How to show Sub-independent Random Variables are uncorrelated.












0














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.










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    0














    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.










    share|cite|improve this question



























      0












      0








      0







      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.










      share|cite|improve this question















      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






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      edited Dec 5 '18 at 0:44

























      asked Dec 3 '18 at 14:12









      Euduardo

      1217




      1217






















          1 Answer
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          1














          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 $$E(XY)=E(X'Y')stackrel{(a)}=E(X')E(Y')stackrel{(b)}=E(X)E(Y).$$






          share|cite|improve this answer





















          • Thanks a lot! It really helped me a lot and gave me insights.
            – Euduardo
            Dec 5 '18 at 3:46











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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1














          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 $$E(XY)=E(X'Y')stackrel{(a)}=E(X')E(Y')stackrel{(b)}=E(X)E(Y).$$






          share|cite|improve this answer





















          • Thanks a lot! It really helped me a lot and gave me insights.
            – Euduardo
            Dec 5 '18 at 3:46
















          1














          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 $$E(XY)=E(X'Y')stackrel{(a)}=E(X')E(Y')stackrel{(b)}=E(X)E(Y).$$






          share|cite|improve this answer





















          • Thanks a lot! It really helped me a lot and gave me insights.
            – Euduardo
            Dec 5 '18 at 3:46














          1












          1








          1






          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 $$E(XY)=E(X'Y')stackrel{(a)}=E(X')E(Y')stackrel{(b)}=E(X)E(Y).$$






          share|cite|improve this answer












          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 $$E(XY)=E(X'Y')stackrel{(a)}=E(X')E(Y')stackrel{(b)}=E(X)E(Y).$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 5 '18 at 1:39









          grand_chat

          20k11225




          20k11225












          • Thanks a lot! It really helped me a lot and gave me insights.
            – Euduardo
            Dec 5 '18 at 3:46


















          • Thanks a lot! It really helped me a lot and gave me insights.
            – Euduardo
            Dec 5 '18 at 3:46
















          Thanks a lot! It really helped me a lot and gave me insights.
          – Euduardo
          Dec 5 '18 at 3:46




          Thanks a lot! It really helped me a lot and gave me insights.
          – Euduardo
          Dec 5 '18 at 3:46


















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