Mixing
Expectation of the product of two random variables is the supremum over all identically distributed copies of one of the factors.
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Let $X in L^1$ and $Y in L^infty$ be two random variables (or, more generally, $X in L^p, , Y in L^q$ for $p, , q$ conjugated). Is it true that
$$ mathbbE left[ X Y right] = sup_ hatY sim Y mathbbE left[ X hatY right]$$
where the supremum is taken over all $hatY$ that have the same distribution as $Y$?
Clearly, the right-hand side is well-defined, but could a-priori be $infty$. Also, the right-hand side is clearly larger than the left.
How could one prove the other direction $geq$?
probability probability-theory random-variables
asked Aug 2 at 7:59
LittleDoe
816
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up vote
0
down vote
favorite
Let $X in L^1$ and $Y in L^infty$ be two random variables (or, more generally, $X in L^p, , Y in L^q$ for $p, , q$ conjugated). Is it true that
$$ mathbbE left[ X Y right] = sup_ hatY sim Y mathbbE left[ X hatY right]$$
where the supremum is taken over all $hatY$ that have the same distribution as $Y$?
Clearly, the right-hand side is well-defined, but could a-priori be $infty$. Also, the right-hand side is clearly larger than the left.
How could one prove the other direction $geq$?
probability probability-theory random-variables
asked Aug 2 at 7:59
LittleDoe
816
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $X in L^1$ and $Y in L^infty$ be two random variables (or, more generally, $X in L^p, , Y in L^q$ for $p, , q$ conjugated). Is it true that
$$ mathbbE left[ X Y right] = sup_ hatY sim Y mathbbE left[ X hatY right]$$
where the supremum is taken over all $hatY$ that have the same distribution as $Y$?
Clearly, the right-hand side is well-defined, but could a-priori be $infty$. Also, the right-hand side is clearly larger than the left.
How could one prove the other direction $geq$?
probability probability-theory random-variables
asked Aug 2 at 7:59
LittleDoe
816
Let $X in L^1$ and $Y in L^infty$ be two random variables (or, more generally, $X in L^p, , Y in L^q$ for $p, , q$ conjugated). Is it true that
$$ mathbbE left[ X Y right] = sup_ hatY sim Y mathbbE left[ X hatY right]$$
where the supremum is taken over all $hatY$ that have the same distribution as $Y$?
Clearly, the right-hand side is well-defined, but could a-priori be $infty$. Also, the right-hand side is clearly larger than the left.
How could one prove the other direction $geq$?
probability probability-theory random-variables
asked Aug 2 at 7:59
LittleDoe
816
asked Aug 2 at 7:59
LittleDoe
816
asked Aug 2 at 7:59
LittleDoe
816
asked Aug 2 at 7:59
LittleDoe
816
816
add a comment |Â
add a comment |Â
1 Answer
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False. Let $Y$ have standard normal distribution and $X=-Y$. Then $EXY=-1$ but $hat Y =X$ has the same distribution as $Y$ so RHS $geq EX^2=1$. This answers the second part. For the first part you need a bounded random variable, so replace normal distribution by any symmetric bounded distribution.
answered Aug 2 at 8:33
Kavi Rama Murthy
19.2k2829
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
False. Let $Y$ have standard normal distribution and $X=-Y$. Then $EXY=-1$ but $hat Y =X$ has the same distribution as $Y$ so RHS $geq EX^2=1$. This answers the second part. For the first part you need a bounded random variable, so replace normal distribution by any symmetric bounded distribution.
answered Aug 2 at 8:33
Kavi Rama Murthy
19.2k2829
add a comment |Â
up vote
1
down vote
accepted
False. Let $Y$ have standard normal distribution and $X=-Y$. Then $EXY=-1$ but $hat Y =X$ has the same distribution as $Y$ so RHS $geq EX^2=1$. This answers the second part. For the first part you need a bounded random variable, so replace normal distribution by any symmetric bounded distribution.
answered Aug 2 at 8:33
Kavi Rama Murthy
19.2k2829
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
False. Let $Y$ have standard normal distribution and $X=-Y$. Then $EXY=-1$ but $hat Y =X$ has the same distribution as $Y$ so RHS $geq EX^2=1$. This answers the second part. For the first part you need a bounded random variable, so replace normal distribution by any symmetric bounded distribution.
answered Aug 2 at 8:33
Kavi Rama Murthy
19.2k2829
False. Let $Y$ have standard normal distribution and $X=-Y$. Then $EXY=-1$ but $hat Y =X$ has the same distribution as $Y$ so RHS $geq EX^2=1$. This answers the second part. For the first part you need a bounded random variable, so replace normal distribution by any symmetric bounded distribution.
answered Aug 2 at 8:33
Kavi Rama Murthy
19.2k2829
answered Aug 2 at 8:33
Kavi Rama Murthy
19.2k2829
answered Aug 2 at 8:33
Kavi Rama Murthy
19.2k2829
answered Aug 2 at 8:33
Kavi Rama Murthy
19.2k2829
19.2k2829
add a comment |Â
add a comment |Â
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