# Thread: Variance-covariance matrix of random vector

1. ## Variance-covariance matrix of random vector

Notation:
Var(Y) is the variance-covariance matrix of a random vector Y
B' is the tranpose of the matrix B.

1) Let A be a m x n matrix of constants, and Y be a n x 1 random vector. Then Var(AY) = A Var(Y) A'

Proof:
Var(AY)
= E[(AY-A E(Y)) (AY-A E(Y))' ]
= E[A(Y-E(Y)) (Y-E(Y))' A' ]
= A E[(Y-E(Y)) (Y-E(Y))'] A'
= A Var(Y) A'

Now, I don't understand the step in red. What theorem is that step using?
I remember a theorem that says if B is a m x n matrix of constants, and X is a n x 1 random vector, then BX is a m x 1 matrix and E(BX) = B E(X), but this theorem doesn't even apply here since it requries X to be a column vector, not a matrix of any dimension.

2) Theorem: Let Y be a n x 1 random vector, and B be a n x 1 vector of constants(nonrandom), then Var(B+Y) = Var(Y).

I don't see why this is true. How can we prove this?
Is it also true that Var(Y+B) = Var(Y) ?

Any help is greatly appreciated!

note: also under discussion in Talk Stats forum

2. Constants can be pulled out of ANY expectation.
Just like in calculus, constants are not part of the integration,
nor the summations in the discrete case.

3. Of course Var(Y+B) = Var(Y), since Y+B=B+Y.

You only need the dimensions to be the same, of Y and B.
And that is assumed when you are asked to add two vectors.