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Thread: Jointly normal distribution

  1. #1
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    Jointly normal distribution

    Let X and Y be jointly normal, with the mean vectors and covariance matrix below:

    $\displaystyle
    \mu =\left(\begin{array}{cc}1\\2\end{array}\right)$ $\displaystyle \Sigma = \left(\begin{array}{cc}2&.5\\.5&1\end{array}\right )$

    Let $\displaystyle Z_{1} = X + Y \mbox{ and } Z_{2} = X - 2Y$

    Find the mean vector and covariance matrix of $\displaystyle Z_{1} \mbox{ and }Z{2}$

    I figured out the following:

    $\displaystyle \mu_{Z_{1}} = \mu_X + \mu_Y = 3,$

    $\displaystyle \mu_{Z_{2}} = \mu_X - 2\mu_Y = -3,$

    $\displaystyle \sigma_{Z_{1}}^{2} = \sigma_{X}^{2} + \sigma_{Y}^{2} + 2\rho\sigma_{X}\sigma_{Y} = 3 + \sqrt{2},$

    $\displaystyle \sigma_{Z_{2}}^{2} = \sigma_{X}^{2} + 4\sigma_{Y}^{2} - 4\rho\sigma_{X}\sigma_{Y} = 6 - 2\sqrt{2}$

    But how do I figure out $\displaystyle Cov(Z_{1}, Z_{2})$?
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  2. #2
    Moo
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    Hello,

    Well, the elements of the covariance matrix are the covariances...
    So we have cov(X,Y)=cov(Y,X)=.5

    And finding cov(Z1,Z2) is then easy, using the fact that the covariance is a bilinear form (cov(ax,y)=a*cov(x,y) , cov(x+y,z)=cov(x,z)+cov(y,z))



    Otherwise, you can note that :
    $\displaystyle \begin{pmatrix}Z_1 \\ Z_2\end{pmatrix}=\begin{pmatrix} 1&1 \\ 1&-2\end{pmatrix} \begin{pmatrix}X\\Y\end{pmatrix}$

    Let's denote $\displaystyle A=\begin{pmatrix} 1&1 \\ 1&-2\end{pmatrix}$


    And we know that $\displaystyle A\begin{pmatrix}X\\Y\end{pmatrix}$ will follow a normal distribution $\displaystyle \mathcal{N}(A\mu,A\Sigma A')$

    See a proof in the attached pdf
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