How can I prove the backwards analog for the dominated convergence theorem for conditional expectationAn application of the General Lebesgue Dominated convergence theoremApplication of dominated convergence theoremGeneralisation of Dominated Convergence TheoremCharacteristic function and dominated convergence theoremDominated Convergence Theorem.Hypothesis of dominated convergence theoremReal Analysis, Folland The Dominated Convergence TheoremA dominated convergence theorem for the conditional expectation $E( cdot mid mathcalF_n)$ where $mathcalF_n$ loses information over timeDurrett exercise 3.2.5: how to apply dominated convergence?How to apply the Lebesgue Dominated Convergence Theorem
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How can I prove the backwards analog for the dominated convergence theorem for conditional expectation
An application of the General Lebesgue Dominated convergence theoremApplication of dominated convergence theoremGeneralisation of Dominated Convergence TheoremCharacteristic function and dominated convergence theoremDominated Convergence Theorem.Hypothesis of dominated convergence theoremReal Analysis, Folland The Dominated Convergence TheoremA dominated convergence theorem for the conditional expectation $E( cdot mid mathcalF_n)$ where $mathcalF_n$ loses information over timeDurrett exercise 3.2.5: how to apply dominated convergence?How to apply the Lebesgue Dominated Convergence Theorem
$begingroup$
Namely:
Suppose $Y_n rightarrow Y_-infty$ a.s. as $n rightarrow infty$ and $|Y_n|leq Z$ a.s. where $EZ <infty$. If $F_n downarrow F_-infty$ then
$E(Y_n|F_n) rightarrow E(Y_-infty|F_-infty)$ a.s.
This is the dominated convergence theorem for conditional expectations (it's from Durrett)
Also $Y_infty = E(X|F_infty)$
probability-theory measure-theory martingales
asked Sep 12 '17 at 18:14
user477465user477465
173113
$endgroup$
This question has an open bounty worth +50
reputation from user477465 ending ending at 2019-03-19 08:17:18Z">in 2 days.
Looking for an answer drawing from credible and/or official sources.
add a comment |
$begingroup$
Namely:
Suppose $Y_n rightarrow Y_-infty$ a.s. as $n rightarrow infty$ and $|Y_n|leq Z$ a.s. where $EZ <infty$. If $F_n downarrow F_-infty$ then
$E(Y_n|F_n) rightarrow E(Y_-infty|F_-infty)$ a.s.
This is the dominated convergence theorem for conditional expectations (it's from Durrett)
Also $Y_infty = E(X|F_infty)$
probability-theory measure-theory martingales
asked Sep 12 '17 at 18:14
user477465user477465
173113
$endgroup$
This question has an open bounty worth +50
reputation from user477465 ending ending at 2019-03-19 08:17:18Z">in 2 days.
Looking for an answer drawing from credible and/or official sources.
$begingroup$
You have Levy's Downward Theorem (check, e.g., David Williams, 'Probability with Martingales'): for a $Win mathcalL_1left(Omega,mathcalF,mathbbPright)$, you have that $M_n:=Eleft[Wleft|mathcalG_nright.right]$ converges almost surely to $M_-inftyinmathcalL_1(Omega)$ with $M_-infty=Eleft(Wleft|mathcalG_-inftyright.right)$. In the proof that you presented, you just need to apply this result to the identity in the second equation of Theorem 4.6.10. Also, the this result needs to be applied in the last equation when you use Jensen's inequality.
$endgroup$
– Mr. X
Mar 12 at 8:54
$begingroup$
In my previous comment, I am assuming $mathcalG_ndownarrow mathcalG_-infty$, of course.
$endgroup$
– Mr. X
Mar 12 at 8:59
$begingroup$
@Mr.X If you have solved this problem, it would be better to post an answer.
$endgroup$
– Saad
Mar 12 at 9:02
add a comment |
$begingroup$
Namely:
Suppose $Y_n rightarrow Y_-infty$ a.s. as $n rightarrow infty$ and $|Y_n|leq Z$ a.s. where $EZ <infty$. If $F_n downarrow F_-infty$ then
$E(Y_n|F_n) rightarrow E(Y_-infty|F_-infty)$ a.s.
This is the dominated convergence theorem for conditional expectations (it's from Durrett)
Also $Y_infty = E(X|F_infty)$
probability-theory measure-theory martingales
asked Sep 12 '17 at 18:14
user477465user477465
173113
$endgroup$
Namely:
Suppose $Y_n rightarrow Y_-infty$ a.s. as $n rightarrow infty$ and $|Y_n|leq Z$ a.s. where $EZ <infty$. If $F_n downarrow F_-infty$ then
$E(Y_n|F_n) rightarrow E(Y_-infty|F_-infty)$ a.s.
This is the dominated convergence theorem for conditional expectations (it's from Durrett)
Also $Y_infty = E(X|F_infty)$
probability-theory measure-theory martingales
probability-theory measure-theory martingales
asked Sep 12 '17 at 18:14
user477465user477465
173113
asked Sep 12 '17 at 18:14
user477465user477465
173113
edited Mar 12 at 8:17
user477465
asked Sep 12 '17 at 18:14
user477465user477465
173113
asked Sep 12 '17 at 18:14
user477465user477465
173113
asked Sep 12 '17 at 18:14
user477465user477465
173113
173113
This question has an open bounty worth +50
reputation from user477465 ending ending at 2019-03-19 08:17:18Z">in 2 days.
Looking for an answer drawing from credible and/or official sources.
This question has an open bounty worth +50
reputation from user477465 ending ending at 2019-03-19 08:17:18Z">in 2 days.
Looking for an answer drawing from credible and/or official sources.
$begingroup$
You have Levy's Downward Theorem (check, e.g., David Williams, 'Probability with Martingales'): for a $Win mathcalL_1left(Omega,mathcalF,mathbbPright)$, you have that $M_n:=Eleft[Wleft|mathcalG_nright.right]$ converges almost surely to $M_-inftyinmathcalL_1(Omega)$ with $M_-infty=Eleft(Wleft|mathcalG_-inftyright.right)$. In the proof that you presented, you just need to apply this result to the identity in the second equation of Theorem 4.6.10. Also, the this result needs to be applied in the last equation when you use Jensen's inequality.
$endgroup$
– Mr. X
Mar 12 at 8:54
$begingroup$
In my previous comment, I am assuming $mathcalG_ndownarrow mathcalG_-infty$, of course.
$endgroup$
– Mr. X
Mar 12 at 8:59
$begingroup$
@Mr.X If you have solved this problem, it would be better to post an answer.
$endgroup$
– Saad
Mar 12 at 9:02
add a comment |
$begingroup$
You have Levy's Downward Theorem (check, e.g., David Williams, 'Probability with Martingales'): for a $Win mathcalL_1left(Omega,mathcalF,mathbbPright)$, you have that $M_n:=Eleft[Wleft|mathcalG_nright.right]$ converges almost surely to $M_-inftyinmathcalL_1(Omega)$ with $M_-infty=Eleft(Wleft|mathcalG_-inftyright.right)$. In the proof that you presented, you just need to apply this result to the identity in the second equation of Theorem 4.6.10. Also, the this result needs to be applied in the last equation when you use Jensen's inequality.
$endgroup$
– Mr. X
Mar 12 at 8:54
$begingroup$
In my previous comment, I am assuming $mathcalG_ndownarrow mathcalG_-infty$, of course.
$endgroup$
– Mr. X
Mar 12 at 8:59
$begingroup$
@Mr.X If you have solved this problem, it would be better to post an answer.
$endgroup$
– Saad
Mar 12 at 9:02
$begingroup$
You have Levy's Downward Theorem (check, e.g., David Williams, 'Probability with Martingales'): for a $Win mathcalL_1left(Omega,mathcalF,mathbbPright)$, you have that $M_n:=Eleft[Wleft|mathcalG_nright.right]$ converges almost surely to $M_-inftyinmathcalL_1(Omega)$ with $M_-infty=Eleft(Wleft|mathcalG_-inftyright.right)$. In the proof that you presented, you just need to apply this result to the identity in the second equation of Theorem 4.6.10. Also, the this result needs to be applied in the last equation when you use Jensen's inequality.
$endgroup$
– Mr. X
Mar 12 at 8:54
$begingroup$
You have Levy's Downward Theorem (check, e.g., David Williams, 'Probability with Martingales'): for a $Win mathcalL_1left(Omega,mathcalF,mathbbPright)$, you have that $M_n:=Eleft[Wleft|mathcalG_nright.right]$ converges almost surely to $M_-inftyinmathcalL_1(Omega)$ with $M_-infty=Eleft(Wleft|mathcalG_-inftyright.right)$. In the proof that you presented, you just need to apply this result to the identity in the second equation of Theorem 4.6.10. Also, the this result needs to be applied in the last equation when you use Jensen's inequality.
$endgroup$
– Mr. X
Mar 12 at 8:54
$begingroup$
In my previous comment, I am assuming $mathcalG_ndownarrow mathcalG_-infty$, of course.
$endgroup$
– Mr. X
Mar 12 at 8:59
$begingroup$
In my previous comment, I am assuming $mathcalG_ndownarrow mathcalG_-infty$, of course.
$endgroup$
– Mr. X
Mar 12 at 8:59
$begingroup$
@Mr.X If you have solved this problem, it would be better to post an answer.
$endgroup$
– Saad
Mar 12 at 9:02
$begingroup$
@Mr.X If you have solved this problem, it would be better to post an answer.
$endgroup$
– Saad
Mar 12 at 9:02
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You have a useful result -- the backwards analog of Theorem 4.6.8. invoked in the proof of Theorem 14.6.10.:
Levy's 'Downward' Theorem [e.g., David Williams, Probability with Martingales]
Let $Win mathcalL_1(Omega,mathcalF,mathbbP)$; let $mathcalG_ndownarrow mathcalG_-infty$ and define $M_noversetDelta=Eleft[Wleft|mathcalG_nright.right]$. Then, $M_nrightarrow M_-infty$, almost surely, with $M_-infty=Eleft[Wleft|mathcalG_-inftyright.right]$.
Now, we just need to adjust the proof of the forward dominated convergence theorem for conditional expectations that you presented to obtain its backwards counter-part.
The first place we can apply this is in the first identity of the second equation of Theorem 4.6.10, i.e., $lim_nrightarrowinftyEleft[W_Nleft|mathcalF_nright.right]=Eleft[W_Nleft|right.mathcalF_-inftyright]$ for all $N$. This holds in view of Levy's Downward Theorem.
The second place where this result should be applied is in the last part, where it is observed that $Eleft[Yleft|mathcalF_nright.right]rightarrow Eleft[Yleft|mathcalF_-inftyright.right]$, more precisely
$left|Eleft[Y_nleft|mathcalF_nright.right]-Eleft[Yleft|mathcalF_nright.right]right|= left|Eleft[Y_nleft|mathcalF_nright.right]-Eleft[Yleft|mathcalF_nright.right]-Eleft[Yleft|mathcalF_-inftyright.right]+Eleft[Yleft|mathcalF_-inftyright.right]right|$
$geq left|Eleft[Y_nleft|mathcalF_nright.right]-Eleft[Yleft|mathcalF_-inftyright.right]right|-left|Eleft[Yleft|mathcalF_-inftyright.right]-Eleft[Yleft|mathcalF_nright.right]right|$
and observe that the last term in the last inequality above converges to zero, almost surely, in light of Levy's Downward Theorem.
Everything else holds as in the original forward version.
Remark. You apply Levy's Downward Theorem to obtain the Downward (or backwards) version of Theorem 14.6.10 just as you apply Levy's Upward Theorem to prove Theorem 14.6.10. (which in the proof to Theorem 14.6.10 is referred to as Theorem 4.6.8). Levy's Downward Theorem replaces Theorem 4.6.8.
answered Mar 12 at 9:46
Mr. XMr. X
46637
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You have a useful result -- the backwards analog of Theorem 4.6.8. invoked in the proof of Theorem 14.6.10.:
Levy's 'Downward' Theorem [e.g., David Williams, Probability with Martingales]
Let $Win mathcalL_1(Omega,mathcalF,mathbbP)$; let $mathcalG_ndownarrow mathcalG_-infty$ and define $M_noversetDelta=Eleft[Wleft|mathcalG_nright.right]$. Then, $M_nrightarrow M_-infty$, almost surely, with $M_-infty=Eleft[Wleft|mathcalG_-inftyright.right]$.
Now, we just need to adjust the proof of the forward dominated convergence theorem for conditional expectations that you presented to obtain its backwards counter-part.
The first place we can apply this is in the first identity of the second equation of Theorem 4.6.10, i.e., $lim_nrightarrowinftyEleft[W_Nleft|mathcalF_nright.right]=Eleft[W_Nleft|right.mathcalF_-inftyright]$ for all $N$. This holds in view of Levy's Downward Theorem.
The second place where this result should be applied is in the last part, where it is observed that $Eleft[Yleft|mathcalF_nright.right]rightarrow Eleft[Yleft|mathcalF_-inftyright.right]$, more precisely
$left|Eleft[Y_nleft|mathcalF_nright.right]-Eleft[Yleft|mathcalF_nright.right]right|= left|Eleft[Y_nleft|mathcalF_nright.right]-Eleft[Yleft|mathcalF_nright.right]-Eleft[Yleft|mathcalF_-inftyright.right]+Eleft[Yleft|mathcalF_-inftyright.right]right|$
$geq left|Eleft[Y_nleft|mathcalF_nright.right]-Eleft[Yleft|mathcalF_-inftyright.right]right|-left|Eleft[Yleft|mathcalF_-inftyright.right]-Eleft[Yleft|mathcalF_nright.right]right|$
and observe that the last term in the last inequality above converges to zero, almost surely, in light of Levy's Downward Theorem.
Everything else holds as in the original forward version.
Remark. You apply Levy's Downward Theorem to obtain the Downward (or backwards) version of Theorem 14.6.10 just as you apply Levy's Upward Theorem to prove Theorem 14.6.10. (which in the proof to Theorem 14.6.10 is referred to as Theorem 4.6.8). Levy's Downward Theorem replaces Theorem 4.6.8.
answered Mar 12 at 9:46
Mr. XMr. X
46637
$endgroup$
add a comment |
$begingroup$
You have a useful result -- the backwards analog of Theorem 4.6.8. invoked in the proof of Theorem 14.6.10.:
Levy's 'Downward' Theorem [e.g., David Williams, Probability with Martingales]
Let $Win mathcalL_1(Omega,mathcalF,mathbbP)$; let $mathcalG_ndownarrow mathcalG_-infty$ and define $M_noversetDelta=Eleft[Wleft|mathcalG_nright.right]$. Then, $M_nrightarrow M_-infty$, almost surely, with $M_-infty=Eleft[Wleft|mathcalG_-inftyright.right]$.
Now, we just need to adjust the proof of the forward dominated convergence theorem for conditional expectations that you presented to obtain its backwards counter-part.
The first place we can apply this is in the first identity of the second equation of Theorem 4.6.10, i.e., $lim_nrightarrowinftyEleft[W_Nleft|mathcalF_nright.right]=Eleft[W_Nleft|right.mathcalF_-inftyright]$ for all $N$. This holds in view of Levy's Downward Theorem.
The second place where this result should be applied is in the last part, where it is observed that $Eleft[Yleft|mathcalF_nright.right]rightarrow Eleft[Yleft|mathcalF_-inftyright.right]$, more precisely
$left|Eleft[Y_nleft|mathcalF_nright.right]-Eleft[Yleft|mathcalF_nright.right]right|= left|Eleft[Y_nleft|mathcalF_nright.right]-Eleft[Yleft|mathcalF_nright.right]-Eleft[Yleft|mathcalF_-inftyright.right]+Eleft[Yleft|mathcalF_-inftyright.right]right|$
$geq left|Eleft[Y_nleft|mathcalF_nright.right]-Eleft[Yleft|mathcalF_-inftyright.right]right|-left|Eleft[Yleft|mathcalF_-inftyright.right]-Eleft[Yleft|mathcalF_nright.right]right|$
and observe that the last term in the last inequality above converges to zero, almost surely, in light of Levy's Downward Theorem.
Everything else holds as in the original forward version.
Remark. You apply Levy's Downward Theorem to obtain the Downward (or backwards) version of Theorem 14.6.10 just as you apply Levy's Upward Theorem to prove Theorem 14.6.10. (which in the proof to Theorem 14.6.10 is referred to as Theorem 4.6.8). Levy's Downward Theorem replaces Theorem 4.6.8.
answered Mar 12 at 9:46
Mr. XMr. X
46637
$endgroup$
add a comment |
$begingroup$
You have a useful result -- the backwards analog of Theorem 4.6.8. invoked in the proof of Theorem 14.6.10.:
Levy's 'Downward' Theorem [e.g., David Williams, Probability with Martingales]
Let $Win mathcalL_1(Omega,mathcalF,mathbbP)$; let $mathcalG_ndownarrow mathcalG_-infty$ and define $M_noversetDelta=Eleft[Wleft|mathcalG_nright.right]$. Then, $M_nrightarrow M_-infty$, almost surely, with $M_-infty=Eleft[Wleft|mathcalG_-inftyright.right]$.
Now, we just need to adjust the proof of the forward dominated convergence theorem for conditional expectations that you presented to obtain its backwards counter-part.
The first place we can apply this is in the first identity of the second equation of Theorem 4.6.10, i.e., $lim_nrightarrowinftyEleft[W_Nleft|mathcalF_nright.right]=Eleft[W_Nleft|right.mathcalF_-inftyright]$ for all $N$. This holds in view of Levy's Downward Theorem.
The second place where this result should be applied is in the last part, where it is observed that $Eleft[Yleft|mathcalF_nright.right]rightarrow Eleft[Yleft|mathcalF_-inftyright.right]$, more precisely
$left|Eleft[Y_nleft|mathcalF_nright.right]-Eleft[Yleft|mathcalF_nright.right]right|= left|Eleft[Y_nleft|mathcalF_nright.right]-Eleft[Yleft|mathcalF_nright.right]-Eleft[Yleft|mathcalF_-inftyright.right]+Eleft[Yleft|mathcalF_-inftyright.right]right|$
$geq left|Eleft[Y_nleft|mathcalF_nright.right]-Eleft[Yleft|mathcalF_-inftyright.right]right|-left|Eleft[Yleft|mathcalF_-inftyright.right]-Eleft[Yleft|mathcalF_nright.right]right|$
and observe that the last term in the last inequality above converges to zero, almost surely, in light of Levy's Downward Theorem.
Everything else holds as in the original forward version.
Remark. You apply Levy's Downward Theorem to obtain the Downward (or backwards) version of Theorem 14.6.10 just as you apply Levy's Upward Theorem to prove Theorem 14.6.10. (which in the proof to Theorem 14.6.10 is referred to as Theorem 4.6.8). Levy's Downward Theorem replaces Theorem 4.6.8.
answered Mar 12 at 9:46
Mr. XMr. X
46637
$endgroup$
You have a useful result -- the backwards analog of Theorem 4.6.8. invoked in the proof of Theorem 14.6.10.:
Levy's 'Downward' Theorem [e.g., David Williams, Probability with Martingales]
Let $Win mathcalL_1(Omega,mathcalF,mathbbP)$; let $mathcalG_ndownarrow mathcalG_-infty$ and define $M_noversetDelta=Eleft[Wleft|mathcalG_nright.right]$. Then, $M_nrightarrow M_-infty$, almost surely, with $M_-infty=Eleft[Wleft|mathcalG_-inftyright.right]$.
Now, we just need to adjust the proof of the forward dominated convergence theorem for conditional expectations that you presented to obtain its backwards counter-part.
The first place we can apply this is in the first identity of the second equation of Theorem 4.6.10, i.e., $lim_nrightarrowinftyEleft[W_Nleft|mathcalF_nright.right]=Eleft[W_Nleft|right.mathcalF_-inftyright]$ for all $N$. This holds in view of Levy's Downward Theorem.
The second place where this result should be applied is in the last part, where it is observed that $Eleft[Yleft|mathcalF_nright.right]rightarrow Eleft[Yleft|mathcalF_-inftyright.right]$, more precisely
$left|Eleft[Y_nleft|mathcalF_nright.right]-Eleft[Yleft|mathcalF_nright.right]right|= left|Eleft[Y_nleft|mathcalF_nright.right]-Eleft[Yleft|mathcalF_nright.right]-Eleft[Yleft|mathcalF_-inftyright.right]+Eleft[Yleft|mathcalF_-inftyright.right]right|$
$geq left|Eleft[Y_nleft|mathcalF_nright.right]-Eleft[Yleft|mathcalF_-inftyright.right]right|-left|Eleft[Yleft|mathcalF_-inftyright.right]-Eleft[Yleft|mathcalF_nright.right]right|$
and observe that the last term in the last inequality above converges to zero, almost surely, in light of Levy's Downward Theorem.
Everything else holds as in the original forward version.
Remark. You apply Levy's Downward Theorem to obtain the Downward (or backwards) version of Theorem 14.6.10 just as you apply Levy's Upward Theorem to prove Theorem 14.6.10. (which in the proof to Theorem 14.6.10 is referred to as Theorem 4.6.8). Levy's Downward Theorem replaces Theorem 4.6.8.
answered Mar 12 at 9:46
Mr. XMr. X
46637
edited Mar 12 at 12:15
answered Mar 12 at 9:46
Mr. XMr. X
46637
answered Mar 12 at 9:46
Mr. XMr. X
46637
answered Mar 12 at 9:46
Mr. XMr. X
46637
46637
add a comment |
add a comment |
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$begingroup$
You have Levy's Downward Theorem (check, e.g., David Williams, 'Probability with Martingales'): for a $Win mathcalL_1left(Omega,mathcalF,mathbbPright)$, you have that $M_n:=Eleft[Wleft|mathcalG_nright.right]$ converges almost surely to $M_-inftyinmathcalL_1(Omega)$ with $M_-infty=Eleft(Wleft|mathcalG_-inftyright.right)$. In the proof that you presented, you just need to apply this result to the identity in the second equation of Theorem 4.6.10. Also, the this result needs to be applied in the last equation when you use Jensen's inequality.
$endgroup$
– Mr. X
Mar 12 at 8:54
$begingroup$
In my previous comment, I am assuming $mathcalG_ndownarrow mathcalG_-infty$, of course.
$endgroup$
– Mr. X
Mar 12 at 8:59
$begingroup$
@Mr.X If you have solved this problem, it would be better to post an answer.
$endgroup$
– Saad
Mar 12 at 9:02