1912 CHAPTER 59. BASIC PROBABILITY

Also, if X∼ Np (m,Σ) then −X∼ Np (−m,Σ) . Furthermore, if X∼ Np (m,Σ) then

E(eit·X)= eit·me−

12 t∗Σt (59.16.32)

Also if a is a constant and X∼ Np (m,Σ) then aX∼ Np(am,a2Σ

).

Proof: Consider E(eit·X) for X∼ Np (m,Σ).

E(eit·X)≡ 1

(2π)p/2 (detΣ)1/2

∫Rp

eit·xe−12 (x−m)∗Σ−1(x−m)dx.

Let R be an orthogonal transformation such that

RΣR∗ = D = diag(σ

21, · · · ,σ2

p).

Then let R(x−m) = y. Then

E(eit·X)= 1

(2π)p/2∏

pi=1 σ i

∫Rp

eit·(R∗y+m)e−12 y∗D−1ydy.

ThereforeE(eit·X)= 1

(2π)p/2∏

pi=1 σ i

∫Rp

eis·(y+Rm)e−12 y∗D−1ydy

where s =Rt. This equals

eit·mp

∏i=1

(∫R

eisiyie− 1

2σ2i

y2idyi

)1√

2πσ i

= eit·mp

∏i=1

(∫R

eisiσ iue−12 u2

du)

1√2π

= eit·mp

∏i=1

e−12 s2

i σ2i

1√2π

∫R

e−12 (u−isiσ i)

2du

= eit·me−12 ∑

pi=1 s2

i σ2i = eit·me−

12 t∗Σt

This proves 59.16.32.Since X1 and X2 are independent, eit·X1 and eit·X2 are also independent. Hence

E(eit·X1+X2

)= E

(eit·X1

)E(eit·X2

).

Thus,

E(eit·X1+X2

)= E

(eit·X1

)E(eit·X2

)= eit·m1e−

12 t∗Σ1teit·m2e−

12 t∗Σ2t

= eit·(m1+m2)e−12 t∗(Σ1+Σ2)t