# Probability almost surely

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##### New member
Let be $X_1, X_2, \dots$ independent random variables. My question is how can we show that $sup_n X_n <\infty$ almost surely $\iff \sum_{n=1}^{\infty} \mathbb{P}(X_n>A)<\infty$ for some positive finite A number.

#### chisigma

##### Well-known member
Of course, setting $\displaystyle \text{sup}_{n} X_{n}=B$, for A>B the 'infinite sum' vanishes. Do You intend to get Your question with the hypothesis A<B?...

Kind regards

$\chi$ $\sigma$

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
Of course, setting $\displaystyle \text{sup}_{n} X_{n}=B$, for A>B the 'infinite sum' vanishes.
$$\sup_n X_n$$ is not a number but a function from the sample space to reals (plus infinity)...

#### girdav

##### Member
Hint: use Borel-Cantelli lemma.

• #### chisigma

##### Well-known member
$$\sup_n X_n$$ is not a number but a function from the sample space to reals (plus infinity)...
I'm afraid that the question has been wrongly expressed and in particular there is confusion between the random variables $X_{n}$ and the probabilities $P_{n}= P\{X_{n}>A\}$...

Kind regards

$\chi$ $\sigma$

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
I don't see any confusion in the question. The sequence $X$ is a function of two arguments, i.e., $X:\mathbb{N}\times\Omega\to\mathbb{R}$. We denote $X(n,\cdot)$ by $X_n$. Then $\sup_nX_n$ is a function $\Omega\ni\omega\mapsto\sup_nX_n(\omega)$. In particular, $X_n$ is not a number; otherwise, $\sup_nX_n$ would also be a number and $\sup_nX_n<\infty$ would be either true or false. As it is, $\sup_nX_n<\infty$ has its own Boolean value for each $\omega\in\Omega$. It is an event (i.e., a subset iof $\Omega$) that holds almost surely, i.e., the probability measure of this event is 1.

• ##### New member
Hint: use Borel-Cantelli lemma.
Thank you for everybody who replied to my post, special thanks to girdav for the hint and to Evgeny.Makarov for 'protecting' my question.
Anyway, I am afriad I still can't solve the problem. To tell the truth I don't really see how the lemma could be used in this case. Which form of the lemma do you mean and how does it give the result?

I would be really grateful if you could help me!

#### girdav

##### Member
I meant this result: http://en.wikipedia.org/wiki/Borel–Cantelli_lemma .

If we assume that for some $A>0$ the series $\sum_{n\geq 0}P(X_n\geq A)$ is convergent then by Borel-Cantelli lemma $P(\limsup_n X_n\geq A)=0$ . Now assume that for all $A$ we have that $\sum_{n\geq 0}P(X_n\geq A)$ is divergent, and apply a converse of Borel-Cantelli lemma, which works for independent random variables.

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• ##### New member
I meant this result:.

If we assume that for some $A>0$ the series $\sum_{n\geq 0}P(X_n\geq A)$ is convergent then by Borel-Cantelli lemma $P(\limsup_n X_n\geq A)=0$ . Now assume that for all $A$ we have that $\sum_{n\geq 0}P(X_n\geq A)$ is divergent, and apply a converse of Borel-Cantelli lemma, which works for independent random variables.
Thank you so much for your quick and clear answer. I really appreciate it!

#### Moo

##### New member
I'm afraid that the question has been wrongly expressed and in particular there is confusion between the random variables $X_{n}$ and the probabilities $P_{n}= P\{X_{n}>A\}$...

Kind regards

$\chi$ $\sigma$
I'm afraid your wording is as wrong as ever. You're always putting some weird notations and it's quite obvious that you're not used to using common things in probability. Hence the probability that you understood wrongly is superior to the probability of the question being wrongly worded.

#### chisigma

##### Well-known member
I'm afraid your wording is as wrong as ever...
I'm afraid... following Dante Alighieri's sentence reported in the signature... that I'm no time to waste with monkeys ...

Kind regards

$\chi$ $\sigma$

#### Moo

##### New member
I'm afraid... following Dante Alighieri's sentence reported in the signature... that I'm no time to waste with monkeys ...

Kind regards

$\chi$ $\sigma$
A monkey that can probably speak a better English than yours, but that wouldn't bother looking for the translation of an Italian sentence no one cares about. And I did mean wording, not working.

Sincerely yours,

Monkey cow.