When is $YE(Ymidmathcal{G}) = E(Y^2midmathcal{G})$?Is it true that $E[|Y|mid{mathcal G}]leq |Y|$ almost...
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When is $YE(Ymidmathcal{G}) = E(Y^2midmathcal{G})$?
Is it true that $E[|Y|mid{mathcal G}]leq |Y|$ almost surely?How to prove that $E(Xmidsigma,(mathcal{G},mathcal{M}))=E(Xmidmathcal{M})$ if $mathcal{G}$ is independent of $X$ and $mathcal{M}$?Does $text E[text E[Xmidmathcal F]midmathcal G]=text E[text E[Xmidmathcal G]midmathcal F]$ hold when $mathcal Gnotsubseteqmathcal F$?Trying to understand $mathbb E[Xmid mathcal F]$ and Martingale conceptProving if $X$ if an $mathcal{F}$ measurable Random variable with $E(X) < infty$, then $E(Xmid mathcal{F}) = X$ a.e.When does $mathbf{E}(XYmidmathcal{A})=Xmathbf{E}(Ymidmathcal{A})$ hold?When “sure” property is equivalent to “almost sure” property?What does $mathbb P{X>xmid mathcal A}$ mean?Product measure : difference between $mathcal Mtimes mathcal M$ and ${Atimes Bmid A,Bin mathcal M}$how to prove $mathbb{E}[UVmid mathcal{B} ] = Umathbb{E}[Vmid mathcal{B} ]$ if $U$ is $mathcal{B}$-measurable?
$begingroup$
I know this is true when $Y in mathcal{G}$, but is there a better (less restrictive) condition that can allow this?
In particular, I want to know when I can say that $E(YE(Ymidmathcal{G})) = E(E(Y^2midmathcal{G}))$, which is, of course, equal to $EY^2$.
probability probability-theory measure-theory
edited Mar 20 at 13:10
StubbornAtom
6,30831440
asked Mar 20 at 13:07
jackson5jackson5
669513
$endgroup$
add a comment |
$begingroup$
I know this is true when $Y in mathcal{G}$, but is there a better (less restrictive) condition that can allow this?
In particular, I want to know when I can say that $E(YE(Ymidmathcal{G})) = E(E(Y^2midmathcal{G}))$, which is, of course, equal to $EY^2$.
probability probability-theory measure-theory
edited Mar 20 at 13:10
StubbornAtom
6,30831440
asked Mar 20 at 13:07
jackson5jackson5
669513
$endgroup$
add a comment |
$begingroup$
I know this is true when $Y in mathcal{G}$, but is there a better (less restrictive) condition that can allow this?
In particular, I want to know when I can say that $E(YE(Ymidmathcal{G})) = E(E(Y^2midmathcal{G}))$, which is, of course, equal to $EY^2$.
probability probability-theory measure-theory
edited Mar 20 at 13:10
StubbornAtom
6,30831440
asked Mar 20 at 13:07
jackson5jackson5
669513
$endgroup$
I know this is true when $Y in mathcal{G}$, but is there a better (less restrictive) condition that can allow this?
In particular, I want to know when I can say that $E(YE(Ymidmathcal{G})) = E(E(Y^2midmathcal{G}))$, which is, of course, equal to $EY^2$.
probability probability-theory measure-theory
probability probability-theory measure-theory
edited Mar 20 at 13:10
StubbornAtom
6,30831440
asked Mar 20 at 13:07
jackson5jackson5
669513
edited Mar 20 at 13:10
StubbornAtom
6,30831440
asked Mar 20 at 13:07
jackson5jackson5
669513
edited Mar 20 at 13:10
StubbornAtom
6,30831440
edited Mar 20 at 13:10
StubbornAtom
6,30831440
edited Mar 20 at 13:10
StubbornAtom
6,30831440
6,30831440
asked Mar 20 at 13:07
jackson5jackson5
669513
asked Mar 20 at 13:07
jackson5jackson5
669513
asked Mar 20 at 13:07
jackson5jackson5
669513
669513
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No.
If I understand correctly your notation, then $Yinmathcal{G}$ means that $Y$ is measurable with respect to $mathcal{G}$. In particular you have $E(Y|mathcal{G}),E(Y^2|mathcal{G})inmathcal{G}$.
But then, from the equation $YE(Y|mathcal{G}) = E(Y^2|mathcal{G})$ you necessarily have that $Yinmathcal{G}$.
answered Mar 20 at 13:29
YankoYanko
8,3842830
$endgroup$
$begingroup$
Yes, that makes perfect sense. Thanks!
$endgroup$
– jackson5
Mar 20 at 13:35
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No.
If I understand correctly your notation, then $Yinmathcal{G}$ means that $Y$ is measurable with respect to $mathcal{G}$. In particular you have $E(Y|mathcal{G}),E(Y^2|mathcal{G})inmathcal{G}$.
But then, from the equation $YE(Y|mathcal{G}) = E(Y^2|mathcal{G})$ you necessarily have that $Yinmathcal{G}$.
answered Mar 20 at 13:29
YankoYanko
8,3842830
$endgroup$
$begingroup$
Yes, that makes perfect sense. Thanks!
$endgroup$
– jackson5
Mar 20 at 13:35
add a comment |
$begingroup$
No.
If I understand correctly your notation, then $Yinmathcal{G}$ means that $Y$ is measurable with respect to $mathcal{G}$. In particular you have $E(Y|mathcal{G}),E(Y^2|mathcal{G})inmathcal{G}$.
But then, from the equation $YE(Y|mathcal{G}) = E(Y^2|mathcal{G})$ you necessarily have that $Yinmathcal{G}$.
answered Mar 20 at 13:29
YankoYanko
8,3842830
$endgroup$
$begingroup$
Yes, that makes perfect sense. Thanks!
$endgroup$
– jackson5
Mar 20 at 13:35
add a comment |
$begingroup$
No.
If I understand correctly your notation, then $Yinmathcal{G}$ means that $Y$ is measurable with respect to $mathcal{G}$. In particular you have $E(Y|mathcal{G}),E(Y^2|mathcal{G})inmathcal{G}$.
But then, from the equation $YE(Y|mathcal{G}) = E(Y^2|mathcal{G})$ you necessarily have that $Yinmathcal{G}$.
answered Mar 20 at 13:29
YankoYanko
8,3842830
$endgroup$
No.
If I understand correctly your notation, then $Yinmathcal{G}$ means that $Y$ is measurable with respect to $mathcal{G}$. In particular you have $E(Y|mathcal{G}),E(Y^2|mathcal{G})inmathcal{G}$.
But then, from the equation $YE(Y|mathcal{G}) = E(Y^2|mathcal{G})$ you necessarily have that $Yinmathcal{G}$.
answered Mar 20 at 13:29
YankoYanko
8,3842830
answered Mar 20 at 13:29
YankoYanko
8,3842830
answered Mar 20 at 13:29
YankoYanko
8,3842830
answered Mar 20 at 13:29
YankoYanko
8,3842830
8,3842830
$begingroup$
Yes, that makes perfect sense. Thanks!
$endgroup$
– jackson5
Mar 20 at 13:35
add a comment |
$begingroup$
Yes, that makes perfect sense. Thanks!
$endgroup$
– jackson5
Mar 20 at 13:35
$begingroup$
Yes, that makes perfect sense. Thanks!
$endgroup$
– jackson5
Mar 20 at 13:35
$begingroup$
Yes, that makes perfect sense. Thanks!
$endgroup$
– jackson5
Mar 20 at 13:35
add a comment |
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