# Strong Law of Large Numbers for not identically distributed Bernoulli variables

Let $$X_{n,j}$$ be independent discrete random variables taking only two values. In particular, $$X_{n,j}=-\mu_{n,j}$$ with probability $$1-\mu_{n,j}$$ and $$X_{n,j}=1-\mu_{n,j}$$ with probability $$\mu_{n,j}$$, where $$0<\mu_{n,j}<1$$. Notice that they all have zero mean, and variance $$\mu_{n,j}(1-\mu_{n,j})$$.

Moreover, suppose that $$\frac 1n \sum_{j=1}^n \mu_{n,j} \to c\in \mathbb R.$$

I would like to conclude that, almost surely, $$Y_n := \frac 1n \sum_{j=1}^n X_{n,j} \to 0$$ but the classical SLLN do not work, since $$X_{n,j}$$ are not equidistributed for fixed $$j$$. The classical Markov/Chebychev estimation yields $$\mathbb P[ |Y_n|\ge \epsilon ]\le \frac 1{\epsilon^2} Var(Y_n) = \frac 1{n^2\epsilon^2} \sum_{j=1}^n \mu_{n,j}(1-\mu_{n,j}) \sim \frac {c-c^2}{n\epsilon^2}$$
that is not summable.

Any idea?

• Haven't look much into it, but it seems like the proof of the bounded fourth moment version of classical SLLN could be adopted here? Feb 12 at 15:02
• I'm trying with Chernoff Bound rn @user10354138 Feb 12 at 15:28

Clearly we have $$\mathbb{E}[X_{n,j}^4]\leq 1$$ and $$\mathbb{E}[X_{n,j}]=0$$ for all $$n,j$$. So \begin{align*} \mathbb{E}[Y_n^4] &= n^{-4}\left(\sum_{j=1}^n\mathbb{E}X_{n,j}^4+6\sum_{1\leq i and hence $$\mathbb{P}(\lvert Y_n\rvert\geq\epsilon)\leq\frac{\mathbb{E}Y_n^4}{\epsilon^4}\leq7\epsilon^{-4}n^{-2}.$$ So $$Y_n\to 0$$ a.s.