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- Feb 13, 2012

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Kind regards

$\chi$ $\sigma$

- Jan 30, 2012

- 2,513

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

- Feb 13, 2012

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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 $P_{n}= P\{X_{n}>A\}$...\(\sup_n X_n\) is not a number but a function from the sample space to reals (plus infinity)...

Kind regards

$\chi$ $\sigma$

- Jan 30, 2012

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Thank you for everybody who replied to my post, special thanks toHint: use Borel-Cantelli lemma.

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!

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.

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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Thank you so much for your quick and clear answer. I really appreciate it!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.

- Feb 12, 2012

- 26

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.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

$\chi$ $\sigma$

- Feb 13, 2012

- 1,704

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

Kind regards

$\chi$ $\sigma$

- Feb 12, 2012

- 26

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

Kind regards

$\chi$ $\sigma$

Sincerely yours,

Monkey cow.

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- Jan 26, 2012

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