Lets say I have 3 RV's that are standard normally distributed and independent : x , y and z
Let X = 4*x + 2*y + 3 and let Y = 5*y + 2*z + 4
How can then the Cov(X,Y) be calculated?
This is what I have done so far to solve this problem: Cov(X,Y) = E[XY] - E[X]E[Y] =
E[(4*x + 2*y + 3)(5*y + 2*z + 4)] - E[4*x + 2*y + 3]E[5*y + 2*z + 4] =
E[20xy+8xz+10y2+4yz+16x+23y+6z+12] - E[4*x + 2*y + 3]E[5*y + 2*z + 4] =
20E[xy]+8E[xz]+10E[y]^2+4E[yz]+16E[x]+23E[y]+6E[z]+12 - (4*E[x] + 2*E[y] + 3 * 5*E[y] + 2*E[z] + 4) =
20 * 0 + 8 * 0 + 10*0^2 +4 * 0 + 16 * 0 +23 * 0 +6 * 0 + 12 - (3 * 4) = 0
But that is not the right answer and I can't figure out why.
UPDATE: 10 * E[y]^2 must be 10 * E[y^2]. It is known that Var(y) = E[y^2] - E[y]^2 = 1 (because N(0 , 1) )and E[y] = 0 which gives 1 = E[y^2] - 0. Hence E[y^2] = 1.
FOR THOSE WHO ARE INTERESTED IN THE ANSWER:
20 * 0 + 8 * 0 + 10 * 1 +4 * 0 + 16 * 0 + 23 * 0 +6 * 0 + 12 - (3 * 4) =
10 + 12 - 12 = 10