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Let $X_0,X_1,\dots$ be a sequence of i.i.d. random variables, with $\mathbb{E}\big[|X_i|\big]=\infty$. Show that: $$ 1.\;\limsup \left(\frac{|X_n|}{n}\right)=\infty\qquad 2.\; \limsup \left(\frac{|X_1+\cdots+X_n|}{n}\right)=\infty$$ almost surely.

For the $1.$ I took arbitrary $R>0$ then: $$\begin{aligned}\infty=\frac{1}{R}\,\mathbb{E}\big[|X_1|\big] &=\frac{1}{R}\int_{\mathbb{R}_{\geq 0}}\mathbb{P}(|X_1|\geq x)\,\mathrm{d}x \\ &=\int_{\mathbb{R}_{\geq 0}}\mathbb{P}(|X_1|\geq Rx)\,\mathrm{d}x \quad (x\mapsto Rx)\\ & =\sum_n\int_n^{n+1}\mathbb{P}(|X_1|\geq Rx)\,\mathrm{d}x\\ &\leq\sum_n\int_n^{n+1}\mathbb{P}(|X_1|\geq Rn)\,\mathrm{d}x\quad \big(\mathbb{P}(|X|>a)\geq\mathbb{P}(|X|>b)\;\text{if}\;a<b\big)\\ &=\sum_n\mathbb{P}(|X_1|\geq Rn)\\& =\sum_n\mathbb{P}(|X_n|\geq Rn)\quad (X_1\sim X_n\;\text{for all}\;n)\end{aligned}$$ We can now use Borel-Cantelli's second lemma to infer $\limsup_n |X_n|/n\geq R$ almost surely, and since $R$ was arbitrary $\limsup |X_n|/n=\infty$ almost surely.

  • Does my proof work?
  • Any hints on the second part? It looks like the strong law, but the modulus puts it on the "wrong" side of the triangle inequality, so I can't apply it.
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  • $\begingroup$ Do you know about 0-1 laws? $\endgroup$
    – saz
    Nov 18, 2016 at 14:33
  • $\begingroup$ @saz No, I just googled it though, but I'm not sure I understand it. How would I approach this with that strategy in mind? $\endgroup$
    – user111064
    Nov 19, 2016 at 0:54
  • $\begingroup$ I realized that there is no need for 0-1-law; see my answer below. $\endgroup$
    – saz
    Nov 20, 2016 at 15:03

1 Answer 1

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Yes, your proof of the first assertion is correct. Just be a little bit careful about null sets; for each $R>0$ there is a null set, say $N_R$, it is important that we can let $R \uparrow \infty$ along a countable sequence.


Regarding the 2nd assertion: Denote by $S_n := \sum_{j=1}^n X_j$ the $n$-th partial sum. Using that

$$|X_n| = |S_n-S_{n-1}| \leq |S_n| + |S_{n-1}|$$

we find

$$\frac{|X_n|}{n} \leq \frac{|S_n|}{n} + \frac{|S_{n-1}|}{n} \leq \frac{|S_n|}{n} + \frac{|S_{n-1}|}{n-1}.$$

This implies

$$\limsup_{n \to \infty} \frac{|X_n|}{n} \leq 2 \limsup_{n \to \infty} \frac{|S_n|}{n}.$$

Since we already know that the left-hand side equals $\infty$ almost surely, this proves

$$\limsup_{n \to \infty} \frac{|S_n|}{n} = \infty \qquad \text{a.s.}$$

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