# Why does $y \text{~} N(X \beta, \sigma^2 V)$ $\implies$ $M^T y \text{~} N(0, \sigma^2 M^T V M)$

Why does $$y \text{~} N(X \beta, \sigma^2 V)$$ $\implies$ $$M^T y \text{~} N(0, \sigma^2 M^T V M)$$?

When $M^T X=0$ and $M^T M = I$.

• multivariate mgf? – BCLC Dec 6 '16 at 17:54
• @BCLC What, where? – mavavilj Dec 6 '16 at 18:00
• mavavilj, I was thinking to just compute E and Var of $M^Ty$ but how do we know it's still normal? – BCLC Dec 6 '16 at 18:01
• affine transforms of normals are normal – bdeonovic Dec 6 '16 at 19:52

$$E[M^Ty] = M^TE[y] = M^TX\beta = 0 \ \beta = 0$$
$$Var[M^Ty] = M^TVar(y)(M^T)^T$$
$$=M^T\sigma^2VM$$