difficulty with proofs; could someone help me with this question that really can not solve? I would also like some indication of material to get through with this kind of question and some hint of material about Markov chain. Thanks in advance.

"A stochastic matrix is called doubly stochastic if its columns sum to 1. Let $X_0 , X_1, \dots$ be a Markov chain on $\{1,\dots, k\}$ with a doubly stochastic transition matrix and initial distribution that is uniform on $\{1, \dots, k\}.$ Show that the distribution of $X_n$ is uniform on $\{1.\dots, k\},$ for all $n \ge 0."$

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$X_1=X_0P$, where $P$ is the transition matrix. As $X_0=[1/k,\ldots,1/k]$, one would have $X_1^i=\frac{1}{k}(P_{2i}+P_{1i}+\ldots+P_{ki})=\frac{1}{k}$ by the double stochastic property (sum of the entries on colums are $1$). The general result follows by induction.

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If a Markov chain with state space $\{1, 2, \dots, k\}$ is ergodic then its stationary distribution is its limiting distribution. As I read your question, it seems you are trying to show that a chain with a doubly stochastic matrix has a uniform stationary distribution on the state space.

You are using the convention that element $p_{ij}$ for $1 \le i,j \le k,$ of the transition matrix $\mathbf{P}$ has $p_{ij} = P(X_n = j | X_{n-1} = i).$ Let vector $\sigma = (\sigma_1, \sigma_2, \dots, \sigma_k).$ If $\sigma$ is uniform, then $\sigma_i = 1/k,$ for $i = 1, \dots, k.$ $$\sigma_i\mathbf{P} = \sum_{j=1}^k \frac 1 k p_{ij} = \frac 1 k\sum_{j=1}p_{ij} = \frac 1 k(1) = \frac 1 k,$$

where the last equality is due to the doubly stochastic property of the transition matrix. The argument for all $i = 1, \dots k$ is the same so $\sigma\mathsf{P} = \sigma,$ and $\sigma$ is a stationalry distribution.

If your question is not covered by this answer and the Answer of @Momo (+1), please explain what part you don't understand, and maybe someone here can help.

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