Kenneth Kuttler

59.17. USE OF CHARACTERISTIC FUNCTIONS TO FIND MOMENTS 1917

and (X2, · · · ,Xp) is also normally distributed with mean m′ and covariance Σp−1. Now from59.16.37, 59.16.36 follows. In case the covariance matrix is diagonal, the above reasoningextends in an obvious way to prove the random variables,

{X1, · · · ,Xp

}are independent.

However, another way to prove this is to use Proposition 59.11.1 on Page 1891 andconsider the characteristic function. Let E (X j) = m j and

P =p

∑j=1

t jX j.

Then since X is normally distributed and the covariance is a diagonal,

D≡

 σ21 0

. . .0 σ2

,

E(eiP) = E

(eit·X)= eit·me−

12 t∗Σt

= exp

∑j=1

it jm j−12

t2j σ

)(59.16.39)

∏j=1

exp(

it jm j−12

t2j σ

)Also,

E(eit jX j

)= E

(exp

(it jX j + ∑

k ̸= ji0Xk

))

= exp(

it jm j−12

t2j σ

)With 59.16.39, this shows

E(eiP)= p

∏j=1

E(eit jX j

)which shows by Proposition 59.11.1 that the random variables,{

X1, · · · ,Xp}

are independent.

59.17 Use Of Characteristic Functions To Find MomentsLet X be a random variable with characteristic function

φ X (t)≡ E (exp(itX))

59.17. USE OF CHARACTERISTIC FUNCTIONS TO FIND MOMENTS 1917and (X2,--- ,X;) is also normally distributed with mean m’ and covariance L,—1. Now from59.16.37, 59.16.36 follows. In case the covariance matrix is diagonal, the above reasoningextends in an obvious way to prove the random variables, {X Iyote Xp} are independent.However, another way to prove this is to use Proposition 59.11.1 on Page 1891 andconsider the characteristic function. Let E (X;) =m, andDpP=" t;X;j=lThen since X is normally distributed and the covariance is a diagonal,ot 0D=0 orE(e?) = E(e#X) =m opt atP= exp [Som 3005) (59.16.39)Also,E(eiXi) = E (ve (« ay+y oxi)KAj1= exp (im _ 503)E (ei) = [YE (e™)j=lWith 59.16.39, this showswhich shows by Proposition 59.11.1 that the random variables,{X1,-++ Xp}are independent. §J59.17. Use Of Characteristic Functions To Find MomentsLet X be a random variable with characteristic functionPx (t) = E (exp (itX))