# Borel-Cantelli argument for maximum of random variables

Let $$(X_n)_{n \geq 1}$$ be a sequence of random variables taking non-integer values, such that for each $$n$$ and each $$i$$, $$\mathbb{P}(X_n \geq i) = 1/i$$. By Borel-Cantelli, I have managed to show that (almost surely) $$$$\limsup_{n \to \infty} \dfrac{\log X_n}{\log n} = 1.$$$$ Next I need to prove that, for $$M_n = \max_{1 \leq k \leq n} X_k$$, we have almost surely: $$$$\lim_{n \to \infty} \dfrac{\log M_n}{\log n} = 1.$$$$ My idea was to fix an $$\epsilon > 0$$, then show that: $$$$\mathbb{P} \left(\limsup_{n \to \infty} \dfrac{\log M_n}{\log n} \leq 1+ \epsilon \right) = 1,$$$$ $$$$\mathbb{P} \left(\liminf_{n \to \infty} \dfrac{\log M_n}{\log n} \geq 1 - \epsilon \right) = 1,$$$$ then take a monotonic intersection over all $$\epsilon \in \mathbb{Q}^+$$ and conclude. However, I have been unable to prove the above equalities, and I am not sure how (and if) the first property that I proved can be useful. Any help appreciated!

Without independence this is obviously wrong. If $$X_n=X_1$$ for all $$n$$ then the result you claim to have proved is clearly false since the $$\lim\ sup$$ is $$0$$, not $$1$$.

Assume independence. The proof for limsup does not require any probability theory.

If $$\lim \sup \frac {a_n} {b_n}=1$$ for some positive increasing sequence $$b_n \to \infty$$ then $$\lim \sup \frac {c_n} {b_n}=1$$ where $$c_n =\max \{a_1,a_2,...,a_n\}$$.

• That answers the question regarding the $\limsup$. How about the $\liminf$? May 6, 2020 at 18:23

I think I have finally found a (probabilistic) solution to this question.

For any $$\alpha > 0$$, we have: $$$$\mathbb{P}[M_n < n^{\alpha}] = \mathbb{P}[X_k < n^{\alpha}]^n \leq (1-n^\alpha)^n \leq \exp(-n^{1-\alpha}).$$$$

Next, we use the fact that $$\sum_n e^{-n^{\epsilon}}$$ converges for any $$\epsilon > 0$$. This means that for any $$\alpha <1$$, the sum of the above probabilities is finite. Therefore, by the first Borel-Cantelli lemma, $$$$\mathbb{P}[M_n \geq n^{\alpha} \text{ eventually}] = \mathbb{P}\left[\liminf_{n \to \infty} \frac{\log M_n}{\log n} \geq \alpha\right] = 1.$$$$ So taking a countable union over rational $$\alpha \in (0,1)$$, we get $$$$\mathbb{P}\left[\liminf_{n \to \infty} \frac{\log M_n}{\log n} \geq 1 \right] = 1.$$$$

On the other hand, $$\limsup \frac{\log M_n}{\log n} \leq \limsup \frac{\log X_n}{\log n}$$. This is because if $$\exists N_1 \in \mathbb{N}$$ such that $$X_n \leq n^\alpha$$ whenever $$n \geq N_1$$, then $$\exists N_2$$ s.t. $$M_n \leq n^\alpha$$ whenever $$n \geq N_2$$. This is accomplished by taking $$M$$ such that $$M^\alpha > \max(X_1, X_2,...,X_{N_1})$$. This shows the $$\limsup$$ is at most $$1$$, so we're done.