# Equivalence of the condition that the supremum of i.i.d. RVs are finite a.s.

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.

(edit)

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 $$\{sup_{n\in\mathbb{N}}{X_n} < \infty\} = \bigcup_{m=1}^{\infty}\bigcap_{n=1}^{\infty}\{X_n \le m\}.$$

• Are you sure the random variables $(X_n)$ are supposed to be i.i.d., or only independent? – Did Dec 10 '18 at 15:20
• I am 100 % sure that they are supposed to be i.i.d., as it was exam problem. – Euduardo Dec 10 '18 at 15:22
• Well then the exam is slightly absurd, but why not. – Did Dec 10 '18 at 15:23
• You're right Did. Now I understand independence is enough, thanks! – Euduardo Dec 11 '18 at 5:09

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 $$P(X_i > M) > 0$$.
• How do you go from $P[X_i >M] >0$ to $P[\sup_n X_n >M] = 1$? – copper.hat Dec 10 '18 at 17:07
• Using the fact that $X_n$ are iid, I see. Thanks. – copper.hat Dec 10 '18 at 18:23