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
edited Nov 21 at 2:10
Henning Makholm
236k16300534
asked Nov 21 at 1:48
Bayesian guy
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.
add a comment |
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
edited Nov 21 at 2:10
Henning Makholm
236k16300534
asked Nov 21 at 1:48
Bayesian guy
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.
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
add a comment |
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
edited Nov 21 at 2:10
Henning Makholm
236k16300534
asked Nov 21 at 1:48
Bayesian guy
47110
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
edited Nov 21 at 2:10
Henning Makholm
236k16300534
asked Nov 21 at 1:48
Bayesian guy
47110
edited Nov 21 at 2:10
Henning Makholm
236k16300534
asked Nov 21 at 1:48
Bayesian guy
47110
edited Nov 21 at 2:10
Henning Makholm
236k16300534
edited Nov 21 at 2:10
Henning Makholm
236k16300534
edited Nov 21 at 2:10
Henning Makholm
236k16300534
236k16300534
asked Nov 21 at 1:48
Bayesian guy
47110
asked Nov 21 at 1:48
Bayesian guy
47110
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
$X_n=-1,X=1$ is a counterexample.
answered Nov 21 at 9:55
Kavi Rama Murthy
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
add a comment |
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.
answered Nov 21 at 9:55
Kavi Rama Murthy
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
add a comment |
up vote
2
down vote
accepted
$X_n=-1,X=1$ is a counterexample.
answered Nov 21 at 9:55
Kavi Rama Murthy
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
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
$X_n=-1,X=1$ is a counterexample.
answered Nov 21 at 9:55
Kavi Rama Murthy
45.4k31853
$X_n=-1,X=1$ is a counterexample.
answered Nov 21 at 9:55
Kavi Rama Murthy
45.4k31853
answered Nov 21 at 9:55
Kavi Rama Murthy
45.4k31853
answered Nov 21 at 9:55
Kavi Rama Murthy
45.4k31853
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
add a comment |
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
add a comment |
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