Mixing
$mathbb{E}[f(X_1+X_2+cdots+X_{n}) mid sigma(X_1,X_2,cdots,X_{n-1})] =mathbb{E}[f(X_1+X_2+cdots+X_{n}) mid...
$begingroup$
$f$ is a bounded measurable functions and the $X_i$'s are independent
to prove the above one must show the following equality : $$I = J$$
where :
$$I=int_{A in sigma(X_1,X_2,cdots,X_{n-1})} mathbb{E}[f(X_1+X_2+cdots+X_n) mid X_1+X_2+cdots+X_{n-1} ]d mathbb{P}$$
$$J=int_{A in sigma(X_1,X_2,cdots,X_{n-1})} f(X_1+X_2+cdots+X_n) d mathbb{P}$$
clearly if $sigma(X_1,X_2,cdots,X_{n-1}) subsetsigma(X_1+X_2+ cdots+X_{n-1})$
then the equality holds
and that inclusion is actually correct for the following argument,
elements of $sigma(X_1,X_2,cdots,X_{n-1})$ are all intersections of elements of $sigma(X_1+X_2+ cdots+X_{n-1})$
is this right ? can someone correct any falseness I said ?
probability-theory measure-theory conditional-expectation
asked Jan 9 at 8:56
rapidracimrapidracim
1,6811319
$endgroup$
add a comment |
$begingroup$
$f$ is a bounded measurable functions and the $X_i$'s are independent
to prove the above one must show the following equality : $$I = J$$
where :
$$I=int_{A in sigma(X_1,X_2,cdots,X_{n-1})} mathbb{E}[f(X_1+X_2+cdots+X_n) mid X_1+X_2+cdots+X_{n-1} ]d mathbb{P}$$
$$J=int_{A in sigma(X_1,X_2,cdots,X_{n-1})} f(X_1+X_2+cdots+X_n) d mathbb{P}$$
clearly if $sigma(X_1,X_2,cdots,X_{n-1}) subsetsigma(X_1+X_2+ cdots+X_{n-1})$
then the equality holds
and that inclusion is actually correct for the following argument,
elements of $sigma(X_1,X_2,cdots,X_{n-1})$ are all intersections of elements of $sigma(X_1+X_2+ cdots+X_{n-1})$
is this right ? can someone correct any falseness I said ?
probability-theory measure-theory conditional-expectation
asked Jan 9 at 8:56
rapidracimrapidracim
1,6811319
$endgroup$
$begingroup$
It is not true that $sigma(X_1,X_2,...,X_n) subset sigma(X_1+X_2+...+X_n)$ except in some trivial cases.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 9:03
$begingroup$
Moer generally, if $X$ is independent of $mathcal G$ and $Y$ is $mathcal G$-measurable, then $E(f(X+Y)midmathcal G)=g(Y)$ where $g(y)=E(f(X+y))$, hence $E(f(X+Y)midmathcal G)=E(f(X+Y)mid Y)$.
$endgroup$
– Did
Jan 20 at 21:58
add a comment |
$begingroup$
$f$ is a bounded measurable functions and the $X_i$'s are independent
to prove the above one must show the following equality : $$I = J$$
where :
$$I=int_{A in sigma(X_1,X_2,cdots,X_{n-1})} mathbb{E}[f(X_1+X_2+cdots+X_n) mid X_1+X_2+cdots+X_{n-1} ]d mathbb{P}$$
$$J=int_{A in sigma(X_1,X_2,cdots,X_{n-1})} f(X_1+X_2+cdots+X_n) d mathbb{P}$$
clearly if $sigma(X_1,X_2,cdots,X_{n-1}) subsetsigma(X_1+X_2+ cdots+X_{n-1})$
then the equality holds
and that inclusion is actually correct for the following argument,
elements of $sigma(X_1,X_2,cdots,X_{n-1})$ are all intersections of elements of $sigma(X_1+X_2+ cdots+X_{n-1})$
is this right ? can someone correct any falseness I said ?
probability-theory measure-theory conditional-expectation
asked Jan 9 at 8:56
rapidracimrapidracim
1,6811319
$endgroup$
$f$ is a bounded measurable functions and the $X_i$'s are independent
to prove the above one must show the following equality : $$I = J$$
where :
$$I=int_{A in sigma(X_1,X_2,cdots,X_{n-1})} mathbb{E}[f(X_1+X_2+cdots+X_n) mid X_1+X_2+cdots+X_{n-1} ]d mathbb{P}$$
$$J=int_{A in sigma(X_1,X_2,cdots,X_{n-1})} f(X_1+X_2+cdots+X_n) d mathbb{P}$$
clearly if $sigma(X_1,X_2,cdots,X_{n-1}) subsetsigma(X_1+X_2+ cdots+X_{n-1})$
then the equality holds
and that inclusion is actually correct for the following argument,
elements of $sigma(X_1,X_2,cdots,X_{n-1})$ are all intersections of elements of $sigma(X_1+X_2+ cdots+X_{n-1})$
is this right ? can someone correct any falseness I said ?
probability-theory measure-theory conditional-expectation
probability-theory measure-theory conditional-expectation
asked Jan 9 at 8:56
rapidracimrapidracim
1,6811319
asked Jan 9 at 8:56
rapidracimrapidracim
1,6811319
asked Jan 9 at 8:56
rapidracimrapidracim
1,6811319
asked Jan 9 at 8:56
rapidracimrapidracim
1,6811319
asked Jan 9 at 8:56
rapidracimrapidracim
1,6811319
1,6811319
$begingroup$
It is not true that $sigma(X_1,X_2,...,X_n) subset sigma(X_1+X_2+...+X_n)$ except in some trivial cases.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 9:03
$begingroup$
Moer generally, if $X$ is independent of $mathcal G$ and $Y$ is $mathcal G$-measurable, then $E(f(X+Y)midmathcal G)=g(Y)$ where $g(y)=E(f(X+y))$, hence $E(f(X+Y)midmathcal G)=E(f(X+Y)mid Y)$.
$endgroup$
– Did
Jan 20 at 21:58
add a comment |
$begingroup$
It is not true that $sigma(X_1,X_2,...,X_n) subset sigma(X_1+X_2+...+X_n)$ except in some trivial cases.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 9:03
$begingroup$
Moer generally, if $X$ is independent of $mathcal G$ and $Y$ is $mathcal G$-measurable, then $E(f(X+Y)midmathcal G)=g(Y)$ where $g(y)=E(f(X+y))$, hence $E(f(X+Y)midmathcal G)=E(f(X+Y)mid Y)$.
$endgroup$
– Did
Jan 20 at 21:58
$begingroup$
It is not true that $sigma(X_1,X_2,...,X_n) subset sigma(X_1+X_2+...+X_n)$ except in some trivial cases.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 9:03
$begingroup$
It is not true that $sigma(X_1,X_2,...,X_n) subset sigma(X_1+X_2+...+X_n)$ except in some trivial cases.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 9:03
$begingroup$
Moer generally, if $X$ is independent of $mathcal G$ and $Y$ is $mathcal G$-measurable, then $E(f(X+Y)midmathcal G)=g(Y)$ where $g(y)=E(f(X+y))$, hence $E(f(X+Y)midmathcal G)=E(f(X+Y)mid Y)$.
$endgroup$
– Did
Jan 20 at 21:58
$begingroup$
Moer generally, if $X$ is independent of $mathcal G$ and $Y$ is $mathcal G$-measurable, then $E(f(X+Y)midmathcal G)=g(Y)$ where $g(y)=E(f(X+y))$, hence $E(f(X+Y)midmathcal G)=E(f(X+Y)mid Y)$.
$endgroup$
– Did
Jan 20 at 21:58
add a comment |
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$begingroup$
It is not true that $sigma(X_1,X_2,...,X_n) subset sigma(X_1+X_2+...+X_n)$ except in some trivial cases.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 9:03
$begingroup$
Moer generally, if $X$ is independent of $mathcal G$ and $Y$ is $mathcal G$-measurable, then $E(f(X+Y)midmathcal G)=g(Y)$ where $g(y)=E(f(X+y))$, hence $E(f(X+Y)midmathcal G)=E(f(X+Y)mid Y)$.
$endgroup$
– Did
Jan 20 at 21:58