Formalizing conditional expectation
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I need help to translate into a conditional exepectation the following problem:
We have an interval of $mathbb{R}$ of size M (say $mathcal{M} = [0, M]$). For each element $x in mathcal{M}$ I define $theta(x)$ which is the score of each element of my interval.
I want these scores to follow a given distribution with cdf $F$. For each subset $mu$ of $mathcal{M}$, I would like to define the average value of $theta$ over this subset.
$$
mathbb{E}left[theta(x) left|x in mu right. right]
$$
How can I calculate this? My first intuition is to do
$$
int_{x in mu(x)}{theta(x) dF(x)}
$$
But it just seems like I am mixing different measures. I would really appreciate some help to formalize this and relate this to the theory of probability.
Thank you for your help.
T.
probability-theory conditional-expectation
asked Dec 7 '18 at 18:05
TochokaTochoka
163
$endgroup$
add a comment |
$begingroup$
I need help to translate into a conditional exepectation the following problem:
We have an interval of $mathbb{R}$ of size M (say $mathcal{M} = [0, M]$). For each element $x in mathcal{M}$ I define $theta(x)$ which is the score of each element of my interval.
I want these scores to follow a given distribution with cdf $F$. For each subset $mu$ of $mathcal{M}$, I would like to define the average value of $theta$ over this subset.
$$
mathbb{E}left[theta(x) left|x in mu right. right]
$$
How can I calculate this? My first intuition is to do
$$
int_{x in mu(x)}{theta(x) dF(x)}
$$
But it just seems like I am mixing different measures. I would really appreciate some help to formalize this and relate this to the theory of probability.
Thank you for your help.
T.
probability-theory conditional-expectation
asked Dec 7 '18 at 18:05
TochokaTochoka
163
$endgroup$
1
$begingroup$
The average value of the function $theta:xmapstotheta(x)$ over the set $musubseteqmathcal M$ of positive measure $|mu|$ is by definition $$frac1{|mu|}int_mutheta(x),dx$$ Nothing probabilistic in there.
$endgroup$
– Did
Dec 7 '18 at 19:22
$begingroup$
$theta$ is a random variable.
$endgroup$
– Tochoka
Dec 7 '18 at 20:57
1
$begingroup$
Yeah I know, and the solution does not use this.
$endgroup$
– Did
Dec 7 '18 at 21:13
add a comment |
$begingroup$
I need help to translate into a conditional exepectation the following problem:
We have an interval of $mathbb{R}$ of size M (say $mathcal{M} = [0, M]$). For each element $x in mathcal{M}$ I define $theta(x)$ which is the score of each element of my interval.
I want these scores to follow a given distribution with cdf $F$. For each subset $mu$ of $mathcal{M}$, I would like to define the average value of $theta$ over this subset.
$$
mathbb{E}left[theta(x) left|x in mu right. right]
$$
How can I calculate this? My first intuition is to do
$$
int_{x in mu(x)}{theta(x) dF(x)}
$$
But it just seems like I am mixing different measures. I would really appreciate some help to formalize this and relate this to the theory of probability.
Thank you for your help.
T.
probability-theory conditional-expectation
asked Dec 7 '18 at 18:05
TochokaTochoka
163
$endgroup$
I need help to translate into a conditional exepectation the following problem:
We have an interval of $mathbb{R}$ of size M (say $mathcal{M} = [0, M]$). For each element $x in mathcal{M}$ I define $theta(x)$ which is the score of each element of my interval.
I want these scores to follow a given distribution with cdf $F$. For each subset $mu$ of $mathcal{M}$, I would like to define the average value of $theta$ over this subset.
$$
mathbb{E}left[theta(x) left|x in mu right. right]
$$
How can I calculate this? My first intuition is to do
$$
int_{x in mu(x)}{theta(x) dF(x)}
$$
But it just seems like I am mixing different measures. I would really appreciate some help to formalize this and relate this to the theory of probability.
Thank you for your help.
T.
probability-theory conditional-expectation
probability-theory conditional-expectation
asked Dec 7 '18 at 18:05
TochokaTochoka
163
asked Dec 7 '18 at 18:05
TochokaTochoka
163
asked Dec 7 '18 at 18:05
TochokaTochoka
163
asked Dec 7 '18 at 18:05
TochokaTochoka
163
asked Dec 7 '18 at 18:05
TochokaTochoka
163
163
1
$begingroup$
The average value of the function $theta:xmapstotheta(x)$ over the set $musubseteqmathcal M$ of positive measure $|mu|$ is by definition $$frac1{|mu|}int_mutheta(x),dx$$ Nothing probabilistic in there.
$endgroup$
– Did
Dec 7 '18 at 19:22
$begingroup$
$theta$ is a random variable.
$endgroup$
– Tochoka
Dec 7 '18 at 20:57
1
$begingroup$
Yeah I know, and the solution does not use this.
$endgroup$
– Did
Dec 7 '18 at 21:13
add a comment |
1
$begingroup$
The average value of the function $theta:xmapstotheta(x)$ over the set $musubseteqmathcal M$ of positive measure $|mu|$ is by definition $$frac1{|mu|}int_mutheta(x),dx$$ Nothing probabilistic in there.
$endgroup$
– Did
Dec 7 '18 at 19:22
$begingroup$
$theta$ is a random variable.
$endgroup$
– Tochoka
Dec 7 '18 at 20:57
1
$begingroup$
Yeah I know, and the solution does not use this.
$endgroup$
– Did
Dec 7 '18 at 21:13
1
1
$begingroup$
The average value of the function $theta:xmapstotheta(x)$ over the set $musubseteqmathcal M$ of positive measure $|mu|$ is by definition $$frac1{|mu|}int_mutheta(x),dx$$ Nothing probabilistic in there.
$endgroup$
– Did
Dec 7 '18 at 19:22
$begingroup$
The average value of the function $theta:xmapstotheta(x)$ over the set $musubseteqmathcal M$ of positive measure $|mu|$ is by definition $$frac1{|mu|}int_mutheta(x),dx$$ Nothing probabilistic in there.
$endgroup$
– Did
Dec 7 '18 at 19:22
$begingroup$
$theta$ is a random variable.
$endgroup$
– Tochoka
Dec 7 '18 at 20:57
$begingroup$
$theta$ is a random variable.
$endgroup$
– Tochoka
Dec 7 '18 at 20:57
1
1
$begingroup$
Yeah I know, and the solution does not use this.
$endgroup$
– Did
Dec 7 '18 at 21:13
$begingroup$
Yeah I know, and the solution does not use this.
$endgroup$
– Did
Dec 7 '18 at 21:13
add a comment |
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1
$begingroup$
The average value of the function $theta:xmapstotheta(x)$ over the set $musubseteqmathcal M$ of positive measure $|mu|$ is by definition $$frac1{|mu|}int_mutheta(x),dx$$ Nothing probabilistic in there.
$endgroup$
– Did
Dec 7 '18 at 19:22
$begingroup$
$theta$ is a random variable.
$endgroup$
– Tochoka
Dec 7 '18 at 20:57
1
$begingroup$
Yeah I know, and the solution does not use this.
$endgroup$
– Did
Dec 7 '18 at 21:13