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I am proving the following :

Suppose $\{X_n : n\in\mathbb{N}\}$ are i.i.d. random variables. Then $P(\sup_{n\in\mathbb{N}}X_n < \infty) = 1$ if and only if $ \sum_{n\in\mathbb{N}}{P(X_n > M)} < \infty$ for some $M>0.$

I guess by the form of the problem, that Borel-Canteli lemma will be used at some point, but I could not thought of a good way to connect it to the problem.

Any idea or hint will be really helpful. Thanks.

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Assume $\sum_{n\in\mathbb{N}}{P(X_n > M)} < \infty.$ In fact, as they are i.i.d., $P(X_n > M) = P(X_1>M)$ for all $n.$ Thus $P(X_1>M) = 0$ and $P(\sup_{n\in\mathbb{N}}X_n > M) \le \sum_{n\in\mathbb{N}}{P(X_n > M)} = 0.$

Conversely, I've realized that $\{sup_{n\in\mathbb{N}}{X_n} < \infty\} = \bigcup_{m=1}^{\infty}\bigcap_{n=1}^{\infty}\{X_n \le m\}.$

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    $\begingroup$ Are you sure the random variables $(X_n)$ are supposed to be i.i.d., or only independent? $\endgroup$ – Did Dec 10 '18 at 15:20
  • $\begingroup$ I am 100 % sure that they are supposed to be i.i.d., as it was exam problem. $\endgroup$ – Euduardo Dec 10 '18 at 15:22
  • $\begingroup$ Well then the exam is slightly absurd, but why not. $\endgroup$ – Did Dec 10 '18 at 15:23
  • $\begingroup$ You're right Did. Now I understand independence is enough, thanks! $\endgroup$ – Euduardo Dec 11 '18 at 5:09
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Suppose $P(\sup_n X_n < \infty) = 1$. Since $\{\sup_n X_n < \infty\} = \bigcup_{M=1}^\infty \{\sup_n X_n \le M\}$, we have $\lim_{M \to \infty} P(\sup_n X_n \le M) = 1$. But if $P(X_i > M) > 0$, $P(\sup_n X_n > M) = 1$. So for some $M$ we must have $P(X_i > M) = 0$.

Conversely, if $P(\sup_n X_n < \infty) < 1$, then for all $M$ we have $P(\sup_n X_n > M) > 0$, and $P(X_i > M) > 0$.

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  • $\begingroup$ How do you go from $P[X_i >M] >0$ to $P[\sup_n X_n >M] = 1$? $\endgroup$ – copper.hat Dec 10 '18 at 17:07
  • $\begingroup$ Second Borel-Cantelli lemma. $\endgroup$ – Robert Israel Dec 10 '18 at 18:21
  • $\begingroup$ Using the fact that $X_n$ are iid, I see. Thanks. $\endgroup$ – copper.hat Dec 10 '18 at 18:23

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