828 CHAPTER 39. STATISTICAL TESTS

It follows from the vanishing of the mixed term that 39.11 equals

n

∑k=1

(Xk− (α +β (tk− t̄)))2

σ2

=n

∑k=1

(Xk−

(α̂ + β̂ (tk− t̄)

))2+(

α̂ + β̂ (tk− t̄))− (α +β (tk− t̄))2

σ2

=1

σ2 ∑k

(α̂ + β̂ (tk− t̄)

)− (α +β (tk− t̄))2 +

nσ̂2

σ2

=1

σ2 ∑k

((α̂−α)+

(β̂ −β

)(tk− t̄)

)2+

nσ̂2

σ2

Consider the mixed term in the first sum.

∑k(α̂−α)

(β̂ −β

)(tk− t̄) = 0

Therefore, from 39.11,n

∑k=1

(Xk− (α +β (tk− t̄)))2

σ2 =

nσ2 (α̂−α)2 +

(β̂ −β

)2

σ2 ∑k(tk− t̄)2 +

nσ̂2

σ2 (39.13)

At this point, we need to consider what we have.

Lemma 39.5.2 Suppose Xi is n(µ i,σ

2i)

and the Xi are independent for i≤ n. Then ∑i aiXi

is n(∑i aiµ i,∑a2

i σ2i).

Proof:Consider the moment generating function.

M (t) ≡ E

(exp

(t ∑

iaiXi

))= E

(n

∏i=1

exp(taiXi)

)

=n

∏i=1

E (exp(taiXi)) =n

∏i=1

e12 t2a2

i σ2i etaiµ i

= e12 t2(∑i a2

i σ2)et ∑i aiµ i ■

Thus α̂ = X̄ is n(

1n ∑

ni=1 (α +β (ti− t̂)) ,∑i

1n2 σ2

)= n

(α, σ2

n

). Also α̂−α

(σ/√

n) =X̄−α

(σ/√

n) isn(0,1) and so the square root of the first term on the right in 39.13 is n(0,1) so that firstterm is X 2 (1). Similarly, consider the second term or rather its square root. This is

β̂ −β(σ/(

∑k (tk− t̄)2)1/2

) (39.14)