Variance Covariance Matrix

#1
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

Ambassador to the humans
#4
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.
 
#5
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.
 

Dason

Ambassador to the humans
#6
They aren't just squaring Y1. It's just taking the variance of Y1 using typical variance rules. Keep in mind that since x1 and x2 are independent the covariance between them is 0
 
#7
They aren't just squaring Y1. It's just taking the variance of Y1 using typical variance rules. Keep in mind that since x1 and x2 are independent the covariance between them is 0
Ok, I have managed to solve the exercise by benefitting from the formula V(X)=E(X^2) - [E(X)]^2, hence I have obtained E(X1^2)=E(X2^2)=1, then I found that V(Y1,Y2)=E(Y1*Y2)=0, and finally I found, after further computations that V(Y1)=V(Y2)=cos^2(ϑ)+sin^2(ϑ)=1. So I found the correct results!