One question:

If we have the vector$\displaystyle x=[x_1,x_2,x_3] ; E(x_1)=8 \ E(x_2)=0, \ E(x_3)=2$:

where

$\displaystyle var(x_1)=4, var(x_2)=1, var(x_3)=4$

and

$\displaystyle cov(x_1,x_2)=-1,cov(x_1,x_3)=2, cov(x_2,x_3)=-2$

what is the variance of the linear combination:

$\displaystyle z=2x_2 + x_3$

*****

I get that

$\displaystyle var(z)=4var(x_2)+var(x_3)+2cov(2x_2,x_3)$

which is equal to

$\displaystyle var(z)=4(1)+4+2(2)(-2)=0$

Am I doing everything right? How can we get that variance? I can't calculate correlation coefficients with a variance of 0, so I am stucked here. I would really appreaciate any sort of help