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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 $M^T M = I$.

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  • $\begingroup$ multivariate mgf? $\endgroup$ – BCLC Dec 6 '16 at 17:54
  • $\begingroup$ @BCLC What, where? $\endgroup$ – mavavilj Dec 6 '16 at 18:00
  • $\begingroup$ mavavilj, I was thinking to just compute E and Var of $M^Ty$ but how do we know it's still normal? $\endgroup$ – BCLC Dec 6 '16 at 18:01
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    $\begingroup$ affine transforms of normals are normal $\endgroup$ – bdeonovic Dec 6 '16 at 19:52
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See 6.2 in The Matrix Cookbook

$$E[M^Ty] = M^TE[y] = M^TX\beta = 0 \ \beta = 0$$

$$Var[M^Ty] = M^TVar(y)(M^T)^T$$

$$=M^T\sigma^2VM$$

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