How to prove this identity for discrete random variables? [closed]
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How to prove that $E(ξE(η|G)) = E(ηE(ξ|G))$, where $ξ$, $G$ and $η$ are random discrete variables. Of course, if both parts exist.
probability-theory random-variables expected-value
asked Nov 20 at 8:02
anykk
595
closed as off-topic by Kavi Rama Murthy, Christopher, Davide Giraudo, Adrian Keister, supinf Nov 20 at 14:10
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Kavi Rama Murthy, Christopher, Davide Giraudo, Adrian Keister, supinf
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
0
down vote
favorite
How to prove that $E(ξE(η|G)) = E(ηE(ξ|G))$, where $ξ$, $G$ and $η$ are random discrete variables. Of course, if both parts exist.
probability-theory random-variables expected-value
asked Nov 20 at 8:02
anykk
595
closed as off-topic by Kavi Rama Murthy, Christopher, Davide Giraudo, Adrian Keister, supinf Nov 20 at 14:10
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Kavi Rama Murthy, Christopher, Davide Giraudo, Adrian Keister, supinf
If this question can be reworded to fit the rules in the help center, please edit the question.
By linearity of expectation, you can change it to $E(xi)E(etamid G) = E(eta)E(ximid G)$. I suspect that that's easier to work with. That being said, what is $G$? I can't imagine that this is true if, for instance, $G$ is $eta=0$.
– Arthur
Nov 20 at 8:10
@Arthur, G is random discrete variable too.
– anykk
Nov 20 at 8:29
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How to prove that $E(ξE(η|G)) = E(ηE(ξ|G))$, where $ξ$, $G$ and $η$ are random discrete variables. Of course, if both parts exist.
probability-theory random-variables expected-value
asked Nov 20 at 8:02
anykk
595
How to prove that $E(ξE(η|G)) = E(ηE(ξ|G))$, where $ξ$, $G$ and $η$ are random discrete variables. Of course, if both parts exist.
probability-theory random-variables expected-value
probability-theory random-variables expected-value
asked Nov 20 at 8:02
anykk
595
asked Nov 20 at 8:02
anykk
595
edited Nov 20 at 8:29
asked Nov 20 at 8:02
anykk
595
asked Nov 20 at 8:02
anykk
595
asked Nov 20 at 8:02
anykk
595
595
closed as off-topic by Kavi Rama Murthy, Christopher, Davide Giraudo, Adrian Keister, supinf Nov 20 at 14:10
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Kavi Rama Murthy, Christopher, Davide Giraudo, Adrian Keister, supinf
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Kavi Rama Murthy, Christopher, Davide Giraudo, Adrian Keister, supinf Nov 20 at 14:10
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Kavi Rama Murthy, Christopher, Davide Giraudo, Adrian Keister, supinf
If this question can be reworded to fit the rules in the help center, please edit the question.
By linearity of expectation, you can change it to $E(xi)E(etamid G) = E(eta)E(ximid G)$. I suspect that that's easier to work with. That being said, what is $G$? I can't imagine that this is true if, for instance, $G$ is $eta=0$.
– Arthur
Nov 20 at 8:10
@Arthur, G is random discrete variable too.
– anykk
Nov 20 at 8:29
add a comment |
By linearity of expectation, you can change it to $E(xi)E(etamid G) = E(eta)E(ximid G)$. I suspect that that's easier to work with. That being said, what is $G$? I can't imagine that this is true if, for instance, $G$ is $eta=0$.
– Arthur
Nov 20 at 8:10
@Arthur, G is random discrete variable too.
– anykk
Nov 20 at 8:29
By linearity of expectation, you can change it to $E(xi)E(etamid G) = E(eta)E(ximid G)$. I suspect that that's easier to work with. That being said, what is $G$? I can't imagine that this is true if, for instance, $G$ is $eta=0$.
– Arthur
Nov 20 at 8:10
By linearity of expectation, you can change it to $E(xi)E(etamid G) = E(eta)E(ximid G)$. I suspect that that's easier to work with. That being said, what is $G$? I can't imagine that this is true if, for instance, $G$ is $eta=0$.
– Arthur
Nov 20 at 8:10
@Arthur, G is random discrete variable too.
– anykk
Nov 20 at 8:29
@Arthur, G is random discrete variable too.
– anykk
Nov 20 at 8:29
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
In this answer $G$ must be looked at as a $sigma$-algebra (eventually generated by a random variable).
Characteristic for $mathbb E[ximid G]$ is that it satisfies:$$int_Axi(omega)P(domega)=int_Amathbb E[ximid G](omega)P(domega)text{ whenever }Atext{ is }Gtext{-measurable}$$
Or equivalently:$$mathbb E[ximathbf1_A]=mathbb E[mathbb E[ximid G]mathbf1_A]text{ whenever }Atext{ is }Gtext{-measurable}$$
This can be expanded to the more general statement that: $$mathbb E[xipsi]=mathbb E[mathbb E[ximid G]psi]text{ whenever }psitext{ is }Gtext{-measurable}$$
Now note that $mathbb E[etamid G]$ is by definition $G$-measurable so that we are allowed to conclude that:$$mathbb E[ximathbb E[etamid G]]=mathbb E[mathbb E[ximid G]mathbb E[etamid G]]tag1$$
Similarly we have:$$mathbb E[etamathbb E[ximid G]]=mathbb E[mathbb E[etamid G]mathbb E[ximid G]]tag2$$
Now note that $(1)$ and $(2)$ have equal RHS.
answered Nov 20 at 8:30
drhab
95.3k543126
Thank you, i'll meditate on it
– anykk
Nov 20 at 8:51
Can you explain me, why the integral and the next line are equivalente? I can't see it. And, how we can expange it to $mathbb E[xi psi]$?
– anykk
Nov 26 at 18:34
By definition $int_AX(omega)P(domega)=int X(omega) mathbf1_A(omega)P(domega)$ and $mathbb E[Xmathbf1_A]$ is just another notation for the RHS. It can be expanded by first proving it for simple functions, then measurable functions that are limits of simple functions. It goes a bit too far to handle this in a comment. I reckon that you must have the disposal of some mathematical material that handles this stuff, don't you?
– drhab
Nov 26 at 18:48
No, I don't. I will learn it later in university. I think, that second proof is more simple, than yours one, but yours is stronger.
– anykk
Nov 26 at 19:26
Then where did you meet the question that you posed? Indeed the other answer (I upvoted it) is more direct. I tried to reveal something of the background of the equalities that you find there.
– drhab
Nov 26 at 19:28
|
show 5 more comments
up vote
2
down vote
From the law of total expectation
begin{align*}
Eleft(xi Eleft(eta|Gright)right) & =Eleft(Eleft(xi Eleft(eta|Gright)|Gright)right)\
& =Eleft(Eleft(eta|Gright)Eleft(xi|Gright)right)\
& =Eleft(Eleft(eta Eleft(xi|Gright)|Gright)right)\
& =Eleft(eta Eleft(xi|Gright)right)
end{align*}
answered Nov 20 at 8:31
hopeless
585
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
In this answer $G$ must be looked at as a $sigma$-algebra (eventually generated by a random variable).
Characteristic for $mathbb E[ximid G]$ is that it satisfies:$$int_Axi(omega)P(domega)=int_Amathbb E[ximid G](omega)P(domega)text{ whenever }Atext{ is }Gtext{-measurable}$$
Or equivalently:$$mathbb E[ximathbf1_A]=mathbb E[mathbb E[ximid G]mathbf1_A]text{ whenever }Atext{ is }Gtext{-measurable}$$
This can be expanded to the more general statement that: $$mathbb E[xipsi]=mathbb E[mathbb E[ximid G]psi]text{ whenever }psitext{ is }Gtext{-measurable}$$
Now note that $mathbb E[etamid G]$ is by definition $G$-measurable so that we are allowed to conclude that:$$mathbb E[ximathbb E[etamid G]]=mathbb E[mathbb E[ximid G]mathbb E[etamid G]]tag1$$
Similarly we have:$$mathbb E[etamathbb E[ximid G]]=mathbb E[mathbb E[etamid G]mathbb E[ximid G]]tag2$$
Now note that $(1)$ and $(2)$ have equal RHS.
answered Nov 20 at 8:30
drhab
95.3k543126
Thank you, i'll meditate on it
– anykk
Nov 20 at 8:51
Can you explain me, why the integral and the next line are equivalente? I can't see it. And, how we can expange it to $mathbb E[xi psi]$?
– anykk
Nov 26 at 18:34
By definition $int_AX(omega)P(domega)=int X(omega) mathbf1_A(omega)P(domega)$ and $mathbb E[Xmathbf1_A]$ is just another notation for the RHS. It can be expanded by first proving it for simple functions, then measurable functions that are limits of simple functions. It goes a bit too far to handle this in a comment. I reckon that you must have the disposal of some mathematical material that handles this stuff, don't you?
– drhab
Nov 26 at 18:48
No, I don't. I will learn it later in university. I think, that second proof is more simple, than yours one, but yours is stronger.
– anykk
Nov 26 at 19:26
Then where did you meet the question that you posed? Indeed the other answer (I upvoted it) is more direct. I tried to reveal something of the background of the equalities that you find there.
– drhab
Nov 26 at 19:28
|
show 5 more comments
up vote
2
down vote
accepted
In this answer $G$ must be looked at as a $sigma$-algebra (eventually generated by a random variable).
Characteristic for $mathbb E[ximid G]$ is that it satisfies:$$int_Axi(omega)P(domega)=int_Amathbb E[ximid G](omega)P(domega)text{ whenever }Atext{ is }Gtext{-measurable}$$
Or equivalently:$$mathbb E[ximathbf1_A]=mathbb E[mathbb E[ximid G]mathbf1_A]text{ whenever }Atext{ is }Gtext{-measurable}$$
This can be expanded to the more general statement that: $$mathbb E[xipsi]=mathbb E[mathbb E[ximid G]psi]text{ whenever }psitext{ is }Gtext{-measurable}$$
Now note that $mathbb E[etamid G]$ is by definition $G$-measurable so that we are allowed to conclude that:$$mathbb E[ximathbb E[etamid G]]=mathbb E[mathbb E[ximid G]mathbb E[etamid G]]tag1$$
Similarly we have:$$mathbb E[etamathbb E[ximid G]]=mathbb E[mathbb E[etamid G]mathbb E[ximid G]]tag2$$
Now note that $(1)$ and $(2)$ have equal RHS.
answered Nov 20 at 8:30
drhab
95.3k543126
Thank you, i'll meditate on it
– anykk
Nov 20 at 8:51
Can you explain me, why the integral and the next line are equivalente? I can't see it. And, how we can expange it to $mathbb E[xi psi]$?
– anykk
Nov 26 at 18:34
By definition $int_AX(omega)P(domega)=int X(omega) mathbf1_A(omega)P(domega)$ and $mathbb E[Xmathbf1_A]$ is just another notation for the RHS. It can be expanded by first proving it for simple functions, then measurable functions that are limits of simple functions. It goes a bit too far to handle this in a comment. I reckon that you must have the disposal of some mathematical material that handles this stuff, don't you?
– drhab
Nov 26 at 18:48
No, I don't. I will learn it later in university. I think, that second proof is more simple, than yours one, but yours is stronger.
– anykk
Nov 26 at 19:26
Then where did you meet the question that you posed? Indeed the other answer (I upvoted it) is more direct. I tried to reveal something of the background of the equalities that you find there.
– drhab
Nov 26 at 19:28
|
show 5 more comments
up vote
2
down vote
accepted
up vote
2
down vote
accepted
In this answer $G$ must be looked at as a $sigma$-algebra (eventually generated by a random variable).
Characteristic for $mathbb E[ximid G]$ is that it satisfies:$$int_Axi(omega)P(domega)=int_Amathbb E[ximid G](omega)P(domega)text{ whenever }Atext{ is }Gtext{-measurable}$$
Or equivalently:$$mathbb E[ximathbf1_A]=mathbb E[mathbb E[ximid G]mathbf1_A]text{ whenever }Atext{ is }Gtext{-measurable}$$
This can be expanded to the more general statement that: $$mathbb E[xipsi]=mathbb E[mathbb E[ximid G]psi]text{ whenever }psitext{ is }Gtext{-measurable}$$
Now note that $mathbb E[etamid G]$ is by definition $G$-measurable so that we are allowed to conclude that:$$mathbb E[ximathbb E[etamid G]]=mathbb E[mathbb E[ximid G]mathbb E[etamid G]]tag1$$
Similarly we have:$$mathbb E[etamathbb E[ximid G]]=mathbb E[mathbb E[etamid G]mathbb E[ximid G]]tag2$$
Now note that $(1)$ and $(2)$ have equal RHS.
answered Nov 20 at 8:30
drhab
95.3k543126
In this answer $G$ must be looked at as a $sigma$-algebra (eventually generated by a random variable).
Characteristic for $mathbb E[ximid G]$ is that it satisfies:$$int_Axi(omega)P(domega)=int_Amathbb E[ximid G](omega)P(domega)text{ whenever }Atext{ is }Gtext{-measurable}$$
Or equivalently:$$mathbb E[ximathbf1_A]=mathbb E[mathbb E[ximid G]mathbf1_A]text{ whenever }Atext{ is }Gtext{-measurable}$$
This can be expanded to the more general statement that: $$mathbb E[xipsi]=mathbb E[mathbb E[ximid G]psi]text{ whenever }psitext{ is }Gtext{-measurable}$$
Now note that $mathbb E[etamid G]$ is by definition $G$-measurable so that we are allowed to conclude that:$$mathbb E[ximathbb E[etamid G]]=mathbb E[mathbb E[ximid G]mathbb E[etamid G]]tag1$$
Similarly we have:$$mathbb E[etamathbb E[ximid G]]=mathbb E[mathbb E[etamid G]mathbb E[ximid G]]tag2$$
Now note that $(1)$ and $(2)$ have equal RHS.
answered Nov 20 at 8:30
drhab
95.3k543126
edited Nov 20 at 8:46
answered Nov 20 at 8:30
drhab
95.3k543126
answered Nov 20 at 8:30
drhab
95.3k543126
answered Nov 20 at 8:30
drhab
95.3k543126
95.3k543126
Thank you, i'll meditate on it
– anykk
Nov 20 at 8:51
Can you explain me, why the integral and the next line are equivalente? I can't see it. And, how we can expange it to $mathbb E[xi psi]$?
– anykk
Nov 26 at 18:34
By definition $int_AX(omega)P(domega)=int X(omega) mathbf1_A(omega)P(domega)$ and $mathbb E[Xmathbf1_A]$ is just another notation for the RHS. It can be expanded by first proving it for simple functions, then measurable functions that are limits of simple functions. It goes a bit too far to handle this in a comment. I reckon that you must have the disposal of some mathematical material that handles this stuff, don't you?
– drhab
Nov 26 at 18:48
No, I don't. I will learn it later in university. I think, that second proof is more simple, than yours one, but yours is stronger.
– anykk
Nov 26 at 19:26
Then where did you meet the question that you posed? Indeed the other answer (I upvoted it) is more direct. I tried to reveal something of the background of the equalities that you find there.
– drhab
Nov 26 at 19:28
|
show 5 more comments
Thank you, i'll meditate on it
– anykk
Nov 20 at 8:51
Can you explain me, why the integral and the next line are equivalente? I can't see it. And, how we can expange it to $mathbb E[xi psi]$?
– anykk
Nov 26 at 18:34
By definition $int_AX(omega)P(domega)=int X(omega) mathbf1_A(omega)P(domega)$ and $mathbb E[Xmathbf1_A]$ is just another notation for the RHS. It can be expanded by first proving it for simple functions, then measurable functions that are limits of simple functions. It goes a bit too far to handle this in a comment. I reckon that you must have the disposal of some mathematical material that handles this stuff, don't you?
– drhab
Nov 26 at 18:48
No, I don't. I will learn it later in university. I think, that second proof is more simple, than yours one, but yours is stronger.
– anykk
Nov 26 at 19:26
Then where did you meet the question that you posed? Indeed the other answer (I upvoted it) is more direct. I tried to reveal something of the background of the equalities that you find there.
– drhab
Nov 26 at 19:28
Thank you, i'll meditate on it
– anykk
Nov 20 at 8:51
Thank you, i'll meditate on it
– anykk
Nov 20 at 8:51
Can you explain me, why the integral and the next line are equivalente? I can't see it. And, how we can expange it to $mathbb E[xi psi]$?
– anykk
Nov 26 at 18:34
Can you explain me, why the integral and the next line are equivalente? I can't see it. And, how we can expange it to $mathbb E[xi psi]$?
– anykk
Nov 26 at 18:34
By definition $int_AX(omega)P(domega)=int X(omega) mathbf1_A(omega)P(domega)$ and $mathbb E[Xmathbf1_A]$ is just another notation for the RHS. It can be expanded by first proving it for simple functions, then measurable functions that are limits of simple functions. It goes a bit too far to handle this in a comment. I reckon that you must have the disposal of some mathematical material that handles this stuff, don't you?
– drhab
Nov 26 at 18:48
By definition $int_AX(omega)P(domega)=int X(omega) mathbf1_A(omega)P(domega)$ and $mathbb E[Xmathbf1_A]$ is just another notation for the RHS. It can be expanded by first proving it for simple functions, then measurable functions that are limits of simple functions. It goes a bit too far to handle this in a comment. I reckon that you must have the disposal of some mathematical material that handles this stuff, don't you?
– drhab
Nov 26 at 18:48
No, I don't. I will learn it later in university. I think, that second proof is more simple, than yours one, but yours is stronger.
– anykk
Nov 26 at 19:26
No, I don't. I will learn it later in university. I think, that second proof is more simple, than yours one, but yours is stronger.
– anykk
Nov 26 at 19:26
Then where did you meet the question that you posed? Indeed the other answer (I upvoted it) is more direct. I tried to reveal something of the background of the equalities that you find there.
– drhab
Nov 26 at 19:28
Then where did you meet the question that you posed? Indeed the other answer (I upvoted it) is more direct. I tried to reveal something of the background of the equalities that you find there.
– drhab
Nov 26 at 19:28
|
show 5 more comments
up vote
2
down vote
From the law of total expectation
begin{align*}
Eleft(xi Eleft(eta|Gright)right) & =Eleft(Eleft(xi Eleft(eta|Gright)|Gright)right)\
& =Eleft(Eleft(eta|Gright)Eleft(xi|Gright)right)\
& =Eleft(Eleft(eta Eleft(xi|Gright)|Gright)right)\
& =Eleft(eta Eleft(xi|Gright)right)
end{align*}
answered Nov 20 at 8:31
hopeless
585
add a comment |
up vote
2
down vote
From the law of total expectation
begin{align*}
Eleft(xi Eleft(eta|Gright)right) & =Eleft(Eleft(xi Eleft(eta|Gright)|Gright)right)\
& =Eleft(Eleft(eta|Gright)Eleft(xi|Gright)right)\
& =Eleft(Eleft(eta Eleft(xi|Gright)|Gright)right)\
& =Eleft(eta Eleft(xi|Gright)right)
end{align*}
answered Nov 20 at 8:31
hopeless
585
add a comment |
up vote
2
down vote
up vote
2
down vote
From the law of total expectation
begin{align*}
Eleft(xi Eleft(eta|Gright)right) & =Eleft(Eleft(xi Eleft(eta|Gright)|Gright)right)\
& =Eleft(Eleft(eta|Gright)Eleft(xi|Gright)right)\
& =Eleft(Eleft(eta Eleft(xi|Gright)|Gright)right)\
& =Eleft(eta Eleft(xi|Gright)right)
end{align*}
answered Nov 20 at 8:31
hopeless
585
From the law of total expectation
begin{align*}
Eleft(xi Eleft(eta|Gright)right) & =Eleft(Eleft(xi Eleft(eta|Gright)|Gright)right)\
& =Eleft(Eleft(eta|Gright)Eleft(xi|Gright)right)\
& =Eleft(Eleft(eta Eleft(xi|Gright)|Gright)right)\
& =Eleft(eta Eleft(xi|Gright)right)
end{align*}
answered Nov 20 at 8:31
hopeless
585
answered Nov 20 at 8:31
hopeless
585
answered Nov 20 at 8:31
hopeless
585
answered Nov 20 at 8:31
hopeless
585
585
add a comment |
add a comment |
By linearity of expectation, you can change it to $E(xi)E(etamid G) = E(eta)E(ximid G)$. I suspect that that's easier to work with. That being said, what is $G$? I can't imagine that this is true if, for instance, $G$ is $eta=0$.
– Arthur
Nov 20 at 8:10
@Arthur, G is random discrete variable too.
– anykk
Nov 20 at 8:29