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The original question is as follows:

$\left\{Y_n\right\}$ are i.i.d random variables. Find the sufficient and neccessary condition for $(\max_{m\le n}Y_m)/n \rightarrow 0$ a.s.

The answer is $EY_1^+ < \infty$. Then $\sum P(Y_n/n > \epsilon)\le EY_1^+< \infty \implies \limsup\left\{Y_n^+/n \le 0 \right\}$ a.s. I can't see why it implies $(\max_{m\le n}Y_m)/n \rightarrow 0$ a.s.

My attempt is as follows:
Let $E=\left\{Y_n^+/n \le 0 \text{ i.o.}\right\}$. Then $P(E)=1$ and for $\omega \in E$, $Y_n^+(\omega)/n \le 0 \text{ i.o.}$ Why will this implies $(\max_{m\le n}Y_m)/n \rightarrow 0$ a.s.? One counter-example I have in mind is let all even number satisfy this condition while all odd number don't.

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    If $\limsup y_n^+/n\leqslant0$, for every $u>0$ there exists $n_u$ such that $y_n^+\leqslant nu$ for every $n\geqslant n_u$. But then $y_n\leqslant y_n^+\leqslant nu$ hence $m_n=\max\limits_{k\leqslant n}y_k$ is such that $m_n/n\leqslant m_{n_u}/n+u$ for every $n\geqslant n_u$, hence $\limsup m_n/n\leqslant0$. Note that there is no way to hope that $\lim m_n/n=0$ only assuming $\limsup y_n^+/n\leqslant0$. – 2017-01-03
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    @Did Thanks for the answer. I need a couple of clarifications: 1. why is your first claim right? 2. It seems at the end you derived $(\limsup m_n/n)\le 0$, which may not be the same as $\limsup {m_n/n \le 0}$. – 2017-01-04
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    I mean $\limsup \{m_n/n \le 0\}$. – 2017-01-04
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    Except that the implication is: $\sum P(Y_n/n > \epsilon)< \infty \implies (\limsup Y_n^+/n) \le 0$ almost surely, not: $\sum P(Y_n/n > \epsilon)< \infty \implies P(\limsup\left\{Y_n^+/n \le 0 \right\})=1$. – 2017-01-04

1 Answers 1

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Let $M_n=\max\limits_{j\le n}Y_j$ and $F$ be the distribution function of $X_1$, then $$ P(M_n\le -n\epsilon)=[F(-n\epsilon)]^n<\alpha^n,\qquad \text{if n is large enough such that }F(-n\epsilon)<\alpha<1.$$ Therefore \begin{eqnarray} P\Bigl(\Bigl\{\dfrac{M_n}n<-\epsilon\Bigr\}\quad \text{i.o.}\Bigr)=0,\qquad \forall \epsilon>0\quad &\Rightarrow& P\Bigl(\liminf_{n\to\infty}\dfrac{M_n}n<-\epsilon\Bigr)=0\qquad \forall \epsilon>0\\ &\Rightarrow& \liminf_{n\to\infty}\dfrac{M_n}n\ge 0\qquad \text{a.s.}\tag{*} \end{eqnarray} Meanwhile, \begin{eqnarray} EY_1^+<\infty &\Rightarrow &\sum_{n=1}^\infty P(Y^+_n>n\epsilon)<\infty\qquad \forall \epsilon>0\\ \text{by Borel-Cantelli Lemma}\quad&\Rightarrow & P\Bigl(\Bigl\{\dfrac{Y_n^+}n >\epsilon\Bigr\}\quad\text{i.o.}\Bigr)=0\qquad \forall \epsilon>0\\ &\Rightarrow & \limsup_{n\to\infty}\dfrac{Y_n^+}n \le \epsilon \quad\text{a.s.}\qquad\forall\epsilon>0\\ &\Rightarrow &\lim_{n\to\infty}\dfrac{Y_n^+}n=0 \quad\text{a.s.}\quad\Bigl(\Leftrightarrow \lim_{n\to\infty}\dfrac{\max\limits_{j\le n}Y_j^+}n=0\Bigr)\\ &\Rightarrow & \limsup_{n\to\infty}\dfrac{Y_n}n\le 0\quad\text{a.s.}\\ &\Rightarrow & \limsup_{n\to\infty}\dfrac{M_n}n\le 0\quad\text{a.s.}\\ \text{by} (*)\qquad &\Rightarrow & \lim_{n\to\infty}\dfrac{M_n}n=0 \quad\text{a.s.} \end{eqnarray}