1914 CHAPTER 59. BASIC PROBABILITY

and the covariance matrix for X is

Σ jk = E((X j−m j)(Xk−mk)

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

formφ X (t) = E

(eit·X)= eit·me−

12 t∗Σt.

Then letting a = (a1, · · · ,ap)

E(

eit ∑pj=1 aiXi

)= E

(eita·X)= eita·me−

12 a∗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 distributed,then 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 characteristic function is given in 59.16.34. I will now relate µ and σ2 to variousquantities involving 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 normally distributed random variable, a ·X is

µ = ∑j

a jm j = a ·m

and its variance isσ

2 = 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.

Then letting s = ta this shows

E(eis·X) = eis·me−

12 s∗E((X−m)(X−m)∗)s

= eis·me−12 s∗Σs

which is the characteristic function of a normally distributed random variable with m 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.