An Inversion of the Law of Large Numbers

Question

Assume $(X_{n})$ are i.i.d. and $E[|X_{1}|]=\infty$.

Can we conclude that $\frac{S_{n}}{n}$ does not converge in probability?

We certainly know that $\frac{S_{n}}{n}$ cannot converge a.s. since the event $$A=\{|X_{n}|>n \; \infty \text{-often}\}$$ has probability $1$ by Borel-Cantelli and thus $|\frac{S_{n}}{n}-\frac{S_{n+1}}{n+1}|$ will be greater than some $\epsilon>0$ infinitely often.

Answer 1

It is possible that in this setting, $S_n/n$ converges in probability to $0$.

Let $(X_i)_{i\geqslant 1}$ be an i.i.d. sequence where $X_i$ is symmetric (that is, $X_i$ and $-X_i$ have the same distribution) and $t\mathbb P\{\lvert X_1\rvert>t\}\to 0$. Then $S_n/n\to 0$ in probability, where $S_n=\sum_{i=1}^nX_i$.

This can be seen by truncation: let $X'_i=X_i\mathbf{1}\{\lvert X_i\rvert\leqslant \delta n\}$ and $X''_i=X_i\mathbf{1}\{\lvert X_i\rvert\gt \delta n\}$. Then it suffices to show the convergences in probability $\sum_{i=1}^nX'_i/n\to 0$ and $\sum_{i=1}^nX''_i/n\to 0$. For the latter, note that $$ \left\{\sum_{i=1}^nX''_i/n\neq 0\right\}\subset \bigcup_{i=1}^n\{\lvert X_i\rvert\gt \delta n\}, $$ take probabilities and use a union bound and $t\mathbb P\{\lvert X_1\rvert>t\}\to 0$. For $Y_n:=\sum_{i=1}^nX'_i/n$, use Chebytchev's inequality. Use independence to control $\mathbb E\left[Y_n^2\right]$ and be reduced to estimate $\mathbb E\left[{X'_{1}}^2\right]$. This can be done by expressing this expectation as an integral of the tail; then we use that $t\mathbb P\{\lvert X_1\rvert>t\}$ is bounded in $t$. More precisely, since the random variables $X'_i$ are uncorrelated and centered,
$$ \mathbb E\left[Y_n^2\right]=\frac 1n\mathbb E\left[{X'_{1}}^2\right]=\frac 1n\int_0^{+\infty}2u\mathbb P\{\lvert X_1\rvert \mathbf{1}\{\lvert X_1\rvert\leqslant \delta n\}>u\}du. $$ The integrand vanishes for $u>\delta n$. Therefore, $$ \mathbb E\left[Y_n^2\right]\leqslant \frac 1n\int_0^{\delta n}2u\mathbb P\{\lvert X_1\rvert >u\}du\leqslant 2\delta \sup_{u>0}u\mathbb P\{\lvert X_1\rvert >u\}. $$

Now, in order to give a counter-example, let $(X_i)_{i\geqslant 1}$ be an i.i.d. sequence where $X_1$ takes integer values: $$ \mathbb P(X_1=k)=\mathbb P(X_1=-k)=\frac c{k^2\log k}, k\geqslant 2, \mathbb P(-1\leqslant X_1\leqslant 1)=0 $$ where $c$ is such that $\sum_{k\geqslant 2}\frac{2c}{k^2\log k}=1$.