What can I say about expected value of expected value of x given y: $E[E[X|Y]]$
$begingroup$
My question is about Y that is discrete and for some random variable X, but if its having a meaning for Y that is continuous then please give that case your attention.
What can I say about $E[E[X|Y]]$?
I know that E[X|Y] is random variable, so It's not trivial case when we calculate expected value of just a number.
And what about $E[E[X|Y]|Y]$? does something like this have a meaning?
if it's, then does for some function $g$ (for simplicity, assuming g with suitable range and continuous) it's true to say that:
$$E[g(Y)E[X|Y]|Y]=g(Y)E[E[X|Y]|Y]$$
Because of a theorem that I seen:
$$E[g(Y)X|Y]=g(Y)E[X|Y]$$
probability means expected-value
asked Dec 14 '18 at 9:17
Mr.OYMr.OY
16610
$endgroup$
add a comment |
$begingroup$
My question is about Y that is discrete and for some random variable X, but if its having a meaning for Y that is continuous then please give that case your attention.
What can I say about $E[E[X|Y]]$?
I know that E[X|Y] is random variable, so It's not trivial case when we calculate expected value of just a number.
And what about $E[E[X|Y]|Y]$? does something like this have a meaning?
if it's, then does for some function $g$ (for simplicity, assuming g with suitable range and continuous) it's true to say that:
$$E[g(Y)E[X|Y]|Y]=g(Y)E[E[X|Y]|Y]$$
Because of a theorem that I seen:
$$E[g(Y)X|Y]=g(Y)E[X|Y]$$
probability means expected-value
asked Dec 14 '18 at 9:17
Mr.OYMr.OY
16610
$endgroup$
1
$begingroup$
In $E[E[Xmid Y]]$ the inner expectation is over $X$ given $Y$ and is a function of $Y$, while the outer expectation is over $Y$ and is a numerical value. So to the extent that $E[E[Xmid Y] mid Y]$ has a meaning, it is the same thing
$endgroup$
– Henry
Dec 14 '18 at 9:25
$begingroup$
@Henry $E[E[X∣Y]∣Y] $ is the same as $E[E[X∣Y]]$? or that you meant that the interpretation of this follows as it was with $E[E[X∣Y]]$?
$endgroup$
– Mr.OY
Dec 14 '18 at 9:35
$begingroup$
If you define $g(y)=E[X mid Y=y]$ then $E[X mid Y] = g(Y)$ and $E[E[X mid Y]] = E[g(Y)]=E[X]$. You can also say that conditioned on $Y=y$ you have $g(Y)=g(y)$ and thus in general and rather obviously $g(Y)=g(Y)$, so in that sense $E[E[X mid Y]mid Y]$ ought to mean $E[g(Y) mid Y] = E[g(Y)]$ and we already know that is $E[X]$
$endgroup$
– Henry
Dec 14 '18 at 16:03
add a comment |
$begingroup$
My question is about Y that is discrete and for some random variable X, but if its having a meaning for Y that is continuous then please give that case your attention.
What can I say about $E[E[X|Y]]$?
I know that E[X|Y] is random variable, so It's not trivial case when we calculate expected value of just a number.
And what about $E[E[X|Y]|Y]$? does something like this have a meaning?
if it's, then does for some function $g$ (for simplicity, assuming g with suitable range and continuous) it's true to say that:
$$E[g(Y)E[X|Y]|Y]=g(Y)E[E[X|Y]|Y]$$
Because of a theorem that I seen:
$$E[g(Y)X|Y]=g(Y)E[X|Y]$$
probability means expected-value
asked Dec 14 '18 at 9:17
Mr.OYMr.OY
16610
$endgroup$
My question is about Y that is discrete and for some random variable X, but if its having a meaning for Y that is continuous then please give that case your attention.
What can I say about $E[E[X|Y]]$?
I know that E[X|Y] is random variable, so It's not trivial case when we calculate expected value of just a number.
And what about $E[E[X|Y]|Y]$? does something like this have a meaning?
if it's, then does for some function $g$ (for simplicity, assuming g with suitable range and continuous) it's true to say that:
$$E[g(Y)E[X|Y]|Y]=g(Y)E[E[X|Y]|Y]$$
Because of a theorem that I seen:
$$E[g(Y)X|Y]=g(Y)E[X|Y]$$
probability means expected-value
probability means expected-value
asked Dec 14 '18 at 9:17
Mr.OYMr.OY
16610
asked Dec 14 '18 at 9:17
Mr.OYMr.OY
16610
edited Dec 14 '18 at 9:22
Mr.OY
asked Dec 14 '18 at 9:17
Mr.OYMr.OY
16610
asked Dec 14 '18 at 9:17
Mr.OYMr.OY
16610
asked Dec 14 '18 at 9:17
Mr.OYMr.OY
16610
16610
1
$begingroup$
In $E[E[Xmid Y]]$ the inner expectation is over $X$ given $Y$ and is a function of $Y$, while the outer expectation is over $Y$ and is a numerical value. So to the extent that $E[E[Xmid Y] mid Y]$ has a meaning, it is the same thing
$endgroup$
– Henry
Dec 14 '18 at 9:25
$begingroup$
@Henry $E[E[X∣Y]∣Y] $ is the same as $E[E[X∣Y]]$? or that you meant that the interpretation of this follows as it was with $E[E[X∣Y]]$?
$endgroup$
– Mr.OY
Dec 14 '18 at 9:35
$begingroup$
If you define $g(y)=E[X mid Y=y]$ then $E[X mid Y] = g(Y)$ and $E[E[X mid Y]] = E[g(Y)]=E[X]$. You can also say that conditioned on $Y=y$ you have $g(Y)=g(y)$ and thus in general and rather obviously $g(Y)=g(Y)$, so in that sense $E[E[X mid Y]mid Y]$ ought to mean $E[g(Y) mid Y] = E[g(Y)]$ and we already know that is $E[X]$
$endgroup$
– Henry
Dec 14 '18 at 16:03
add a comment |
1
$begingroup$
In $E[E[Xmid Y]]$ the inner expectation is over $X$ given $Y$ and is a function of $Y$, while the outer expectation is over $Y$ and is a numerical value. So to the extent that $E[E[Xmid Y] mid Y]$ has a meaning, it is the same thing
$endgroup$
– Henry
Dec 14 '18 at 9:25
$begingroup$
@Henry $E[E[X∣Y]∣Y] $ is the same as $E[E[X∣Y]]$? or that you meant that the interpretation of this follows as it was with $E[E[X∣Y]]$?
$endgroup$
– Mr.OY
Dec 14 '18 at 9:35
$begingroup$
If you define $g(y)=E[X mid Y=y]$ then $E[X mid Y] = g(Y)$ and $E[E[X mid Y]] = E[g(Y)]=E[X]$. You can also say that conditioned on $Y=y$ you have $g(Y)=g(y)$ and thus in general and rather obviously $g(Y)=g(Y)$, so in that sense $E[E[X mid Y]mid Y]$ ought to mean $E[g(Y) mid Y] = E[g(Y)]$ and we already know that is $E[X]$
$endgroup$
– Henry
Dec 14 '18 at 16:03
1
1
$begingroup$
In $E[E[Xmid Y]]$ the inner expectation is over $X$ given $Y$ and is a function of $Y$, while the outer expectation is over $Y$ and is a numerical value. So to the extent that $E[E[Xmid Y] mid Y]$ has a meaning, it is the same thing
$endgroup$
– Henry
Dec 14 '18 at 9:25
$begingroup$
In $E[E[Xmid Y]]$ the inner expectation is over $X$ given $Y$ and is a function of $Y$, while the outer expectation is over $Y$ and is a numerical value. So to the extent that $E[E[Xmid Y] mid Y]$ has a meaning, it is the same thing
$endgroup$
– Henry
Dec 14 '18 at 9:25
$begingroup$
@Henry $E[E[X∣Y]∣Y] $ is the same as $E[E[X∣Y]]$? or that you meant that the interpretation of this follows as it was with $E[E[X∣Y]]$?
$endgroup$
– Mr.OY
Dec 14 '18 at 9:35
$begingroup$
@Henry $E[E[X∣Y]∣Y] $ is the same as $E[E[X∣Y]]$? or that you meant that the interpretation of this follows as it was with $E[E[X∣Y]]$?
$endgroup$
– Mr.OY
Dec 14 '18 at 9:35
$begingroup$
If you define $g(y)=E[X mid Y=y]$ then $E[X mid Y] = g(Y)$ and $E[E[X mid Y]] = E[g(Y)]=E[X]$. You can also say that conditioned on $Y=y$ you have $g(Y)=g(y)$ and thus in general and rather obviously $g(Y)=g(Y)$, so in that sense $E[E[X mid Y]mid Y]$ ought to mean $E[g(Y) mid Y] = E[g(Y)]$ and we already know that is $E[X]$
$endgroup$
– Henry
Dec 14 '18 at 16:03
$begingroup$
If you define $g(y)=E[X mid Y=y]$ then $E[X mid Y] = g(Y)$ and $E[E[X mid Y]] = E[g(Y)]=E[X]$. You can also say that conditioned on $Y=y$ you have $g(Y)=g(y)$ and thus in general and rather obviously $g(Y)=g(Y)$, so in that sense $E[E[X mid Y]mid Y]$ ought to mean $E[g(Y) mid Y] = E[g(Y)]$ and we already know that is $E[X]$
$endgroup$
– Henry
Dec 14 '18 at 16:03
add a comment |
2 Answers
2
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$begingroup$
If $X$ is measurable wrt $sigma(Y)$ i.e. the $sigma$-algebra generated by random variable $Y$ then $mathbb E[Xmid Y]=X$.
This can be applied on $Z:=mathbb E[Xmid Y]$ because $mathbb E[Xmid Y]$ is by definition measurable wrt $sigma(Y)$.
This results in $mathbb E[Zmid Y]=Z$ or equivalently: $$mathbb E[mathbb E[Xmid Y]mid Y]=mathbb E[Xmid Y]$$
answered Dec 14 '18 at 11:58
drhabdrhab
101k545136
$endgroup$
add a comment |
$begingroup$
It is a basic fact about conditional expectations that $E(E(X|Y))=EX$.
answered Dec 14 '18 at 9:19
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
If $X$ is measurable wrt $sigma(Y)$ i.e. the $sigma$-algebra generated by random variable $Y$ then $mathbb E[Xmid Y]=X$.
This can be applied on $Z:=mathbb E[Xmid Y]$ because $mathbb E[Xmid Y]$ is by definition measurable wrt $sigma(Y)$.
This results in $mathbb E[Zmid Y]=Z$ or equivalently: $$mathbb E[mathbb E[Xmid Y]mid Y]=mathbb E[Xmid Y]$$
answered Dec 14 '18 at 11:58
drhabdrhab
101k545136
$endgroup$
add a comment |
$begingroup$
If $X$ is measurable wrt $sigma(Y)$ i.e. the $sigma$-algebra generated by random variable $Y$ then $mathbb E[Xmid Y]=X$.
This can be applied on $Z:=mathbb E[Xmid Y]$ because $mathbb E[Xmid Y]$ is by definition measurable wrt $sigma(Y)$.
This results in $mathbb E[Zmid Y]=Z$ or equivalently: $$mathbb E[mathbb E[Xmid Y]mid Y]=mathbb E[Xmid Y]$$
answered Dec 14 '18 at 11:58
drhabdrhab
101k545136
$endgroup$
add a comment |
$begingroup$
If $X$ is measurable wrt $sigma(Y)$ i.e. the $sigma$-algebra generated by random variable $Y$ then $mathbb E[Xmid Y]=X$.
This can be applied on $Z:=mathbb E[Xmid Y]$ because $mathbb E[Xmid Y]$ is by definition measurable wrt $sigma(Y)$.
This results in $mathbb E[Zmid Y]=Z$ or equivalently: $$mathbb E[mathbb E[Xmid Y]mid Y]=mathbb E[Xmid Y]$$
answered Dec 14 '18 at 11:58
drhabdrhab
101k545136
$endgroup$
If $X$ is measurable wrt $sigma(Y)$ i.e. the $sigma$-algebra generated by random variable $Y$ then $mathbb E[Xmid Y]=X$.
This can be applied on $Z:=mathbb E[Xmid Y]$ because $mathbb E[Xmid Y]$ is by definition measurable wrt $sigma(Y)$.
This results in $mathbb E[Zmid Y]=Z$ or equivalently: $$mathbb E[mathbb E[Xmid Y]mid Y]=mathbb E[Xmid Y]$$
answered Dec 14 '18 at 11:58
drhabdrhab
101k545136
answered Dec 14 '18 at 11:58
drhabdrhab
101k545136
answered Dec 14 '18 at 11:58
drhabdrhab
101k545136
answered Dec 14 '18 at 11:58
drhabdrhab
101k545136
101k545136
add a comment |
add a comment |
$begingroup$
It is a basic fact about conditional expectations that $E(E(X|Y))=EX$.
answered Dec 14 '18 at 9:19
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
$endgroup$
add a comment |
$begingroup$
It is a basic fact about conditional expectations that $E(E(X|Y))=EX$.
answered Dec 14 '18 at 9:19
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
$endgroup$
add a comment |
$begingroup$
It is a basic fact about conditional expectations that $E(E(X|Y))=EX$.
answered Dec 14 '18 at 9:19
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
$endgroup$
It is a basic fact about conditional expectations that $E(E(X|Y))=EX$.
answered Dec 14 '18 at 9:19
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
answered Dec 14 '18 at 9:19
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
answered Dec 14 '18 at 9:19
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
answered Dec 14 '18 at 9:19
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
60.6k42161
add a comment |
add a comment |
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$begingroup$
In $E[E[Xmid Y]]$ the inner expectation is over $X$ given $Y$ and is a function of $Y$, while the outer expectation is over $Y$ and is a numerical value. So to the extent that $E[E[Xmid Y] mid Y]$ has a meaning, it is the same thing
$endgroup$
– Henry
Dec 14 '18 at 9:25
$begingroup$
@Henry $E[E[X∣Y]∣Y] $ is the same as $E[E[X∣Y]]$? or that you meant that the interpretation of this follows as it was with $E[E[X∣Y]]$?
$endgroup$
– Mr.OY
Dec 14 '18 at 9:35
$begingroup$
If you define $g(y)=E[X mid Y=y]$ then $E[X mid Y] = g(Y)$ and $E[E[X mid Y]] = E[g(Y)]=E[X]$. You can also say that conditioned on $Y=y$ you have $g(Y)=g(y)$ and thus in general and rather obviously $g(Y)=g(Y)$, so in that sense $E[E[X mid Y]mid Y]$ ought to mean $E[g(Y) mid Y] = E[g(Y)]$ and we already know that is $E[X]$
$endgroup$
– Henry
Dec 14 '18 at 16:03