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.
  • $\begingroup$ Do you know about 0-1 laws? $\endgroup$ – saz Nov 18 '16 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 '16 at 0:54
  • $\begingroup$ I realized that there is no need for 0-1-law; see my answer below. $\endgroup$ – saz Nov 20 '16 at 15:03

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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.