Prove or disprove: If $X | U$ is independent of $Y | V$, then $E[XY|U,V] = E[X|U] cdot E[Y|V]$.
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I may have missed something basic here. Suppose $U$ and $V$ are continuous random variables such that $E[X|U]$ and $E[Y|V]$ makes sense for some random variables $X$ and $Y$. If $X|U$ is independent of $Y|V$, does it follow that
$$
E[XY|U,V] = E[X|U]cdot E[Y|V] ?
$$
Any simpler explanation is greatly appreciated.
conditional-expectation conditional-probability
asked Aug 26 at 5:38
venrey
12811
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up vote
1
down vote
favorite
I may have missed something basic here. Suppose $U$ and $V$ are continuous random variables such that $E[X|U]$ and $E[Y|V]$ makes sense for some random variables $X$ and $Y$. If $X|U$ is independent of $Y|V$, does it follow that
$$
E[XY|U,V] = E[X|U]cdot E[Y|V] ?
$$
Any simpler explanation is greatly appreciated.
conditional-expectation conditional-probability
asked Aug 26 at 5:38
venrey
12811
Can you state precisely what you mean by $X|U$ and $Y|V$ being independent? $X|U$ is not a random variable. You can talk about two random variables being independent when conditioned on a given sigma field, but you are conditioning on two different random variables. As far as I know such a concept does not exist.
– Kavi Rama Murthy
Aug 26 at 12:02
@KaviRamaMurthy I have a side question Kavi. ""Let U be a uniform random variable on (0, 1), and suppose that the conditional distribution of X, given that U = p, is binomial with parameters n and p"" Then is it acceptable to write $X|U = p sim Binomial(n,p)$ therefore $X|U$ is a Binomial RV? Or does $X|U$ still not make sense? I know it can make sense to say that $E[X|U]$ is a RV.
– HJ_beginner
Aug 26 at 17:20
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I may have missed something basic here. Suppose $U$ and $V$ are continuous random variables such that $E[X|U]$ and $E[Y|V]$ makes sense for some random variables $X$ and $Y$. If $X|U$ is independent of $Y|V$, does it follow that
$$
E[XY|U,V] = E[X|U]cdot E[Y|V] ?
$$
Any simpler explanation is greatly appreciated.
conditional-expectation conditional-probability
asked Aug 26 at 5:38
venrey
12811
I may have missed something basic here. Suppose $U$ and $V$ are continuous random variables such that $E[X|U]$ and $E[Y|V]$ makes sense for some random variables $X$ and $Y$. If $X|U$ is independent of $Y|V$, does it follow that
$$
E[XY|U,V] = E[X|U]cdot E[Y|V] ?
$$
Any simpler explanation is greatly appreciated.
conditional-expectation conditional-probability
asked Aug 26 at 5:38
venrey
12811
asked Aug 26 at 5:38
venrey
12811
asked Aug 26 at 5:38
venrey
12811
asked Aug 26 at 5:38
venrey
12811
12811
Can you state precisely what you mean by $X|U$ and $Y|V$ being independent? $X|U$ is not a random variable. You can talk about two random variables being independent when conditioned on a given sigma field, but you are conditioning on two different random variables. As far as I know such a concept does not exist.
– Kavi Rama Murthy
Aug 26 at 12:02
@KaviRamaMurthy I have a side question Kavi. ""Let U be a uniform random variable on (0, 1), and suppose that the conditional distribution of X, given that U = p, is binomial with parameters n and p"" Then is it acceptable to write $X|U = p sim Binomial(n,p)$ therefore $X|U$ is a Binomial RV? Or does $X|U$ still not make sense? I know it can make sense to say that $E[X|U]$ is a RV.
– HJ_beginner
Aug 26 at 17:20
add a comment |Â
Can you state precisely what you mean by $X|U$ and $Y|V$ being independent? $X|U$ is not a random variable. You can talk about two random variables being independent when conditioned on a given sigma field, but you are conditioning on two different random variables. As far as I know such a concept does not exist.
– Kavi Rama Murthy
Aug 26 at 12:02
@KaviRamaMurthy I have a side question Kavi. ""Let U be a uniform random variable on (0, 1), and suppose that the conditional distribution of X, given that U = p, is binomial with parameters n and p"" Then is it acceptable to write $X|U = p sim Binomial(n,p)$ therefore $X|U$ is a Binomial RV? Or does $X|U$ still not make sense? I know it can make sense to say that $E[X|U]$ is a RV.
– HJ_beginner
Aug 26 at 17:20
Can you state precisely what you mean by $X|U$ and $Y|V$ being independent? $X|U$ is not a random variable. You can talk about two random variables being independent when conditioned on a given sigma field, but you are conditioning on two different random variables. As far as I know such a concept does not exist.
– Kavi Rama Murthy
Aug 26 at 12:02
Can you state precisely what you mean by $X|U$ and $Y|V$ being independent? $X|U$ is not a random variable. You can talk about two random variables being independent when conditioned on a given sigma field, but you are conditioning on two different random variables. As far as I know such a concept does not exist.
– Kavi Rama Murthy
Aug 26 at 12:02
@KaviRamaMurthy I have a side question Kavi. ""Let U be a uniform random variable on (0, 1), and suppose that the conditional distribution of X, given that U = p, is binomial with parameters n and p"" Then is it acceptable to write $X|U = p sim Binomial(n,p)$ therefore $X|U$ is a Binomial RV? Or does $X|U$ still not make sense? I know it can make sense to say that $E[X|U]$ is a RV.
– HJ_beginner
Aug 26 at 17:20
@KaviRamaMurthy I have a side question Kavi. ""Let U be a uniform random variable on (0, 1), and suppose that the conditional distribution of X, given that U = p, is binomial with parameters n and p"" Then is it acceptable to write $X|U = p sim Binomial(n,p)$ therefore $X|U$ is a Binomial RV? Or does $X|U$ still not make sense? I know it can make sense to say that $E[X|U]$ is a RV.
– HJ_beginner
Aug 26 at 17:20
add a comment |Â
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Can you state precisely what you mean by $X|U$ and $Y|V$ being independent? $X|U$ is not a random variable. You can talk about two random variables being independent when conditioned on a given sigma field, but you are conditioning on two different random variables. As far as I know such a concept does not exist.
– Kavi Rama Murthy
Aug 26 at 12:02
@KaviRamaMurthy I have a side question Kavi. ""Let U be a uniform random variable on (0, 1), and suppose that the conditional distribution of X, given that U = p, is binomial with parameters n and p"" Then is it acceptable to write $X|U = p sim Binomial(n,p)$ therefore $X|U$ is a Binomial RV? Or does $X|U$ still not make sense? I know it can make sense to say that $E[X|U]$ is a RV.
– HJ_beginner
Aug 26 at 17:20