If $E|X_n|to E|X|$ then $X_n overset{1}{rightarrow} X$ [closed]











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If $E|X_n|to E|X|$ then $X_n overset{1}{rightarrow} X$. I can't find a counterexample to prove that this is not necessarily true. Any hints?






probability







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closed as off-topic by Batominovski, user10354138, user26857, Lee David Chung Lin, Gibbs Nov 22 at 11:09


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Batominovski, user10354138, user26857, Lee David Chung Lin, Gibbs

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 1




    Hint: Try $X_n,Xgeq 0$ and $mathbb{E}X_n=mathbb{E}X$.
    – user10354138
    Nov 21 at 1:52










  • What does $X_noverset{1}{to}X$ mean? Convergence in $L^1$?
    – Batominovski
    Nov 21 at 2:13










  • @Batominovski That's right.
    – Bayesian guy
    Nov 21 at 2:16












  • Forget about the subscript $n$ and the (implied) series. If you have just two r.v.s and $E[Y]=E[X]$, does that imply $Y=X$ in any sense? Surely not.
    – antkam
    Nov 21 at 20:07















up vote
-1
down vote

favorite












If $E|X_n|to E|X|$ then $X_n overset{1}{rightarrow} X$. I can't find a counterexample to prove that this is not necessarily true. Any hints?






probability







share|cite|improve this question















closed as off-topic by Batominovski, user10354138, user26857, Lee David Chung Lin, Gibbs Nov 22 at 11:09


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Batominovski, user10354138, user26857, Lee David Chung Lin, Gibbs

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 1




    Hint: Try $X_n,Xgeq 0$ and $mathbb{E}X_n=mathbb{E}X$.
    – user10354138
    Nov 21 at 1:52










  • What does $X_noverset{1}{to}X$ mean? Convergence in $L^1$?
    – Batominovski
    Nov 21 at 2:13










  • @Batominovski That's right.
    – Bayesian guy
    Nov 21 at 2:16












  • Forget about the subscript $n$ and the (implied) series. If you have just two r.v.s and $E[Y]=E[X]$, does that imply $Y=X$ in any sense? Surely not.
    – antkam
    Nov 21 at 20:07













up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











If $E|X_n|to E|X|$ then $X_n overset{1}{rightarrow} X$. I can't find a counterexample to prove that this is not necessarily true. Any hints?






probability







share|cite|improve this question















If $E|X_n|to E|X|$ then $X_n overset{1}{rightarrow} X$. I can't find a counterexample to prove that this is not necessarily true. Any hints?






probability




probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 21 at 2:10









Henning Makholm

236k16300534




236k16300534










asked Nov 21 at 1:48









Bayesian guy

47110




47110




closed as off-topic by Batominovski, user10354138, user26857, Lee David Chung Lin, Gibbs Nov 22 at 11:09


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Batominovski, user10354138, user26857, Lee David Chung Lin, Gibbs

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Batominovski, user10354138, user26857, Lee David Chung Lin, Gibbs Nov 22 at 11:09


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Batominovski, user10354138, user26857, Lee David Chung Lin, Gibbs

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    Hint: Try $X_n,Xgeq 0$ and $mathbb{E}X_n=mathbb{E}X$.
    – user10354138
    Nov 21 at 1:52










  • What does $X_noverset{1}{to}X$ mean? Convergence in $L^1$?
    – Batominovski
    Nov 21 at 2:13










  • @Batominovski That's right.
    – Bayesian guy
    Nov 21 at 2:16












  • Forget about the subscript $n$ and the (implied) series. If you have just two r.v.s and $E[Y]=E[X]$, does that imply $Y=X$ in any sense? Surely not.
    – antkam
    Nov 21 at 20:07














  • 1




    Hint: Try $X_n,Xgeq 0$ and $mathbb{E}X_n=mathbb{E}X$.
    – user10354138
    Nov 21 at 1:52










  • What does $X_noverset{1}{to}X$ mean? Convergence in $L^1$?
    – Batominovski
    Nov 21 at 2:13










  • @Batominovski That's right.
    – Bayesian guy
    Nov 21 at 2:16












  • Forget about the subscript $n$ and the (implied) series. If you have just two r.v.s and $E[Y]=E[X]$, does that imply $Y=X$ in any sense? Surely not.
    – antkam
    Nov 21 at 20:07








1




1




Hint: Try $X_n,Xgeq 0$ and $mathbb{E}X_n=mathbb{E}X$.
– user10354138
Nov 21 at 1:52




Hint: Try $X_n,Xgeq 0$ and $mathbb{E}X_n=mathbb{E}X$.
– user10354138
Nov 21 at 1:52












What does $X_noverset{1}{to}X$ mean? Convergence in $L^1$?
– Batominovski
Nov 21 at 2:13




What does $X_noverset{1}{to}X$ mean? Convergence in $L^1$?
– Batominovski
Nov 21 at 2:13












@Batominovski That's right.
– Bayesian guy
Nov 21 at 2:16






@Batominovski That's right.
– Bayesian guy
Nov 21 at 2:16














Forget about the subscript $n$ and the (implied) series. If you have just two r.v.s and $E[Y]=E[X]$, does that imply $Y=X$ in any sense? Surely not.
– antkam
Nov 21 at 20:07




Forget about the subscript $n$ and the (implied) series. If you have just two r.v.s and $E[Y]=E[X]$, does that imply $Y=X$ in any sense? Surely not.
– antkam
Nov 21 at 20:07










1 Answer
1






active

oldest

votes

















up vote
2
down vote



accepted










$X_n=-1,X=1$ is a counterexample.






share|cite|improve this answer





















  • Thanks. And what about the case without the absolute value, is it still false?
    – Bayesian guy
    Nov 21 at 18:44






  • 1




    @Bayesianguy Take $X$ standard normal and $X_n=-X$ for all $n$.
    – Kavi Rama Murthy
    Nov 21 at 23:08


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










$X_n=-1,X=1$ is a counterexample.






share|cite|improve this answer





















  • Thanks. And what about the case without the absolute value, is it still false?
    – Bayesian guy
    Nov 21 at 18:44






  • 1




    @Bayesianguy Take $X$ standard normal and $X_n=-X$ for all $n$.
    – Kavi Rama Murthy
    Nov 21 at 23:08















up vote
2
down vote



accepted










$X_n=-1,X=1$ is a counterexample.






share|cite|improve this answer





















  • Thanks. And what about the case without the absolute value, is it still false?
    – Bayesian guy
    Nov 21 at 18:44






  • 1




    @Bayesianguy Take $X$ standard normal and $X_n=-X$ for all $n$.
    – Kavi Rama Murthy
    Nov 21 at 23:08













up vote
2
down vote



accepted







up vote
2
down vote



accepted






$X_n=-1,X=1$ is a counterexample.






share|cite|improve this answer












$X_n=-1,X=1$ is a counterexample.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 21 at 9:55









Kavi Rama Murthy

45.4k31853




45.4k31853












  • Thanks. And what about the case without the absolute value, is it still false?
    – Bayesian guy
    Nov 21 at 18:44






  • 1




    @Bayesianguy Take $X$ standard normal and $X_n=-X$ for all $n$.
    – Kavi Rama Murthy
    Nov 21 at 23:08


















  • Thanks. And what about the case without the absolute value, is it still false?
    – Bayesian guy
    Nov 21 at 18:44






  • 1




    @Bayesianguy Take $X$ standard normal and $X_n=-X$ for all $n$.
    – Kavi Rama Murthy
    Nov 21 at 23:08
















Thanks. And what about the case without the absolute value, is it still false?
– Bayesian guy
Nov 21 at 18:44




Thanks. And what about the case without the absolute value, is it still false?
– Bayesian guy
Nov 21 at 18:44




1




1




@Bayesianguy Take $X$ standard normal and $X_n=-X$ for all $n$.
– Kavi Rama Murthy
Nov 21 at 23:08




@Bayesianguy Take $X$ standard normal and $X_n=-X$ for all $n$.
– Kavi Rama Murthy
Nov 21 at 23:08



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