How to show Sub-independent Random Variables are uncorrelated.
$begingroup$
I want to prove the following:
If two RVs $X, Y$ are sub-independent, i.e., $phi_{X+Y}(t) = phi_X(t)phi_Y(t), tinmathbb{R}$ then $X, Y$are uncorrelated. Keep $Cov(X,Y) = E(XY)-E(X)E(Y) = 0$ in mind, I'm considering the way to connect $E(X)$ with characteristic function $phi_X(t)$. In fact, $phi_{X}'(0) = iE(X), phi_{Y}'(0) = iE(Y).$ Finding $E(XY)$ will end the proof, but I could not proceed further.
Any idea or tips would be really helpful.
random-variables independence correlation characteristic-functions
asked Dec 3 '18 at 14:12
EuduardoEuduardo
1638
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add a comment |
$begingroup$
I want to prove the following:
If two RVs $X, Y$ are sub-independent, i.e., $phi_{X+Y}(t) = phi_X(t)phi_Y(t), tinmathbb{R}$ then $X, Y$are uncorrelated. Keep $Cov(X,Y) = E(XY)-E(X)E(Y) = 0$ in mind, I'm considering the way to connect $E(X)$ with characteristic function $phi_X(t)$. In fact, $phi_{X}'(0) = iE(X), phi_{Y}'(0) = iE(Y).$ Finding $E(XY)$ will end the proof, but I could not proceed further.
Any idea or tips would be really helpful.
random-variables independence correlation characteristic-functions
asked Dec 3 '18 at 14:12
EuduardoEuduardo
1638
$endgroup$
add a comment |
$begingroup$
I want to prove the following:
If two RVs $X, Y$ are sub-independent, i.e., $phi_{X+Y}(t) = phi_X(t)phi_Y(t), tinmathbb{R}$ then $X, Y$are uncorrelated. Keep $Cov(X,Y) = E(XY)-E(X)E(Y) = 0$ in mind, I'm considering the way to connect $E(X)$ with characteristic function $phi_X(t)$. In fact, $phi_{X}'(0) = iE(X), phi_{Y}'(0) = iE(Y).$ Finding $E(XY)$ will end the proof, but I could not proceed further.
Any idea or tips would be really helpful.
random-variables independence correlation characteristic-functions
asked Dec 3 '18 at 14:12
EuduardoEuduardo
1638
$endgroup$
I want to prove the following:
If two RVs $X, Y$ are sub-independent, i.e., $phi_{X+Y}(t) = phi_X(t)phi_Y(t), tinmathbb{R}$ then $X, Y$are uncorrelated. Keep $Cov(X,Y) = E(XY)-E(X)E(Y) = 0$ in mind, I'm considering the way to connect $E(X)$ with characteristic function $phi_X(t)$. In fact, $phi_{X}'(0) = iE(X), phi_{Y}'(0) = iE(Y).$ Finding $E(XY)$ will end the proof, but I could not proceed further.
Any idea or tips would be really helpful.
random-variables independence correlation characteristic-functions
random-variables independence correlation characteristic-functions
asked Dec 3 '18 at 14:12
EuduardoEuduardo
1638
asked Dec 3 '18 at 14:12
EuduardoEuduardo
1638
edited Dec 5 '18 at 0:44
Euduardo
asked Dec 3 '18 at 14:12
EuduardoEuduardo
1638
asked Dec 3 '18 at 14:12
EuduardoEuduardo
1638
asked Dec 3 '18 at 14:12
EuduardoEuduardo
1638
1638
add a comment |
add a comment |
1 Answer
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The trick is to introduce auxiliary variables $X'$ and $Y'$ with known joint distribution. Define $X'$ and $Y'$ to be independent but with the same marginal distributions as $X$ and $Y$ respectively. Then
$$
phi_{X'+Y'}(t)stackrel{(a)}=phi_{X'}(t)phi_{Y'}(t) stackrel{(b)}=phi_X(t)phi_Y(t)stackrel{(c)}=phi_{X+Y}(t);
$$
in (a) we use independence of $X'$ and $Y'$, in (b) we use equality of marginal distributions; and (c) is the definition of sub-independence. Therefore, by uniqueness of characteristic functions, $X'+Y'$ has the same distribution as $X+Y$. In particular,
$$
E(X'+Y')^2=E(X+Y)^2
$$
(assuming second moments exist), which in turn implies $$E(XY)=E(X'Y')stackrel{(a)}=E(X')E(Y')stackrel{(b)}=E(X)E(Y).$$
answered Dec 5 '18 at 1:39
grand_chatgrand_chat
20.3k11326
$endgroup$
$begingroup$
Thanks a lot! It really helped me a lot and gave me insights.
$endgroup$
– Euduardo
Dec 5 '18 at 3:46
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
The trick is to introduce auxiliary variables $X'$ and $Y'$ with known joint distribution. Define $X'$ and $Y'$ to be independent but with the same marginal distributions as $X$ and $Y$ respectively. Then
$$
phi_{X'+Y'}(t)stackrel{(a)}=phi_{X'}(t)phi_{Y'}(t) stackrel{(b)}=phi_X(t)phi_Y(t)stackrel{(c)}=phi_{X+Y}(t);
$$
in (a) we use independence of $X'$ and $Y'$, in (b) we use equality of marginal distributions; and (c) is the definition of sub-independence. Therefore, by uniqueness of characteristic functions, $X'+Y'$ has the same distribution as $X+Y$. In particular,
$$
E(X'+Y')^2=E(X+Y)^2
$$
(assuming second moments exist), which in turn implies $$E(XY)=E(X'Y')stackrel{(a)}=E(X')E(Y')stackrel{(b)}=E(X)E(Y).$$
answered Dec 5 '18 at 1:39
grand_chatgrand_chat
20.3k11326
$endgroup$
$begingroup$
Thanks a lot! It really helped me a lot and gave me insights.
$endgroup$
– Euduardo
Dec 5 '18 at 3:46
add a comment |
$begingroup$
The trick is to introduce auxiliary variables $X'$ and $Y'$ with known joint distribution. Define $X'$ and $Y'$ to be independent but with the same marginal distributions as $X$ and $Y$ respectively. Then
$$
phi_{X'+Y'}(t)stackrel{(a)}=phi_{X'}(t)phi_{Y'}(t) stackrel{(b)}=phi_X(t)phi_Y(t)stackrel{(c)}=phi_{X+Y}(t);
$$
in (a) we use independence of $X'$ and $Y'$, in (b) we use equality of marginal distributions; and (c) is the definition of sub-independence. Therefore, by uniqueness of characteristic functions, $X'+Y'$ has the same distribution as $X+Y$. In particular,
$$
E(X'+Y')^2=E(X+Y)^2
$$
(assuming second moments exist), which in turn implies $$E(XY)=E(X'Y')stackrel{(a)}=E(X')E(Y')stackrel{(b)}=E(X)E(Y).$$
answered Dec 5 '18 at 1:39
grand_chatgrand_chat
20.3k11326
$endgroup$
$begingroup$
Thanks a lot! It really helped me a lot and gave me insights.
$endgroup$
– Euduardo
Dec 5 '18 at 3:46
add a comment |
$begingroup$
The trick is to introduce auxiliary variables $X'$ and $Y'$ with known joint distribution. Define $X'$ and $Y'$ to be independent but with the same marginal distributions as $X$ and $Y$ respectively. Then
$$
phi_{X'+Y'}(t)stackrel{(a)}=phi_{X'}(t)phi_{Y'}(t) stackrel{(b)}=phi_X(t)phi_Y(t)stackrel{(c)}=phi_{X+Y}(t);
$$
in (a) we use independence of $X'$ and $Y'$, in (b) we use equality of marginal distributions; and (c) is the definition of sub-independence. Therefore, by uniqueness of characteristic functions, $X'+Y'$ has the same distribution as $X+Y$. In particular,
$$
E(X'+Y')^2=E(X+Y)^2
$$
(assuming second moments exist), which in turn implies $$E(XY)=E(X'Y')stackrel{(a)}=E(X')E(Y')stackrel{(b)}=E(X)E(Y).$$
answered Dec 5 '18 at 1:39
grand_chatgrand_chat
20.3k11326
$endgroup$
The trick is to introduce auxiliary variables $X'$ and $Y'$ with known joint distribution. Define $X'$ and $Y'$ to be independent but with the same marginal distributions as $X$ and $Y$ respectively. Then
$$
phi_{X'+Y'}(t)stackrel{(a)}=phi_{X'}(t)phi_{Y'}(t) stackrel{(b)}=phi_X(t)phi_Y(t)stackrel{(c)}=phi_{X+Y}(t);
$$
in (a) we use independence of $X'$ and $Y'$, in (b) we use equality of marginal distributions; and (c) is the definition of sub-independence. Therefore, by uniqueness of characteristic functions, $X'+Y'$ has the same distribution as $X+Y$. In particular,
$$
E(X'+Y')^2=E(X+Y)^2
$$
(assuming second moments exist), which in turn implies $$E(XY)=E(X'Y')stackrel{(a)}=E(X')E(Y')stackrel{(b)}=E(X)E(Y).$$
answered Dec 5 '18 at 1:39
grand_chatgrand_chat
20.3k11326
answered Dec 5 '18 at 1:39
grand_chatgrand_chat
20.3k11326
answered Dec 5 '18 at 1:39
grand_chatgrand_chat
20.3k11326
answered Dec 5 '18 at 1:39
grand_chatgrand_chat
20.3k11326
20.3k11326
$begingroup$
Thanks a lot! It really helped me a lot and gave me insights.
$endgroup$
– Euduardo
Dec 5 '18 at 3:46
add a comment |
$begingroup$
Thanks a lot! It really helped me a lot and gave me insights.
$endgroup$
– Euduardo
Dec 5 '18 at 3:46
$begingroup$
Thanks a lot! It really helped me a lot and gave me insights.
$endgroup$
– Euduardo
Dec 5 '18 at 3:46
$begingroup$
Thanks a lot! It really helped me a lot and gave me insights.
$endgroup$
– Euduardo
Dec 5 '18 at 3:46
add a comment |
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