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Thread: Variance(U) formula if U=aX+bY+cZ+d

  1. #1
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    Variance(U) formula if U=aX+bY+cZ+d

    I don't find this formula anywhere. Wikipedia has the case where:
    Var(aX+bY)=a^2*Var(X)+a^2*Var(Y)+2abCov(X,Y).
    How to derive formula for my case?
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  2. #2
    Member Last_Singularity's Avatar
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    Quote Originally Posted by totalnewbie View Post
    I don't find this formula anywhere. Wikipedia has the case where:
    Var(aX+bY)=a^2*Var(X)+a^2*Var(Y)+2abCov(X,Y).
    How to derive formula for my case?
    So you want to find $\displaystyle var(aX+bY+cZ+d)$

    Denote
    $\displaystyle \alpha = aX + bY$
    $\displaystyle \beta = cZ+d$

    Your problem then becomes $\displaystyle var(\alpha + \beta)$

    Of course, from the formula given on wikipedia, you know how to expand: $\displaystyle var(\alpha + \beta)$
    $\displaystyle =var(\alpha) + var(\beta) + 2 cov(\alpha,\beta)$
    $\displaystyle =var(aX + bY) + var(cZ+d) + 2 cov(aX + bY,cZ+d)$

    At this point, you can further simplify the first two variance terms. But for the last one, you need to know that
    $\displaystyle cov(aX+bY,cW+dV)$$\displaystyle = (ac)cov(X,W) + (ad)cov(X,V) + (bc)cov(Y,W) + (bd)cov(Y,V)$

    Can you finish the expansion from here?
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