I have a sequence of random variables $(X_i)_{i \geq 1}$, not necessary equaly distributed. They have moment of any order : $\forall k \geq 1,~\mathbb{E}(X_i^k) < \infty$ and $\mathbb{E}[X_i] = 0$. Morevover I know that $\mathbb{E}[X_i X_j] = 0$ for $i \neq j$.
Is there a way to obtain a weak law of large number for
$$\frac{1}{n} \sum_{i = 1}^n (|X_i| - \mathbb{E}[|X_i|]) \to 0, ~ \text{when } n \to \infty$$
where the convergence is, at least, in probablity ?
The rv's $X_i$ are not supposed to be positive or negatives.
Do you have any counter-example ? Or any proof of it ? Markov inequality seems not to work because of the absolute value.
Is there any further hypothesis to assume in order to get this LLN ?
Thanks,
Let $(\varepsilon_i)$ be an i.i.d. sequence of random variables taking values $1$ and $-1$ with equal probability and let $Y$ be a random variable which is independent of $(\varepsilon_i)$ and such that $\mathbb E\left[\lvert Y\rvert^k\right]$ is finite for each $k\geqslant 1$. Let $X_i=\varepsilon_iY$. Then all the assumptions are satisfied and $$ \frac{1}{n} \sum_{i = 1}^n (|X_i| - \mathbb{E}[|X_i|]) \to 0 $$ is equivalent to $$ \frac{1}{n} \sum_{i = 1}^n (|Y| - \mathbb{E}[|Y|]) \to 0. $$ Since we have freedom on the choice of $Y$ as long as it is independent of $(\varepsilon_i)$, we can even take $Y$ such that $\lvert Y\vert$ is not constant.