1. ## variance cov matrix

One question:

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

where

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

and

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

what is the variance of the linear combination:

$z=2x_2 + x_3$

*****

I get that

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

which is equal to

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

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

2. yes that variance is zero
also note that the correlation between X2 and X3 is -1.

3. thnk you!

So I am supposed to calculate the correlation between Z and another linear combination. What am I supposed to do if var(z)=0? Just say that it is impossible to calculate?

4. what is this other linear combination?
and note anyone can make this absurd.
we need to follow the cauchy-schwartz inequality
that's why I calculated your correlation
you can make the covariance between X2 and X3 equal -100 which makes this impossible

5. the other linear combinations are

w=x_1+x_2+x_3
y=x_1-x_3

yahh, I guess that it is a problem in the question. Thank you!