Mixing
Non-trivial examples of $E[X|X^2] = X$ and $E[Y|Y^2] = 0$.
$begingroup$
I want to find a few examples of non-trivial random variables $X, Y$ such that $E[X|X^2] = X$ and $E[Y|Y^2] = 0$.
From what i gather for $E[X|X^2] = X$ i need to find a function such that if $X=u$ then $f(-u)=0$. Then for $E[Y|Y^2] = 0$ if $Y=u$ then $f(u)=f(-u)$.
conditional-expectation expected-value
asked Jan 28 at 16:39
Student7549Student7549
164
$endgroup$
add a comment |
$begingroup$
I want to find a few examples of non-trivial random variables $X, Y$ such that $E[X|X^2] = X$ and $E[Y|Y^2] = 0$.
From what i gather for $E[X|X^2] = X$ i need to find a function such that if $X=u$ then $f(-u)=0$. Then for $E[Y|Y^2] = 0$ if $Y=u$ then $f(u)=f(-u)$.
conditional-expectation expected-value
asked Jan 28 at 16:39
Student7549Student7549
164
$endgroup$
2
$begingroup$
Any random variable $X$ with pdf $f_X(x) = 0$ for any $x < 0$ satisfies the first property, e.g. exponential, chi-squared, uniform on a positive interval. For $Y$, any distribution that is symmetric around $0$ works, e.g. normal, Cauchy.
$endgroup$
– Alex
Jan 28 at 16:56
add a comment |
$begingroup$
I want to find a few examples of non-trivial random variables $X, Y$ such that $E[X|X^2] = X$ and $E[Y|Y^2] = 0$.
From what i gather for $E[X|X^2] = X$ i need to find a function such that if $X=u$ then $f(-u)=0$. Then for $E[Y|Y^2] = 0$ if $Y=u$ then $f(u)=f(-u)$.
conditional-expectation expected-value
asked Jan 28 at 16:39
Student7549Student7549
164
$endgroup$
I want to find a few examples of non-trivial random variables $X, Y$ such that $E[X|X^2] = X$ and $E[Y|Y^2] = 0$.
From what i gather for $E[X|X^2] = X$ i need to find a function such that if $X=u$ then $f(-u)=0$. Then for $E[Y|Y^2] = 0$ if $Y=u$ then $f(u)=f(-u)$.
conditional-expectation expected-value
conditional-expectation expected-value
asked Jan 28 at 16:39
Student7549Student7549
164
asked Jan 28 at 16:39
Student7549Student7549
164
edited Jan 28 at 16:48
Student7549
asked Jan 28 at 16:39
Student7549Student7549
164
asked Jan 28 at 16:39
Student7549Student7549
164
asked Jan 28 at 16:39
Student7549Student7549
164
164
2
$begingroup$
Any random variable $X$ with pdf $f_X(x) = 0$ for any $x < 0$ satisfies the first property, e.g. exponential, chi-squared, uniform on a positive interval. For $Y$, any distribution that is symmetric around $0$ works, e.g. normal, Cauchy.
$endgroup$
– Alex
Jan 28 at 16:56
add a comment |
2
$begingroup$
Any random variable $X$ with pdf $f_X(x) = 0$ for any $x < 0$ satisfies the first property, e.g. exponential, chi-squared, uniform on a positive interval. For $Y$, any distribution that is symmetric around $0$ works, e.g. normal, Cauchy.
$endgroup$
– Alex
Jan 28 at 16:56
2
2
$begingroup$
Any random variable $X$ with pdf $f_X(x) = 0$ for any $x < 0$ satisfies the first property, e.g. exponential, chi-squared, uniform on a positive interval. For $Y$, any distribution that is symmetric around $0$ works, e.g. normal, Cauchy.
$endgroup$
– Alex
Jan 28 at 16:56
$begingroup$
Any random variable $X$ with pdf $f_X(x) = 0$ for any $x < 0$ satisfies the first property, e.g. exponential, chi-squared, uniform on a positive interval. For $Y$, any distribution that is symmetric around $0$ works, e.g. normal, Cauchy.
$endgroup$
– Alex
Jan 28 at 16:56
add a comment |
2 Answers
2
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$begingroup$
Assume that $X$ is any positive random variable. Then $X=sqrt{X^2}$ is $sigma(X^2)$-measurable so the conditional expectation of $X$ given $X^2$ is $X$.
Let $Y$ be any $L^2$ symmetric random variable. Then, for any bounded function $g$, $Yg(Y^2)$ is symmetric so has mean $0$, which is the mean value of $0cdot g(Y^2)$, and $0$ is $sigma(Y^2)$-measurable, so the conditional expectation of $Y$ given $Y^2$ is zero.
answered Jan 28 at 16:54
MindlackMindlack
4,910211
$endgroup$
add a comment |
$begingroup$
if you look at
A problem in Jun Shao mathematical statistics:exercises and solutions P33
example of conditions that $E(X|X^2)=0$
$E(X|X^2=t)=sqrt(t) bigg(frac{f(sqrt(t))-f(-sqrt(t))}{f(sqrt(t))+f(-sqrt(t))} bigg)$
$E(X|X^2=t)=0$ hence $f(sqrt(t))-f(-sqrt(t))=0$
so it is enough
$f(t)=f(-t)$ so you need to choose a symmetric distribution around Zero.
example of conditions that $E(X|X^2)=X$
$E(X|X^2=t)=sqrt(t) $ if
$bigg(frac{f(sqrt(t))-f(-sqrt(t))}{f(sqrt(t))+f(-sqrt(t))} bigg)=1$
so $2f(-sqrt(t))=0$ so $X$ need to be positive.
answered 2 days ago
masoudmasoud
1035
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Assume that $X$ is any positive random variable. Then $X=sqrt{X^2}$ is $sigma(X^2)$-measurable so the conditional expectation of $X$ given $X^2$ is $X$.
Let $Y$ be any $L^2$ symmetric random variable. Then, for any bounded function $g$, $Yg(Y^2)$ is symmetric so has mean $0$, which is the mean value of $0cdot g(Y^2)$, and $0$ is $sigma(Y^2)$-measurable, so the conditional expectation of $Y$ given $Y^2$ is zero.
answered Jan 28 at 16:54
MindlackMindlack
4,910211
$endgroup$
add a comment |
$begingroup$
Assume that $X$ is any positive random variable. Then $X=sqrt{X^2}$ is $sigma(X^2)$-measurable so the conditional expectation of $X$ given $X^2$ is $X$.
Let $Y$ be any $L^2$ symmetric random variable. Then, for any bounded function $g$, $Yg(Y^2)$ is symmetric so has mean $0$, which is the mean value of $0cdot g(Y^2)$, and $0$ is $sigma(Y^2)$-measurable, so the conditional expectation of $Y$ given $Y^2$ is zero.
answered Jan 28 at 16:54
MindlackMindlack
4,910211
$endgroup$
add a comment |
$begingroup$
Assume that $X$ is any positive random variable. Then $X=sqrt{X^2}$ is $sigma(X^2)$-measurable so the conditional expectation of $X$ given $X^2$ is $X$.
Let $Y$ be any $L^2$ symmetric random variable. Then, for any bounded function $g$, $Yg(Y^2)$ is symmetric so has mean $0$, which is the mean value of $0cdot g(Y^2)$, and $0$ is $sigma(Y^2)$-measurable, so the conditional expectation of $Y$ given $Y^2$ is zero.
answered Jan 28 at 16:54
MindlackMindlack
4,910211
$endgroup$
Assume that $X$ is any positive random variable. Then $X=sqrt{X^2}$ is $sigma(X^2)$-measurable so the conditional expectation of $X$ given $X^2$ is $X$.
Let $Y$ be any $L^2$ symmetric random variable. Then, for any bounded function $g$, $Yg(Y^2)$ is symmetric so has mean $0$, which is the mean value of $0cdot g(Y^2)$, and $0$ is $sigma(Y^2)$-measurable, so the conditional expectation of $Y$ given $Y^2$ is zero.
answered Jan 28 at 16:54
MindlackMindlack
4,910211
answered Jan 28 at 16:54
MindlackMindlack
4,910211
answered Jan 28 at 16:54
MindlackMindlack
4,910211
answered Jan 28 at 16:54
MindlackMindlack
4,910211
4,910211
add a comment |
add a comment |
$begingroup$
if you look at
A problem in Jun Shao mathematical statistics:exercises and solutions P33
example of conditions that $E(X|X^2)=0$
$E(X|X^2=t)=sqrt(t) bigg(frac{f(sqrt(t))-f(-sqrt(t))}{f(sqrt(t))+f(-sqrt(t))} bigg)$
$E(X|X^2=t)=0$ hence $f(sqrt(t))-f(-sqrt(t))=0$
so it is enough
$f(t)=f(-t)$ so you need to choose a symmetric distribution around Zero.
example of conditions that $E(X|X^2)=X$
$E(X|X^2=t)=sqrt(t) $ if
$bigg(frac{f(sqrt(t))-f(-sqrt(t))}{f(sqrt(t))+f(-sqrt(t))} bigg)=1$
so $2f(-sqrt(t))=0$ so $X$ need to be positive.
answered 2 days ago
masoudmasoud
1035
$endgroup$
add a comment |
$begingroup$
if you look at
A problem in Jun Shao mathematical statistics:exercises and solutions P33
example of conditions that $E(X|X^2)=0$
$E(X|X^2=t)=sqrt(t) bigg(frac{f(sqrt(t))-f(-sqrt(t))}{f(sqrt(t))+f(-sqrt(t))} bigg)$
$E(X|X^2=t)=0$ hence $f(sqrt(t))-f(-sqrt(t))=0$
so it is enough
$f(t)=f(-t)$ so you need to choose a symmetric distribution around Zero.
example of conditions that $E(X|X^2)=X$
$E(X|X^2=t)=sqrt(t) $ if
$bigg(frac{f(sqrt(t))-f(-sqrt(t))}{f(sqrt(t))+f(-sqrt(t))} bigg)=1$
so $2f(-sqrt(t))=0$ so $X$ need to be positive.
answered 2 days ago
masoudmasoud
1035
$endgroup$
add a comment |
$begingroup$
if you look at
A problem in Jun Shao mathematical statistics:exercises and solutions P33
example of conditions that $E(X|X^2)=0$
$E(X|X^2=t)=sqrt(t) bigg(frac{f(sqrt(t))-f(-sqrt(t))}{f(sqrt(t))+f(-sqrt(t))} bigg)$
$E(X|X^2=t)=0$ hence $f(sqrt(t))-f(-sqrt(t))=0$
so it is enough
$f(t)=f(-t)$ so you need to choose a symmetric distribution around Zero.
example of conditions that $E(X|X^2)=X$
$E(X|X^2=t)=sqrt(t) $ if
$bigg(frac{f(sqrt(t))-f(-sqrt(t))}{f(sqrt(t))+f(-sqrt(t))} bigg)=1$
so $2f(-sqrt(t))=0$ so $X$ need to be positive.
answered 2 days ago
masoudmasoud
1035
$endgroup$
if you look at
A problem in Jun Shao mathematical statistics:exercises and solutions P33
example of conditions that $E(X|X^2)=0$
$E(X|X^2=t)=sqrt(t) bigg(frac{f(sqrt(t))-f(-sqrt(t))}{f(sqrt(t))+f(-sqrt(t))} bigg)$
$E(X|X^2=t)=0$ hence $f(sqrt(t))-f(-sqrt(t))=0$
so it is enough
$f(t)=f(-t)$ so you need to choose a symmetric distribution around Zero.
example of conditions that $E(X|X^2)=X$
$E(X|X^2=t)=sqrt(t) $ if
$bigg(frac{f(sqrt(t))-f(-sqrt(t))}{f(sqrt(t))+f(-sqrt(t))} bigg)=1$
so $2f(-sqrt(t))=0$ so $X$ need to be positive.
answered 2 days ago
masoudmasoud
1035
answered 2 days ago
masoudmasoud
1035
answered 2 days ago
masoudmasoud
1035
answered 2 days ago
masoudmasoud
1035
1035
add a comment |
add a comment |
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$begingroup$
Any random variable $X$ with pdf $f_X(x) = 0$ for any $x < 0$ satisfies the first property, e.g. exponential, chi-squared, uniform on a positive interval. For $Y$, any distribution that is symmetric around $0$ works, e.g. normal, Cauchy.
$endgroup$
– Alex
Jan 28 at 16:56