# How can I prove the asymptotic equipartition property (AEP) for an identically distributed markov chain?

 Hi, everyone. I am recently reading the lecture note of EE376a : Information Theory course from Stanford University. This note introduces that we can prove the Asymptotic Equipartition Property (or the Shannon-McMillan-Breiman Theorem) for the case that the given stochastic process $$\left\{ X_n \right\}$$ is a time-invariant discrete-time Markov chain with a finite state space $$\mathcal{X}$$ such that every $$X_n$$ is identically distributed over $$\mathcal{X}$$ with the distributuion $$p$$. The following is the statement. :

$$-\frac{1}{n} \log p(X_1, \cdots, X_n) \rightarrow H(X_2|X_1)$$

in probability as $$n \rightarrow \infty$$. The note introduces the proof of this statement by using Weak Law of Large Numbers for weak dependency. :

Let $$\{ Y_n \}$$ be a sequence of identically distributed random variables such that $$\lim_{n \rightarrow \infty} \frac{1}{n^2} \sum_{i=1}^{n} \sum_{j=1}^{n} Cov[Y_i, Y_j] = 0.$$ Then, $$\frac{1}{n} \sum_{i=1}^{n} Y_i \rightarrow \mathbb{E}[Y_1]$$ in probability as $$n \rightarrow \infty$$.

Thus, we let $$Y_k := \log p(X_k |X_{k-1})$$ for $$k \geq 2$$ and $$Y_1 := \log p(X_1)$$. Therefore, it suffices to verify that $$\lim_{n \rightarrow \infty} \frac{1}{n^2} \sum_{i=1}^{n} \sum_{j=1}^{n} Cov[Y_i, Y_j] = 0 \ \cdots (\star)$$ and the statement comes from the WLLN for weak dependency. Now, here is my question. How can I prove the part $$(\star)$$? Please, share some good ideas for this problem.