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?
Best Answer
The fourth moment bound works.
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<j\leq n}\mathbb{E}X_i^2X_j^2\right)\\ &\leq n^{-3}+6n^{-2}\leq 7n^{-2} \end{align*} 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.