762 CHAPTER 28. THE NORMAL DISTRIBUTION

which is the characteristic function of a random variable which is N (−m,Σ) . Theorem27.1.4 again implies −X ∼ N (−m,Σ) . Finally consider the last claim. You apply what isknown aboutX with t replaced with at and then massage things. This gives the character-istic function for aX is given by

E (exp(i t·aX)) = exp(i t·am)exp(−1

2t∗Σa2t

)which is the characteristic function of a normal random vector having mean am and co-variance a2Σ. ■

28.2 Linear CombinationsFollowing [44] a random vector has a generalized normal distribution if its characteristicfunction is given as eit·me−

12 t∗Σt where Σ is symmetric and has nonnegative eigenvalues.

For a random real valued variable, m is scalar and so is Σ so the characteristic functionof such a generalized normally distributed random variable is eitµ e−

12 t2σ2

. These gener-alized normal distributions do not require Σ to be invertible, only that the eigenvalues benonnegative. In one dimension this would correspond the characteristic function of a diracmeasure having point mass 1 at µ. In higher dimensions, it could be a mixture of suchthings with more familiar things. I won’t try very hard to distinguish between general-ized normal distributions and normal distributions in which the covariance matrix has allpositive eigenvalues. These generalized normal distributions are discussed more a littlelater.

Here are some other interesting results about normal distributions found in [44]. Thenext theorem has to do with the question whether a random vector is normally distributedin the above generalized sense.

Theorem 28.2.1 LetX = (X1, · · · ,Xp) where each Xi is a real valued random vari-able. Then X is normally distributed in the above generalized sense if and only if everylinear combination, ∑

pj=1 aiXi is normally distributed. In this case the mean ofX is

m= (E (X1) , · · · ,E (Xp))

and the covariance matrix forX is

Σ jk = E((X j−m j)(Xk−mk)

∗) .Proof: Suppose first X is normally distributed. Then its characteristic function is of

the formφX (t) = E

(eit·X)= eit·me−

12 t∗Σt.

Then letting a= (a1, · · · ,ap)

E(

eit ∑pj=1 aiXi

)= E

(eita·X)= eita·me−

12a∗Σat2

which is the characteristic function of a normally distributed random variable with meana ·m and variance σ2 = a∗Σa. This proves half of the theorem. If X is normally dis-tributed, then every linear combination is normally distributed.