Expectation and the equal in distribution
$begingroup$
Is the following statement true?
If $X$ and $Y$ are 2 random variables and $text{ }$$f$ is any bounded function then $mathbb{E}(f(X))=mathbb{E}(f(Y))$ if and only if $X$ and $Y$ have equal distribution?
Are there some counterexamples of this statement?
real-analysis probability probability-theory
asked Jan 20 at 1:08
PTTS_guyPTTS_guy
134
$endgroup$
add a comment |
$begingroup$
Is the following statement true?
If $X$ and $Y$ are 2 random variables and $text{ }$$f$ is any bounded function then $mathbb{E}(f(X))=mathbb{E}(f(Y))$ if and only if $X$ and $Y$ have equal distribution?
Are there some counterexamples of this statement?
real-analysis probability probability-theory
asked Jan 20 at 1:08
PTTS_guyPTTS_guy
134
$endgroup$
add a comment |
$begingroup$
Is the following statement true?
If $X$ and $Y$ are 2 random variables and $text{ }$$f$ is any bounded function then $mathbb{E}(f(X))=mathbb{E}(f(Y))$ if and only if $X$ and $Y$ have equal distribution?
Are there some counterexamples of this statement?
real-analysis probability probability-theory
asked Jan 20 at 1:08
PTTS_guyPTTS_guy
134
$endgroup$
Is the following statement true?
If $X$ and $Y$ are 2 random variables and $text{ }$$f$ is any bounded function then $mathbb{E}(f(X))=mathbb{E}(f(Y))$ if and only if $X$ and $Y$ have equal distribution?
Are there some counterexamples of this statement?
real-analysis probability probability-theory
real-analysis probability probability-theory
asked Jan 20 at 1:08
PTTS_guyPTTS_guy
134
asked Jan 20 at 1:08
PTTS_guyPTTS_guy
134
asked Jan 20 at 1:08
PTTS_guyPTTS_guy
134
asked Jan 20 at 1:08
PTTS_guyPTTS_guy
134
asked Jan 20 at 1:08
PTTS_guyPTTS_guy
134
134
add a comment |
add a comment |
1 Answer
1
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$begingroup$
If $P_X=P_Y$, then clearly the expectations are equal. For the opposite direction, let $f(x):=1_{A}(x)$, where $Ain mathcal{B}(mathbb{R})$. Then
$$
P_X(A)=mathsf{E}f(X)=mathsf{E}f(Y)=P_Y(A).
$$
answered Jan 20 at 1:26
d.k.o.d.k.o.
9,553628
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add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
If $P_X=P_Y$, then clearly the expectations are equal. For the opposite direction, let $f(x):=1_{A}(x)$, where $Ain mathcal{B}(mathbb{R})$. Then
$$
P_X(A)=mathsf{E}f(X)=mathsf{E}f(Y)=P_Y(A).
$$
answered Jan 20 at 1:26
d.k.o.d.k.o.
9,553628
$endgroup$
add a comment |
$begingroup$
If $P_X=P_Y$, then clearly the expectations are equal. For the opposite direction, let $f(x):=1_{A}(x)$, where $Ain mathcal{B}(mathbb{R})$. Then
$$
P_X(A)=mathsf{E}f(X)=mathsf{E}f(Y)=P_Y(A).
$$
answered Jan 20 at 1:26
d.k.o.d.k.o.
9,553628
$endgroup$
add a comment |
$begingroup$
If $P_X=P_Y$, then clearly the expectations are equal. For the opposite direction, let $f(x):=1_{A}(x)$, where $Ain mathcal{B}(mathbb{R})$. Then
$$
P_X(A)=mathsf{E}f(X)=mathsf{E}f(Y)=P_Y(A).
$$
answered Jan 20 at 1:26
d.k.o.d.k.o.
9,553628
$endgroup$
If $P_X=P_Y$, then clearly the expectations are equal. For the opposite direction, let $f(x):=1_{A}(x)$, where $Ain mathcal{B}(mathbb{R})$. Then
$$
P_X(A)=mathsf{E}f(X)=mathsf{E}f(Y)=P_Y(A).
$$
answered Jan 20 at 1:26
d.k.o.d.k.o.
9,553628
answered Jan 20 at 1:26
d.k.o.d.k.o.
9,553628
answered Jan 20 at 1:26
d.k.o.d.k.o.
9,553628
answered Jan 20 at 1:26
d.k.o.d.k.o.
9,553628
9,553628
add a comment |
add a comment |
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