Proving $mathbb{D}(xi|mathcal{D})=mathbb{E}(xi^2|mathcal{D}) -(mathbb{E}(xi|mathcal{D}))^2 $ [closed]
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The conditional dispersion of random variable $xi$ according $sigma$-algebra $mathcal{D}$ is a random variable: $mathbb{D}(xi|mathcal{D})=mathbb{E}((xi - mathbb{E}(xi|mathcal{D}))^2|mathcal{D}).$ I need to show that next to equalities is correct
$mathbb{D}(xi|mathcal{D})=mathbb{E}(xi^2|mathcal{D}) -(mathbb{E}(xi|mathcal{D}))^2 $
and
$mathbb{D}xi=mathbb{E}mathbb{D}(xi|mathcal{D})+mathbb{D}mathbb{E}(xi|mathcal{D})$.
So how to begin and what properties to use?
probability probability-theory conditional-expectation random
asked Jan 2 at 20:54
AtstovasAtstovas
1139
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closed as off-topic by Did, Davide Giraudo, José Carlos Santos, Cesareo, Leucippus Jan 3 at 0:40
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Davide Giraudo, José Carlos Santos, Cesareo, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
The conditional dispersion of random variable $xi$ according $sigma$-algebra $mathcal{D}$ is a random variable: $mathbb{D}(xi|mathcal{D})=mathbb{E}((xi - mathbb{E}(xi|mathcal{D}))^2|mathcal{D}).$ I need to show that next to equalities is correct
$mathbb{D}(xi|mathcal{D})=mathbb{E}(xi^2|mathcal{D}) -(mathbb{E}(xi|mathcal{D}))^2 $
and
$mathbb{D}xi=mathbb{E}mathbb{D}(xi|mathcal{D})+mathbb{D}mathbb{E}(xi|mathcal{D})$.
So how to begin and what properties to use?
probability probability-theory conditional-expectation random
asked Jan 2 at 20:54
AtstovasAtstovas
1139
$endgroup$
closed as off-topic by Did, Davide Giraudo, José Carlos Santos, Cesareo, Leucippus Jan 3 at 0:40
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Davide Giraudo, José Carlos Santos, Cesareo, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
The conditional dispersion of random variable $xi$ according $sigma$-algebra $mathcal{D}$ is a random variable: $mathbb{D}(xi|mathcal{D})=mathbb{E}((xi - mathbb{E}(xi|mathcal{D}))^2|mathcal{D}).$ I need to show that next to equalities is correct
$mathbb{D}(xi|mathcal{D})=mathbb{E}(xi^2|mathcal{D}) -(mathbb{E}(xi|mathcal{D}))^2 $
and
$mathbb{D}xi=mathbb{E}mathbb{D}(xi|mathcal{D})+mathbb{D}mathbb{E}(xi|mathcal{D})$.
So how to begin and what properties to use?
probability probability-theory conditional-expectation random
asked Jan 2 at 20:54
AtstovasAtstovas
1139
$endgroup$
The conditional dispersion of random variable $xi$ according $sigma$-algebra $mathcal{D}$ is a random variable: $mathbb{D}(xi|mathcal{D})=mathbb{E}((xi - mathbb{E}(xi|mathcal{D}))^2|mathcal{D}).$ I need to show that next to equalities is correct
$mathbb{D}(xi|mathcal{D})=mathbb{E}(xi^2|mathcal{D}) -(mathbb{E}(xi|mathcal{D}))^2 $
and
$mathbb{D}xi=mathbb{E}mathbb{D}(xi|mathcal{D})+mathbb{D}mathbb{E}(xi|mathcal{D})$.
So how to begin and what properties to use?
probability probability-theory conditional-expectation random
probability probability-theory conditional-expectation random
asked Jan 2 at 20:54
AtstovasAtstovas
1139
asked Jan 2 at 20:54
AtstovasAtstovas
1139
asked Jan 2 at 20:54
AtstovasAtstovas
1139
asked Jan 2 at 20:54
AtstovasAtstovas
1139
asked Jan 2 at 20:54
AtstovasAtstovas
1139
1139
closed as off-topic by Did, Davide Giraudo, José Carlos Santos, Cesareo, Leucippus Jan 3 at 0:40
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Davide Giraudo, José Carlos Santos, Cesareo, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Did, Davide Giraudo, José Carlos Santos, Cesareo, Leucippus Jan 3 at 0:40
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Davide Giraudo, José Carlos Santos, Cesareo, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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First, expand out the square:
$$ mathbb D(xi | mathcal D) = mathbb E left( xi^2 - 2 xi mathbb E(xi | mathcal D) + mathbb E (xi | mathcal D)^2 mid mathcal D right).$$
Then use:
- Linearity:
$$ mathbb E(X_1 + X_2 | mathcal D) = mathbb E(X_1 | mathcal D) + mathbb E(X_2 | mathcal D), mathbb E(aX|mathcal D)=amathbb E(X|mathcal D).$$
- Pulling out known factors: if $X$ is $mathcal D$-measurable, then $$ mathbb E(XY| mathcal D) = X mathbb E(Y | mathcal D).$$
(And remember, $mathbb E(xi | mathcal D)$ is $mathcal D$-measurable, by definition. Hence $mathbb E(xi | mathcal D)^2$ is $mathcal D$-measurable too.)
answered Jan 2 at 21:19
Kenny WongKenny Wong
19.2k21441
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$begingroup$
Can you check if I did it right? $ mathbb D(xi | mathcal D) = mathbb E left( xi^2 - 2 xi mathbb E(xi | mathcal D) + mathbb E (xi | mathcal D)^2 mid mathcal D right)=mathbb{E}(xi^2|mathcal{D})-2xi mathbb{E}(xi |mathcal{D})+mathbb{E}(xi|mathcal{D})^2=mathbb{E}(xi^2|mathcal{D})-mathbb{E}(xi|mathcal{D})^2?$
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– Atstovas
Jan 3 at 10:17
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@Atstovas Should be $mathbb E(xi^2 | mathcal D) - 2 mathbb E(xi | mathcal D)^2 + mathbb E(xi | mathcal D)^2$.
$endgroup$
– Kenny Wong
Jan 3 at 11:54
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Thank you! But I don’t know how to do the second equality
$endgroup$
– Atstovas
Jan 3 at 12:00
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First, expand out the square:
$$ mathbb D(xi | mathcal D) = mathbb E left( xi^2 - 2 xi mathbb E(xi | mathcal D) + mathbb E (xi | mathcal D)^2 mid mathcal D right).$$
Then use:
- Linearity:
$$ mathbb E(X_1 + X_2 | mathcal D) = mathbb E(X_1 | mathcal D) + mathbb E(X_2 | mathcal D), mathbb E(aX|mathcal D)=amathbb E(X|mathcal D).$$
- Pulling out known factors: if $X$ is $mathcal D$-measurable, then $$ mathbb E(XY| mathcal D) = X mathbb E(Y | mathcal D).$$
(And remember, $mathbb E(xi | mathcal D)$ is $mathcal D$-measurable, by definition. Hence $mathbb E(xi | mathcal D)^2$ is $mathcal D$-measurable too.)
answered Jan 2 at 21:19
Kenny WongKenny Wong
19.2k21441
$endgroup$
$begingroup$
Can you check if I did it right? $ mathbb D(xi | mathcal D) = mathbb E left( xi^2 - 2 xi mathbb E(xi | mathcal D) + mathbb E (xi | mathcal D)^2 mid mathcal D right)=mathbb{E}(xi^2|mathcal{D})-2xi mathbb{E}(xi |mathcal{D})+mathbb{E}(xi|mathcal{D})^2=mathbb{E}(xi^2|mathcal{D})-mathbb{E}(xi|mathcal{D})^2?$
$endgroup$
– Atstovas
Jan 3 at 10:17
$begingroup$
@Atstovas Should be $mathbb E(xi^2 | mathcal D) - 2 mathbb E(xi | mathcal D)^2 + mathbb E(xi | mathcal D)^2$.
$endgroup$
– Kenny Wong
Jan 3 at 11:54
$begingroup$
Thank you! But I don’t know how to do the second equality
$endgroup$
– Atstovas
Jan 3 at 12:00
add a comment |
$begingroup$
First, expand out the square:
$$ mathbb D(xi | mathcal D) = mathbb E left( xi^2 - 2 xi mathbb E(xi | mathcal D) + mathbb E (xi | mathcal D)^2 mid mathcal D right).$$
Then use:
- Linearity:
$$ mathbb E(X_1 + X_2 | mathcal D) = mathbb E(X_1 | mathcal D) + mathbb E(X_2 | mathcal D), mathbb E(aX|mathcal D)=amathbb E(X|mathcal D).$$
- Pulling out known factors: if $X$ is $mathcal D$-measurable, then $$ mathbb E(XY| mathcal D) = X mathbb E(Y | mathcal D).$$
(And remember, $mathbb E(xi | mathcal D)$ is $mathcal D$-measurable, by definition. Hence $mathbb E(xi | mathcal D)^2$ is $mathcal D$-measurable too.)
answered Jan 2 at 21:19
Kenny WongKenny Wong
19.2k21441
$endgroup$
$begingroup$
Can you check if I did it right? $ mathbb D(xi | mathcal D) = mathbb E left( xi^2 - 2 xi mathbb E(xi | mathcal D) + mathbb E (xi | mathcal D)^2 mid mathcal D right)=mathbb{E}(xi^2|mathcal{D})-2xi mathbb{E}(xi |mathcal{D})+mathbb{E}(xi|mathcal{D})^2=mathbb{E}(xi^2|mathcal{D})-mathbb{E}(xi|mathcal{D})^2?$
$endgroup$
– Atstovas
Jan 3 at 10:17
$begingroup$
@Atstovas Should be $mathbb E(xi^2 | mathcal D) - 2 mathbb E(xi | mathcal D)^2 + mathbb E(xi | mathcal D)^2$.
$endgroup$
– Kenny Wong
Jan 3 at 11:54
$begingroup$
Thank you! But I don’t know how to do the second equality
$endgroup$
– Atstovas
Jan 3 at 12:00
add a comment |
$begingroup$
First, expand out the square:
$$ mathbb D(xi | mathcal D) = mathbb E left( xi^2 - 2 xi mathbb E(xi | mathcal D) + mathbb E (xi | mathcal D)^2 mid mathcal D right).$$
Then use:
- Linearity:
$$ mathbb E(X_1 + X_2 | mathcal D) = mathbb E(X_1 | mathcal D) + mathbb E(X_2 | mathcal D), mathbb E(aX|mathcal D)=amathbb E(X|mathcal D).$$
- Pulling out known factors: if $X$ is $mathcal D$-measurable, then $$ mathbb E(XY| mathcal D) = X mathbb E(Y | mathcal D).$$
(And remember, $mathbb E(xi | mathcal D)$ is $mathcal D$-measurable, by definition. Hence $mathbb E(xi | mathcal D)^2$ is $mathcal D$-measurable too.)
answered Jan 2 at 21:19
Kenny WongKenny Wong
19.2k21441
$endgroup$
First, expand out the square:
$$ mathbb D(xi | mathcal D) = mathbb E left( xi^2 - 2 xi mathbb E(xi | mathcal D) + mathbb E (xi | mathcal D)^2 mid mathcal D right).$$
Then use:
- Linearity:
$$ mathbb E(X_1 + X_2 | mathcal D) = mathbb E(X_1 | mathcal D) + mathbb E(X_2 | mathcal D), mathbb E(aX|mathcal D)=amathbb E(X|mathcal D).$$
- Pulling out known factors: if $X$ is $mathcal D$-measurable, then $$ mathbb E(XY| mathcal D) = X mathbb E(Y | mathcal D).$$
(And remember, $mathbb E(xi | mathcal D)$ is $mathcal D$-measurable, by definition. Hence $mathbb E(xi | mathcal D)^2$ is $mathcal D$-measurable too.)
answered Jan 2 at 21:19
Kenny WongKenny Wong
19.2k21441
edited Jan 2 at 21:45
answered Jan 2 at 21:19
Kenny WongKenny Wong
19.2k21441
answered Jan 2 at 21:19
Kenny WongKenny Wong
19.2k21441
answered Jan 2 at 21:19
Kenny WongKenny Wong
19.2k21441
19.2k21441
$begingroup$
Can you check if I did it right? $ mathbb D(xi | mathcal D) = mathbb E left( xi^2 - 2 xi mathbb E(xi | mathcal D) + mathbb E (xi | mathcal D)^2 mid mathcal D right)=mathbb{E}(xi^2|mathcal{D})-2xi mathbb{E}(xi |mathcal{D})+mathbb{E}(xi|mathcal{D})^2=mathbb{E}(xi^2|mathcal{D})-mathbb{E}(xi|mathcal{D})^2?$
$endgroup$
– Atstovas
Jan 3 at 10:17
$begingroup$
@Atstovas Should be $mathbb E(xi^2 | mathcal D) - 2 mathbb E(xi | mathcal D)^2 + mathbb E(xi | mathcal D)^2$.
$endgroup$
– Kenny Wong
Jan 3 at 11:54
$begingroup$
Thank you! But I don’t know how to do the second equality
$endgroup$
– Atstovas
Jan 3 at 12:00
add a comment |
$begingroup$
Can you check if I did it right? $ mathbb D(xi | mathcal D) = mathbb E left( xi^2 - 2 xi mathbb E(xi | mathcal D) + mathbb E (xi | mathcal D)^2 mid mathcal D right)=mathbb{E}(xi^2|mathcal{D})-2xi mathbb{E}(xi |mathcal{D})+mathbb{E}(xi|mathcal{D})^2=mathbb{E}(xi^2|mathcal{D})-mathbb{E}(xi|mathcal{D})^2?$
$endgroup$
– Atstovas
Jan 3 at 10:17
$begingroup$
@Atstovas Should be $mathbb E(xi^2 | mathcal D) - 2 mathbb E(xi | mathcal D)^2 + mathbb E(xi | mathcal D)^2$.
$endgroup$
– Kenny Wong
Jan 3 at 11:54
$begingroup$
Thank you! But I don’t know how to do the second equality
$endgroup$
– Atstovas
Jan 3 at 12:00
$begingroup$
Can you check if I did it right? $ mathbb D(xi | mathcal D) = mathbb E left( xi^2 - 2 xi mathbb E(xi | mathcal D) + mathbb E (xi | mathcal D)^2 mid mathcal D right)=mathbb{E}(xi^2|mathcal{D})-2xi mathbb{E}(xi |mathcal{D})+mathbb{E}(xi|mathcal{D})^2=mathbb{E}(xi^2|mathcal{D})-mathbb{E}(xi|mathcal{D})^2?$
$endgroup$
– Atstovas
Jan 3 at 10:17
$begingroup$
Can you check if I did it right? $ mathbb D(xi | mathcal D) = mathbb E left( xi^2 - 2 xi mathbb E(xi | mathcal D) + mathbb E (xi | mathcal D)^2 mid mathcal D right)=mathbb{E}(xi^2|mathcal{D})-2xi mathbb{E}(xi |mathcal{D})+mathbb{E}(xi|mathcal{D})^2=mathbb{E}(xi^2|mathcal{D})-mathbb{E}(xi|mathcal{D})^2?$
$endgroup$
– Atstovas
Jan 3 at 10:17
$begingroup$
@Atstovas Should be $mathbb E(xi^2 | mathcal D) - 2 mathbb E(xi | mathcal D)^2 + mathbb E(xi | mathcal D)^2$.
$endgroup$
– Kenny Wong
Jan 3 at 11:54
$begingroup$
@Atstovas Should be $mathbb E(xi^2 | mathcal D) - 2 mathbb E(xi | mathcal D)^2 + mathbb E(xi | mathcal D)^2$.
$endgroup$
– Kenny Wong
Jan 3 at 11:54
$begingroup$
Thank you! But I don’t know how to do the second equality
$endgroup$
– Atstovas
Jan 3 at 12:00
$begingroup$
Thank you! But I don’t know how to do the second equality
$endgroup$
– Atstovas
Jan 3 at 12:00
add a comment |
