Suppose $X_n rightarrow X$ a.s. and for each $n$, $X_n perp textit F$. Then is it true that $X perp textit...
For random variables $X_n$ and $X$, suppose $X_n rightarrow X$ a.s. and for each $n$, $X_n perp mathcal F$ i.e. independent with $mathcal F$.
My question is:
Is it true that $X perp mathcal F$ ?
I know that I can assume $X_n rightarrow X$ pointwise since measure zero sets can be ignored when considering independence. Then, I know $X$ is $sigma (X_n : n geq1)$ measurable. By assumption, $sigma (X_n) perp mathcal F$ for all $n$. I tried to make use of Dynkin's lemma by constructing a pi system which generates $sigma (X_n : n geq1)$ and is independent with $mathcal F$ , but it gets me nowhere.
Any hint is appreciated.
probability random-variables independence probability-limit-theorems
asked Nov 23 at 13:28
izimath
366110
add a comment |
For random variables $X_n$ and $X$, suppose $X_n rightarrow X$ a.s. and for each $n$, $X_n perp mathcal F$ i.e. independent with $mathcal F$.
My question is:
Is it true that $X perp mathcal F$ ?
I know that I can assume $X_n rightarrow X$ pointwise since measure zero sets can be ignored when considering independence. Then, I know $X$ is $sigma (X_n : n geq1)$ measurable. By assumption, $sigma (X_n) perp mathcal F$ for all $n$. I tried to make use of Dynkin's lemma by constructing a pi system which generates $sigma (X_n : n geq1)$ and is independent with $mathcal F$ , but it gets me nowhere.
Any hint is appreciated.
probability random-variables independence probability-limit-theorems
asked Nov 23 at 13:28
izimath
366110
Why can't you do something like this: $E[XF] = E[lim_n X_n F] = lim_n E[X_n F] = lim_n E[X_n] E[F] = E[X] E[F]$?
– mathworker21
Nov 23 at 13:32
@mathworker21 I think you are tring to use bounded convergence theorem to $1_B (X_n)$ for some borel set $B$, but $1_B (X_n)$ does not converge to $1_B (X)$
– izimath
Nov 23 at 13:46
add a comment |
For random variables $X_n$ and $X$, suppose $X_n rightarrow X$ a.s. and for each $n$, $X_n perp mathcal F$ i.e. independent with $mathcal F$.
My question is:
Is it true that $X perp mathcal F$ ?
I know that I can assume $X_n rightarrow X$ pointwise since measure zero sets can be ignored when considering independence. Then, I know $X$ is $sigma (X_n : n geq1)$ measurable. By assumption, $sigma (X_n) perp mathcal F$ for all $n$. I tried to make use of Dynkin's lemma by constructing a pi system which generates $sigma (X_n : n geq1)$ and is independent with $mathcal F$ , but it gets me nowhere.
Any hint is appreciated.
probability random-variables independence probability-limit-theorems
asked Nov 23 at 13:28
izimath
366110
For random variables $X_n$ and $X$, suppose $X_n rightarrow X$ a.s. and for each $n$, $X_n perp mathcal F$ i.e. independent with $mathcal F$.
My question is:
Is it true that $X perp mathcal F$ ?
I know that I can assume $X_n rightarrow X$ pointwise since measure zero sets can be ignored when considering independence. Then, I know $X$ is $sigma (X_n : n geq1)$ measurable. By assumption, $sigma (X_n) perp mathcal F$ for all $n$. I tried to make use of Dynkin's lemma by constructing a pi system which generates $sigma (X_n : n geq1)$ and is independent with $mathcal F$ , but it gets me nowhere.
Any hint is appreciated.
probability random-variables independence probability-limit-theorems
probability random-variables independence probability-limit-theorems
asked Nov 23 at 13:28
izimath
366110
asked Nov 23 at 13:28
izimath
366110
edited Nov 23 at 14:13
asked Nov 23 at 13:28
izimath
366110
asked Nov 23 at 13:28
izimath
366110
asked Nov 23 at 13:28
izimath
366110
366110
Why can't you do something like this: $E[XF] = E[lim_n X_n F] = lim_n E[X_n F] = lim_n E[X_n] E[F] = E[X] E[F]$?
– mathworker21
Nov 23 at 13:32
@mathworker21 I think you are tring to use bounded convergence theorem to $1_B (X_n)$ for some borel set $B$, but $1_B (X_n)$ does not converge to $1_B (X)$
– izimath
Nov 23 at 13:46
add a comment |
Why can't you do something like this: $E[XF] = E[lim_n X_n F] = lim_n E[X_n F] = lim_n E[X_n] E[F] = E[X] E[F]$?
– mathworker21
Nov 23 at 13:32
@mathworker21 I think you are tring to use bounded convergence theorem to $1_B (X_n)$ for some borel set $B$, but $1_B (X_n)$ does not converge to $1_B (X)$
– izimath
Nov 23 at 13:46
Why can't you do something like this: $E[XF] = E[lim_n X_n F] = lim_n E[X_n F] = lim_n E[X_n] E[F] = E[X] E[F]$?
– mathworker21
Nov 23 at 13:32
Why can't you do something like this: $E[XF] = E[lim_n X_n F] = lim_n E[X_n F] = lim_n E[X_n] E[F] = E[X] E[F]$?
– mathworker21
Nov 23 at 13:32
@mathworker21 I think you are tring to use bounded convergence theorem to $1_B (X_n)$ for some borel set $B$, but $1_B (X_n)$ does not converge to $1_B (X)$
– izimath
Nov 23 at 13:46
@mathworker21 I think you are tring to use bounded convergence theorem to $1_B (X_n)$ for some borel set $B$, but $1_B (X_n)$ does not converge to $1_B (X)$
– izimath
Nov 23 at 13:46
add a comment |
1 Answer
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We can follow these steps:
Using the dominated convergence theorem, show that for all continuous and bounded function $varphicolonmathbb Rtomathbb R$ and each $Finmathcal F$, the following equality holds:
$$tag{*} mathbb Eleft[varphi(X)mathbf 1_Fright]= mathbb Eleft[varphi(X) right]mathbb P(F).$$Approximation the indicator function of a closed set $S$ by continuous bounded functions, using $d(cdot,S)$ (distance to a set). The monotone convergence theorem allows to extend (*) when $varphi$ is the indicator function of $S$.
- This shows that $mathbb P({Xin S}cap F)=mathbb P(Xin S)mathbb P(F)$ for all closed sets, and using complements, this holds also when $S$ is open. Then we can conclude by the Dynkin's lemma.
answered Nov 23 at 14:10
Davide Giraudo
125k16150259
In the second step, can we use a closed interval $S$, just for simplicity?
– izimath
Nov 23 at 14:23
Yes, and after we have to work the collection of complements of closed intervals instead of the class of open set. So I think the argument can also work.
– Davide Giraudo
Nov 23 at 14:42
Thanks very much. What do you think about the approach that I was trying to do? I don't see a way through.
– izimath
Nov 23 at 14:47
Well, I guess $sigma (X_n : n geq1)$ is too big to be independent with $mathcal F$.
– izimath
Nov 23 at 14:54
add a comment |
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1 Answer
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1 Answer
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oldest
votes
We can follow these steps:
Using the dominated convergence theorem, show that for all continuous and bounded function $varphicolonmathbb Rtomathbb R$ and each $Finmathcal F$, the following equality holds:
$$tag{*} mathbb Eleft[varphi(X)mathbf 1_Fright]= mathbb Eleft[varphi(X) right]mathbb P(F).$$Approximation the indicator function of a closed set $S$ by continuous bounded functions, using $d(cdot,S)$ (distance to a set). The monotone convergence theorem allows to extend (*) when $varphi$ is the indicator function of $S$.
- This shows that $mathbb P({Xin S}cap F)=mathbb P(Xin S)mathbb P(F)$ for all closed sets, and using complements, this holds also when $S$ is open. Then we can conclude by the Dynkin's lemma.
answered Nov 23 at 14:10
Davide Giraudo
125k16150259
In the second step, can we use a closed interval $S$, just for simplicity?
– izimath
Nov 23 at 14:23
Yes, and after we have to work the collection of complements of closed intervals instead of the class of open set. So I think the argument can also work.
– Davide Giraudo
Nov 23 at 14:42
Thanks very much. What do you think about the approach that I was trying to do? I don't see a way through.
– izimath
Nov 23 at 14:47
Well, I guess $sigma (X_n : n geq1)$ is too big to be independent with $mathcal F$.
– izimath
Nov 23 at 14:54
add a comment |
We can follow these steps:
Using the dominated convergence theorem, show that for all continuous and bounded function $varphicolonmathbb Rtomathbb R$ and each $Finmathcal F$, the following equality holds:
$$tag{*} mathbb Eleft[varphi(X)mathbf 1_Fright]= mathbb Eleft[varphi(X) right]mathbb P(F).$$Approximation the indicator function of a closed set $S$ by continuous bounded functions, using $d(cdot,S)$ (distance to a set). The monotone convergence theorem allows to extend (*) when $varphi$ is the indicator function of $S$.
- This shows that $mathbb P({Xin S}cap F)=mathbb P(Xin S)mathbb P(F)$ for all closed sets, and using complements, this holds also when $S$ is open. Then we can conclude by the Dynkin's lemma.
answered Nov 23 at 14:10
Davide Giraudo
125k16150259
In the second step, can we use a closed interval $S$, just for simplicity?
– izimath
Nov 23 at 14:23
Yes, and after we have to work the collection of complements of closed intervals instead of the class of open set. So I think the argument can also work.
– Davide Giraudo
Nov 23 at 14:42
Thanks very much. What do you think about the approach that I was trying to do? I don't see a way through.
– izimath
Nov 23 at 14:47
Well, I guess $sigma (X_n : n geq1)$ is too big to be independent with $mathcal F$.
– izimath
Nov 23 at 14:54
add a comment |
We can follow these steps:
Using the dominated convergence theorem, show that for all continuous and bounded function $varphicolonmathbb Rtomathbb R$ and each $Finmathcal F$, the following equality holds:
$$tag{*} mathbb Eleft[varphi(X)mathbf 1_Fright]= mathbb Eleft[varphi(X) right]mathbb P(F).$$Approximation the indicator function of a closed set $S$ by continuous bounded functions, using $d(cdot,S)$ (distance to a set). The monotone convergence theorem allows to extend (*) when $varphi$ is the indicator function of $S$.
- This shows that $mathbb P({Xin S}cap F)=mathbb P(Xin S)mathbb P(F)$ for all closed sets, and using complements, this holds also when $S$ is open. Then we can conclude by the Dynkin's lemma.
answered Nov 23 at 14:10
Davide Giraudo
125k16150259
We can follow these steps:
Using the dominated convergence theorem, show that for all continuous and bounded function $varphicolonmathbb Rtomathbb R$ and each $Finmathcal F$, the following equality holds:
$$tag{*} mathbb Eleft[varphi(X)mathbf 1_Fright]= mathbb Eleft[varphi(X) right]mathbb P(F).$$Approximation the indicator function of a closed set $S$ by continuous bounded functions, using $d(cdot,S)$ (distance to a set). The monotone convergence theorem allows to extend (*) when $varphi$ is the indicator function of $S$.
- This shows that $mathbb P({Xin S}cap F)=mathbb P(Xin S)mathbb P(F)$ for all closed sets, and using complements, this holds also when $S$ is open. Then we can conclude by the Dynkin's lemma.
answered Nov 23 at 14:10
Davide Giraudo
125k16150259
answered Nov 23 at 14:10
Davide Giraudo
125k16150259
answered Nov 23 at 14:10
Davide Giraudo
125k16150259
answered Nov 23 at 14:10
Davide Giraudo
125k16150259
125k16150259
In the second step, can we use a closed interval $S$, just for simplicity?
– izimath
Nov 23 at 14:23
Yes, and after we have to work the collection of complements of closed intervals instead of the class of open set. So I think the argument can also work.
– Davide Giraudo
Nov 23 at 14:42
Thanks very much. What do you think about the approach that I was trying to do? I don't see a way through.
– izimath
Nov 23 at 14:47
Well, I guess $sigma (X_n : n geq1)$ is too big to be independent with $mathcal F$.
– izimath
Nov 23 at 14:54
add a comment |
In the second step, can we use a closed interval $S$, just for simplicity?
– izimath
Nov 23 at 14:23
Yes, and after we have to work the collection of complements of closed intervals instead of the class of open set. So I think the argument can also work.
– Davide Giraudo
Nov 23 at 14:42
Thanks very much. What do you think about the approach that I was trying to do? I don't see a way through.
– izimath
Nov 23 at 14:47
Well, I guess $sigma (X_n : n geq1)$ is too big to be independent with $mathcal F$.
– izimath
Nov 23 at 14:54
In the second step, can we use a closed interval $S$, just for simplicity?
– izimath
Nov 23 at 14:23
In the second step, can we use a closed interval $S$, just for simplicity?
– izimath
Nov 23 at 14:23
Yes, and after we have to work the collection of complements of closed intervals instead of the class of open set. So I think the argument can also work.
– Davide Giraudo
Nov 23 at 14:42
Yes, and after we have to work the collection of complements of closed intervals instead of the class of open set. So I think the argument can also work.
– Davide Giraudo
Nov 23 at 14:42
Thanks very much. What do you think about the approach that I was trying to do? I don't see a way through.
– izimath
Nov 23 at 14:47
Thanks very much. What do you think about the approach that I was trying to do? I don't see a way through.
– izimath
Nov 23 at 14:47
Well, I guess $sigma (X_n : n geq1)$ is too big to be independent with $mathcal F$.
– izimath
Nov 23 at 14:54
Well, I guess $sigma (X_n : n geq1)$ is too big to be independent with $mathcal F$.
– izimath
Nov 23 at 14:54
add a comment |
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Why can't you do something like this: $E[XF] = E[lim_n X_n F] = lim_n E[X_n F] = lim_n E[X_n] E[F] = E[X] E[F]$?
– mathworker21
Nov 23 at 13:32
@mathworker21 I think you are tring to use bounded convergence theorem to $1_B (X_n)$ for some borel set $B$, but $1_B (X_n)$ does not converge to $1_B (X)$
– izimath
Nov 23 at 13:46