I want to prove that $cov(AX+a,BY+b) = Acov(X,Y)B^t$ for any matrices $A,B$ and vectors $a,b$ if $X$ and $Y$ are normal distributed vectors, where $m$ is the mean of $X$ and $n$ the mean of $Y$
Is this proof correct? I am basically only using the linearity of the expectation $E$. It seems that I didn't even use properties of the normal distribution, so this should be true for any distributions with the given properties, right?
$$\begin{align*} cov(AX+a,BY+b) &= E[(AX+a - Am-a)(BY+b-Bn-b)^t] \\ &= E[(AX- Am)(BY-Bn)^t]\\ &= E[AXY^tB^t]-E[AXn^tB^t]-E[AmY^tB^t]+E[Amn^tB^t]\\ &= A\cdot E[XY^t]B^t-A\cdot E[X]n^tB^t-Am\cdot E[Y^t]B^t+Amn^tB^t\\ &=A(E[XY^t]-E[X]n^t-mE[Y^t]+mn^t)B^t\\ &=A(E[(X-m)(Y-n)^t]B^t\\ &=Acov(X,Y)B^t \end{align*}$$
Yes this is correct, and you're right it's true for any distribution.