Show that $X perp!!!perp Y iff forall f , text{bounded and measurable} , mathbb{E}[f(X)mid Y] = E[f(X)]$
$begingroup$
I could only prove one direction, let $A in sigma(Y)$
$$int_Amathbb{E}[f(X)mid Y] dmathbb{P} =int_Af(X) dmathbb{P} = mathbb{E}[f(X) mathbb{1}_A] = mathbb{E}[f(X)] mathbb{E}[ mathbb{1}_A] = int_A mathbb{E}[f(X)]dmathbb{P}$$
for third equality I used independence and every constant is measurable. so...
how to do the reverse implication though ?
probability-theory conditional-expectation independence
edited Jan 8 at 18:33
Davide Giraudo
128k17156268
asked Jan 8 at 18:08
rapidracimrapidracim
1,7291419
$endgroup$
add a comment |
$begingroup$
I could only prove one direction, let $A in sigma(Y)$
$$int_Amathbb{E}[f(X)mid Y] dmathbb{P} =int_Af(X) dmathbb{P} = mathbb{E}[f(X) mathbb{1}_A] = mathbb{E}[f(X)] mathbb{E}[ mathbb{1}_A] = int_A mathbb{E}[f(X)]dmathbb{P}$$
for third equality I used independence and every constant is measurable. so...
how to do the reverse implication though ?
probability-theory conditional-expectation independence
edited Jan 8 at 18:33
Davide Giraudo
128k17156268
asked Jan 8 at 18:08
rapidracimrapidracim
1,7291419
$endgroup$
add a comment |
$begingroup$
I could only prove one direction, let $A in sigma(Y)$
$$int_Amathbb{E}[f(X)mid Y] dmathbb{P} =int_Af(X) dmathbb{P} = mathbb{E}[f(X) mathbb{1}_A] = mathbb{E}[f(X)] mathbb{E}[ mathbb{1}_A] = int_A mathbb{E}[f(X)]dmathbb{P}$$
for third equality I used independence and every constant is measurable. so...
how to do the reverse implication though ?
probability-theory conditional-expectation independence
edited Jan 8 at 18:33
Davide Giraudo
128k17156268
asked Jan 8 at 18:08
rapidracimrapidracim
1,7291419
$endgroup$
I could only prove one direction, let $A in sigma(Y)$
$$int_Amathbb{E}[f(X)mid Y] dmathbb{P} =int_Af(X) dmathbb{P} = mathbb{E}[f(X) mathbb{1}_A] = mathbb{E}[f(X)] mathbb{E}[ mathbb{1}_A] = int_A mathbb{E}[f(X)]dmathbb{P}$$
for third equality I used independence and every constant is measurable. so...
how to do the reverse implication though ?
probability-theory conditional-expectation independence
probability-theory conditional-expectation independence
edited Jan 8 at 18:33
Davide Giraudo
128k17156268
asked Jan 8 at 18:08
rapidracimrapidracim
1,7291419
edited Jan 8 at 18:33
Davide Giraudo
128k17156268
asked Jan 8 at 18:08
rapidracimrapidracim
1,7291419
edited Jan 8 at 18:33
Davide Giraudo
128k17156268
edited Jan 8 at 18:33
Davide Giraudo
128k17156268
edited Jan 8 at 18:33
Davide Giraudo
128k17156268
128k17156268
asked Jan 8 at 18:08
rapidracimrapidracim
1,7291419
asked Jan 8 at 18:08
rapidracimrapidracim
1,7291419
asked Jan 8 at 18:08
rapidracimrapidracim
1,7291419
1,7291419
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
For the opposite implication, let $A$ and $B$ be two Borel subsets of the real line. Using the assumption where $f$ is the indicator function of $A$, we get
$$tag{*}
mathbb Eleft[mathbf 1_{A}left(Xright)mid Yright]=mathbb Pleft(Xin Aright).
$$
By using the definition of conditional expectation with ${Yin B}insigma(B)$,
we get that
$$
mathbb Eleft[mathbb Eleft[mathbf 1_{A}left(Xright)mid Yright]mathbf 1_Bleft(Yright)right]=mathbb Eleft[ mathbf 1_{A}left(Xright)mathbf 1_Bleft(Yright)right]$$
and replacing $mathbb Eleft[mathbf 1_{A}left(Xright)mid Yright]$ by $mathbb Pleft(Xin Aright)$ gives the independence between $X$ and $Y$.
Observe also that a similar criterion can be established in order to show the independence between $X$ and a $sigma$-algebra $mathcal G$, namely, that for each measurable bounded function $f$, the equality
$$
mathbb Eleft[f(X)midmathcal Gright]=mathbb Eleft[f(X) right]
$$
holds almost surely.
answered Jan 8 at 18:19
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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$begingroup$
For the opposite implication, let $A$ and $B$ be two Borel subsets of the real line. Using the assumption where $f$ is the indicator function of $A$, we get
$$tag{*}
mathbb Eleft[mathbf 1_{A}left(Xright)mid Yright]=mathbb Pleft(Xin Aright).
$$
By using the definition of conditional expectation with ${Yin B}insigma(B)$,
we get that
$$
mathbb Eleft[mathbb Eleft[mathbf 1_{A}left(Xright)mid Yright]mathbf 1_Bleft(Yright)right]=mathbb Eleft[ mathbf 1_{A}left(Xright)mathbf 1_Bleft(Yright)right]$$
and replacing $mathbb Eleft[mathbf 1_{A}left(Xright)mid Yright]$ by $mathbb Pleft(Xin Aright)$ gives the independence between $X$ and $Y$.
Observe also that a similar criterion can be established in order to show the independence between $X$ and a $sigma$-algebra $mathcal G$, namely, that for each measurable bounded function $f$, the equality
$$
mathbb Eleft[f(X)midmathcal Gright]=mathbb Eleft[f(X) right]
$$
holds almost surely.
answered Jan 8 at 18:19
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
add a comment |
$begingroup$
For the opposite implication, let $A$ and $B$ be two Borel subsets of the real line. Using the assumption where $f$ is the indicator function of $A$, we get
$$tag{*}
mathbb Eleft[mathbf 1_{A}left(Xright)mid Yright]=mathbb Pleft(Xin Aright).
$$
By using the definition of conditional expectation with ${Yin B}insigma(B)$,
we get that
$$
mathbb Eleft[mathbb Eleft[mathbf 1_{A}left(Xright)mid Yright]mathbf 1_Bleft(Yright)right]=mathbb Eleft[ mathbf 1_{A}left(Xright)mathbf 1_Bleft(Yright)right]$$
and replacing $mathbb Eleft[mathbf 1_{A}left(Xright)mid Yright]$ by $mathbb Pleft(Xin Aright)$ gives the independence between $X$ and $Y$.
Observe also that a similar criterion can be established in order to show the independence between $X$ and a $sigma$-algebra $mathcal G$, namely, that for each measurable bounded function $f$, the equality
$$
mathbb Eleft[f(X)midmathcal Gright]=mathbb Eleft[f(X) right]
$$
holds almost surely.
answered Jan 8 at 18:19
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
add a comment |
$begingroup$
For the opposite implication, let $A$ and $B$ be two Borel subsets of the real line. Using the assumption where $f$ is the indicator function of $A$, we get
$$tag{*}
mathbb Eleft[mathbf 1_{A}left(Xright)mid Yright]=mathbb Pleft(Xin Aright).
$$
By using the definition of conditional expectation with ${Yin B}insigma(B)$,
we get that
$$
mathbb Eleft[mathbb Eleft[mathbf 1_{A}left(Xright)mid Yright]mathbf 1_Bleft(Yright)right]=mathbb Eleft[ mathbf 1_{A}left(Xright)mathbf 1_Bleft(Yright)right]$$
and replacing $mathbb Eleft[mathbf 1_{A}left(Xright)mid Yright]$ by $mathbb Pleft(Xin Aright)$ gives the independence between $X$ and $Y$.
Observe also that a similar criterion can be established in order to show the independence between $X$ and a $sigma$-algebra $mathcal G$, namely, that for each measurable bounded function $f$, the equality
$$
mathbb Eleft[f(X)midmathcal Gright]=mathbb Eleft[f(X) right]
$$
holds almost surely.
answered Jan 8 at 18:19
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
For the opposite implication, let $A$ and $B$ be two Borel subsets of the real line. Using the assumption where $f$ is the indicator function of $A$, we get
$$tag{*}
mathbb Eleft[mathbf 1_{A}left(Xright)mid Yright]=mathbb Pleft(Xin Aright).
$$
By using the definition of conditional expectation with ${Yin B}insigma(B)$,
we get that
$$
mathbb Eleft[mathbb Eleft[mathbf 1_{A}left(Xright)mid Yright]mathbf 1_Bleft(Yright)right]=mathbb Eleft[ mathbf 1_{A}left(Xright)mathbf 1_Bleft(Yright)right]$$
and replacing $mathbb Eleft[mathbf 1_{A}left(Xright)mid Yright]$ by $mathbb Pleft(Xin Aright)$ gives the independence between $X$ and $Y$.
Observe also that a similar criterion can be established in order to show the independence between $X$ and a $sigma$-algebra $mathcal G$, namely, that for each measurable bounded function $f$, the equality
$$
mathbb Eleft[f(X)midmathcal Gright]=mathbb Eleft[f(X) right]
$$
holds almost surely.
answered Jan 8 at 18:19
Davide GiraudoDavide Giraudo
128k17156268
edited Jan 8 at 18:35
answered Jan 8 at 18:19
Davide GiraudoDavide Giraudo
128k17156268
answered Jan 8 at 18:19
Davide GiraudoDavide Giraudo
128k17156268
answered Jan 8 at 18:19
Davide GiraudoDavide Giraudo
128k17156268
128k17156268
add a comment |
add a comment |
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