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Using the law of iterated expectations with multiple variables

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0

0

$begingroup$

I'm given that:

$E[epsilon_i | dist_i, range_i] = 0$

How can I apply the law of iterated expectations to find that

$E[epsilon_i|dist_i] =0$

is also true?

probability conditional-expectation expected-value

share|cite|improve this question

edited Feb 27 at 2:22

mitmath514

asked Feb 27 at 1:37

mitmath514mitmath514

213

$endgroup$

add a comment | 

0

0

$begingroup$

I'm given that:

$E[epsilon_i | dist_i, range_i] = 0$

How can I apply the law of iterated expectations to find that

$E[epsilon_i|dist_i] =0$

is also true?

probability conditional-expectation expected-value

share|cite|improve this question

edited Feb 27 at 2:22

mitmath514

asked Feb 27 at 1:37

mitmath514mitmath514

213

$endgroup$

add a comment | 

$begingroup$

I'm given that:

$E[epsilon_i | dist_i, range_i] = 0$

How can I apply the law of iterated expectations to find that

$E[epsilon_i|dist_i] =0$

is also true?

probability conditional-expectation expected-value

share|cite|improve this question

edited Feb 27 at 2:22

mitmath514

asked Feb 27 at 1:37

mitmath514mitmath514

213

$endgroup$

share|cite|improve this question

edited Feb 27 at 2:22

mitmath514

asked Feb 27 at 1:37

mitmath514mitmath514

213

share|cite|improve this question

edited Feb 27 at 2:22

mitmath514

asked Feb 27 at 1:37

mitmath514mitmath514

213

asked Feb 27 at 1:37

mitmath514mitmath514

213

asked Feb 27 at 1:37

mitmath514mitmath514

213

1 Answer
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oldest

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0

$begingroup$

Tower property

if $F_1 subset F_2$ so $E(E(Z|F_2)|F_1)=E(Z|F_1)$

by taking $F_1=sigma(X)$ and $F_2=sigma(X,Y)$ so $E(E(Z|sigma(X,Y))|sigma(X))=E(Z|sigma(X))$

on the other hand

$E(E(Z|X,Y)|X)=E(Z|X)$

choose $(Z=epsilon_i,X=dist_i,Y=range_i)$

$E[Z|X]=Ebigg(E[Z | X, Y])|X bigg) =E(0|X)=0$

(just for hint $E[Z=epsilon_i|X=dist_i]=E(E[Z=epsilon_i | X=dist_i, Y=range_i])|X=dist_i) =E(0)=0$ )

share|cite|improve this answer

answered Mar 29 at 2:10

masoudmasoud

3128

$endgroup$

add a comment | 

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0

$begingroup$

Tower property

if $F_1 subset F_2$ so $E(E(Z|F_2)|F_1)=E(Z|F_1)$

by taking $F_1=sigma(X)$ and $F_2=sigma(X,Y)$ so $E(E(Z|sigma(X,Y))|sigma(X))=E(Z|sigma(X))$

on the other hand

$E(E(Z|X,Y)|X)=E(Z|X)$

choose $(Z=epsilon_i,X=dist_i,Y=range_i)$

$E[Z|X]=Ebigg(E[Z | X, Y])|X bigg) =E(0|X)=0$

(just for hint $E[Z=epsilon_i|X=dist_i]=E(E[Z=epsilon_i | X=dist_i, Y=range_i])|X=dist_i) =E(0)=0$ )

share|cite|improve this answer

answered Mar 29 at 2:10

masoudmasoud

3128

$endgroup$

add a comment | 

0

$begingroup$

Tower property

if $F_1 subset F_2$ so $E(E(Z|F_2)|F_1)=E(Z|F_1)$

by taking $F_1=sigma(X)$ and $F_2=sigma(X,Y)$ so $E(E(Z|sigma(X,Y))|sigma(X))=E(Z|sigma(X))$

on the other hand

$E(E(Z|X,Y)|X)=E(Z|X)$

choose $(Z=epsilon_i,X=dist_i,Y=range_i)$

$E[Z|X]=Ebigg(E[Z | X, Y])|X bigg) =E(0|X)=0$

(just for hint $E[Z=epsilon_i|X=dist_i]=E(E[Z=epsilon_i | X=dist_i, Y=range_i])|X=dist_i) =E(0)=0$ )

share|cite|improve this answer

answered Mar 29 at 2:10

masoudmasoud

3128

$endgroup$

add a comment | 

$begingroup$

Tower property

if $F_1 subset F_2$ so $E(E(Z|F_2)|F_1)=E(Z|F_1)$

by taking $F_1=sigma(X)$ and $F_2=sigma(X,Y)$ so $E(E(Z|sigma(X,Y))|sigma(X))=E(Z|sigma(X))$

on the other hand

$E(E(Z|X,Y)|X)=E(Z|X)$

choose $(Z=epsilon_i,X=dist_i,Y=range_i)$

$E[Z|X]=Ebigg(E[Z | X, Y])|X bigg) =E(0|X)=0$

(just for hint $E[Z=epsilon_i|X=dist_i]=E(E[Z=epsilon_i | X=dist_i, Y=range_i])|X=dist_i) =E(0)=0$ )

share|cite|improve this answer

answered Mar 29 at 2:10

masoudmasoud

3128

$endgroup$

share|cite|improve this answer

answered Mar 29 at 2:10

masoudmasoud

3128

answered Mar 29 at 2:10

masoudmasoud

3128

answered Mar 29 at 2:10

masoudmasoud

3128