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












1












$begingroup$


I am proving the following :



Suppose ${X_n : ninmathbb{N}}$ are i.i.d. random variables. Then $P(sup_{ninmathbb{N}}X_n < infty) = 1$ if and only if $ sum_{ninmathbb{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 $sum_{ninmathbb{N}}{P(X_n > M)} < infty.$ 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 $P(sup_{ninmathbb{N}}X_n > M) le sum_{ninmathbb{N}}{P(X_n > M)} = 0.$



Conversely, I've realized that ${sup_{ninmathbb{N}}{X_n} < infty} = bigcup_{m=1}^{infty}bigcap_{n=1}^{infty}{X_n le m}.$






probability-theory measure-theory random-variables independence borel-cantelli-lemmas







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Are you sure the random variables $(X_n)$ are supposed to be i.i.d., or only independent?
    $endgroup$
    – Did
    Dec 10 '18 at 15:20










  • $begingroup$
    I am 100 % sure that they are supposed to be i.i.d., as it was exam problem.
    $endgroup$
    – Euduardo
    Dec 10 '18 at 15:22










  • $begingroup$
    Well then the exam is slightly absurd, but why not.
    $endgroup$
    – Did
    Dec 10 '18 at 15:23










  • $begingroup$
    You're right Did. Now I understand independence is enough, thanks!
    $endgroup$
    – Euduardo
    Dec 11 '18 at 5:09
















1












$begingroup$


I am proving the following :



Suppose ${X_n : ninmathbb{N}}$ are i.i.d. random variables. Then $P(sup_{ninmathbb{N}}X_n < infty) = 1$ if and only if $ sum_{ninmathbb{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 $sum_{ninmathbb{N}}{P(X_n > M)} < infty.$ 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 $P(sup_{ninmathbb{N}}X_n > M) le sum_{ninmathbb{N}}{P(X_n > M)} = 0.$



Conversely, I've realized that ${sup_{ninmathbb{N}}{X_n} < infty} = bigcup_{m=1}^{infty}bigcap_{n=1}^{infty}{X_n le m}.$






probability-theory measure-theory random-variables independence borel-cantelli-lemmas







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Are you sure the random variables $(X_n)$ are supposed to be i.i.d., or only independent?
    $endgroup$
    – Did
    Dec 10 '18 at 15:20










  • $begingroup$
    I am 100 % sure that they are supposed to be i.i.d., as it was exam problem.
    $endgroup$
    – Euduardo
    Dec 10 '18 at 15:22










  • $begingroup$
    Well then the exam is slightly absurd, but why not.
    $endgroup$
    – Did
    Dec 10 '18 at 15:23










  • $begingroup$
    You're right Did. Now I understand independence is enough, thanks!
    $endgroup$
    – Euduardo
    Dec 11 '18 at 5:09














1












1








1





$begingroup$


I am proving the following :



Suppose ${X_n : ninmathbb{N}}$ are i.i.d. random variables. Then $P(sup_{ninmathbb{N}}X_n < infty) = 1$ if and only if $ sum_{ninmathbb{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 $sum_{ninmathbb{N}}{P(X_n > M)} < infty.$ 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 $P(sup_{ninmathbb{N}}X_n > M) le sum_{ninmathbb{N}}{P(X_n > M)} = 0.$



Conversely, I've realized that ${sup_{ninmathbb{N}}{X_n} < infty} = bigcup_{m=1}^{infty}bigcap_{n=1}^{infty}{X_n le m}.$






probability-theory measure-theory random-variables independence borel-cantelli-lemmas







share|cite|improve this question











$endgroup$




I am proving the following :



Suppose ${X_n : ninmathbb{N}}$ are i.i.d. random variables. Then $P(sup_{ninmathbb{N}}X_n < infty) = 1$ if and only if $ sum_{ninmathbb{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 $sum_{ninmathbb{N}}{P(X_n > M)} < infty.$ 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 $P(sup_{ninmathbb{N}}X_n > M) le sum_{ninmathbb{N}}{P(X_n > M)} = 0.$



Conversely, I've realized that ${sup_{ninmathbb{N}}{X_n} < infty} = bigcup_{m=1}^{infty}bigcap_{n=1}^{infty}{X_n le m}.$






probability-theory measure-theory random-variables independence borel-cantelli-lemmas




probability-theory measure-theory random-variables independence borel-cantelli-lemmas






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 15:10







Euduardo

















asked Dec 10 '18 at 14:15









EuduardoEuduardo

1638




1638








  • 1




    $begingroup$
    Are you sure the random variables $(X_n)$ are supposed to be i.i.d., or only independent?
    $endgroup$
    – Did
    Dec 10 '18 at 15:20










  • $begingroup$
    I am 100 % sure that they are supposed to be i.i.d., as it was exam problem.
    $endgroup$
    – Euduardo
    Dec 10 '18 at 15:22










  • $begingroup$
    Well then the exam is slightly absurd, but why not.
    $endgroup$
    – Did
    Dec 10 '18 at 15:23










  • $begingroup$
    You're right Did. Now I understand independence is enough, thanks!
    $endgroup$
    – Euduardo
    Dec 11 '18 at 5:09














  • 1




    $begingroup$
    Are you sure the random variables $(X_n)$ are supposed to be i.i.d., or only independent?
    $endgroup$
    – Did
    Dec 10 '18 at 15:20










  • $begingroup$
    I am 100 % sure that they are supposed to be i.i.d., as it was exam problem.
    $endgroup$
    – Euduardo
    Dec 10 '18 at 15:22










  • $begingroup$
    Well then the exam is slightly absurd, but why not.
    $endgroup$
    – Did
    Dec 10 '18 at 15:23










  • $begingroup$
    You're right Did. Now I understand independence is enough, thanks!
    $endgroup$
    – Euduardo
    Dec 11 '18 at 5:09








1




1




$begingroup$
Are you sure the random variables $(X_n)$ are supposed to be i.i.d., or only independent?
$endgroup$
– Did
Dec 10 '18 at 15:20




$begingroup$
Are you sure the random variables $(X_n)$ are supposed to be i.i.d., or only independent?
$endgroup$
– Did
Dec 10 '18 at 15:20












$begingroup$
I am 100 % sure that they are supposed to be i.i.d., as it was exam problem.
$endgroup$
– Euduardo
Dec 10 '18 at 15:22




$begingroup$
I am 100 % sure that they are supposed to be i.i.d., as it was exam problem.
$endgroup$
– Euduardo
Dec 10 '18 at 15:22












$begingroup$
Well then the exam is slightly absurd, but why not.
$endgroup$
– Did
Dec 10 '18 at 15:23




$begingroup$
Well then the exam is slightly absurd, but why not.
$endgroup$
– Did
Dec 10 '18 at 15:23












$begingroup$
You're right Did. Now I understand independence is enough, thanks!
$endgroup$
– Euduardo
Dec 11 '18 at 5:09




$begingroup$
You're right Did. Now I understand independence is enough, thanks!
$endgroup$
– Euduardo
Dec 11 '18 at 5:09










1 Answer
1






active

oldest

votes


















3












$begingroup$

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 $P(X_i > M) > 0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How do you go from $P[X_i >M] >0$ to $P[sup_n X_n >M] = 1$?
    $endgroup$
    – copper.hat
    Dec 10 '18 at 17:07










  • $begingroup$
    Second Borel-Cantelli lemma.
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 18:21










  • $begingroup$
    Using the fact that $X_n$ are iid, I see. Thanks.
    $endgroup$
    – copper.hat
    Dec 10 '18 at 18:23











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1 Answer
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1 Answer
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active

oldest

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active

oldest

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active

oldest

votes









3












$begingroup$

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 $P(X_i > M) > 0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How do you go from $P[X_i >M] >0$ to $P[sup_n X_n >M] = 1$?
    $endgroup$
    – copper.hat
    Dec 10 '18 at 17:07










  • $begingroup$
    Second Borel-Cantelli lemma.
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 18:21










  • $begingroup$
    Using the fact that $X_n$ are iid, I see. Thanks.
    $endgroup$
    – copper.hat
    Dec 10 '18 at 18:23
















3












$begingroup$

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 $P(X_i > M) > 0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How do you go from $P[X_i >M] >0$ to $P[sup_n X_n >M] = 1$?
    $endgroup$
    – copper.hat
    Dec 10 '18 at 17:07










  • $begingroup$
    Second Borel-Cantelli lemma.
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 18:21










  • $begingroup$
    Using the fact that $X_n$ are iid, I see. Thanks.
    $endgroup$
    – copper.hat
    Dec 10 '18 at 18:23














3












3








3





$begingroup$

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 $P(X_i > M) > 0$.






share|cite|improve this answer









$endgroup$



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 $P(X_i > M) > 0$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 10 '18 at 15:11









Robert IsraelRobert Israel

324k23214468




324k23214468












  • $begingroup$
    How do you go from $P[X_i >M] >0$ to $P[sup_n X_n >M] = 1$?
    $endgroup$
    – copper.hat
    Dec 10 '18 at 17:07










  • $begingroup$
    Second Borel-Cantelli lemma.
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 18:21










  • $begingroup$
    Using the fact that $X_n$ are iid, I see. Thanks.
    $endgroup$
    – copper.hat
    Dec 10 '18 at 18:23


















  • $begingroup$
    How do you go from $P[X_i >M] >0$ to $P[sup_n X_n >M] = 1$?
    $endgroup$
    – copper.hat
    Dec 10 '18 at 17:07










  • $begingroup$
    Second Borel-Cantelli lemma.
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 18:21










  • $begingroup$
    Using the fact that $X_n$ are iid, I see. Thanks.
    $endgroup$
    – copper.hat
    Dec 10 '18 at 18:23
















$begingroup$
How do you go from $P[X_i >M] >0$ to $P[sup_n X_n >M] = 1$?
$endgroup$
– copper.hat
Dec 10 '18 at 17:07




$begingroup$
How do you go from $P[X_i >M] >0$ to $P[sup_n X_n >M] = 1$?
$endgroup$
– copper.hat
Dec 10 '18 at 17:07












$begingroup$
Second Borel-Cantelli lemma.
$endgroup$
– Robert Israel
Dec 10 '18 at 18:21




$begingroup$
Second Borel-Cantelli lemma.
$endgroup$
– Robert Israel
Dec 10 '18 at 18:21












$begingroup$
Using the fact that $X_n$ are iid, I see. Thanks.
$endgroup$
– copper.hat
Dec 10 '18 at 18:23




$begingroup$
Using the fact that $X_n$ are iid, I see. Thanks.
$endgroup$
– copper.hat
Dec 10 '18 at 18:23


















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