Conditional Exspectation of Conditional Exspectation
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Picture of the Task here
I got this task to solve and i am very dissapointet of myself that i can't solve this.
I will write only m instead of Municipality and F instead of FederlState.
For a) i got $$mathbb{E}[wagevert m]=2 mathbb{1}_{m=1}+mathbb{1}_{m=2}+2.5mathbb{1}_{m=3}+2.5mathbb{1}_{m=4}. $$
For b) i have $$ mathbb{E}[vert F]=frac{9}{4} mathbb{1}_{F=1}+2 mathbb{1}_{F=2}.
$$
So nowc)
For the first one i have $$ mathbb{E}[mathbb{E} [wage vert m]vert F] overset{tower; property}{=} mathbb{E}[wagevert F]=frac{9}{4} mathbb{1}_{F=1}+2 mathbb{1}_{F=2},$$
since $sigma(F)subset sigma(m)$.
For the second one i am not quite sure what to do. My idea was $$ mathbb{E} [mathbb{E} [wagevert F]vert m]=mathbb{E} [frac{9}{4} mathbb{1}_{F=1}+2 mathbb{1}_{F=2}vert m]=frac{9}{4} mathbb{1}_{min {1,3}}+2 mathbb{1}_{min{2,4}},$$ what is quite the same as before, so i think i made a mistke, but don't know where.
conditional-expectation
asked Dec 3 '18 at 10:15
Flo GeysFlo Geys
33
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add a comment |
$begingroup$
Picture of the Task here
I got this task to solve and i am very dissapointet of myself that i can't solve this.
I will write only m instead of Municipality and F instead of FederlState.
For a) i got $$mathbb{E}[wagevert m]=2 mathbb{1}_{m=1}+mathbb{1}_{m=2}+2.5mathbb{1}_{m=3}+2.5mathbb{1}_{m=4}. $$
For b) i have $$ mathbb{E}[vert F]=frac{9}{4} mathbb{1}_{F=1}+2 mathbb{1}_{F=2}.
$$
So nowc)
For the first one i have $$ mathbb{E}[mathbb{E} [wage vert m]vert F] overset{tower; property}{=} mathbb{E}[wagevert F]=frac{9}{4} mathbb{1}_{F=1}+2 mathbb{1}_{F=2},$$
since $sigma(F)subset sigma(m)$.
For the second one i am not quite sure what to do. My idea was $$ mathbb{E} [mathbb{E} [wagevert F]vert m]=mathbb{E} [frac{9}{4} mathbb{1}_{F=1}+2 mathbb{1}_{F=2}vert m]=frac{9}{4} mathbb{1}_{min {1,3}}+2 mathbb{1}_{min{2,4}},$$ what is quite the same as before, so i think i made a mistke, but don't know where.
conditional-expectation
asked Dec 3 '18 at 10:15
Flo GeysFlo Geys
33
$endgroup$
add a comment |
$begingroup$
Picture of the Task here
I got this task to solve and i am very dissapointet of myself that i can't solve this.
I will write only m instead of Municipality and F instead of FederlState.
For a) i got $$mathbb{E}[wagevert m]=2 mathbb{1}_{m=1}+mathbb{1}_{m=2}+2.5mathbb{1}_{m=3}+2.5mathbb{1}_{m=4}. $$
For b) i have $$ mathbb{E}[vert F]=frac{9}{4} mathbb{1}_{F=1}+2 mathbb{1}_{F=2}.
$$
So nowc)
For the first one i have $$ mathbb{E}[mathbb{E} [wage vert m]vert F] overset{tower; property}{=} mathbb{E}[wagevert F]=frac{9}{4} mathbb{1}_{F=1}+2 mathbb{1}_{F=2},$$
since $sigma(F)subset sigma(m)$.
For the second one i am not quite sure what to do. My idea was $$ mathbb{E} [mathbb{E} [wagevert F]vert m]=mathbb{E} [frac{9}{4} mathbb{1}_{F=1}+2 mathbb{1}_{F=2}vert m]=frac{9}{4} mathbb{1}_{min {1,3}}+2 mathbb{1}_{min{2,4}},$$ what is quite the same as before, so i think i made a mistke, but don't know where.
conditional-expectation
asked Dec 3 '18 at 10:15
Flo GeysFlo Geys
33
$endgroup$
Picture of the Task here
I got this task to solve and i am very dissapointet of myself that i can't solve this.
I will write only m instead of Municipality and F instead of FederlState.
For a) i got $$mathbb{E}[wagevert m]=2 mathbb{1}_{m=1}+mathbb{1}_{m=2}+2.5mathbb{1}_{m=3}+2.5mathbb{1}_{m=4}. $$
For b) i have $$ mathbb{E}[vert F]=frac{9}{4} mathbb{1}_{F=1}+2 mathbb{1}_{F=2}.
$$
So nowc)
For the first one i have $$ mathbb{E}[mathbb{E} [wage vert m]vert F] overset{tower; property}{=} mathbb{E}[wagevert F]=frac{9}{4} mathbb{1}_{F=1}+2 mathbb{1}_{F=2},$$
since $sigma(F)subset sigma(m)$.
For the second one i am not quite sure what to do. My idea was $$ mathbb{E} [mathbb{E} [wagevert F]vert m]=mathbb{E} [frac{9}{4} mathbb{1}_{F=1}+2 mathbb{1}_{F=2}vert m]=frac{9}{4} mathbb{1}_{min {1,3}}+2 mathbb{1}_{min{2,4}},$$ what is quite the same as before, so i think i made a mistke, but don't know where.
conditional-expectation
conditional-expectation
asked Dec 3 '18 at 10:15
Flo GeysFlo Geys
33
asked Dec 3 '18 at 10:15
Flo GeysFlo Geys
33
edited Dec 3 '18 at 10:32
Flo Geys
asked Dec 3 '18 at 10:15
Flo GeysFlo Geys
33
asked Dec 3 '18 at 10:15
Flo GeysFlo Geys
33
asked Dec 3 '18 at 10:15
Flo GeysFlo Geys
33
33
add a comment |
add a comment |
1 Answer
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"...what is quite the same as before, so I think I made a mistake..."
No, you did not.
Let $X,Y,Z$ be random variables with $sigma(Y)subseteqsigma(Z)$ or equivalently $Y=kleft(Zright)$ for some function $k$.
Observe that for any suitable function $g$ we have: $$mathbb{E}left(mathbb{E}left[mathbb{E}left[Xmid Zright]mid Yright]gleft(Yright)right)=mathbb{E}left(mathbb{E}left[Xmid Zright]gleft(Yright)right)=mathbb{E}left(mathbb{E}left[Xmid Zright]gleft(kleft(Zright)right)right)=$$$$mathbb{E}left(Xgleft(kleft(Zright)right)right)=mathbb{E}left(Xgleft(Yright)right)$$
This allows the conclusion that $mathbb{E}left[mathbb{E}left[Xmid Zright]mid Yright]=mathbb{E}left[Xmid Yright]$
Setting $X=text{wage}$, $Y=F$ and $Z=m$ we find: $$mathbb{E}left[mathbb{E}left[text{wage}mid mright]mid Fright]=mathbb{E}left[text{wage}mid Fright]$$
answered Dec 3 '18 at 12:48
drhabdrhab
99.1k544130
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$begingroup$
"...what is quite the same as before, so I think I made a mistake..."
No, you did not.
Let $X,Y,Z$ be random variables with $sigma(Y)subseteqsigma(Z)$ or equivalently $Y=kleft(Zright)$ for some function $k$.
Observe that for any suitable function $g$ we have: $$mathbb{E}left(mathbb{E}left[mathbb{E}left[Xmid Zright]mid Yright]gleft(Yright)right)=mathbb{E}left(mathbb{E}left[Xmid Zright]gleft(Yright)right)=mathbb{E}left(mathbb{E}left[Xmid Zright]gleft(kleft(Zright)right)right)=$$$$mathbb{E}left(Xgleft(kleft(Zright)right)right)=mathbb{E}left(Xgleft(Yright)right)$$
This allows the conclusion that $mathbb{E}left[mathbb{E}left[Xmid Zright]mid Yright]=mathbb{E}left[Xmid Yright]$
Setting $X=text{wage}$, $Y=F$ and $Z=m$ we find: $$mathbb{E}left[mathbb{E}left[text{wage}mid mright]mid Fright]=mathbb{E}left[text{wage}mid Fright]$$
answered Dec 3 '18 at 12:48
drhabdrhab
99.1k544130
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add a comment |
$begingroup$
"...what is quite the same as before, so I think I made a mistake..."
No, you did not.
Let $X,Y,Z$ be random variables with $sigma(Y)subseteqsigma(Z)$ or equivalently $Y=kleft(Zright)$ for some function $k$.
Observe that for any suitable function $g$ we have: $$mathbb{E}left(mathbb{E}left[mathbb{E}left[Xmid Zright]mid Yright]gleft(Yright)right)=mathbb{E}left(mathbb{E}left[Xmid Zright]gleft(Yright)right)=mathbb{E}left(mathbb{E}left[Xmid Zright]gleft(kleft(Zright)right)right)=$$$$mathbb{E}left(Xgleft(kleft(Zright)right)right)=mathbb{E}left(Xgleft(Yright)right)$$
This allows the conclusion that $mathbb{E}left[mathbb{E}left[Xmid Zright]mid Yright]=mathbb{E}left[Xmid Yright]$
Setting $X=text{wage}$, $Y=F$ and $Z=m$ we find: $$mathbb{E}left[mathbb{E}left[text{wage}mid mright]mid Fright]=mathbb{E}left[text{wage}mid Fright]$$
answered Dec 3 '18 at 12:48
drhabdrhab
99.1k544130
$endgroup$
add a comment |
$begingroup$
"...what is quite the same as before, so I think I made a mistake..."
No, you did not.
Let $X,Y,Z$ be random variables with $sigma(Y)subseteqsigma(Z)$ or equivalently $Y=kleft(Zright)$ for some function $k$.
Observe that for any suitable function $g$ we have: $$mathbb{E}left(mathbb{E}left[mathbb{E}left[Xmid Zright]mid Yright]gleft(Yright)right)=mathbb{E}left(mathbb{E}left[Xmid Zright]gleft(Yright)right)=mathbb{E}left(mathbb{E}left[Xmid Zright]gleft(kleft(Zright)right)right)=$$$$mathbb{E}left(Xgleft(kleft(Zright)right)right)=mathbb{E}left(Xgleft(Yright)right)$$
This allows the conclusion that $mathbb{E}left[mathbb{E}left[Xmid Zright]mid Yright]=mathbb{E}left[Xmid Yright]$
Setting $X=text{wage}$, $Y=F$ and $Z=m$ we find: $$mathbb{E}left[mathbb{E}left[text{wage}mid mright]mid Fright]=mathbb{E}left[text{wage}mid Fright]$$
answered Dec 3 '18 at 12:48
drhabdrhab
99.1k544130
$endgroup$
"...what is quite the same as before, so I think I made a mistake..."
No, you did not.
Let $X,Y,Z$ be random variables with $sigma(Y)subseteqsigma(Z)$ or equivalently $Y=kleft(Zright)$ for some function $k$.
Observe that for any suitable function $g$ we have: $$mathbb{E}left(mathbb{E}left[mathbb{E}left[Xmid Zright]mid Yright]gleft(Yright)right)=mathbb{E}left(mathbb{E}left[Xmid Zright]gleft(Yright)right)=mathbb{E}left(mathbb{E}left[Xmid Zright]gleft(kleft(Zright)right)right)=$$$$mathbb{E}left(Xgleft(kleft(Zright)right)right)=mathbb{E}left(Xgleft(Yright)right)$$
This allows the conclusion that $mathbb{E}left[mathbb{E}left[Xmid Zright]mid Yright]=mathbb{E}left[Xmid Yright]$
Setting $X=text{wage}$, $Y=F$ and $Z=m$ we find: $$mathbb{E}left[mathbb{E}left[text{wage}mid mright]mid Fright]=mathbb{E}left[text{wage}mid Fright]$$
answered Dec 3 '18 at 12:48
drhabdrhab
99.1k544130
answered Dec 3 '18 at 12:48
drhabdrhab
99.1k544130
answered Dec 3 '18 at 12:48
drhabdrhab
99.1k544130
answered Dec 3 '18 at 12:48
drhabdrhab
99.1k544130
99.1k544130
add a comment |
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