Weak law of large numbers for epsilon sequence that tends to 0

Question

Assume a $S_n=\sum_{i=1}^n X_i$ fulfills a weak law of large numbers, i.e. for every $\epsilon>0$, $$P\left(|\tfrac{1}{n}S_n - \mu| > \epsilon\right) \rightarrow 0$$ Does there exist a sequence $(\epsilon_n)_n$ with $\epsilon_n \rightarrow 0$ such that $$P\left(|\tfrac{1}{n}S_n - \mu| > \epsilon_n\right) \rightarrow 0$$ holds? Or do I need to make further assumptions about the $X_i$s?

Answer 1

Yes, and it really has nothing to do with law of large numbers.

Lemma: Suppose $Y_n$ are random variables with $\mathbb{P}(Y_n>\epsilon)\to 0$ for all $\epsilon>0$, then there exists a sequence $\epsilon_n\downarrow 0$ such that $\mathbb{P}(Y_n>\epsilon_n)\to 0$.

Proof: For every $m\geq 1$, there exists $N_m\in\mathbb{N}$ such that $\mathbb{P}(Y_n>2^{-m})<\frac1m$ for all $n\geq N_m$. So WLOG $N_{m+1}>N_m$, and choose $\epsilon_{N_m}=\epsilon_{N_m+1}=\dots=\epsilon_{N_{m+1}-1}=2^{-m}$. Then $\mathbb{P}(Y_n>\epsilon_n)<\frac1m$ for all $n\geq N_m$, hence the desired result.

Answer 2

For convergence in probability, by Chebyshev's inequality, an upper bound on the probability is $$ P(|\frac{S_n}{n} - \mu|>\varepsilon_n) <\frac{Var S_n}{n^2\varepsilon_n} = \frac{\sigma^2}{n \varepsilon_n} $$ Since second moment is finite, as long as $$ n \varepsilon_n \to \infty $$ the ratio converges to $\mu$.