800 CHAPTER 39. STATISTICAL TESTS

Thus the first term in 39.2 is the moment generating function of X̄ . Some computationsshow that the second term is the moment generating function of the vector(

X1− X̄ · · · Xn− X̄)

Indeed,

E

(exp

n

∑k=1

tk (Xk− X̄)

)= E

(exp

(∑k

tkXk−∑k

tk1n ∑

jX j

))

= E

(exp

(∑k

tkXk−∑j

t j1n ∑

kXk

))= E

(exp

(∑

j∑k

tk− t j

nXk

))

= ∏j

E

(exp

(∑k

tk− t j

nXk

))= ∏

j∏

kE(

exp(

tk− t j

nXk

))

= ∏j

∏k

(exp

((tk− t j

nµ

)+

12

(tk− t j

n

)2

σ2

))

= exp

(∑

j∑k

12

(tk− t j

n

)2

σ2

)

Therefore, by Proposition 38.9.7, X̄ and this random vector are linearly independent. ■The above proposition leads to something interesting, the distribution of nS2/σ2. Let

the Xk be independent and normally distributed with mean µ and variance σ2. Then

S2 ≡ 1n

n

∑k=1

(Xk− X̄)2

The distribution of nS2/σ2 = ∑nk=1

(Xk−X̄)2

σ2 will be considered. If we know this, then sinceS2 is experimentally determined, it will follow that we could estimate σ2. First note that

Xk− X̄ = Xk−µ +µ− X̄

and so(Xk− X̄)

2= (Xk−µ)2−2(Xk−µ)(X̄−µ)+(X̄−µ)

2

2n

∑k=1

(Xk−µ)(X̄−µ) = 2n(X̄−µ)(X̄−µ)

Therefore,n

∑k=1

(Xk− X̄)2+n(X̄−µ)

2=

n

∑k=1

(Xk−µ)2

Thenn

∑k=1

(Xk− X̄)2

σ2 +n(X̄−µ)

2

σ2 =n

∑k=1

(Xk−µ)2

σ2