Convergence of conditional expectations equivalence

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Suppose that as sequence of random variables $X_n$ and $X$ are defined on $(Omega,mathcalF,P)$ and $mathcalGsubset mathcalF$.
I know that for any $Gin mathcalG$,

$$
E[f(X_n)1_G]to E[f(X)1_G]
$$

for some bounded function $f$. Does it imply that

$$
E[f(X_n)mid mathcalG]to E[f(X)mid mathcalG] quadtexta.s.?
$$

---

The other way is trivial by an application of the dominated convergence theorem, i.e.

$$
lim_n E[f(X_n)1_G]=E[lim_n E[f(X_n)midmathcalG]1_G]= E[f(X)1_G].
$$

probability-theory conditional-expectation

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edited Jul 28 at 0:27

asked Jul 27 at 23:46

akm47

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up vote
1
down vote

favorite

Suppose that as sequence of random variables $X_n$ and $X$ are defined on $(Omega,mathcalF,P)$ and $mathcalGsubset mathcalF$.
I know that for any $Gin mathcalG$,

$$
E[f(X_n)1_G]to E[f(X)1_G]
$$

for some bounded function $f$. Does it imply that

$$
E[f(X_n)mid mathcalG]to E[f(X)mid mathcalG] quadtexta.s.?
$$

---

The other way is trivial by an application of the dominated convergence theorem, i.e.

$$
lim_n E[f(X_n)1_G]=E[lim_n E[f(X_n)midmathcalG]1_G]= E[f(X)1_G].
$$

probability-theory conditional-expectation

share|cite|improve this question

edited Jul 28 at 0:27

asked Jul 27 at 23:46

akm47

184

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Suppose that as sequence of random variables $X_n$ and $X$ are defined on $(Omega,mathcalF,P)$ and $mathcalGsubset mathcalF$.
I know that for any $Gin mathcalG$,

$$
E[f(X_n)1_G]to E[f(X)1_G]
$$

for some bounded function $f$. Does it imply that

$$
E[f(X_n)mid mathcalG]to E[f(X)mid mathcalG] quadtexta.s.?
$$

---

The other way is trivial by an application of the dominated convergence theorem, i.e.

$$
lim_n E[f(X_n)1_G]=E[lim_n E[f(X_n)midmathcalG]1_G]= E[f(X)1_G].
$$

probability-theory conditional-expectation

share|cite|improve this question

edited Jul 28 at 0:27

asked Jul 27 at 23:46

akm47

184

Suppose that as sequence of random variables $X_n$ and $X$ are defined on $(Omega,mathcalF,P)$ and $mathcalGsubset mathcalF$.
I know that for any $Gin mathcalG$,

$$
E[f(X_n)1_G]to E[f(X)1_G]
$$

for some bounded function $f$. Does it imply that

$$
E[f(X_n)mid mathcalG]to E[f(X)mid mathcalG] quadtexta.s.?
$$

---

The other way is trivial by an application of the dominated convergence theorem, i.e.

$$
lim_n E[f(X_n)1_G]=E[lim_n E[f(X_n)midmathcalG]1_G]= E[f(X)1_G].
$$

probability-theory conditional-expectation

share|cite|improve this question

edited Jul 28 at 0:27

asked Jul 27 at 23:46

akm47

184

share|cite|improve this question

share|cite|improve this question

edited Jul 28 at 0:27

edited Jul 28 at 0:27

edited Jul 28 at 0:27

asked Jul 27 at 23:46

akm47

184

asked Jul 27 at 23:46

akm47

184

asked Jul 27 at 23:46

akm47

184

184

add a comment | 

add a comment | 

1 Answer
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No. Take $mathcal G =mathcal F$. Let $X_n to X$ in probability. Then the hypothesis is satisfied for any bounded measurable $f$, But well known examples show that $f(X_n)$ need not converge almost surely. If you want $mathcal G$ to be properly contained in $mathcal F$ take the latter to be Lebesgue measurable sets in $(0,1)$, the former to be Borel sigma algebra, $F$ to be the identity function and consider the standard example of $X_n$ converging in probability but not almost surely.

share|cite|improve this answer

edited Jul 28 at 1:52

answered Jul 28 at 0:25

Kavi Rama Murthy

19.9k2829

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm aware of this example, so I particularly specified that $mathcalG$ is a strict subset of $mathcalF$. Does it change anything?
  – akm47
  Jul 28 at 0:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @akm47 I have edited the answer. I should point out that not every Mathematician uses $subset$ for proper subset. You should have added $mathcal G neq mathcal F$ in the statement.
  – Kavi Rama Murthy
  Jul 28 at 1:54
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

No. Take $mathcal G =mathcal F$. Let $X_n to X$ in probability. Then the hypothesis is satisfied for any bounded measurable $f$, But well known examples show that $f(X_n)$ need not converge almost surely. If you want $mathcal G$ to be properly contained in $mathcal F$ take the latter to be Lebesgue measurable sets in $(0,1)$, the former to be Borel sigma algebra, $F$ to be the identity function and consider the standard example of $X_n$ converging in probability but not almost surely.

share|cite|improve this answer

edited Jul 28 at 1:52

answered Jul 28 at 0:25

Kavi Rama Murthy

19.9k2829

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm aware of this example, so I particularly specified that $mathcalG$ is a strict subset of $mathcalF$. Does it change anything?
  – akm47
  Jul 28 at 0:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @akm47 I have edited the answer. I should point out that not every Mathematician uses $subset$ for proper subset. You should have added $mathcal G neq mathcal F$ in the statement.
  – Kavi Rama Murthy
  Jul 28 at 1:54
  

  
  
  
  

add a comment | 

up vote
2
down vote

No. Take $mathcal G =mathcal F$. Let $X_n to X$ in probability. Then the hypothesis is satisfied for any bounded measurable $f$, But well known examples show that $f(X_n)$ need not converge almost surely. If you want $mathcal G$ to be properly contained in $mathcal F$ take the latter to be Lebesgue measurable sets in $(0,1)$, the former to be Borel sigma algebra, $F$ to be the identity function and consider the standard example of $X_n$ converging in probability but not almost surely.

share|cite|improve this answer

edited Jul 28 at 1:52

answered Jul 28 at 0:25

Kavi Rama Murthy

19.9k2829

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm aware of this example, so I particularly specified that $mathcalG$ is a strict subset of $mathcalF$. Does it change anything?
  – akm47
  Jul 28 at 0:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @akm47 I have edited the answer. I should point out that not every Mathematician uses $subset$ for proper subset. You should have added $mathcal G neq mathcal F$ in the statement.
  – Kavi Rama Murthy
  Jul 28 at 1:54
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

No. Take $mathcal G =mathcal F$. Let $X_n to X$ in probability. Then the hypothesis is satisfied for any bounded measurable $f$, But well known examples show that $f(X_n)$ need not converge almost surely. If you want $mathcal G$ to be properly contained in $mathcal F$ take the latter to be Lebesgue measurable sets in $(0,1)$, the former to be Borel sigma algebra, $F$ to be the identity function and consider the standard example of $X_n$ converging in probability but not almost surely.

share|cite|improve this answer

edited Jul 28 at 1:52

answered Jul 28 at 0:25

Kavi Rama Murthy

19.9k2829

No. Take $mathcal G =mathcal F$. Let $X_n to X$ in probability. Then the hypothesis is satisfied for any bounded measurable $f$, But well known examples show that $f(X_n)$ need not converge almost surely. If you want $mathcal G$ to be properly contained in $mathcal F$ take the latter to be Lebesgue measurable sets in $(0,1)$, the former to be Borel sigma algebra, $F$ to be the identity function and consider the standard example of $X_n$ converging in probability but not almost surely.

share|cite|improve this answer

edited Jul 28 at 1:52

answered Jul 28 at 0:25

Kavi Rama Murthy

19.9k2829

share|cite|improve this answer

share|cite|improve this answer

edited Jul 28 at 1:52

edited Jul 28 at 1:52

edited Jul 28 at 1:52

answered Jul 28 at 0:25

Kavi Rama Murthy

19.9k2829

answered Jul 28 at 0:25

Kavi Rama Murthy

19.9k2829

answered Jul 28 at 0:25

Kavi Rama Murthy

19.9k2829

19.9k2829

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm aware of this example, so I particularly specified that $mathcalG$ is a strict subset of $mathcalF$. Does it change anything?
  – akm47
  Jul 28 at 0:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @akm47 I have edited the answer. I should point out that not every Mathematician uses $subset$ for proper subset. You should have added $mathcal G neq mathcal F$ in the statement.
  – Kavi Rama Murthy
  Jul 28 at 1:54
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm aware of this example, so I particularly specified that $mathcalG$ is a strict subset of $mathcalF$. Does it change anything?
  – akm47
  Jul 28 at 0:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @akm47 I have edited the answer. I should point out that not every Mathematician uses $subset$ for proper subset. You should have added $mathcal G neq mathcal F$ in the statement.
  – Kavi Rama Murthy
  Jul 28 at 1:54
  

  
  
  
  

I'm aware of this example, so I particularly specified that $mathcalG$ is a strict subset of $mathcalF$. Does it change anything?
– akm47
Jul 28 at 0:30

I'm aware of this example, so I particularly specified that $mathcalG$ is a strict subset of $mathcalF$. Does it change anything?
– akm47
Jul 28 at 0:30

@akm47 I have edited the answer. I should point out that not every Mathematician uses $subset$ for proper subset. You should have added $mathcal G neq mathcal F$ in the statement.
– Kavi Rama Murthy
Jul 28 at 1:54

@akm47 I have edited the answer. I should point out that not every Mathematician uses $subset$ for proper subset. You should have added $mathcal G neq mathcal F$ in the statement.
– Kavi Rama Murthy
Jul 28 at 1:54

add a comment | 

 

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