59.20. GENERALIZED MULTIVARIATE NORMAL 1935

Also, it suffices to consider E(Xq

k

). Differentiate both sides. Using the repeated index

summation convention,

e−12 t∗Σt (−Σnk jt j

)=∫Rp

ixkeit·xdµ

Now differentiate again.

e−12 t∗Σt (−Σnk jt j

)(−Σnk jt j

)+(−Σnkk) =−

∫Rp

x2keit·xdλ Xn

Next let t = 0 to conclude that E(X2

nk

)= Σnkk. Of course you can continue differentiating

as long as desired and obtain E(X2m

nk

)is equal to some polynomial formula involving Σnkk

and these are given to converge to Σkk. Therefore, for any q > 1,{E (|Xn|q)} is boundedand so from the above, ∫

Ω

|X|q dP≤ lim infn→∞

2qE (|Xn|q)< ∞

So yes, X is indeed in Lq (Ω,Rp) for every q. What about the covariance?From the definition of the characteristic function,

e−12 t∗Σt =

∫Rp

eit·xdµ

and so taking the derivative with respect to tk of both sides,

e−12 t∗Σt (−Σk jt j

)=∫Rp

ixkeit·xdµ

Now differentiate with respect to tl on both sides.

e−12 t∗Σt (−Σliti)

(−Σk jt j

)+ e−

12 t∗Σt (−Σkl)

=∫Rp

ixk (ixl)eit·xdµ =−∫Rp

xkxleit·xdµ

Now let t = 0 to obtainΣkl =

∫Rp

xkxleit·xdµ = E (XkXl)

If m ̸= 0, the same kind of argument holds with a little more details. This proves thefollowing theorem.

Theorem 59.20.2 Let Σ be nonnegative and self adjoint p× p matrix. Then there exists arandom variable X whose distribution measure λ X has characteristic function

eit·me−12 t∗Σt

AlsoE((X−m)(X−m)∗

)= Σ

that isE((X−m)i (X−m) j

)= Σi j

This is generalized normally distributed random variable.