How I can proof that if $\bf{Y}$ is a random vector with distribution multivariate normal $N_p(\bf{\mu},\bf{\Sigma}$) each $Y_i\sim N(\mu_i,\sigma_i^2)$.

I know that $$M_\bf{Y}(\bf{t})=\exp\{\bf{\mu}'\bf{t}+\frac{1}{2}\bf{t}'\bf{\Sigma}t\}$$

If I take $\bf{t}=(0,\dots,0,1,\dots,0)=t_i$ then $$M_\bf{Y}(\bf{t})=\exp\{\mu_i+\frac{1}{2}t_i^2\sigma_i^2\}=M_{Y_i}(t_i)=M_\bf{Y}(0,\dots,0,1,\dots,0)$$

and $M_{Y_i}(t_i)$ is the moment generating function of a normal random variable so $Y_i\sim N(\mu_i,\sigma_i^2)$.

Is it valid? I mean it proofs anything?


1 Answer 1


Yes, your method is correct. The m.g.f. for a vector random variable $X$, if it is finite in a neighborhood of the origin, has packed in it the m.g.f. of any linear function of $X$. In your case, the $i$-th coordinate $X_i$ is a linear function of $X$. Suppose $X$ has m.g.f. $\psi(t) = E\exp(\langle t, X\rangle)$ and that $Y=AX$ for some matrix $A$. Then $\phi(u) = E\exp(\langle u, Y\rangle) = E\exp( \langle A' u, X\rangle) = \psi(A' u)$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .