1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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 $\phi(u) = E\exp(\langle u, Y\rangle) = E\exp( \langle A' u, X\rangle) = \psi(A' u)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.