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

• Jan 5th 2009, 06:27 AM
totalnewbie
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?
• Jan 5th 2009, 08:06 AM
Last_Singularity
Quote:

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 $var(aX+bY+cZ+d)$

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

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

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

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

Can you finish the expansion from here?