# Advanced Probability Theory

Let $(X_n)_{n \geq 1}$ be a sequence of i.i.d. random variables. Show that: $$\frac{X_n}{n} \rightarrow 0 \text{ P-a.s} \Leftrightarrow E[|X_1|] < \infty$$

I thought of using the Borel-Cantelli lemma's but i don't know hot to proceed..

• Isn't that a strong law? Don't you also need that E[X]=0 because the sample mean converges to E[X]? – Jimmy R. Mar 25 '14 at 13:17
• @Stefanos No, it is not. He doesn't want to show that $\frac1n\sum_{k=1}^n X_k\to0$. He only wants to show that $\frac{X_n}n\to 0$. – Cm7F7Bb Mar 25 '14 at 13:19
• @V.C. Ok, yes... true – Jimmy R. Mar 25 '14 at 13:26

Indeed Borel-Cantelli, twice.

First note that for every positive $c$ and every real number $x$, $$c\sum_{n=1}^\infty\mathbf 1_{|x|\geqslant cn}\leqslant|x|\leqslant c\sum_{n=0}^\infty\mathbf 1_{|x|\geqslant cn},$$ hence, as soon as $(X_n)$ is identically distributed, $$cU(c)\leqslant E(|X_1|)\leqslant c+cU(c),\qquad U(c)=\sum_{n=1}^\infty P(|X_n|\geqslant cn).$$ In particular, here is a result of independent interest:

Assume that the sequence $(X_n)$ is identically distributed. Then the expectation $E(|X_1|)$ is finite if and only if, for every positive $c$, the series $\sum\limits_nP(|X_n|\geqslant cn)$ converges.

Now to the proof of the exercise.

If $E(|X_1|)$ is finite, then the series $U(c)$ converges hence the so-called easy part of Borel-Cantelli lemma shows that the event $\limsup\limits_{n\to\infty}\,[|X_n|\geqslant cn]$ has probability $0$, thus, almost surely $\limsup\limits_{n\to\infty}\,|X_n|/n\leqslant c$. This holds for every $c\gt0$ hence $\lim\limits_{n\to\infty}\,|X_n|/n=0$ almost surely, which implies that $\lim\limits_{n\to\infty}\,X_n/n=0$ almost surely. This part does not use the independence.

On the other hand, if $E(|X_1|)$ is infinite, then the series $U(c)$ diverges and the sequence $(X_n)$ is independent hence the so-called difficult part of Borel-Cantelli lemma shows that the event $\limsup\limits_{n\to\infty}\,[|X_n|\geqslant cn]$ has probability $1$, thus, almost surely $\limsup\limits_{n\to\infty}\,|X_n|/n\geqslant c$. This holds for every $c\gt0$ hence $\limsup\limits_{n\to\infty}\,|X_n|/n=+\infty$ almost surely, which implies that the sequence of general term $X_n/n$ diverges almost surely.

Replace $X$ by $|X|$: we can then assume that $X>0$.

• if $EX_1<\infty$: $$\frac 1n X_n = \frac 1n X_0 + \frac 1n \sum_{k=1}^n (X_k - X_{k-1}) \to E(X_1 - X_{0}) = 0$$a.s., because of the $LLN$.

• if $\frac {X_n}n\to 0$ a.s.: $$\frac 1n \sum_{k=1}^n (X_k1_{X_k<N} - X_{k-1}1_{X_{k-1}<N})\to EX1_{X<N}$$ Now RHS $\to EX$ via monotonic convergence. $$\frac 1n \sum_{k=1}^n (X_k1_{X_k\ge N} - X_{k-1}1_{X_{k-1}\ge N}) = o(1) + \frac 1n X_n1_{X_n\ge N}\to 0$$ because $0\le\frac 1n X_n1_{X_n\ge N} \le \frac1n {X_n}$. Now make the sum and take the limit $n\to\infty$ and then $N\to\infty$, and you are done.