Prove nontrivial conditional expectation relation [closed]
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I wish to prove that for non-factorizable probability distribution function of two variables $W(x,y)ne f(x)g(y)$ the conditional expectation value $E[y^2|x]$ is not constant (that is, has some $x$ dependence). It can be assumed that in general $W(x,y)ne W(x,-y)$. Beside some general considerations I don't have a direction (this is not hw question, but arised as part of research). If it helps, $W(x,y)$ and its derivatives can be considered continuous.
probability functions
edited Jan 10 at 15:47
Did
249k23228466
asked Jan 10 at 14:59
AlexanderAlexander
1385
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closed as off-topic by Namaste, max_zorn, mrtaurho, José Carlos Santos, Chris Custer Jan 23 at 9:00
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, max_zorn, mrtaurho, José Carlos Santos, Chris Custer
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I wish to prove that for non-factorizable probability distribution function of two variables $W(x,y)ne f(x)g(y)$ the conditional expectation value $E[y^2|x]$ is not constant (that is, has some $x$ dependence). It can be assumed that in general $W(x,y)ne W(x,-y)$. Beside some general considerations I don't have a direction (this is not hw question, but arised as part of research). If it helps, $W(x,y)$ and its derivatives can be considered continuous.
probability functions
edited Jan 10 at 15:47
Did
249k23228466
asked Jan 10 at 14:59
AlexanderAlexander
1385
$endgroup$
closed as off-topic by Namaste, max_zorn, mrtaurho, José Carlos Santos, Chris Custer Jan 23 at 9:00
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, max_zorn, mrtaurho, José Carlos Santos, Chris Custer
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Please do not modify significantly your question after answers addressing it are posted.
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– Did
Jan 10 at 15:48
add a comment |
$begingroup$
I wish to prove that for non-factorizable probability distribution function of two variables $W(x,y)ne f(x)g(y)$ the conditional expectation value $E[y^2|x]$ is not constant (that is, has some $x$ dependence). It can be assumed that in general $W(x,y)ne W(x,-y)$. Beside some general considerations I don't have a direction (this is not hw question, but arised as part of research). If it helps, $W(x,y)$ and its derivatives can be considered continuous.
probability functions
edited Jan 10 at 15:47
Did
249k23228466
asked Jan 10 at 14:59
AlexanderAlexander
1385
$endgroup$
I wish to prove that for non-factorizable probability distribution function of two variables $W(x,y)ne f(x)g(y)$ the conditional expectation value $E[y^2|x]$ is not constant (that is, has some $x$ dependence). It can be assumed that in general $W(x,y)ne W(x,-y)$. Beside some general considerations I don't have a direction (this is not hw question, but arised as part of research). If it helps, $W(x,y)$ and its derivatives can be considered continuous.
probability functions
probability functions
edited Jan 10 at 15:47
Did
249k23228466
asked Jan 10 at 14:59
AlexanderAlexander
1385
edited Jan 10 at 15:47
Did
249k23228466
asked Jan 10 at 14:59
AlexanderAlexander
1385
edited Jan 10 at 15:47
Did
249k23228466
edited Jan 10 at 15:47
Did
249k23228466
edited Jan 10 at 15:47
Did
249k23228466
249k23228466
asked Jan 10 at 14:59
AlexanderAlexander
1385
asked Jan 10 at 14:59
AlexanderAlexander
1385
asked Jan 10 at 14:59
AlexanderAlexander
1385
1385
closed as off-topic by Namaste, max_zorn, mrtaurho, José Carlos Santos, Chris Custer Jan 23 at 9:00
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, max_zorn, mrtaurho, José Carlos Santos, Chris Custer
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Namaste, max_zorn, mrtaurho, José Carlos Santos, Chris Custer Jan 23 at 9:00
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, max_zorn, mrtaurho, José Carlos Santos, Chris Custer
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Please do not modify significantly your question after answers addressing it are posted.
$endgroup$
– Did
Jan 10 at 15:48
add a comment |
$begingroup$
Please do not modify significantly your question after answers addressing it are posted.
$endgroup$
– Did
Jan 10 at 15:48
$begingroup$
Please do not modify significantly your question after answers addressing it are posted.
$endgroup$
– Did
Jan 10 at 15:48
$begingroup$
Please do not modify significantly your question after answers addressing it are posted.
$endgroup$
– Did
Jan 10 at 15:48
add a comment |
1 Answer
1
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oldest
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The result does not hold. Here is an example such that $E(y^2mid x)$ is constant but $(x,y)$ is not independent.
Choose some bounded independent random variables $x$ and $z$ such that $E(z)=0$, say, with $x$ uniform on $(0,1)$ and $z$ uniform on $(-1,1)$, and define $$y=sqrt{1+zx}$$ Then $(x,y)$ is not independent since, for example,
$$P(y^2geqslanttfrac32mid x)=left(1-tfrac1{2x}right)mathbf 1_{x>1/2}$$ depends on $x$,
but $$E(y^2mid x)=E(1+zxmid x)=1+E(z)x=1$$ does not depend on $x$.
answered Jan 10 at 15:35
DidDid
249k23228466
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$begingroup$
Thank you! Now we can proceed further - what are sufficient conditions for it to depend on $x$?
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– Alexander
Jan 10 at 15:43
1
$begingroup$
Sorry but this has no nontautological answer. So, no, we cannot "proceed further"...
$endgroup$
– Did
Jan 10 at 15:46
$begingroup$
Can you add the combined PDF $W(x,y)$ please?
$endgroup$
– Alexander
Jan 10 at 15:52
$begingroup$
What do you find difficult in writing down the joint PDF of $(x,y)$, starting from the PDF of $x$ (which is explicit) and the conditional PDF of $y^2$ conditionally on $x$?
$endgroup$
– Did
Jan 10 at 15:59
1
$begingroup$
Seeing me practice won't help you, then.
$endgroup$
– Did
Jan 10 at 16:01
|
show 2 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The result does not hold. Here is an example such that $E(y^2mid x)$ is constant but $(x,y)$ is not independent.
Choose some bounded independent random variables $x$ and $z$ such that $E(z)=0$, say, with $x$ uniform on $(0,1)$ and $z$ uniform on $(-1,1)$, and define $$y=sqrt{1+zx}$$ Then $(x,y)$ is not independent since, for example,
$$P(y^2geqslanttfrac32mid x)=left(1-tfrac1{2x}right)mathbf 1_{x>1/2}$$ depends on $x$,
but $$E(y^2mid x)=E(1+zxmid x)=1+E(z)x=1$$ does not depend on $x$.
answered Jan 10 at 15:35
DidDid
249k23228466
$endgroup$
$begingroup$
Thank you! Now we can proceed further - what are sufficient conditions for it to depend on $x$?
$endgroup$
– Alexander
Jan 10 at 15:43
1
$begingroup$
Sorry but this has no nontautological answer. So, no, we cannot "proceed further"...
$endgroup$
– Did
Jan 10 at 15:46
$begingroup$
Can you add the combined PDF $W(x,y)$ please?
$endgroup$
– Alexander
Jan 10 at 15:52
$begingroup$
What do you find difficult in writing down the joint PDF of $(x,y)$, starting from the PDF of $x$ (which is explicit) and the conditional PDF of $y^2$ conditionally on $x$?
$endgroup$
– Did
Jan 10 at 15:59
1
$begingroup$
Seeing me practice won't help you, then.
$endgroup$
– Did
Jan 10 at 16:01
|
show 2 more comments
$begingroup$
The result does not hold. Here is an example such that $E(y^2mid x)$ is constant but $(x,y)$ is not independent.
Choose some bounded independent random variables $x$ and $z$ such that $E(z)=0$, say, with $x$ uniform on $(0,1)$ and $z$ uniform on $(-1,1)$, and define $$y=sqrt{1+zx}$$ Then $(x,y)$ is not independent since, for example,
$$P(y^2geqslanttfrac32mid x)=left(1-tfrac1{2x}right)mathbf 1_{x>1/2}$$ depends on $x$,
but $$E(y^2mid x)=E(1+zxmid x)=1+E(z)x=1$$ does not depend on $x$.
answered Jan 10 at 15:35
DidDid
249k23228466
$endgroup$
$begingroup$
Thank you! Now we can proceed further - what are sufficient conditions for it to depend on $x$?
$endgroup$
– Alexander
Jan 10 at 15:43
1
$begingroup$
Sorry but this has no nontautological answer. So, no, we cannot "proceed further"...
$endgroup$
– Did
Jan 10 at 15:46
$begingroup$
Can you add the combined PDF $W(x,y)$ please?
$endgroup$
– Alexander
Jan 10 at 15:52
$begingroup$
What do you find difficult in writing down the joint PDF of $(x,y)$, starting from the PDF of $x$ (which is explicit) and the conditional PDF of $y^2$ conditionally on $x$?
$endgroup$
– Did
Jan 10 at 15:59
1
$begingroup$
Seeing me practice won't help you, then.
$endgroup$
– Did
Jan 10 at 16:01
|
show 2 more comments
$begingroup$
The result does not hold. Here is an example such that $E(y^2mid x)$ is constant but $(x,y)$ is not independent.
Choose some bounded independent random variables $x$ and $z$ such that $E(z)=0$, say, with $x$ uniform on $(0,1)$ and $z$ uniform on $(-1,1)$, and define $$y=sqrt{1+zx}$$ Then $(x,y)$ is not independent since, for example,
$$P(y^2geqslanttfrac32mid x)=left(1-tfrac1{2x}right)mathbf 1_{x>1/2}$$ depends on $x$,
but $$E(y^2mid x)=E(1+zxmid x)=1+E(z)x=1$$ does not depend on $x$.
answered Jan 10 at 15:35
DidDid
249k23228466
$endgroup$
The result does not hold. Here is an example such that $E(y^2mid x)$ is constant but $(x,y)$ is not independent.
Choose some bounded independent random variables $x$ and $z$ such that $E(z)=0$, say, with $x$ uniform on $(0,1)$ and $z$ uniform on $(-1,1)$, and define $$y=sqrt{1+zx}$$ Then $(x,y)$ is not independent since, for example,
$$P(y^2geqslanttfrac32mid x)=left(1-tfrac1{2x}right)mathbf 1_{x>1/2}$$ depends on $x$,
but $$E(y^2mid x)=E(1+zxmid x)=1+E(z)x=1$$ does not depend on $x$.
answered Jan 10 at 15:35
DidDid
249k23228466
answered Jan 10 at 15:35
DidDid
249k23228466
answered Jan 10 at 15:35
DidDid
249k23228466
answered Jan 10 at 15:35
DidDid
249k23228466
249k23228466
$begingroup$
Thank you! Now we can proceed further - what are sufficient conditions for it to depend on $x$?
$endgroup$
– Alexander
Jan 10 at 15:43
1
$begingroup$
Sorry but this has no nontautological answer. So, no, we cannot "proceed further"...
$endgroup$
– Did
Jan 10 at 15:46
$begingroup$
Can you add the combined PDF $W(x,y)$ please?
$endgroup$
– Alexander
Jan 10 at 15:52
$begingroup$
What do you find difficult in writing down the joint PDF of $(x,y)$, starting from the PDF of $x$ (which is explicit) and the conditional PDF of $y^2$ conditionally on $x$?
$endgroup$
– Did
Jan 10 at 15:59
1
$begingroup$
Seeing me practice won't help you, then.
$endgroup$
– Did
Jan 10 at 16:01
|
show 2 more comments
$begingroup$
Thank you! Now we can proceed further - what are sufficient conditions for it to depend on $x$?
$endgroup$
– Alexander
Jan 10 at 15:43
1
$begingroup$
Sorry but this has no nontautological answer. So, no, we cannot "proceed further"...
$endgroup$
– Did
Jan 10 at 15:46
$begingroup$
Can you add the combined PDF $W(x,y)$ please?
$endgroup$
– Alexander
Jan 10 at 15:52
$begingroup$
What do you find difficult in writing down the joint PDF of $(x,y)$, starting from the PDF of $x$ (which is explicit) and the conditional PDF of $y^2$ conditionally on $x$?
$endgroup$
– Did
Jan 10 at 15:59
1
$begingroup$
Seeing me practice won't help you, then.
$endgroup$
– Did
Jan 10 at 16:01
$begingroup$
Thank you! Now we can proceed further - what are sufficient conditions for it to depend on $x$?
$endgroup$
– Alexander
Jan 10 at 15:43
$begingroup$
Thank you! Now we can proceed further - what are sufficient conditions for it to depend on $x$?
$endgroup$
– Alexander
Jan 10 at 15:43
1
1
$begingroup$
Sorry but this has no nontautological answer. So, no, we cannot "proceed further"...
$endgroup$
– Did
Jan 10 at 15:46
$begingroup$
Sorry but this has no nontautological answer. So, no, we cannot "proceed further"...
$endgroup$
– Did
Jan 10 at 15:46
$begingroup$
Can you add the combined PDF $W(x,y)$ please?
$endgroup$
– Alexander
Jan 10 at 15:52
$begingroup$
Can you add the combined PDF $W(x,y)$ please?
$endgroup$
– Alexander
Jan 10 at 15:52
$begingroup$
What do you find difficult in writing down the joint PDF of $(x,y)$, starting from the PDF of $x$ (which is explicit) and the conditional PDF of $y^2$ conditionally on $x$?
$endgroup$
– Did
Jan 10 at 15:59
$begingroup$
What do you find difficult in writing down the joint PDF of $(x,y)$, starting from the PDF of $x$ (which is explicit) and the conditional PDF of $y^2$ conditionally on $x$?
$endgroup$
– Did
Jan 10 at 15:59
1
1
$begingroup$
Seeing me practice won't help you, then.
$endgroup$
– Did
Jan 10 at 16:01
$begingroup$
Seeing me practice won't help you, then.
$endgroup$
– Did
Jan 10 at 16:01
|
show 2 more comments
$begingroup$
Please do not modify significantly your question after answers addressing it are posted.
$endgroup$
– Did
Jan 10 at 15:48