I have a 2D random variable $X$ and $EX=(1,0)^T$ and
$ VX= \begin{bmatrix} 4 && 4 \\ 4 && 16 \end{bmatrix} $
If $Y_1=X_1+2$ and $Y_2=X_2-X_1+1$, then how do I find the variance matrix for $Y$?
I have tried the following where I emitted the constants as my guess is they don't affect variance: $Cov(Y_1,Y1)=Cov(X_1,X_1)=Var(X_1)=4$
$Cov(Y_2,Y_2) =Cov(X_2-X_1,X_2-X_1)\\ =Cov(X_2,X_2-X_1)-Cov(X_1,X_2-X_1)\\ =Cov(X_2,X_2)-Cov(X_2,X_1)-Cov(X_1,X_2)+Cov(X_1,X_1)\\ =16-4-4+4=12 $
Variance= $\begin{bmatrix} 4 && 0\\ 0 && 12 \end{bmatrix} $
But the result seems wrong because now they are no longer correlated just because of addition and/or subtraction
Guide:
Find a $2\times2$ matrix $A$ such that $Y=v+AX$ for some vector $v$ and apply the general rule: $$\mathsf{V}(v+AX)=A\mathsf{V}(X)A^T$$
Btw: I think that your result is correct.
It might well be that $X_1$ and $X_2-X_1$ are uncorrelated while $X_1$ and $X_2$ are not uncorrelated.