How do we show that where $(X_i)_{i\geq1}$ is a stochastic process with finite state space $X$, that if one of the following limits exist, then also other exist and all three coincide?
$$\lim_{n \to \infty}\frac{H(X_1,...,X_n)}{n}$$ $$\lim_{n \to \infty}\frac{H(X_{k+1},...,X_{k+n})}{n}$$ $$\lim_{n \to \infty}\frac{H(X_{k+1},...,X_{k+n} | X_1,...,X_k)}{n}$$
I understand how we are able to prove that for a stationary process $H(X) = H'(X)$, where $H'(X) = \sum^{n}_{i = 1}\frac{H(X_i|X_{i-1},...,X_1)}{n}$ , but am not sure how to approach this problem.
The answer is for stochastic process $\left\lbrace X_i \right\rbrace$ which is stationary.
\begin{eqnarray} H_1 &= &\lim_{n \to \infty}\frac{H(X_1,...,X_n)}{n} \\ H_2 &= &\lim_{n \to \infty}\frac{H(X_{k+1},...,X_{k+n})}{n} \\ H_3 &= &\lim_{n \to \infty}\frac{H(X_{k+1},...,X_{k+n} | X_1,...,X_k)}{n} \end{eqnarray}
The core of proof is $H_0 = \lim_{n \to \infty} H(X_n|X_{n-1}, \cdots, X_1)$.
1) $H_0$ does exist
$$0 < H(X_{n+1} | X_n, \cdots, X_1) \leq H(X_{n+1} | X_n, \cdots, X_2) = H(X_n | X_{n-1}, \cdots, X_1)$$ where the first inequality is from the fact that entropy is non negative, the second one is because adding $X_1$ reduces uncertainty, and the equality is from stationary property.
$H(X_n|X_{n-1}, \cdots, X_1)$ is a non negative decreasing function wrt $n$, thus $H_0$ exists.
2) $H_1= H_0$
By the chain rule, $$\frac{1}{n} H(X_1,...,X_n) = \frac{1}{n}\sum^{n}_{i = 1}H(X_i|X_{i-1},...,X_1)$$
Use Cesàro mean with proven $H_0$ is the limit of $H(X_n|X_{n-1},...,X_1)$, we have $H_1 = H_0$. (I can add the proof of Cesàro mean if you want).
2) $H_2= H_1$
By definition of stationary process, $H(X_{k+1},...,X_{k+n}) = H(X_1,...,X_n)$.
3) $H_3= H_0$
$$\frac{H(X_{k+1},...,X_{k+n} | X_1,...,X_k)}{n} = \frac{1}{n}\sum^{k+n}_{i = k+1}H(X_i|X_{i-1},...,X_1)$$
Always Cesàro mean, when $n \to \infty$, the limit of LHS is $H_3$, and the limit of RHS is $H_0$.
Conclusion,
all roads lead to $H_0$.