# Showing that $\limsup_{n\to \infty} \frac{\max\{X_1,\ldots, X_n\}}{n} \leq 1$ almost surely

Let $$(X_n)_{n \geq 1}$$ be a sequence of i.i.d. random variables such that $$X_1 \geq 0$$ a.s. and $$\sum_{n=1}^{\infty} n \mathbb P (X_1 > n) < \infty$$ .

I have to show that $$\limsup_{n \to \infty} \frac{\max\{X_1,\ldots, X_n\}}{n} \leq 1 \quad \text{a.s.}$$

Here's my attempt:

Let $$\limsup X_n / n < \infty$$ with probability 1. Let the nonnegative iid random variables be $$Y_n = \max[X_n, 0]$$ for all $$n \in \{1, 2, \ldots\}$$. Then with prob. 1 we have $$\limsup Y_n / n < \infty$$ and: $$\limsup \frac{1}{n} \sum X_i \leq \lim \frac{1}{n} \sum Y_i = E[Y]$$ (by LLN) where $$Y = Y_1$$. Since $$E[Y] = 1 \rightarrow \limsup_{n \to \infty} \frac{\max\{X_1, \ldots, X_n\}}{n} \leq E[Y] = 1$$ a.s.

Is this possible and correct? Any comment and correction is appreciated.

• Yu said that $X_1\geq 0$ a.s, then shouldn't $X_n=Y_n$ a.s? May 24 '21 at 13:58
• You are not told that $E[X_i]=1$ for all $i$. Also your last line does not follow, the one that seeks to relate the previous lines to $\max[X_1, ..., X_n]$. This is not a law of large numbers problem. Also I think you need to show the inequality for $\limsup_{n\rightarrow\infty} \frac{\max[X_1, ..., X_n]}{n}$, which is not the same as $\lim_{n\rightarrow\infty} \sup \frac{\max[X_1, ..., X_n]}{n}$. May 24 '21 at 14:06
• Instead, you should start by computing a bound on $P[\max[X_1, ..., X_n]/n>1]$. May 24 '21 at 14:07

This can be proved by application of the 1st Borel-Cantelli lemma which states that: If $$(A_n)_{n\in\mathbb{N}}$$ is a (not necessarily independent) sequence of events such that $$\sum_{n\in\mathbb{N}}\mathbb{P}(A_n)<\infty$$, then $$\mathbb{P}(A_n\ happens\ infinitely\ often)=0$$. We will write $$A_n\ i.o.$$ to mean $$A_n\ happens\ infinitely\ often$$.

Now consider that

\begin{align} limsup\frac{max\{X_1, ..., X_n\}}{n}\leq 1\ \ a.s. & \Leftrightarrow \mathbb{P}(limsup\frac{max\{X_1, ..., X_n\}}{n} > 1)=0 \\ & \Leftrightarrow\mathbb{P}(\forall m>N,\ sup_{n>m}\frac{max\{X_1, ..., X_n\}}{n}>1)=0 \\ & \Leftrightarrow\mathbb{P}(\frac{max\{X_1, ..., X_n\}}{n}>1\ \ \ i.o.)=0 \end{align} This final implication can be seen from the fact that if $$\frac{max\{X_1, ..., X_n\}}{n}>1$$ does not happen for infinitely many $$n$$ (say that the largest $$n$$ for which this is true is $$n'$$), then $$sup_{n>n'}\frac{max\{X_1, ..., X_n\}}{n}\leq 1$$.

Hence applying the Borel-Cantelli lemma, we are left to show that $$\sum_{n\in\mathbb{N}}\mathbb{P}(\frac{max\{X_1, ..., X_n\}}{n}>1)<\infty$$.

We can calculate that: \begin{align} \mathbb{P}(\frac{max\{X_1, ..., X_n\}}{n}>1)&=\mathbb{P}(max\{X_1, ..., X_n\}>n)\\ &=\mathbb{P}(\{X_1>n\}\cup \{X_2>n\}\cup ... \cup \{X_n>n\})\\ &=\mathbb{P}(X_1>n)+\mathbb{P}(X_2>n)+...+\mathbb{P}(X_n>n)\\ &=n\mathbb{P}(X_1>n) \end{align} since the $$X_i$$s are i.i.d

And we see exactly why the conditions were given as they were, since the given condition $$\sum_{n\in\mathbb{N}}n\mathbb{P}(X_1>n)<\infty$$ implies that $$\sum_{n\in\mathbb{N}}\mathbb{P}(\frac{max\{X_1, ..., X_n\}}{n}>1)<\infty$$, and so we conclude, by the Borel-Cantelli lemma, that $$limsup\frac{max\{X_1, ..., X_n\}}{n}\leq 1$$ almost surely

• Thanks a lot Jacob! =) Your answer helped a lot! May 27 '21 at 7:31