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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- Jan 5th 2009, 06:27 AMtotalnewbieVariance(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? - Jan 5th 2009, 08:06 AMLast_Singularity
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?