Iterated Expectation when Prior is a Function — what are the constraints?
$begingroup$
I have the expression
$$E[E[Y|g(X)]] = E[Y]$$
and am trying to determine for what functions $g$ this holds true.
The expression is similar to iterated expectation. I believe the only valid answer is $g(X)=f(X|Y)$ because anything else will prevent the term $f(Y|X)f(X)=f(Y)$ from forming during the development of the iterate expectation theorem. Is this correct?
probability-theory
asked Jan 2 at 19:23
AvedisAvedis
647
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add a comment |
$begingroup$
I have the expression
$$E[E[Y|g(X)]] = E[Y]$$
and am trying to determine for what functions $g$ this holds true.
The expression is similar to iterated expectation. I believe the only valid answer is $g(X)=f(X|Y)$ because anything else will prevent the term $f(Y|X)f(X)=f(Y)$ from forming during the development of the iterate expectation theorem. Is this correct?
probability-theory
asked Jan 2 at 19:23
AvedisAvedis
647
$endgroup$
2
$begingroup$
If $E|Y|<infty$, then $E[E[Y|g(X)]] = E[Y]$ holds for any (Borel measurable) $g$ and $X$. So we don't need any additional conditions.
$endgroup$
– Song
Jan 2 at 19:56
2
$begingroup$
The point is that $E[E[Y|Z]] = E[Y]$ for any random variable $Z$, as long as $E[Y]$ is defined. So in particular this works for $Z = g(X)$ as long as that is a random variable.
$endgroup$
– Robert Israel
Jan 2 at 20:23
add a comment |
$begingroup$
I have the expression
$$E[E[Y|g(X)]] = E[Y]$$
and am trying to determine for what functions $g$ this holds true.
The expression is similar to iterated expectation. I believe the only valid answer is $g(X)=f(X|Y)$ because anything else will prevent the term $f(Y|X)f(X)=f(Y)$ from forming during the development of the iterate expectation theorem. Is this correct?
probability-theory
asked Jan 2 at 19:23
AvedisAvedis
647
$endgroup$
I have the expression
$$E[E[Y|g(X)]] = E[Y]$$
and am trying to determine for what functions $g$ this holds true.
The expression is similar to iterated expectation. I believe the only valid answer is $g(X)=f(X|Y)$ because anything else will prevent the term $f(Y|X)f(X)=f(Y)$ from forming during the development of the iterate expectation theorem. Is this correct?
probability-theory
probability-theory
asked Jan 2 at 19:23
AvedisAvedis
647
asked Jan 2 at 19:23
AvedisAvedis
647
asked Jan 2 at 19:23
AvedisAvedis
647
asked Jan 2 at 19:23
AvedisAvedis
647
asked Jan 2 at 19:23
AvedisAvedis
647
647
2
$begingroup$
If $E|Y|<infty$, then $E[E[Y|g(X)]] = E[Y]$ holds for any (Borel measurable) $g$ and $X$. So we don't need any additional conditions.
$endgroup$
– Song
Jan 2 at 19:56
2
$begingroup$
The point is that $E[E[Y|Z]] = E[Y]$ for any random variable $Z$, as long as $E[Y]$ is defined. So in particular this works for $Z = g(X)$ as long as that is a random variable.
$endgroup$
– Robert Israel
Jan 2 at 20:23
add a comment |
2
$begingroup$
If $E|Y|<infty$, then $E[E[Y|g(X)]] = E[Y]$ holds for any (Borel measurable) $g$ and $X$. So we don't need any additional conditions.
$endgroup$
– Song
Jan 2 at 19:56
2
$begingroup$
The point is that $E[E[Y|Z]] = E[Y]$ for any random variable $Z$, as long as $E[Y]$ is defined. So in particular this works for $Z = g(X)$ as long as that is a random variable.
$endgroup$
– Robert Israel
Jan 2 at 20:23
2
2
$begingroup$
If $E|Y|<infty$, then $E[E[Y|g(X)]] = E[Y]$ holds for any (Borel measurable) $g$ and $X$. So we don't need any additional conditions.
$endgroup$
– Song
Jan 2 at 19:56
$begingroup$
If $E|Y|<infty$, then $E[E[Y|g(X)]] = E[Y]$ holds for any (Borel measurable) $g$ and $X$. So we don't need any additional conditions.
$endgroup$
– Song
Jan 2 at 19:56
2
2
$begingroup$
The point is that $E[E[Y|Z]] = E[Y]$ for any random variable $Z$, as long as $E[Y]$ is defined. So in particular this works for $Z = g(X)$ as long as that is a random variable.
$endgroup$
– Robert Israel
Jan 2 at 20:23
$begingroup$
The point is that $E[E[Y|Z]] = E[Y]$ for any random variable $Z$, as long as $E[Y]$ is defined. So in particular this works for $Z = g(X)$ as long as that is a random variable.
$endgroup$
– Robert Israel
Jan 2 at 20:23
add a comment |
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$begingroup$
If $E|Y|<infty$, then $E[E[Y|g(X)]] = E[Y]$ holds for any (Borel measurable) $g$ and $X$. So we don't need any additional conditions.
$endgroup$
– Song
Jan 2 at 19:56
2
$begingroup$
The point is that $E[E[Y|Z]] = E[Y]$ for any random variable $Z$, as long as $E[Y]$ is defined. So in particular this works for $Z = g(X)$ as long as that is a random variable.
$endgroup$
– Robert Israel
Jan 2 at 20:23