# Variance Covariance Matrix

#### hellostudent

##### New Member
Hello, I'm trying to solve this exercise My procedure is the following:

Let's write the data:

E(X1)=0; V(X1)=1E(X2)=0; V(X2)=1X= [X1,X2]Y=[Y1,Y2]Y= A * X

Considering the random vector X = [X1,X2], we know that E(X)=[E(X1),E(X2)]=[0,0]

we also know that if Y = A * X, **E(Y)=E(A*X)=A*E(X) = A*[0,0] = [0,0]

So the mean of Y is 0 (and this result is correct according to the solutions given by my exercise sheet).But then, I'm failing to find a way to compute the variance-covariance matrix

Can I state that if X1 and X2 are independent, are Y1 and Y2, independent too?

I state this considering that if (Y1,Y2)=A(X1,X2), (Y1,Y2)=(AX1,AX2), so if X1 and X2 are independent, Y1 and Y2, that are respectively X1 and X2 multiplied by the same matrix A (so by the same value), they are independent too.

I benefit from this deduction to compute the covariance Cov(X1,X2). In fact, I know that since E(Y) = [0,0], E(Y1)=E(Y2)=0and, since Y1 and Y2 are independent, E(Y1*Y2)=E(Y1)*E(Y2)=0, and (X1,X2)=E(X1*X2)*E(X1)*E(X2)=0*0*0=0

So I know that two terms of the variance-covariance matrix are 0 and 0 (Cov(X1,X2) and Cov(X2,X1).

Some of my classmates arrived to the results with this procedure, of which I did not understand the steps: Do you know how to help me to continue?

#### Dason

Tell us what parts of your classmates derivation confuses you

#### Dason

They used variance rules and the previously established
Y1=cos(ϑ)*X1-sin(ϑ)*X2

Keep in mind that the cos and sin terms are just constants here.

#### hellostudent

##### New Member
If Y1=cos(ϑ)*X1-sin(ϑ)*X2, why is V(Y1)=cos^2(ϑ)*V(X1)+sin^2*V(X2)?
It seems that they squared the two terms of the binomial, but it's not a square of Y1 because there isn't the double product -2*cos(ϑ)*X1*sin(ϑ)*X2. I don't know what's the variance rule that I'm ignoring.