Consider a discrete Markov source, {\mathscr{X} \ = \ X_i\}_{i=1}^{\infty}} ona finite alphabet set. Let the initial distribution be {Q} and the transition probability for the {n^{th}} step be {P_n}. When can we say that {\mathscr{X}} is stationary?
Clearly, the source has to be time invariant and thefore we need {P_n = P, \forall n}. For {\mathscr{X}} to be stationary, we need
\displaystyle f(X_1) = f(X_2) = \ \ldots \ f(X_n)
etc, where {f()} is the distribution. But {f(X_n) = QP^n}. Thus {Q = QP} guarantees that all {X_n}‘s have the same distribution. Now, consider, say, {f(X_1;X_2;X_3)} and {f(X_2;X_3;X_4)}.
\displaystyle f(X_1;X_2;X_3) = f(X_1)f(X_2/X_1)f(X_3/X_2).
\displaystyle f(X_2;X_3;X_4) = f(X_2)f(X_3/X_2)f(X_4/X_3).
It is clear that for the two joint distributions to be equal, {f(X_1) = f(X_2)} is enough and therefore {Q = QP} is sufficient.

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