Mixing
If $E|X_n|to E|X|$ then $X_n overset{1}{rightarrow} X$ [closed]
-1
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 '18 at 2:10
Henning Makholm
238k16303538
asked Nov 21 '18 at 1:48
Bayesian guy
47110
closed as off-topic by Batominovski, user10354138, user26857, Lee David Chung Lin, Gibbs Nov 22 '18 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 |
-1
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 '18 at 2:10
Henning Makholm
238k16303538
asked Nov 21 '18 at 1:48
Bayesian guy
47110
closed as off-topic by Batominovski, user10354138, user26857, Lee David Chung Lin, Gibbs Nov 22 '18 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 '18 at 1:52
What does $X_noverset{1}{to}X$ mean? Convergence in $L^1$?
– Batominovski
Nov 21 '18 at 2:13
@Batominovski That's right.
– Bayesian guy
Nov 21 '18 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 '18 at 20:07
add a comment |
-1
-1
-1
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 '18 at 2:10
Henning Makholm
238k16303538
asked Nov 21 '18 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 '18 at 2:10
Henning Makholm
238k16303538
asked Nov 21 '18 at 1:48
Bayesian guy
47110
edited Nov 21 '18 at 2:10
Henning Makholm
238k16303538
asked Nov 21 '18 at 1:48
Bayesian guy
47110
edited Nov 21 '18 at 2:10
Henning Makholm
238k16303538
edited Nov 21 '18 at 2:10
Henning Makholm
238k16303538
edited Nov 21 '18 at 2:10
Henning Makholm
238k16303538
238k16303538
asked Nov 21 '18 at 1:48
Bayesian guy
47110
asked Nov 21 '18 at 1:48
Bayesian guy
47110
asked Nov 21 '18 at 1:48
Bayesian guy
47110
47110
closed as off-topic by Batominovski, user10354138, user26857, Lee David Chung Lin, Gibbs Nov 22 '18 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 '18 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 '18 at 1:52
What does $X_noverset{1}{to}X$ mean? Convergence in $L^1$?
– Batominovski
Nov 21 '18 at 2:13
@Batominovski That's right.
– Bayesian guy
Nov 21 '18 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 '18 at 20:07
add a comment |
1
Hint: Try $X_n,Xgeq 0$ and $mathbb{E}X_n=mathbb{E}X$.
– user10354138
Nov 21 '18 at 1:52
What does $X_noverset{1}{to}X$ mean? Convergence in $L^1$?
– Batominovski
Nov 21 '18 at 2:13
@Batominovski That's right.
– Bayesian guy
Nov 21 '18 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 '18 at 20:07
1
1
Hint: Try $X_n,Xgeq 0$ and $mathbb{E}X_n=mathbb{E}X$.
– user10354138
Nov 21 '18 at 1:52
Hint: Try $X_n,Xgeq 0$ and $mathbb{E}X_n=mathbb{E}X$.
– user10354138
Nov 21 '18 at 1:52
What does $X_noverset{1}{to}X$ mean? Convergence in $L^1$?
– Batominovski
Nov 21 '18 at 2:13
What does $X_noverset{1}{to}X$ mean? Convergence in $L^1$?
– Batominovski
Nov 21 '18 at 2:13
@Batominovski That's right.
– Bayesian guy
Nov 21 '18 at 2:16
@Batominovski That's right.
– Bayesian guy
Nov 21 '18 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 '18 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 '18 at 20:07
add a comment |
1 Answer
1
active
oldest
votes
2
$X_n=-1,X=1$ is a counterexample.
answered Nov 21 '18 at 9:55
Kavi Rama Murthy
51k31854
Thanks. And what about the case without the absolute value, is it still false?
– Bayesian guy
Nov 21 '18 at 18:44
1
@Bayesianguy Take $X$ standard normal and $X_n=-X$ for all $n$.
– Kavi Rama Murthy
Nov 21 '18 at 23:08
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
$X_n=-1,X=1$ is a counterexample.
answered Nov 21 '18 at 9:55
Kavi Rama Murthy
51k31854
Thanks. And what about the case without the absolute value, is it still false?
– Bayesian guy
Nov 21 '18 at 18:44
1
@Bayesianguy Take $X$ standard normal and $X_n=-X$ for all $n$.
– Kavi Rama Murthy
Nov 21 '18 at 23:08
add a comment |
2
$X_n=-1,X=1$ is a counterexample.
answered Nov 21 '18 at 9:55
Kavi Rama Murthy
51k31854
Thanks. And what about the case without the absolute value, is it still false?
– Bayesian guy
Nov 21 '18 at 18:44
1
@Bayesianguy Take $X$ standard normal and $X_n=-X$ for all $n$.
– Kavi Rama Murthy
Nov 21 '18 at 23:08
add a comment |
2
2
2
$X_n=-1,X=1$ is a counterexample.
answered Nov 21 '18 at 9:55
Kavi Rama Murthy
51k31854
$X_n=-1,X=1$ is a counterexample.
answered Nov 21 '18 at 9:55
Kavi Rama Murthy
51k31854
answered Nov 21 '18 at 9:55
Kavi Rama Murthy
51k31854
answered Nov 21 '18 at 9:55
Kavi Rama Murthy
51k31854
answered Nov 21 '18 at 9:55
Kavi Rama Murthy
51k31854
51k31854
Thanks. And what about the case without the absolute value, is it still false?
– Bayesian guy
Nov 21 '18 at 18:44
1
@Bayesianguy Take $X$ standard normal and $X_n=-X$ for all $n$.
– Kavi Rama Murthy
Nov 21 '18 at 23:08
add a comment |
Thanks. And what about the case without the absolute value, is it still false?
– Bayesian guy
Nov 21 '18 at 18:44
1
@Bayesianguy Take $X$ standard normal and $X_n=-X$ for all $n$.
– Kavi Rama Murthy
Nov 21 '18 at 23:08
Thanks. And what about the case without the absolute value, is it still false?
– Bayesian guy
Nov 21 '18 at 18:44
Thanks. And what about the case without the absolute value, is it still false?
– Bayesian guy
Nov 21 '18 at 18:44
1
1
@Bayesianguy Take $X$ standard normal and $X_n=-X$ for all $n$.
– Kavi Rama Murthy
Nov 21 '18 at 23:08
@Bayesianguy Take $X$ standard normal and $X_n=-X$ for all $n$.
– Kavi Rama Murthy
Nov 21 '18 at 23:08
add a comment |
1
Hint: Try $X_n,Xgeq 0$ and $mathbb{E}X_n=mathbb{E}X$.
– user10354138
Nov 21 '18 at 1:52
What does $X_noverset{1}{to}X$ mean? Convergence in $L^1$?
– Batominovski
Nov 21 '18 at 2:13
@Batominovski That's right.
– Bayesian guy
Nov 21 '18 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 '18 at 20:07