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

1. ## 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?

2. Originally Posted by totalnewbie
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?