The Equivalent Condition of the Weak Law of Large Numbers When Random Variables Are Uniformly Bounded

Question

When the random variable $\{X_n,n\ge1\}$ satisfies the uniformly bounded condition, why does $$ \frac{1}{n^2}\operatorname{Var}\left(\sum_{k=1}^{n}X_k\right)\rightarrow0 $$ become a necessary and sufficient condition for the establishment of the weak law of large numbers? From Chebyshev's law of large numbers, we can easily use the above conditions to deduce that the law of large numbers holds. So why is this condition necessary?

Answer 1

Suppose that $(X_n)$ are uniformly bounded and let $Y_n:=\sum_{i=1}^n (X_i-\mathbb E[X_i])/n$. Then $(Y_n)_{n\geqslant 1}$ is also uniformly bounded and so is $(Y_n^2)_{n\geqslant 1}$. In particular, $(Y_n^2)$ is uniformly integrable.

The weak law of large numbers is, by definition, the convergence in probability of $Y_n\to 0$, which is equivalent to $Y_n^2\to 0$ in probability.

Now we use the following fact: convergence in $\mathbb L^1$ of $(Y_n^2)$ to $0$ is equivalent to (convergence in probability of $(Y_n^2)_n$ to $0$ and uniform integrability of $(Y_n^2)_n$), see here.

An other way to see this is that if $\lvert X_n\rvert\leqslant C$ almost surely, then letting $Y_n:=\sum_{i=1}^n (X_i-\mathbb E[X_i])/n$ , we have $\lvert Y_n\rvert\leqslant C$. If $Y_n\to 0$ in probability, then $$ \operatorname{Var}(Y_n)\leqslant \mathbb E\left[Y_n^2\right]=\mathbb E\left[Y_n^2\mathbf{1}_{\{\lvert Y_n\rvert>\delta\}}\right]+\mathbb E\left[Y_n^2\mathbf{1}_{\{\lvert Y_n\rvert\leqslant\delta\}}\right] \leqslant C^2\mathbb P\left(\lvert Y_n\rvert>\delta\right)+\delta^2. $$