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Thread: Help with bivariate normal dist question

  1. #1
    Senior Member Danneedshelp's Avatar
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    Help with bivariate normal dist question

    Q: Consider $\displaystyle U_{1}=Y_{1}+Y_{2}$ and $\displaystyle U_{2}=Y_{1}-Y_{2}$. Assume $\displaystyle U_{1}$ and $\displaystyle U_{2}$ have a bivariate normal distribution. Show that $\displaystyle U_{1}$ and $\displaystyle U_{2}$ are independent.

    My work: Suppose $\displaystyle U_{1}=Y_{1}+Y_{2}$ and $\displaystyle U_{2}=Y_{1}-Y_{2}$ have a bivariate normal distribution. To show $\displaystyle U_{1}$ and $\displaystyle U_{2}$ are independent we must show $\displaystyle Cov(U_{1},U_{2})=\sum_{i}\sum_{j}\\a_{i}b_{j}Cov(U _{1},U_{2})=0$.

    I am having trouble with the latter computation:

    $\displaystyle Cov(U_{1},U_{2})=\sum_{i}\sum_{j}\\a_{i}b_{j}Cov(U _{1},U_{2})
    =a_{1}b_{1}Cov(Y_{1},Y_{1})+$
    $\displaystyle a_{1}b_{2}Cov(Y_{1},Y_{2})+
    a_{2}b_{2}Cov(Y_{2},Y_{2})$

    How do I treat $\displaystyle Cov(Y_{1},Y_{1})$ given $\displaystyle U_{1}=Y_{1}+Y_{2}$ and $\displaystyle U_{2}=Y_{1}-Y_{2}$?

    All in all, I'm pretty stuck. Any help would be greatly appreciated.

    Thanks
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  2. #2
    MHF Contributor matheagle's Avatar
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    $\displaystyle Cov(Y_{1},Y_{1})=V(Y_1)$

    you can prove that via the definition or the short cut formula

    Use $\displaystyle Cov(X,Y)=E(XY)-E(X)E(Y)$ and replace X with Y.

    $\displaystyle Cov(Y,Y)=E(YY)-E(Y)E(Y)=E(Y^2)-(E(Y))^2=V(Y)$
    Last edited by matheagle; Nov 23rd 2009 at 11:56 PM.
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