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Let $(X_n)_{n\in\mathbb N}$ be a sequence of independent random variables. Show that $\sup_nX_n<\infty$ almost surely iff there exists $A>0$ such that, $\sum_nP(X_n>A)<\infty$

By Borel-Cantelli we have

$$\sum_nP(X_n>A)=\infty\Longleftrightarrow P(\limsup_nX_n>A)=1$$

Hence $\forall A>0$;

$\sum_nP(X_n>A)=\infty\Rightarrow 1=P(\limsup_nX_n>A)<P(\sup_nX_n>A)$

Can you give a hint for the other direction, Thanks.

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    $\begingroup$ Since you have independent events, the "iff" in your first display can be strengthened: if $\sum_n P(X_n > A) < \infty$, then we not only have $P(\limsup X_n > A) < 1$, we actually have $P(\limsup X_n > A) = 0$. This is often called the second Borel-Cantelli lemma, or the Borel zero-one law. $\endgroup$ Commented Jun 4, 2014 at 20:17
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    $\begingroup$ @Nate Eldredge Thanks a lot, but is it not called Kolmogorov zero-one law ? $\endgroup$
    – OBDA
    Commented Jun 4, 2014 at 20:20
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    $\begingroup$ No, the Kolmogorov zero-one law is a different (though related) result. $\endgroup$ Commented Jun 4, 2014 at 20:26
  • $\begingroup$ @NateEldredge Borel zero-one law is the BCL2 rather than BCL2 and BCL1? $\endgroup$
    – BCLC
    Commented Mar 21, 2018 at 17:21

1 Answer 1

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I'll assume the random variables are real-valued.

'direction 1'

Let $A_n^c := \{X_n > A\}$. By BCL1, we have

$$P(\limsup A_n^C) = 0$$

$$\to P(\liminf A_n) = 1$$

$$\to P(\lim A_n) = 1$$

$$\to \lim P(A_n) = 1$$

$$\to P(\bigcap_{n=1}^{\infty} A_n) = 1$$

$$\to \prod_{n=1}^{\infty} P(A_n) = 1 \ \text{Why?}$$

$$\to \forall n \in \mathbb N, P(A_n) = 1$$

$$\to \forall n \in \mathbb N, P\{X_n \le A\} = 1$$

$$\to P( \sup_{n \ge 1} (X_n) < \infty) = 1$$

'direction 2'

Show the contrapositive:

By BCL2, we have

$$P(\limsup A_n^C) = 1 \ (\forall A > 0)$$

$$\to P(\liminf A_n) = 0$$

$$\to \lim_{m \to \infty} P(\bigcap_{n=m}^{\infty} A_n) = 0$$

$$\to \lim_{m \to \infty} \prod_{n=m}^{\infty} P(A_n) = 0$$

$$\to \sup_{m \ge 1} \prod_{n=m}^{\infty} P(A_n) = 0$$

$\to \forall m \ge 1, \exists n \ge m$ s.t. $P(A_n) = 0$

$$\to P(X_n \le A) = 0 \tag{*}$$

Now suppose on the contrary that $P[\sup X_n < \infty] = 1$.

$$\to \sup[X_1, X_2, ...] < \infty \ \text{a.s.}$$

$\to \exists N \ge 0$ s.t.

$$P(\bigcap_{n=1}^{\infty} X_n \le N ) = 1$$

$$\to \prod_{n=1}^{\infty} P(X_n \le N ) = 1$$

$$\to P(X_n \le N ) = 1 \forall n \ge 1$$

Choose $A = N$ in $(*)$. ↯

$$\therefore, \forall A > 0, \sum_{n=1}^{\infty} P(X_n > A) =\infty \to P[\sup X_n < \infty] < 1$$

QED

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