Mixing
Conditional expectation of asymptotically independent random variables
$begingroup$
Suppose that $W_n to W_{infty}$ a.s. where $W_{infty}$ is independent of random variable $V$. Moreover, suppose that $E[|V|]<infty$.
Is it true that
begin{align}
lim_{n toinfty} E[V|W_n]= E[V|W_infty]=E[V] text{ a.s.}
end{align}
The last equality is of course trivial. Therefore, I am looking the proof of the first. This looks like some kind of continuity result, and I am actually not sure if it holds. If the result doesn't hold, I would like to know what extra conditions can be added for it to hold.
probability-theory conditional-expectation
asked Feb 1 at 15:05
BobyBoby
1,0691930
$endgroup$
add a comment |
$begingroup$
Suppose that $W_n to W_{infty}$ a.s. where $W_{infty}$ is independent of random variable $V$. Moreover, suppose that $E[|V|]<infty$.
Is it true that
begin{align}
lim_{n toinfty} E[V|W_n]= E[V|W_infty]=E[V] text{ a.s.}
end{align}
The last equality is of course trivial. Therefore, I am looking the proof of the first. This looks like some kind of continuity result, and I am actually not sure if it holds. If the result doesn't hold, I would like to know what extra conditions can be added for it to hold.
probability-theory conditional-expectation
asked Feb 1 at 15:05
BobyBoby
1,0691930
$endgroup$
add a comment |
$begingroup$
Suppose that $W_n to W_{infty}$ a.s. where $W_{infty}$ is independent of random variable $V$. Moreover, suppose that $E[|V|]<infty$.
Is it true that
begin{align}
lim_{n toinfty} E[V|W_n]= E[V|W_infty]=E[V] text{ a.s.}
end{align}
The last equality is of course trivial. Therefore, I am looking the proof of the first. This looks like some kind of continuity result, and I am actually not sure if it holds. If the result doesn't hold, I would like to know what extra conditions can be added for it to hold.
probability-theory conditional-expectation
asked Feb 1 at 15:05
BobyBoby
1,0691930
$endgroup$
Suppose that $W_n to W_{infty}$ a.s. where $W_{infty}$ is independent of random variable $V$. Moreover, suppose that $E[|V|]<infty$.
Is it true that
begin{align}
lim_{n toinfty} E[V|W_n]= E[V|W_infty]=E[V] text{ a.s.}
end{align}
The last equality is of course trivial. Therefore, I am looking the proof of the first. This looks like some kind of continuity result, and I am actually not sure if it holds. If the result doesn't hold, I would like to know what extra conditions can be added for it to hold.
probability-theory conditional-expectation
probability-theory conditional-expectation
asked Feb 1 at 15:05
BobyBoby
1,0691930
asked Feb 1 at 15:05
BobyBoby
1,0691930
edited Feb 1 at 19:00
Boby
asked Feb 1 at 15:05
BobyBoby
1,0691930
asked Feb 1 at 15:05
BobyBoby
1,0691930
asked Feb 1 at 15:05
BobyBoby
1,0691930
1,0691930
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Let $V$ be any integrable random variable, $Y=V$ and $W_n=Y/n$. Then $W_nto 0=:W_infty$ which is independent of $V$ but the $sigma$-algebra generated by $W_n$ is the same as that generated by $V$ hence $mathbb Eleft[Vmid W_nright]=V$ and we get a counter-example for any non-degenerated $V$.
answered Feb 2 at 11:02
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
$begingroup$
Thank you for the counterexample. Do you think we can prove this under some additional assumptions?
$endgroup$
– Boby
Feb 2 at 16:18
$begingroup$
The only non trivial thing I have in head is when the sequence of $sigma$-algebras $(sigmaleft(W_nright))_n$ is non-increasing.
$endgroup$
– Davide Giraudo
Feb 2 at 17:50
$begingroup$
You mean when it is a filtration?
$endgroup$
– Boby
Feb 3 at 2:06
$begingroup$
A reversed filtration, yes.
$endgroup$
– Davide Giraudo
Feb 3 at 10:26
$begingroup$
That would be the martingale convergence, right?
$endgroup$
– Boby
Feb 3 at 14:59
|
show 3 more comments
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1 Answer
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1 Answer
1
active
oldest
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$begingroup$
Let $V$ be any integrable random variable, $Y=V$ and $W_n=Y/n$. Then $W_nto 0=:W_infty$ which is independent of $V$ but the $sigma$-algebra generated by $W_n$ is the same as that generated by $V$ hence $mathbb Eleft[Vmid W_nright]=V$ and we get a counter-example for any non-degenerated $V$.
answered Feb 2 at 11:02
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
$begingroup$
Thank you for the counterexample. Do you think we can prove this under some additional assumptions?
$endgroup$
– Boby
Feb 2 at 16:18
$begingroup$
The only non trivial thing I have in head is when the sequence of $sigma$-algebras $(sigmaleft(W_nright))_n$ is non-increasing.
$endgroup$
– Davide Giraudo
Feb 2 at 17:50
$begingroup$
You mean when it is a filtration?
$endgroup$
– Boby
Feb 3 at 2:06
$begingroup$
A reversed filtration, yes.
$endgroup$
– Davide Giraudo
Feb 3 at 10:26
$begingroup$
That would be the martingale convergence, right?
$endgroup$
– Boby
Feb 3 at 14:59
|
show 3 more comments
$begingroup$
Let $V$ be any integrable random variable, $Y=V$ and $W_n=Y/n$. Then $W_nto 0=:W_infty$ which is independent of $V$ but the $sigma$-algebra generated by $W_n$ is the same as that generated by $V$ hence $mathbb Eleft[Vmid W_nright]=V$ and we get a counter-example for any non-degenerated $V$.
answered Feb 2 at 11:02
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
$begingroup$
Thank you for the counterexample. Do you think we can prove this under some additional assumptions?
$endgroup$
– Boby
Feb 2 at 16:18
$begingroup$
The only non trivial thing I have in head is when the sequence of $sigma$-algebras $(sigmaleft(W_nright))_n$ is non-increasing.
$endgroup$
– Davide Giraudo
Feb 2 at 17:50
$begingroup$
You mean when it is a filtration?
$endgroup$
– Boby
Feb 3 at 2:06
$begingroup$
A reversed filtration, yes.
$endgroup$
– Davide Giraudo
Feb 3 at 10:26
$begingroup$
That would be the martingale convergence, right?
$endgroup$
– Boby
Feb 3 at 14:59
|
show 3 more comments
$begingroup$
Let $V$ be any integrable random variable, $Y=V$ and $W_n=Y/n$. Then $W_nto 0=:W_infty$ which is independent of $V$ but the $sigma$-algebra generated by $W_n$ is the same as that generated by $V$ hence $mathbb Eleft[Vmid W_nright]=V$ and we get a counter-example for any non-degenerated $V$.
answered Feb 2 at 11:02
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
Let $V$ be any integrable random variable, $Y=V$ and $W_n=Y/n$. Then $W_nto 0=:W_infty$ which is independent of $V$ but the $sigma$-algebra generated by $W_n$ is the same as that generated by $V$ hence $mathbb Eleft[Vmid W_nright]=V$ and we get a counter-example for any non-degenerated $V$.
answered Feb 2 at 11:02
Davide GiraudoDavide Giraudo
128k17156268
edited Feb 9 at 13:10
answered Feb 2 at 11:02
Davide GiraudoDavide Giraudo
128k17156268
answered Feb 2 at 11:02
Davide GiraudoDavide Giraudo
128k17156268
answered Feb 2 at 11:02
Davide GiraudoDavide Giraudo
128k17156268
128k17156268
$begingroup$
Thank you for the counterexample. Do you think we can prove this under some additional assumptions?
$endgroup$
– Boby
Feb 2 at 16:18
$begingroup$
The only non trivial thing I have in head is when the sequence of $sigma$-algebras $(sigmaleft(W_nright))_n$ is non-increasing.
$endgroup$
– Davide Giraudo
Feb 2 at 17:50
$begingroup$
You mean when it is a filtration?
$endgroup$
– Boby
Feb 3 at 2:06
$begingroup$
A reversed filtration, yes.
$endgroup$
– Davide Giraudo
Feb 3 at 10:26
$begingroup$
That would be the martingale convergence, right?
$endgroup$
– Boby
Feb 3 at 14:59
|
show 3 more comments
$begingroup$
Thank you for the counterexample. Do you think we can prove this under some additional assumptions?
$endgroup$
– Boby
Feb 2 at 16:18
$begingroup$
The only non trivial thing I have in head is when the sequence of $sigma$-algebras $(sigmaleft(W_nright))_n$ is non-increasing.
$endgroup$
– Davide Giraudo
Feb 2 at 17:50
$begingroup$
You mean when it is a filtration?
$endgroup$
– Boby
Feb 3 at 2:06
$begingroup$
A reversed filtration, yes.
$endgroup$
– Davide Giraudo
Feb 3 at 10:26
$begingroup$
That would be the martingale convergence, right?
$endgroup$
– Boby
Feb 3 at 14:59
$begingroup$
Thank you for the counterexample. Do you think we can prove this under some additional assumptions?
$endgroup$
– Boby
Feb 2 at 16:18
$begingroup$
Thank you for the counterexample. Do you think we can prove this under some additional assumptions?
$endgroup$
– Boby
Feb 2 at 16:18
$begingroup$
The only non trivial thing I have in head is when the sequence of $sigma$-algebras $(sigmaleft(W_nright))_n$ is non-increasing.
$endgroup$
– Davide Giraudo
Feb 2 at 17:50
$begingroup$
The only non trivial thing I have in head is when the sequence of $sigma$-algebras $(sigmaleft(W_nright))_n$ is non-increasing.
$endgroup$
– Davide Giraudo
Feb 2 at 17:50
$begingroup$
You mean when it is a filtration?
$endgroup$
– Boby
Feb 3 at 2:06
$begingroup$
You mean when it is a filtration?
$endgroup$
– Boby
Feb 3 at 2:06
$begingroup$
A reversed filtration, yes.
$endgroup$
– Davide Giraudo
Feb 3 at 10:26
$begingroup$
A reversed filtration, yes.
$endgroup$
– Davide Giraudo
Feb 3 at 10:26
$begingroup$
That would be the martingale convergence, right?
$endgroup$
– Boby
Feb 3 at 14:59
$begingroup$
That would be the martingale convergence, right?
$endgroup$
– Boby
Feb 3 at 14:59
|
show 3 more comments
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