use indicator function to prove independence
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I would like to know why for $forall A in mathcal{F}$ $$E[1_{A}e^{itX}] = P(A)E[e^{itX}]$$ will imply the independence between $X$ and $mathcal{F}$. Thanks in advance!
probability characteristic-functions
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VanDDF
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I would like to know why for $forall A in mathcal{F}$ $$E[1_{A}e^{itX}] = P(A)E[e^{itX}]$$ will imply the independence between $X$ and $mathcal{F}$. Thanks in advance!
probability characteristic-functions
asked yesterday
VanDDF
212
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up vote
0
down vote
favorite
up vote
0
down vote
favorite
I would like to know why for $forall A in mathcal{F}$ $$E[1_{A}e^{itX}] = P(A)E[e^{itX}]$$ will imply the independence between $X$ and $mathcal{F}$. Thanks in advance!
probability characteristic-functions
asked yesterday
VanDDF
212
I would like to know why for $forall A in mathcal{F}$ $$E[1_{A}e^{itX}] = P(A)E[e^{itX}]$$ will imply the independence between $X$ and $mathcal{F}$. Thanks in advance!
probability characteristic-functions
probability characteristic-functions
asked yesterday
VanDDF
212
asked yesterday
VanDDF
212
asked yesterday
VanDDF
212
asked yesterday
VanDDF
212
asked yesterday
VanDDF
212
212
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add a comment |
3 Answers
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Use simple function approximation to prove that $Ee^{itY}e^{itX}=Ee^{itY}Ee^{itX}$ for all $t$ for any $mathcal F$ measurable $Y$. This proves that $X$ and $Y$ are independent for any $mathcal F$ measurable $Y$. QED.
answered yesterday
Kavi Rama Murthy
38.6k31747
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Hint: let $Omega$ denote the probability space and $F$ the cdf of $X$. Since $$int_A exp itx dF(x)int_Omega dF(y)=int_A dF(x)int_Omegaexp ity dF(y)$$(where I've inserted a factor of $1$ on the left-hand side),$$int_Omega dF(y)int_A dF(x) [exp itx-exp ity]=0.$$
answered yesterday
J.G.
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Note that for $X= 1_A$, $Ain mathcal{F}$, it holds that$$E[e^{i(sX+tY)}] = E[(e^{is}1_A + 1_{A^c})e^{itY}]= (e^{is}P(A)+P(A^c))E[e^{itY}] = E[e^{isX}]E[e^{itY}].$$
Thus $1_A$ and $Y$ are independent for all $Ainmathcal{F}$, and we conclude that $Y$ and $mathcal{F}$ are independent.
answered yesterday
Song
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Use simple function approximation to prove that $Ee^{itY}e^{itX}=Ee^{itY}Ee^{itX}$ for all $t$ for any $mathcal F$ measurable $Y$. This proves that $X$ and $Y$ are independent for any $mathcal F$ measurable $Y$. QED.
answered yesterday
Kavi Rama Murthy
38.6k31747
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up vote
1
down vote
Use simple function approximation to prove that $Ee^{itY}e^{itX}=Ee^{itY}Ee^{itX}$ for all $t$ for any $mathcal F$ measurable $Y$. This proves that $X$ and $Y$ are independent for any $mathcal F$ measurable $Y$. QED.
answered yesterday
Kavi Rama Murthy
38.6k31747
add a comment |
up vote
1
down vote
up vote
1
down vote
Use simple function approximation to prove that $Ee^{itY}e^{itX}=Ee^{itY}Ee^{itX}$ for all $t$ for any $mathcal F$ measurable $Y$. This proves that $X$ and $Y$ are independent for any $mathcal F$ measurable $Y$. QED.
answered yesterday
Kavi Rama Murthy
38.6k31747
Use simple function approximation to prove that $Ee^{itY}e^{itX}=Ee^{itY}Ee^{itX}$ for all $t$ for any $mathcal F$ measurable $Y$. This proves that $X$ and $Y$ are independent for any $mathcal F$ measurable $Y$. QED.
answered yesterday
Kavi Rama Murthy
38.6k31747
answered yesterday
Kavi Rama Murthy
38.6k31747
answered yesterday
Kavi Rama Murthy
38.6k31747
answered yesterday
Kavi Rama Murthy
38.6k31747
38.6k31747
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add a comment |
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0
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Hint: let $Omega$ denote the probability space and $F$ the cdf of $X$. Since $$int_A exp itx dF(x)int_Omega dF(y)=int_A dF(x)int_Omegaexp ity dF(y)$$(where I've inserted a factor of $1$ on the left-hand side),$$int_Omega dF(y)int_A dF(x) [exp itx-exp ity]=0.$$
answered yesterday
J.G.
17.6k11829
add a comment |
up vote
0
down vote
Hint: let $Omega$ denote the probability space and $F$ the cdf of $X$. Since $$int_A exp itx dF(x)int_Omega dF(y)=int_A dF(x)int_Omegaexp ity dF(y)$$(where I've inserted a factor of $1$ on the left-hand side),$$int_Omega dF(y)int_A dF(x) [exp itx-exp ity]=0.$$
answered yesterday
J.G.
17.6k11829
add a comment |
up vote
0
down vote
up vote
0
down vote
Hint: let $Omega$ denote the probability space and $F$ the cdf of $X$. Since $$int_A exp itx dF(x)int_Omega dF(y)=int_A dF(x)int_Omegaexp ity dF(y)$$(where I've inserted a factor of $1$ on the left-hand side),$$int_Omega dF(y)int_A dF(x) [exp itx-exp ity]=0.$$
answered yesterday
J.G.
17.6k11829
Hint: let $Omega$ denote the probability space and $F$ the cdf of $X$. Since $$int_A exp itx dF(x)int_Omega dF(y)=int_A dF(x)int_Omegaexp ity dF(y)$$(where I've inserted a factor of $1$ on the left-hand side),$$int_Omega dF(y)int_A dF(x) [exp itx-exp ity]=0.$$
answered yesterday
J.G.
17.6k11829
answered yesterday
J.G.
17.6k11829
answered yesterday
J.G.
17.6k11829
answered yesterday
J.G.
17.6k11829
17.6k11829
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add a comment |
up vote
0
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Note that for $X= 1_A$, $Ain mathcal{F}$, it holds that$$E[e^{i(sX+tY)}] = E[(e^{is}1_A + 1_{A^c})e^{itY}]= (e^{is}P(A)+P(A^c))E[e^{itY}] = E[e^{isX}]E[e^{itY}].$$
Thus $1_A$ and $Y$ are independent for all $Ainmathcal{F}$, and we conclude that $Y$ and $mathcal{F}$ are independent.
answered yesterday
Song
3816
add a comment |
up vote
0
down vote
Note that for $X= 1_A$, $Ain mathcal{F}$, it holds that$$E[e^{i(sX+tY)}] = E[(e^{is}1_A + 1_{A^c})e^{itY}]= (e^{is}P(A)+P(A^c))E[e^{itY}] = E[e^{isX}]E[e^{itY}].$$
Thus $1_A$ and $Y$ are independent for all $Ainmathcal{F}$, and we conclude that $Y$ and $mathcal{F}$ are independent.
answered yesterday
Song
3816
add a comment |
up vote
0
down vote
up vote
0
down vote
Note that for $X= 1_A$, $Ain mathcal{F}$, it holds that$$E[e^{i(sX+tY)}] = E[(e^{is}1_A + 1_{A^c})e^{itY}]= (e^{is}P(A)+P(A^c))E[e^{itY}] = E[e^{isX}]E[e^{itY}].$$
Thus $1_A$ and $Y$ are independent for all $Ainmathcal{F}$, and we conclude that $Y$ and $mathcal{F}$ are independent.
answered yesterday
Song
3816
Note that for $X= 1_A$, $Ain mathcal{F}$, it holds that$$E[e^{i(sX+tY)}] = E[(e^{is}1_A + 1_{A^c})e^{itY}]= (e^{is}P(A)+P(A^c))E[e^{itY}] = E[e^{isX}]E[e^{itY}].$$
Thus $1_A$ and $Y$ are independent for all $Ainmathcal{F}$, and we conclude that $Y$ and $mathcal{F}$ are independent.
answered yesterday
Song
3816
answered yesterday
Song
3816
answered yesterday
Song
3816
answered yesterday
Song
3816
3816
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