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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.

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  • $\begingroup$ Yu said that $X_1\geq 0$ a.s, then shouldn't $X_n=Y_n$ a.s? $\endgroup$ May 24 '21 at 13:58
  • $\begingroup$ 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 $\lim_{n\rightarrow\infty} \sup \frac{\max[X_1, ..., X_n]}{n}$. $\endgroup$
    – Michael
    May 24 '21 at 14:06
  • $\begingroup$ Instead, you should start by computing a bound on $P[\max[X_1, ..., X_n]/n>1]$. $\endgroup$
    – Michael
    May 24 '21 at 14:07
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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

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    $\begingroup$ Thanks a lot Jacob! =) Your answer helped a lot! $\endgroup$
    – Stanisla
    May 27 '21 at 7:31

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