Sequence of Independent random variables

Question

Suppose $\{X_n, n \geq 1\}$ are independent random variables. Show $$P[\sup_{n} X_n < \infty]=1$$ iff $$\sum_{n}P[X_n > M]< \infty \text{ for some }M.$$ Proof so far: $\Longleftarrow$ If $\sum_{n}P[X_n > M]< \infty$ for some $M$, then $P\{[X_n > M] \text{ i.o}\}=0$. So $$P \{\liminf[X_n \leq M]\}=1-P \{\limsup[X_n > M]\}=1$$ So we have a sequence of $X_{n},X_{n+1},....$ all less than or equal to $M$. So the intersection of such sets would be the $ \sup_{k\geq n} X_k \leq M <\infty$. Hence $P[\sup_{n} X_n < \infty]=1$

Answer 1

Note: I think the proof below allows the random variables to take values in $(\overline{\mathbb R}, \mathscr B(\overline{\mathbb R}))$, in w/c case @zhoraster's claim may not hold.

If anything below assumes the random variables take values only in $(\mathbb R, \mathscr B(\mathbb R))$, pls point it out.

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'if'

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

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

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

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

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

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

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

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

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

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

'only if'

Show the contrapositive:

By BCL2, we have

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

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

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

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

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

$\to \forall \epsilon > 0, \exists, m \ge 1$ s.t.

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

$\exists n \ge m$ s.t. $P(A_n) = 0$

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

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 $M = N$ in $(*)$. ↯

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

QED