relations between expectation conditioned on subsets and conditional expectation
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Let $mathbb{E}[|Y|]< infty$. Set $mathbb{E}[Y |B] = frac{mathbb{E}[Y mathbb{1}_B]}{P(B)}$ for $B$ with $P(B) > 0$. Suppose that $mathbb{E}[Y | X in [a,b]] in [a,b]$ for all $a<b$ with $mathbb{P}(X in [a,b] ) >0$ holds. My question is whether we can show that this implies that $mathbb{P} ( mathbb{E}[Y |X] in [a,b] | X in [a,b]) =1$, without first showing that $Y$ and $X$ are a.s. equal.
probability conditional-expectation
asked Nov 16 at 20:24
boukkoun
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Let $mathbb{E}[|Y|]< infty$. Set $mathbb{E}[Y |B] = frac{mathbb{E}[Y mathbb{1}_B]}{P(B)}$ for $B$ with $P(B) > 0$. Suppose that $mathbb{E}[Y | X in [a,b]] in [a,b]$ for all $a<b$ with $mathbb{P}(X in [a,b] ) >0$ holds. My question is whether we can show that this implies that $mathbb{P} ( mathbb{E}[Y |X] in [a,b] | X in [a,b]) =1$, without first showing that $Y$ and $X$ are a.s. equal.
probability conditional-expectation
asked Nov 16 at 20:24
boukkoun
676
Definition of $E[Y|X]$ gives this immediately.
– Kavi Rama Murthy
Nov 16 at 23:45
add a comment |
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favorite
up vote
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favorite
Let $mathbb{E}[|Y|]< infty$. Set $mathbb{E}[Y |B] = frac{mathbb{E}[Y mathbb{1}_B]}{P(B)}$ for $B$ with $P(B) > 0$. Suppose that $mathbb{E}[Y | X in [a,b]] in [a,b]$ for all $a<b$ with $mathbb{P}(X in [a,b] ) >0$ holds. My question is whether we can show that this implies that $mathbb{P} ( mathbb{E}[Y |X] in [a,b] | X in [a,b]) =1$, without first showing that $Y$ and $X$ are a.s. equal.
probability conditional-expectation
asked Nov 16 at 20:24
boukkoun
676
Let $mathbb{E}[|Y|]< infty$. Set $mathbb{E}[Y |B] = frac{mathbb{E}[Y mathbb{1}_B]}{P(B)}$ for $B$ with $P(B) > 0$. Suppose that $mathbb{E}[Y | X in [a,b]] in [a,b]$ for all $a<b$ with $mathbb{P}(X in [a,b] ) >0$ holds. My question is whether we can show that this implies that $mathbb{P} ( mathbb{E}[Y |X] in [a,b] | X in [a,b]) =1$, without first showing that $Y$ and $X$ are a.s. equal.
probability conditional-expectation
probability conditional-expectation
asked Nov 16 at 20:24
boukkoun
676
asked Nov 16 at 20:24
boukkoun
676
edited Nov 21 at 6:15
asked Nov 16 at 20:24
boukkoun
676
asked Nov 16 at 20:24
boukkoun
676
asked Nov 16 at 20:24
boukkoun
676
676
Definition of $E[Y|X]$ gives this immediately.
– Kavi Rama Murthy
Nov 16 at 23:45
add a comment |
Definition of $E[Y|X]$ gives this immediately.
– Kavi Rama Murthy
Nov 16 at 23:45
Definition of $E[Y|X]$ gives this immediately.
– Kavi Rama Murthy
Nov 16 at 23:45
Definition of $E[Y|X]$ gives this immediately.
– Kavi Rama Murthy
Nov 16 at 23:45
add a comment |
2 Answers
2
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Yes, it is direct if you abstract a little and call the r.v. $Z_{ab}:=E[Y|Xin [a,b]]$.
This r.v. is defined in $Omega_{ab}subseteq Omega$, where $Omega_{ab}={omegainOmega: X(omega)in [a,b]}$.
Your initial statement "it is given that..." can be expressed as $Z_{ab}in [a,b]$ for all $omegainOmega_{ab}$, which directly implies that the desired probability is one.
answered Nov 16 at 21:22
Rostov
487210
Thanks for your answer! I have 2 concerns though: $mathbb{E}[Y|Xin[a,b]]$ is a constant for each $a$ and $b$, there is nothing random left there. Moreover you would need to relate $mathbb{E}[Y|X]$ in a rigorous way to $Z_{ab}$ in some way. The problem is that it seems very intuitive that the statement I cited holds, and in fact it does hold. So what I was looking for is a rigorous proof for it.
– boukkoun
Nov 17 at 22:29
add a comment |
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$$mathbb{E}[Y|B]=frac{mathbb{E}[Y1_B]}{mathbb{P}(B)}= int_{Omega} Y(omega)frac{1_B(omega)mathbb{P}(domega)}{mathbb{P}(B)} =int_{Omega} Y(omega)mathbb{P}(domega|B).$$
Let $Xsim Unif(0,2)$ and $Ysim Unif(-2,0)$ independent. Then
$$ mathbb{E}[Y|Xin [0,2]]=mathbb{E}[Y]=-1notin [0,2].$$
answered Nov 17 at 5:54
Daniel Camarena Perez
58728
We assume that $mathbb{E}[Y|X in [a,b]] in [a,b]$, of course this is not true in general.
– boukkoun
Nov 17 at 22:21
I am sorry if my wording was not clear, I changed it now.
– boukkoun
Nov 17 at 22:32
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Yes, it is direct if you abstract a little and call the r.v. $Z_{ab}:=E[Y|Xin [a,b]]$.
This r.v. is defined in $Omega_{ab}subseteq Omega$, where $Omega_{ab}={omegainOmega: X(omega)in [a,b]}$.
Your initial statement "it is given that..." can be expressed as $Z_{ab}in [a,b]$ for all $omegainOmega_{ab}$, which directly implies that the desired probability is one.
answered Nov 16 at 21:22
Rostov
487210
Thanks for your answer! I have 2 concerns though: $mathbb{E}[Y|Xin[a,b]]$ is a constant for each $a$ and $b$, there is nothing random left there. Moreover you would need to relate $mathbb{E}[Y|X]$ in a rigorous way to $Z_{ab}$ in some way. The problem is that it seems very intuitive that the statement I cited holds, and in fact it does hold. So what I was looking for is a rigorous proof for it.
– boukkoun
Nov 17 at 22:29
add a comment |
up vote
0
down vote
Yes, it is direct if you abstract a little and call the r.v. $Z_{ab}:=E[Y|Xin [a,b]]$.
This r.v. is defined in $Omega_{ab}subseteq Omega$, where $Omega_{ab}={omegainOmega: X(omega)in [a,b]}$.
Your initial statement "it is given that..." can be expressed as $Z_{ab}in [a,b]$ for all $omegainOmega_{ab}$, which directly implies that the desired probability is one.
answered Nov 16 at 21:22
Rostov
487210
Thanks for your answer! I have 2 concerns though: $mathbb{E}[Y|Xin[a,b]]$ is a constant for each $a$ and $b$, there is nothing random left there. Moreover you would need to relate $mathbb{E}[Y|X]$ in a rigorous way to $Z_{ab}$ in some way. The problem is that it seems very intuitive that the statement I cited holds, and in fact it does hold. So what I was looking for is a rigorous proof for it.
– boukkoun
Nov 17 at 22:29
add a comment |
up vote
0
down vote
up vote
0
down vote
Yes, it is direct if you abstract a little and call the r.v. $Z_{ab}:=E[Y|Xin [a,b]]$.
This r.v. is defined in $Omega_{ab}subseteq Omega$, where $Omega_{ab}={omegainOmega: X(omega)in [a,b]}$.
Your initial statement "it is given that..." can be expressed as $Z_{ab}in [a,b]$ for all $omegainOmega_{ab}$, which directly implies that the desired probability is one.
answered Nov 16 at 21:22
Rostov
487210
Yes, it is direct if you abstract a little and call the r.v. $Z_{ab}:=E[Y|Xin [a,b]]$.
This r.v. is defined in $Omega_{ab}subseteq Omega$, where $Omega_{ab}={omegainOmega: X(omega)in [a,b]}$.
Your initial statement "it is given that..." can be expressed as $Z_{ab}in [a,b]$ for all $omegainOmega_{ab}$, which directly implies that the desired probability is one.
answered Nov 16 at 21:22
Rostov
487210
answered Nov 16 at 21:22
Rostov
487210
answered Nov 16 at 21:22
Rostov
487210
answered Nov 16 at 21:22
Rostov
487210
487210
Thanks for your answer! I have 2 concerns though: $mathbb{E}[Y|Xin[a,b]]$ is a constant for each $a$ and $b$, there is nothing random left there. Moreover you would need to relate $mathbb{E}[Y|X]$ in a rigorous way to $Z_{ab}$ in some way. The problem is that it seems very intuitive that the statement I cited holds, and in fact it does hold. So what I was looking for is a rigorous proof for it.
– boukkoun
Nov 17 at 22:29
add a comment |
Thanks for your answer! I have 2 concerns though: $mathbb{E}[Y|Xin[a,b]]$ is a constant for each $a$ and $b$, there is nothing random left there. Moreover you would need to relate $mathbb{E}[Y|X]$ in a rigorous way to $Z_{ab}$ in some way. The problem is that it seems very intuitive that the statement I cited holds, and in fact it does hold. So what I was looking for is a rigorous proof for it.
– boukkoun
Nov 17 at 22:29
Thanks for your answer! I have 2 concerns though: $mathbb{E}[Y|Xin[a,b]]$ is a constant for each $a$ and $b$, there is nothing random left there. Moreover you would need to relate $mathbb{E}[Y|X]$ in a rigorous way to $Z_{ab}$ in some way. The problem is that it seems very intuitive that the statement I cited holds, and in fact it does hold. So what I was looking for is a rigorous proof for it.
– boukkoun
Nov 17 at 22:29
Thanks for your answer! I have 2 concerns though: $mathbb{E}[Y|Xin[a,b]]$ is a constant for each $a$ and $b$, there is nothing random left there. Moreover you would need to relate $mathbb{E}[Y|X]$ in a rigorous way to $Z_{ab}$ in some way. The problem is that it seems very intuitive that the statement I cited holds, and in fact it does hold. So what I was looking for is a rigorous proof for it.
– boukkoun
Nov 17 at 22:29
add a comment |
up vote
0
down vote
$$mathbb{E}[Y|B]=frac{mathbb{E}[Y1_B]}{mathbb{P}(B)}= int_{Omega} Y(omega)frac{1_B(omega)mathbb{P}(domega)}{mathbb{P}(B)} =int_{Omega} Y(omega)mathbb{P}(domega|B).$$
Let $Xsim Unif(0,2)$ and $Ysim Unif(-2,0)$ independent. Then
$$ mathbb{E}[Y|Xin [0,2]]=mathbb{E}[Y]=-1notin [0,2].$$
answered Nov 17 at 5:54
Daniel Camarena Perez
58728
We assume that $mathbb{E}[Y|X in [a,b]] in [a,b]$, of course this is not true in general.
– boukkoun
Nov 17 at 22:21
I am sorry if my wording was not clear, I changed it now.
– boukkoun
Nov 17 at 22:32
add a comment |
up vote
0
down vote
$$mathbb{E}[Y|B]=frac{mathbb{E}[Y1_B]}{mathbb{P}(B)}= int_{Omega} Y(omega)frac{1_B(omega)mathbb{P}(domega)}{mathbb{P}(B)} =int_{Omega} Y(omega)mathbb{P}(domega|B).$$
Let $Xsim Unif(0,2)$ and $Ysim Unif(-2,0)$ independent. Then
$$ mathbb{E}[Y|Xin [0,2]]=mathbb{E}[Y]=-1notin [0,2].$$
answered Nov 17 at 5:54
Daniel Camarena Perez
58728
We assume that $mathbb{E}[Y|X in [a,b]] in [a,b]$, of course this is not true in general.
– boukkoun
Nov 17 at 22:21
I am sorry if my wording was not clear, I changed it now.
– boukkoun
Nov 17 at 22:32
add a comment |
up vote
0
down vote
up vote
0
down vote
$$mathbb{E}[Y|B]=frac{mathbb{E}[Y1_B]}{mathbb{P}(B)}= int_{Omega} Y(omega)frac{1_B(omega)mathbb{P}(domega)}{mathbb{P}(B)} =int_{Omega} Y(omega)mathbb{P}(domega|B).$$
Let $Xsim Unif(0,2)$ and $Ysim Unif(-2,0)$ independent. Then
$$ mathbb{E}[Y|Xin [0,2]]=mathbb{E}[Y]=-1notin [0,2].$$
answered Nov 17 at 5:54
Daniel Camarena Perez
58728
$$mathbb{E}[Y|B]=frac{mathbb{E}[Y1_B]}{mathbb{P}(B)}= int_{Omega} Y(omega)frac{1_B(omega)mathbb{P}(domega)}{mathbb{P}(B)} =int_{Omega} Y(omega)mathbb{P}(domega|B).$$
Let $Xsim Unif(0,2)$ and $Ysim Unif(-2,0)$ independent. Then
$$ mathbb{E}[Y|Xin [0,2]]=mathbb{E}[Y]=-1notin [0,2].$$
answered Nov 17 at 5:54
Daniel Camarena Perez
58728
answered Nov 17 at 5:54
Daniel Camarena Perez
58728
answered Nov 17 at 5:54
Daniel Camarena Perez
58728
answered Nov 17 at 5:54
Daniel Camarena Perez
58728
58728
We assume that $mathbb{E}[Y|X in [a,b]] in [a,b]$, of course this is not true in general.
– boukkoun
Nov 17 at 22:21
I am sorry if my wording was not clear, I changed it now.
– boukkoun
Nov 17 at 22:32
add a comment |
We assume that $mathbb{E}[Y|X in [a,b]] in [a,b]$, of course this is not true in general.
– boukkoun
Nov 17 at 22:21
I am sorry if my wording was not clear, I changed it now.
– boukkoun
Nov 17 at 22:32
We assume that $mathbb{E}[Y|X in [a,b]] in [a,b]$, of course this is not true in general.
– boukkoun
Nov 17 at 22:21
We assume that $mathbb{E}[Y|X in [a,b]] in [a,b]$, of course this is not true in general.
– boukkoun
Nov 17 at 22:21
I am sorry if my wording was not clear, I changed it now.
– boukkoun
Nov 17 at 22:32
I am sorry if my wording was not clear, I changed it now.
– boukkoun
Nov 17 at 22:32
add a comment |
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Definition of $E[Y|X]$ gives this immediately.
– Kavi Rama Murthy
Nov 16 at 23:45