1
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$ Feb 12 at 15:02
  • $\begingroup$ I'm trying with Chernoff Bound rn @user10354138 $\endgroup$
    – Exodd
    Feb 12 at 15:28
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thanks! I also verified that Chernoff Bound works here $\endgroup$
    – Exodd
    Feb 12 at 15:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.