28.2. LINEAR COMBINATIONS 763

Conversely, suppose every linear combination is normally distributed. Next suppose∑

pj=1 a jX j = a ·X is normally distributed with mean µ and variance σ2 so that its charac-

teristic function is given as eitµ e−12 t2σ2

. I will now relate µ and σ2 to various quantitiesinvolving the X j. Letting m j = E (X j) ,m= (m1, · · · ,mp)

∗

µ =p

∑j=1

a jE (X j) =p

∑j=1

a jm j, σ2 = E

( p

∑j=1

a jX j−p

∑j=1

a jm j

)2

= E

( p

∑j=1

a j (X j−m j)

)2= ∑

j,ka jakE ((X j−m j)(Xk−mk))

It follows the mean of the random variable, a ·X is µ = ∑ j a jm j = a ·m and its varianceis

σ2 = a∗E

((X−m)(X−m)∗

)a

Therefore, E(eita·X) = eitµ e−

12 t2σ2

= eita·me−12 t2a∗E((X−m)(X−m)∗)a. Letting s = ta

this showsE(eis·X)= eis·me−

12s∗E((X−m)(X−m)∗)s = eis·me−

12s∗Σs

which is the characteristic function of a normally distributed random variable withm givenabove and Σ given by

Σ jk = E ((X j−m j)(Xk−mk)) .

By assumption, a is completely arbitrary and so it follows that s is also. Hence, X isnormally distributed as claimed. ■

Corollary 28.2.2 Let X = (X1, · · · ,Xp) ,Y = (Y1, · · · ,Yp) where each Xi,Yi is a realvalued random variable. Suppose also that for every a∈Rp, a ·X and a ·Y are both nor-mally distributed with the same mean and variance. Then X and Y are both multivariatenormal random vectors with the same mean and variance.

Proof: In the Proof of Theorem 28.2.1 the proof implies that the characteristic functionsof a ·X and a ·Y are both of the form eitme−

12 σ2t2

. Then as in the proof of that theorem, itmust be the case that m = ∑

pj=1 a jm j where E (Xi) = mi = E (Yi) and

σ2 = a∗E

((X−m)(X−m)∗

)a= a∗E

((Y −m)(Y −m)∗

)a

and this last equation must hold for every a. Therefore,

E((X−m)(X−m)∗

)= E

((Y −m)(Y −m)∗

)≡ Σ

and so the characteristic function of both X and Y is eis·me−12s∗Σs as in the proof of

Theorem 28.2.1. ■

Theorem 28.2.3 Suppose X = (X1, · · · ,Xp) is normally distributed with mean mand covariance Σ. Then if X1 is uncorrelated with any of the Xi, meaning

E ((X1−m1)(X j−m j)) = 0 for j > 1,

then X1 and (X2, · · · ,Xp) are both normally distributed and the two random vectors areindependent. Here m j ≡ E (X j) . More generally, if the covariance matrix is a diagonalmatrix, the random variables,

{X1, · · · ,Xp

}are independent.