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Let $X_1,\cdots,X_n$ be sequence of positive, iid random variables such that $\mathbb{E} X_1 <\infty$. How can I show that

$$\frac{1}{n}\max\left(X_1,X_2,\cdots,X_n\right)\to 0 \text{ in probability}$$

I can prove it assuming $\mathbb{E}X_1^2 < \infty$, but I don't know how to prove it with the given assumption.

Note: While almost surely convergence implies convergence in probability, I am looking for a solution that does not take that route, and does not use Borel-Cantelli lemma.

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  • $\begingroup$ Are the $X_i$ positive? $\endgroup$ – Will M. Jun 24 '18 at 0:30
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    $\begingroup$ Doesn't he say that? $\endgroup$ – mathworker21 Jun 24 '18 at 0:30
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First, \begin{eqnarray} P\{|\max|\geq n\delta\}=1-[1-P(|X_1|\geq n\delta)]^n. \end{eqnarray} Next we estimate $P(|X_1|\geq n\delta)$. Since $E|X_1|<+\infty$, $$E\{|X_1|;|X_1|\geq n\delta\}:=\alpha(n)\to0,\text{as}\ n\to\infty.$$ So we have $$P(|X_1|\geq n\delta)\leq\frac{\alpha(n)}{n\delta}.$$

Thus, $$1-[1-P(|X_1|\geq n\delta)]^n\leq 1-[1-\frac{\alpha(n)}{n\delta}]^{\frac{n\delta}{\alpha(n)}\frac{\alpha(n)}{\delta}}\to 0.$$

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  • $\begingroup$ What is $E \left\{ \left|X_1\right|; \left|X_1 \right| \ge n\delta \right\}$? $\endgroup$ – Yuki Kawabata Jun 24 '18 at 12:16
  • $\begingroup$ It means the integration of $|X_1|$ on the set $\{|X_1|\geq n\delta\}$. $\endgroup$ – XIAODA QU Jun 24 '18 at 12:23
  • $\begingroup$ And why do we have the inequality after "so we have"? $\endgroup$ – Yuki Kawabata Jun 24 '18 at 13:50
  • $\begingroup$ That’s the Markov inequality $\endgroup$ – Galton Jun 24 '18 at 15:49
  • $\begingroup$ @YukiKawabata Just consider $n\delta$ as a function on $\{|X_1|\geq n\delta\}$, whose integration over $\{|X_1|\geq n\delta\}$ is smaller than $|X_1|$'s, similar to Markov inequality. $\endgroup$ – XIAODA QU Jun 25 '18 at 8:22
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Hint:

Each random variable has an identical cumulative distribution function $\phi(x)$. The probability that $$\max(X_1, ..., X_n) > x$$ is equal to 1 minus the probability that all $X_i$ are less than $x$, which is $\phi(x)^n$.

That is, the cdf for the max is $1-\phi(x)^n$. Now realize that $0<\phi(x)<1$ and think about for a fixed $x$ what the limit is. Can you now use the cdf to show convergence to zero in probability?

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    $\begingroup$ But here the $x$ you have depends on $n$, right? $\endgroup$ – mathworker21 Jun 24 '18 at 0:34
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Take some $\lambda > 0$. Then $P(\max(X_1,\dots,X_n) > n\lambda) = P(\cup_j \{X_j > \lambda n\}) = 1-P(\cap_j \{X_j \le \lambda n\})$

$= 1-(P(X_1 \le \lambda n))^n = 1-(1-P(X_1 > \lambda n))^n$. So we just need to show that $(1-P(X_1 > \lambda n))^n \to 1$. But the LHS is $e^{n\log(1-P(X_1 > \lambda n))}$, so it suffices to show $n\log(1-P(X_1 > \lambda n)) \to 0$. Note $-n\log(1-P(X_1 > \lambda n)) = n\log(\frac{1}{1-P(X_1 > \lambda n)}) \asymp nP(X_1 > \lambda n)$. But since $X_1 \in L^1$, $nP(X_1 > \lambda n) \to 0$, as desired.

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  • $\begingroup$ First, what is $\asymp$? Second, why $n P\left(X_1 > \lambda n\right) \to 0$? Markov inequality only concludes that for any $n$, $nP \left(X_1 > \lambda n\right) < E X_1/\lambda$. $\endgroup$ – Yuki Kawabata Jun 24 '18 at 0:58
  • $\begingroup$ It just means that there exist constants $c,C > 0$ (depending just on $X_1,\lambda$) so that for all $n$, $c\log(\frac{1}{1-P(X_1 > \lambda n)}) \le P(X_1 > \lambda n) \le C\log(\frac{1}{1-P(X_1 > \lambda n)})$. $\endgroup$ – mathworker21 Jun 24 '18 at 1:17
  • $\begingroup$ And exactly, I did not use Markov's inequality. It is too weak for this purpose. I used the fact that $||X_1||_1 = \int_0^\infty \alpha P(X_1 > \alpha)d\alpha$. Since $||X_1||_1 < \infty$, it must be that $\alpha P(X_1 > \alpha) \to 0$ as $\alpha \to +\infty$. $\endgroup$ – mathworker21 Jun 24 '18 at 1:18

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