Almost sure convergence and Borell - Cantelli Lemma 2












0












$begingroup$


Suppose we have the following random variable:



$X_n = n$ with probability $frac{1}{n}$ and $0$ with probability $1-frac{1}{n}$. We we can define this variable on the probability space $([0,1], mathcal{B}[0,1], lambda)$ where $lambda$ is the Lebesgue measure. Also, we have independence for all $ngeq1$.



It is said that this $X_n$ converges to $0$ almost surely (several sources). However, once we check this by the Borel - Cantelli Lemma 2, we get that $sum_{n=1}^infty P(X_n=n)=infty$. Given that the events are independent, we know that $X_n=n$ infinitely often.



What could be the reason I am receiving this contradiction?






convergence borel-cantelli-lemmas







share|cite|improve this question









$endgroup$












  • $begingroup$
    The source is correct, and never asserts that $(X_n)$ is independent (a hypothesis which you seem to have invented by yourself and which does not hold in the setting of the notes you are referring to).
    $endgroup$
    – Did
    Dec 17 '18 at 10:41










  • $begingroup$
    @Did I used the independence, because they do claim it in the same note later. It may have been a typo in the note. However, could you elaborate why the independence does not hold in this setting?
    $endgroup$
    – Ovi
    Dec 17 '18 at 10:52










  • $begingroup$
    Because they explicitely build $(X_n)$ as $X_n=mathbf 1_{(0,1/n)}$ on $[0,1]$ endowed with its Borel sigma-field and the Lebesgue measure, and these are not independent. "they do claim it in the same note later" Sure they do, but for completely different sequences $(X_n)$. No typo here.
    $endgroup$
    – Did
    Dec 17 '18 at 11:44










  • $begingroup$
    @Did Thank you, I completely got it now!
    $endgroup$
    – Ovi
    Dec 17 '18 at 11:54
















0












$begingroup$


Suppose we have the following random variable:



$X_n = n$ with probability $frac{1}{n}$ and $0$ with probability $1-frac{1}{n}$. We we can define this variable on the probability space $([0,1], mathcal{B}[0,1], lambda)$ where $lambda$ is the Lebesgue measure. Also, we have independence for all $ngeq1$.



It is said that this $X_n$ converges to $0$ almost surely (several sources). However, once we check this by the Borel - Cantelli Lemma 2, we get that $sum_{n=1}^infty P(X_n=n)=infty$. Given that the events are independent, we know that $X_n=n$ infinitely often.



What could be the reason I am receiving this contradiction?






convergence borel-cantelli-lemmas







share|cite|improve this question









$endgroup$












  • $begingroup$
    The source is correct, and never asserts that $(X_n)$ is independent (a hypothesis which you seem to have invented by yourself and which does not hold in the setting of the notes you are referring to).
    $endgroup$
    – Did
    Dec 17 '18 at 10:41










  • $begingroup$
    @Did I used the independence, because they do claim it in the same note later. It may have been a typo in the note. However, could you elaborate why the independence does not hold in this setting?
    $endgroup$
    – Ovi
    Dec 17 '18 at 10:52










  • $begingroup$
    Because they explicitely build $(X_n)$ as $X_n=mathbf 1_{(0,1/n)}$ on $[0,1]$ endowed with its Borel sigma-field and the Lebesgue measure, and these are not independent. "they do claim it in the same note later" Sure they do, but for completely different sequences $(X_n)$. No typo here.
    $endgroup$
    – Did
    Dec 17 '18 at 11:44










  • $begingroup$
    @Did Thank you, I completely got it now!
    $endgroup$
    – Ovi
    Dec 17 '18 at 11:54














0












0








0





$begingroup$


Suppose we have the following random variable:



$X_n = n$ with probability $frac{1}{n}$ and $0$ with probability $1-frac{1}{n}$. We we can define this variable on the probability space $([0,1], mathcal{B}[0,1], lambda)$ where $lambda$ is the Lebesgue measure. Also, we have independence for all $ngeq1$.



It is said that this $X_n$ converges to $0$ almost surely (several sources). However, once we check this by the Borel - Cantelli Lemma 2, we get that $sum_{n=1}^infty P(X_n=n)=infty$. Given that the events are independent, we know that $X_n=n$ infinitely often.



What could be the reason I am receiving this contradiction?






convergence borel-cantelli-lemmas







share|cite|improve this question









$endgroup$




Suppose we have the following random variable:



$X_n = n$ with probability $frac{1}{n}$ and $0$ with probability $1-frac{1}{n}$. We we can define this variable on the probability space $([0,1], mathcal{B}[0,1], lambda)$ where $lambda$ is the Lebesgue measure. Also, we have independence for all $ngeq1$.



It is said that this $X_n$ converges to $0$ almost surely (several sources). However, once we check this by the Borel - Cantelli Lemma 2, we get that $sum_{n=1}^infty P(X_n=n)=infty$. Given that the events are independent, we know that $X_n=n$ infinitely often.



What could be the reason I am receiving this contradiction?






convergence borel-cantelli-lemmas




convergence borel-cantelli-lemmas






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 17 '18 at 10:24









OviOvi

63




63












  • $begingroup$
    The source is correct, and never asserts that $(X_n)$ is independent (a hypothesis which you seem to have invented by yourself and which does not hold in the setting of the notes you are referring to).
    $endgroup$
    – Did
    Dec 17 '18 at 10:41










  • $begingroup$
    @Did I used the independence, because they do claim it in the same note later. It may have been a typo in the note. However, could you elaborate why the independence does not hold in this setting?
    $endgroup$
    – Ovi
    Dec 17 '18 at 10:52










  • $begingroup$
    Because they explicitely build $(X_n)$ as $X_n=mathbf 1_{(0,1/n)}$ on $[0,1]$ endowed with its Borel sigma-field and the Lebesgue measure, and these are not independent. "they do claim it in the same note later" Sure they do, but for completely different sequences $(X_n)$. No typo here.
    $endgroup$
    – Did
    Dec 17 '18 at 11:44










  • $begingroup$
    @Did Thank you, I completely got it now!
    $endgroup$
    – Ovi
    Dec 17 '18 at 11:54


















  • $begingroup$
    The source is correct, and never asserts that $(X_n)$ is independent (a hypothesis which you seem to have invented by yourself and which does not hold in the setting of the notes you are referring to).
    $endgroup$
    – Did
    Dec 17 '18 at 10:41










  • $begingroup$
    @Did I used the independence, because they do claim it in the same note later. It may have been a typo in the note. However, could you elaborate why the independence does not hold in this setting?
    $endgroup$
    – Ovi
    Dec 17 '18 at 10:52










  • $begingroup$
    Because they explicitely build $(X_n)$ as $X_n=mathbf 1_{(0,1/n)}$ on $[0,1]$ endowed with its Borel sigma-field and the Lebesgue measure, and these are not independent. "they do claim it in the same note later" Sure they do, but for completely different sequences $(X_n)$. No typo here.
    $endgroup$
    – Did
    Dec 17 '18 at 11:44










  • $begingroup$
    @Did Thank you, I completely got it now!
    $endgroup$
    – Ovi
    Dec 17 '18 at 11:54
















$begingroup$
The source is correct, and never asserts that $(X_n)$ is independent (a hypothesis which you seem to have invented by yourself and which does not hold in the setting of the notes you are referring to).
$endgroup$
– Did
Dec 17 '18 at 10:41




$begingroup$
The source is correct, and never asserts that $(X_n)$ is independent (a hypothesis which you seem to have invented by yourself and which does not hold in the setting of the notes you are referring to).
$endgroup$
– Did
Dec 17 '18 at 10:41












$begingroup$
@Did I used the independence, because they do claim it in the same note later. It may have been a typo in the note. However, could you elaborate why the independence does not hold in this setting?
$endgroup$
– Ovi
Dec 17 '18 at 10:52




$begingroup$
@Did I used the independence, because they do claim it in the same note later. It may have been a typo in the note. However, could you elaborate why the independence does not hold in this setting?
$endgroup$
– Ovi
Dec 17 '18 at 10:52












$begingroup$
Because they explicitely build $(X_n)$ as $X_n=mathbf 1_{(0,1/n)}$ on $[0,1]$ endowed with its Borel sigma-field and the Lebesgue measure, and these are not independent. "they do claim it in the same note later" Sure they do, but for completely different sequences $(X_n)$. No typo here.
$endgroup$
– Did
Dec 17 '18 at 11:44




$begingroup$
Because they explicitely build $(X_n)$ as $X_n=mathbf 1_{(0,1/n)}$ on $[0,1]$ endowed with its Borel sigma-field and the Lebesgue measure, and these are not independent. "they do claim it in the same note later" Sure they do, but for completely different sequences $(X_n)$. No typo here.
$endgroup$
– Did
Dec 17 '18 at 11:44












$begingroup$
@Did Thank you, I completely got it now!
$endgroup$
– Ovi
Dec 17 '18 at 11:54




$begingroup$
@Did Thank you, I completely got it now!
$endgroup$
– Ovi
Dec 17 '18 at 11:54










1 Answer
1






active

oldest

votes


















0












$begingroup$

This is an answer to the question as posted here. The random variables are not defined explicitly in this question. It is just mentioned that they are defined on $(0,1)$ with Lebesgue measure. Explicit definition was given in the comments after I posted this answer.



I don't know what those sources are but we can only conclude that $X_n to 0$ in probability. It need not converge almost surely.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the comment! Here is the link: nptel.ac.in/courses/108106083/lecture28_Convergence.pdf). Page 3, Example. I also think that, clearly, we can only talk about convergence in probability.
    $endgroup$
    – Ovi
    Dec 17 '18 at 10:39










  • $begingroup$
    See comment on main.
    $endgroup$
    – Did
    Dec 17 '18 at 10:41











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

This is an answer to the question as posted here. The random variables are not defined explicitly in this question. It is just mentioned that they are defined on $(0,1)$ with Lebesgue measure. Explicit definition was given in the comments after I posted this answer.



I don't know what those sources are but we can only conclude that $X_n to 0$ in probability. It need not converge almost surely.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the comment! Here is the link: nptel.ac.in/courses/108106083/lecture28_Convergence.pdf). Page 3, Example. I also think that, clearly, we can only talk about convergence in probability.
    $endgroup$
    – Ovi
    Dec 17 '18 at 10:39










  • $begingroup$
    See comment on main.
    $endgroup$
    – Did
    Dec 17 '18 at 10:41
















0












$begingroup$

This is an answer to the question as posted here. The random variables are not defined explicitly in this question. It is just mentioned that they are defined on $(0,1)$ with Lebesgue measure. Explicit definition was given in the comments after I posted this answer.



I don't know what those sources are but we can only conclude that $X_n to 0$ in probability. It need not converge almost surely.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the comment! Here is the link: nptel.ac.in/courses/108106083/lecture28_Convergence.pdf). Page 3, Example. I also think that, clearly, we can only talk about convergence in probability.
    $endgroup$
    – Ovi
    Dec 17 '18 at 10:39










  • $begingroup$
    See comment on main.
    $endgroup$
    – Did
    Dec 17 '18 at 10:41














0












0








0





$begingroup$

This is an answer to the question as posted here. The random variables are not defined explicitly in this question. It is just mentioned that they are defined on $(0,1)$ with Lebesgue measure. Explicit definition was given in the comments after I posted this answer.



I don't know what those sources are but we can only conclude that $X_n to 0$ in probability. It need not converge almost surely.






share|cite|improve this answer











$endgroup$



This is an answer to the question as posted here. The random variables are not defined explicitly in this question. It is just mentioned that they are defined on $(0,1)$ with Lebesgue measure. Explicit definition was given in the comments after I posted this answer.



I don't know what those sources are but we can only conclude that $X_n to 0$ in probability. It need not converge almost surely.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 17 '18 at 11:48

























answered Dec 17 '18 at 10:34









Kavi Rama MurthyKavi Rama Murthy

66.7k52867




66.7k52867












  • $begingroup$
    Thanks for the comment! Here is the link: nptel.ac.in/courses/108106083/lecture28_Convergence.pdf). Page 3, Example. I also think that, clearly, we can only talk about convergence in probability.
    $endgroup$
    – Ovi
    Dec 17 '18 at 10:39










  • $begingroup$
    See comment on main.
    $endgroup$
    – Did
    Dec 17 '18 at 10:41


















  • $begingroup$
    Thanks for the comment! Here is the link: nptel.ac.in/courses/108106083/lecture28_Convergence.pdf). Page 3, Example. I also think that, clearly, we can only talk about convergence in probability.
    $endgroup$
    – Ovi
    Dec 17 '18 at 10:39










  • $begingroup$
    See comment on main.
    $endgroup$
    – Did
    Dec 17 '18 at 10:41
















$begingroup$
Thanks for the comment! Here is the link: nptel.ac.in/courses/108106083/lecture28_Convergence.pdf). Page 3, Example. I also think that, clearly, we can only talk about convergence in probability.
$endgroup$
– Ovi
Dec 17 '18 at 10:39




$begingroup$
Thanks for the comment! Here is the link: nptel.ac.in/courses/108106083/lecture28_Convergence.pdf). Page 3, Example. I also think that, clearly, we can only talk about convergence in probability.
$endgroup$
– Ovi
Dec 17 '18 at 10:39












$begingroup$
See comment on main.
$endgroup$
– Did
Dec 17 '18 at 10:41




$begingroup$
See comment on main.
$endgroup$
– Did
Dec 17 '18 at 10:41


















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