On computing a conditional expectation $E[Y|X]$ where $Y$ is not known.
$begingroup$
Preface:
This question is based on the answer to this question given in the comments.
The problem:
Consider the Lebesgue probability space on the interval $[0,1)$. (I.e. the state space is $Ω = [0, 1)$, the $sigma$-field is the set of Lebesgue measurable sets and the measure is the Lebesgue measure.) We define the random variable $X$ as:
$$X(w)=begin{cases} 2w &, 0leq w < 1/2 \ 2w−1 &, 1/2leq w<1 end{cases}$$
Compute the conditional expectation $E(Y |X)$ where $Y : [0, 1) to mathbb{R}$ is a measurable function.
My question:
Why can we write $$E(Ymid X)(omega)=frac{Y(omega)+Y(omega+frac12)mathbf 1_{2omega<1}+Y(omega-frac12)mathbf 1_{2omegageqslant1}}2$$ ? What theorem are we appealing to?
EDIT: I think an answer was given to a similar question here.
probability probability-theory expected-value
asked Oct 7 '18 at 21:33
MonoliteMonolite
1,5412925
$endgroup$
add a comment |
$begingroup$
Preface:
This question is based on the answer to this question given in the comments.
The problem:
Consider the Lebesgue probability space on the interval $[0,1)$. (I.e. the state space is $Ω = [0, 1)$, the $sigma$-field is the set of Lebesgue measurable sets and the measure is the Lebesgue measure.) We define the random variable $X$ as:
$$X(w)=begin{cases} 2w &, 0leq w < 1/2 \ 2w−1 &, 1/2leq w<1 end{cases}$$
Compute the conditional expectation $E(Y |X)$ where $Y : [0, 1) to mathbb{R}$ is a measurable function.
My question:
Why can we write $$E(Ymid X)(omega)=frac{Y(omega)+Y(omega+frac12)mathbf 1_{2omega<1}+Y(omega-frac12)mathbf 1_{2omegageqslant1}}2$$ ? What theorem are we appealing to?
EDIT: I think an answer was given to a similar question here.
probability probability-theory expected-value
asked Oct 7 '18 at 21:33
MonoliteMonolite
1,5412925
$endgroup$
$begingroup$
The title does not agree with the equation you have written Are you computing $E(X|Y)$ or $E(Y|X)$?
$endgroup$
– Kavi Rama Murthy
Oct 7 '18 at 23:41
1
$begingroup$
@KaviRamaMurthy from the linked question it is $E(Y mid X)$
$endgroup$
– Henry
Oct 7 '18 at 23:43
$begingroup$
@KaviRamaMurthy I will edit, sorry for the typo.
$endgroup$
– Monolite
Oct 8 '18 at 10:39
1
$begingroup$
@Did Maybe you could take a look at this question since it comes from a comment of yours. In any case thanks for all the answers and your activity on mathstack!
$endgroup$
– Monolite
Oct 23 '18 at 22:43
add a comment |
$begingroup$
Preface:
This question is based on the answer to this question given in the comments.
The problem:
Consider the Lebesgue probability space on the interval $[0,1)$. (I.e. the state space is $Ω = [0, 1)$, the $sigma$-field is the set of Lebesgue measurable sets and the measure is the Lebesgue measure.) We define the random variable $X$ as:
$$X(w)=begin{cases} 2w &, 0leq w < 1/2 \ 2w−1 &, 1/2leq w<1 end{cases}$$
Compute the conditional expectation $E(Y |X)$ where $Y : [0, 1) to mathbb{R}$ is a measurable function.
My question:
Why can we write $$E(Ymid X)(omega)=frac{Y(omega)+Y(omega+frac12)mathbf 1_{2omega<1}+Y(omega-frac12)mathbf 1_{2omegageqslant1}}2$$ ? What theorem are we appealing to?
EDIT: I think an answer was given to a similar question here.
probability probability-theory expected-value
asked Oct 7 '18 at 21:33
MonoliteMonolite
1,5412925
$endgroup$
Preface:
This question is based on the answer to this question given in the comments.
The problem:
Consider the Lebesgue probability space on the interval $[0,1)$. (I.e. the state space is $Ω = [0, 1)$, the $sigma$-field is the set of Lebesgue measurable sets and the measure is the Lebesgue measure.) We define the random variable $X$ as:
$$X(w)=begin{cases} 2w &, 0leq w < 1/2 \ 2w−1 &, 1/2leq w<1 end{cases}$$
Compute the conditional expectation $E(Y |X)$ where $Y : [0, 1) to mathbb{R}$ is a measurable function.
My question:
Why can we write $$E(Ymid X)(omega)=frac{Y(omega)+Y(omega+frac12)mathbf 1_{2omega<1}+Y(omega-frac12)mathbf 1_{2omegageqslant1}}2$$ ? What theorem are we appealing to?
EDIT: I think an answer was given to a similar question here.
probability probability-theory expected-value
probability probability-theory expected-value
asked Oct 7 '18 at 21:33
MonoliteMonolite
1,5412925
asked Oct 7 '18 at 21:33
MonoliteMonolite
1,5412925
edited Dec 2 '18 at 16:03
Monolite
asked Oct 7 '18 at 21:33
MonoliteMonolite
1,5412925
asked Oct 7 '18 at 21:33
MonoliteMonolite
1,5412925
asked Oct 7 '18 at 21:33
MonoliteMonolite
1,5412925
1,5412925
$begingroup$
The title does not agree with the equation you have written Are you computing $E(X|Y)$ or $E(Y|X)$?
$endgroup$
– Kavi Rama Murthy
Oct 7 '18 at 23:41
1
$begingroup$
@KaviRamaMurthy from the linked question it is $E(Y mid X)$
$endgroup$
– Henry
Oct 7 '18 at 23:43
$begingroup$
@KaviRamaMurthy I will edit, sorry for the typo.
$endgroup$
– Monolite
Oct 8 '18 at 10:39
1
$begingroup$
@Did Maybe you could take a look at this question since it comes from a comment of yours. In any case thanks for all the answers and your activity on mathstack!
$endgroup$
– Monolite
Oct 23 '18 at 22:43
add a comment |
$begingroup$
The title does not agree with the equation you have written Are you computing $E(X|Y)$ or $E(Y|X)$?
$endgroup$
– Kavi Rama Murthy
Oct 7 '18 at 23:41
1
$begingroup$
@KaviRamaMurthy from the linked question it is $E(Y mid X)$
$endgroup$
– Henry
Oct 7 '18 at 23:43
$begingroup$
@KaviRamaMurthy I will edit, sorry for the typo.
$endgroup$
– Monolite
Oct 8 '18 at 10:39
1
$begingroup$
@Did Maybe you could take a look at this question since it comes from a comment of yours. In any case thanks for all the answers and your activity on mathstack!
$endgroup$
– Monolite
Oct 23 '18 at 22:43
$begingroup$
The title does not agree with the equation you have written Are you computing $E(X|Y)$ or $E(Y|X)$?
$endgroup$
– Kavi Rama Murthy
Oct 7 '18 at 23:41
$begingroup$
The title does not agree with the equation you have written Are you computing $E(X|Y)$ or $E(Y|X)$?
$endgroup$
– Kavi Rama Murthy
Oct 7 '18 at 23:41
1
1
$begingroup$
@KaviRamaMurthy from the linked question it is $E(Y mid X)$
$endgroup$
– Henry
Oct 7 '18 at 23:43
$begingroup$
@KaviRamaMurthy from the linked question it is $E(Y mid X)$
$endgroup$
– Henry
Oct 7 '18 at 23:43
$begingroup$
@KaviRamaMurthy I will edit, sorry for the typo.
$endgroup$
– Monolite
Oct 8 '18 at 10:39
$begingroup$
@KaviRamaMurthy I will edit, sorry for the typo.
$endgroup$
– Monolite
Oct 8 '18 at 10:39
1
1
$begingroup$
@Did Maybe you could take a look at this question since it comes from a comment of yours. In any case thanks for all the answers and your activity on mathstack!
$endgroup$
– Monolite
Oct 23 '18 at 22:43
$begingroup$
@Did Maybe you could take a look at this question since it comes from a comment of yours. In any case thanks for all the answers and your activity on mathstack!
$endgroup$
– Monolite
Oct 23 '18 at 22:43
add a comment |
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$begingroup$
The title does not agree with the equation you have written Are you computing $E(X|Y)$ or $E(Y|X)$?
$endgroup$
– Kavi Rama Murthy
Oct 7 '18 at 23:41
1
$begingroup$
@KaviRamaMurthy from the linked question it is $E(Y mid X)$
$endgroup$
– Henry
Oct 7 '18 at 23:43
$begingroup$
@KaviRamaMurthy I will edit, sorry for the typo.
$endgroup$
– Monolite
Oct 8 '18 at 10:39
1
$begingroup$
@Did Maybe you could take a look at this question since it comes from a comment of yours. In any case thanks for all the answers and your activity on mathstack!
$endgroup$
– Monolite
Oct 23 '18 at 22:43