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I have following question and stuck at the 1st one.

$\left\{X_n\right\}$ are i.i.d. random variables with non-zero finite mean. Let $S_n=X_1+X_2+\dots+X_n$. Prove
1. $|X_n|/n \xrightarrow{a.s.}0$;
2. $(\max_{1\le k \le n}|X_k|)/n \xrightarrow{a.s.} 0$
3. $(\max_{1\le k \le n}|X_k|)/(1+S_n) \xrightarrow{a.s.} 0$

I tried to use Borel-Cantelli lemma and Markov inequality, but failed.
1st Try: It's equivalent to show for any $\epsilon > 0$, we have $P(|X_n|/n > \epsilon \text{ i.o.})=0$

$$\sum_{n=1}^{\infty} P(|X_n|/n > \epsilon) \le \sum_{n=1}^{\infty}E[|X_n|]/n\epsilon=\infty$$ I figured it didn't work probably because i.i.d. condition was not used.
2nd Try: $$\begin{align}P(|X_n|/n > \epsilon \text{ i.o.}) &= \lim_{m \rightarrow \infty}P(\bigcup_{n=m}^\infty |X_n|/n > \epsilon) \\ &= \lim_{m \rightarrow \infty}1-P((\bigcup_{n=m}^\infty |X_n|/n > \epsilon)^c) \\ &=\lim_{m \rightarrow \infty}1-\prod_{n=m}^\infty P(|X_1|/n \le \epsilon)\end{align}$$ Stuck again.

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2 Answers 2

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The Markov inequaltiy is too weak in this case. Instead, use the fact that the $X_n$ are all identically distributed and the integral comparison test.

$$\sum \limits_{n = 1}^\infty P(|X_n|/n > \varepsilon) = \sum \limits_{n = 1}^\infty P(|X_1|/\varepsilon > n) \le 1 + \int_0^\infty P(|X_1|/\varepsilon > x) \, dx$$

Now note that for any nonnegative random variable $Y$ the equation $$\begin{align*} \int_0^\infty P(Y > x) \, dx &= \int_0^\infty \int_0^\infty I\{y > x\} \, dP^Y(y) \, dx = \int_0^\infty \int_0^\infty I\{y > x\} \, dx \, dP^Y(y)\\ &= \int_0^\infty y \, dP^Y(y) = E[Y] \end{align*}$$ holds.

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  • $\begingroup$ Any idea on 2nd one? By a similar argument, I arrived $$\sum_{n=1}^{\infty} 1-P(|X_1|/\epsilon \le n)^n$$ How to show its convergence? $\endgroup$
    – Lei Hao
    Commented Nov 29, 2016 at 4:49
  • $\begingroup$ I don't know how to solve the other two. $\endgroup$
    – Dominik
    Commented Nov 29, 2016 at 7:25
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    $\begingroup$ The second one is very easy after you've proved the first one. What you need is just observing that $\lim\limits_{n\to\infty}nP(|X_1|>n\epsilon)=0$. For the third one, apply strong law of large numbers and invoke the consequence in 2. $\endgroup$ Commented Nov 29, 2016 at 8:22
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I guess following is the answer. $$\begin{align}P(|X_n|/n > \epsilon \text{ i.o.}) &= \lim_{m \rightarrow \infty}P(\bigcup_{n=m}^\infty |X_n|/n > \epsilon)\\ &= \lim_{m \rightarrow \infty}P(\bigcup_{n=m}^\infty |X_1|/n > \epsilon)\\ &= \lim_{m \rightarrow \infty}P(|X_1|/m > \epsilon) \\ & \le \lim_{m \rightarrow \infty} E(|X_1|)/n \epsilon \rightarrow 0\end{align}$$

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  • $\begingroup$ How could you arrive at the second equality? $\endgroup$ Commented Nov 28, 2016 at 11:16
  • $\begingroup$ Since $X_n$ are i.i.d, $$\left\{\bigcup_{n=m}^\infty |X_n|/n > \epsilon \right\}=\left\{\bigcup_{n=m}^\infty |X_1|/n > \epsilon \right\}=\left\{ |X_1|/m > \epsilon \right\}$$, the last equality is due to the fact $$ \left \{ |X_1| / n > \epsilon \right \} \subset \left\{|X_1|/m > \epsilon \right\}$$ for any n > m $\endgroup$
    – Lei Hao
    Commented Nov 28, 2016 at 12:36
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    $\begingroup$ In your comment the first equality is wrong. We do not have $\{|X_n|/n>\epsilon\}=\{|X_1|/n>\epsilon\}$. $\endgroup$ Commented Nov 28, 2016 at 14:23
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    $\begingroup$ The same distribution has nothing to do with $\{|X_n|/n>\epsilon\}=\{|X_1|/n>\epsilon\}$. $\endgroup$ Commented Nov 29, 2016 at 3:42
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    $\begingroup$ So how do you get $P(\bigcup_{n=m}^\infty |X_n|/n > \epsilon) = P(\bigcup_{n=m}^\infty |X_1|/n > \epsilon)$? You've never proved this is true. Of course, you can't make it since it's just not true. $\endgroup$ Commented Nov 29, 2016 at 8:11

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