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question about the notation of conditional expectation
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question about the notation of conditional expectation
Intuition behind Conditional ExpectationIntuition behind conditional expectation when sigma algebra isn't generated by a partitionA question on Conditional Expectation from BreimanExpectation and Variance of Conditional Sum (using formal definition of conditional expectation)Difference between conditional expectation and conditional probabiltyConditional expectation of a random variable conditional on a function of the random variableNotation for Conditional ExpectationWhat is the expectation of $X$ given $X$
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$begingroup$
Let $X$ be a standard normal random variable. We need to compute the integral
$$E(X~|X>0).$$
In the book, they give the answer $1over sqrt2pi$. Then I am confused with this conditional expectation notation.
Since from my understanding, it should be
$$int_0^infty 2cdot xover sqrt2pi e^-x^2over 2, dx$$
which gives the answer $sqrt2overpi$. I add a factor "2" here. Can anyone tell me this conditon expectation $E(X|X>0)$ notation meaning? Since it is not standard, we usually conditioning on a $sigma$-algebra, rather than a set.
probability probability-theory conditional-expectation conditional-probability
asked 8 hours ago
Qing SUNQing SUN
1005 bronze badges
$endgroup$
add a comment
|
$begingroup$
Let $X$ be a standard normal random variable. We need to compute the integral
$$E(X~|X>0).$$
In the book, they give the answer $1over sqrt2pi$. Then I am confused with this conditional expectation notation.
Since from my understanding, it should be
$$int_0^infty 2cdot xover sqrt2pi e^-x^2over 2, dx$$
which gives the answer $sqrt2overpi$. I add a factor "2" here. Can anyone tell me this conditon expectation $E(X|X>0)$ notation meaning? Since it is not standard, we usually conditioning on a $sigma$-algebra, rather than a set.
probability probability-theory conditional-expectation conditional-probability
asked 8 hours ago
Qing SUNQing SUN
1005 bronze badges
$endgroup$
$begingroup$
You are right as far as I can tell.
$endgroup$
– Will M.
6 hours ago
$begingroup$
In my experience, when one sees $E(Amid B),$ usually $B$ is a set. It seems unusual to me to put an entire sigma algebra in that place in the formula. But that doesn't rule out the possibility that your book uses the notation $E(Amid B)$ in a non-standard way when $B$ is a set. It might be instructive to see whether that use of the notation occurs in any other examples or exercises and what results are given for those occurrences.
$endgroup$
– David K
6 hours ago
add a comment
|
$begingroup$
Let $X$ be a standard normal random variable. We need to compute the integral
$$E(X~|X>0).$$
In the book, they give the answer $1over sqrt2pi$. Then I am confused with this conditional expectation notation.
Since from my understanding, it should be
$$int_0^infty 2cdot xover sqrt2pi e^-x^2over 2, dx$$
which gives the answer $sqrt2overpi$. I add a factor "2" here. Can anyone tell me this conditon expectation $E(X|X>0)$ notation meaning? Since it is not standard, we usually conditioning on a $sigma$-algebra, rather than a set.
probability probability-theory conditional-expectation conditional-probability
asked 8 hours ago
Qing SUNQing SUN
1005 bronze badges
$endgroup$
Let $X$ be a standard normal random variable. We need to compute the integral
$$E(X~|X>0).$$
In the book, they give the answer $1over sqrt2pi$. Then I am confused with this conditional expectation notation.
Since from my understanding, it should be
$$int_0^infty 2cdot xover sqrt2pi e^-x^2over 2, dx$$
which gives the answer $sqrt2overpi$. I add a factor "2" here. Can anyone tell me this conditon expectation $E(X|X>0)$ notation meaning? Since it is not standard, we usually conditioning on a $sigma$-algebra, rather than a set.
probability probability-theory conditional-expectation conditional-probability
probability probability-theory conditional-expectation conditional-probability
asked 8 hours ago
Qing SUNQing SUN
1005 bronze badges
asked 8 hours ago
Qing SUNQing SUN
1005 bronze badges
asked 8 hours ago
Qing SUNQing SUN
1005 bronze badges
asked 8 hours ago
Qing SUNQing SUN
1005 bronze badges
asked 8 hours ago
Qing SUNQing SUN
1005 bronze badges
1005 bronze badges
$begingroup$
You are right as far as I can tell.
$endgroup$
– Will M.
6 hours ago
$begingroup$
In my experience, when one sees $E(Amid B),$ usually $B$ is a set. It seems unusual to me to put an entire sigma algebra in that place in the formula. But that doesn't rule out the possibility that your book uses the notation $E(Amid B)$ in a non-standard way when $B$ is a set. It might be instructive to see whether that use of the notation occurs in any other examples or exercises and what results are given for those occurrences.
$endgroup$
– David K
6 hours ago
add a comment
|
$begingroup$
You are right as far as I can tell.
$endgroup$
– Will M.
6 hours ago
$begingroup$
In my experience, when one sees $E(Amid B),$ usually $B$ is a set. It seems unusual to me to put an entire sigma algebra in that place in the formula. But that doesn't rule out the possibility that your book uses the notation $E(Amid B)$ in a non-standard way when $B$ is a set. It might be instructive to see whether that use of the notation occurs in any other examples or exercises and what results are given for those occurrences.
$endgroup$
– David K
6 hours ago
$begingroup$
You are right as far as I can tell.
$endgroup$
– Will M.
6 hours ago
$begingroup$
You are right as far as I can tell.
$endgroup$
– Will M.
6 hours ago
$begingroup$
In my experience, when one sees $E(Amid B),$ usually $B$ is a set. It seems unusual to me to put an entire sigma algebra in that place in the formula. But that doesn't rule out the possibility that your book uses the notation $E(Amid B)$ in a non-standard way when $B$ is a set. It might be instructive to see whether that use of the notation occurs in any other examples or exercises and what results are given for those occurrences.
$endgroup$
– David K
6 hours ago
$begingroup$
In my experience, when one sees $E(Amid B),$ usually $B$ is a set. It seems unusual to me to put an entire sigma algebra in that place in the formula. But that doesn't rule out the possibility that your book uses the notation $E(Amid B)$ in a non-standard way when $B$ is a set. It might be instructive to see whether that use of the notation occurs in any other examples or exercises and what results are given for those occurrences.
$endgroup$
– David K
6 hours ago
add a comment
|
1 Answer
1
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votes
$begingroup$
By the wikipedia page for conditional expectation, we should have
$$E(X|A) = fracE(Xmathbf1_A)P(A)$$
In particular, $E(X|X>0) = sqrtfrac2pi$ when $X$ is a standard normal.
Related, I believe I have seen the notation $E(X;A) := E(Xmathbf1_A),$ which would give an answer of $E(X;X>0) = frac1sqrt2pi$
answered 8 hours ago
Brian MoehringBrian Moehring
3,6773 silver badges14 bronze badges
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1 Answer
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1 Answer
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$begingroup$
By the wikipedia page for conditional expectation, we should have
$$E(X|A) = fracE(Xmathbf1_A)P(A)$$
In particular, $E(X|X>0) = sqrtfrac2pi$ when $X$ is a standard normal.
Related, I believe I have seen the notation $E(X;A) := E(Xmathbf1_A),$ which would give an answer of $E(X;X>0) = frac1sqrt2pi$
answered 8 hours ago
Brian MoehringBrian Moehring
3,6773 silver badges14 bronze badges
$endgroup$
add a comment
|
$begingroup$
By the wikipedia page for conditional expectation, we should have
$$E(X|A) = fracE(Xmathbf1_A)P(A)$$
In particular, $E(X|X>0) = sqrtfrac2pi$ when $X$ is a standard normal.
Related, I believe I have seen the notation $E(X;A) := E(Xmathbf1_A),$ which would give an answer of $E(X;X>0) = frac1sqrt2pi$
answered 8 hours ago
Brian MoehringBrian Moehring
3,6773 silver badges14 bronze badges
$endgroup$
add a comment
|
$begingroup$
By the wikipedia page for conditional expectation, we should have
$$E(X|A) = fracE(Xmathbf1_A)P(A)$$
In particular, $E(X|X>0) = sqrtfrac2pi$ when $X$ is a standard normal.
Related, I believe I have seen the notation $E(X;A) := E(Xmathbf1_A),$ which would give an answer of $E(X;X>0) = frac1sqrt2pi$
answered 8 hours ago
Brian MoehringBrian Moehring
3,6773 silver badges14 bronze badges
$endgroup$
By the wikipedia page for conditional expectation, we should have
$$E(X|A) = fracE(Xmathbf1_A)P(A)$$
In particular, $E(X|X>0) = sqrtfrac2pi$ when $X$ is a standard normal.
Related, I believe I have seen the notation $E(X;A) := E(Xmathbf1_A),$ which would give an answer of $E(X;X>0) = frac1sqrt2pi$
answered 8 hours ago
Brian MoehringBrian Moehring
3,6773 silver badges14 bronze badges
edited 7 hours ago
answered 8 hours ago
Brian MoehringBrian Moehring
3,6773 silver badges14 bronze badges
answered 8 hours ago
Brian MoehringBrian Moehring
3,6773 silver badges14 bronze badges
answered 8 hours ago
Brian MoehringBrian Moehring
3,6773 silver badges14 bronze badges
3,6773 silver badges14 bronze badges
add a comment
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$begingroup$
You are right as far as I can tell.
$endgroup$
– Will M.
6 hours ago
$begingroup$
In my experience, when one sees $E(Amid B),$ usually $B$ is a set. It seems unusual to me to put an entire sigma algebra in that place in the formula. But that doesn't rule out the possibility that your book uses the notation $E(Amid B)$ in a non-standard way when $B$ is a set. It might be instructive to see whether that use of the notation occurs in any other examples or exercises and what results are given for those occurrences.
$endgroup$
– David K
6 hours ago