I was wondering if the Weak Law of Large Numbers also holds true for dependent identically distributed random variables with constant mean and bounded variance.
Let $X_1,X_2\dots X_n$ be a sequence of dependent but identically distributed random variables with $E(X_i)=\mu$ and $Var(X_i) \le B_i$. Then is it true that, $$n^{-1}\sum\limits_{i=1}^nX_i \to \mu, \text{ in probability }$$
If we were to use the Chebyshev's inequality for some $\epsilon >0 $, $$P \left( \left | {n^{-1}\sum\limits_{i=1}^nX_i -\mu } \right | \ge \epsilon \right) \le \frac{Var \left(n^{-1}\sum\limits_{i=1}^nX_i \right )}{\epsilon^2}= \frac{\sum\limits_{i,j}Cov(X_i,X_j)}{n^2\epsilon^2}$$
But by Cauchy Schwarz, $$Cov(X_i,X_j)\le \sqrt{Var(X_i)}\sqrt{Var(X_j)}\le \sqrt{B_iB_j}$$ and this ensures, $$0 \le P \left( \left | {n^{-1}\sum\limits_{i=1}^nX_i -\mu } \right | \ge \epsilon \right)\le \frac{\sqrt{B_iB_j}}{n^2\epsilon^2}$$ and hence by the Sandwich Theorem, we have convergence in probability.
Are these arguments valid? Is my conclusion on assuming constant mean and bounded variance are all that is needed to ensure WLLN to be true for dependent sequence of random variables?
Thanks