39.5. LINEAR REGRESSION 827

and soα̂ =

1n ∑

kXk ≡ X̄

Then alson

∑k=1

Xk (tk− t̄)−β ∑k(tk− t̄)2 = 0

and so

β̂ =∑

nk=1 Xk (tk− t̄)

∑k (tk− t̄)2 =∑

nk=1 (Xk− X̄)(tk− t̄)

∑k (tk− t̄)2

because ∑nk=1 X̄ (tk− t̄) = 0. It remains to find the maximum likelihood estimate for σ2.

Using 39.10,

nσ−

n

∑k=1

(Xk−

(α̂ + β̂ (tk− t̄)

))2

σ3 = 0

σ̂2 =

1n

n

∑k=1

(Xk−

(α̂ + β̂ (tk− t̄)

))2

Now considern

∑k=1

(Xk− (α +β (tk− t̄)))2

σ2 (39.11)

I will add in α̂ + β̂ (tk− t̄) and subtract it and then write this as a sum of quadratic forms.First of all, note that it is the sum of the squares of independent random variables in n(0,1)and so it is X 2 (n). It equals

n

∑k=1

(Xk−

(α̂ + β̂ (tk− t̄)

)+((

α̂ + β̂ (tk− t̄))− (α +β (tk− t̄))

))2

σ2 (39.12)

This will be expanded. I need to consider the mixed term in which I will use the abovedescriptions of α̂ and β̂ .

∑k

(Xk−

(α̂ + β̂ (tk− t̄)

))((α̂ + β̂ (tk− t̄)

)− (α +β (tk− t̄))

)= ∑

k

[(Xk− X̄)−

(β̂ (tk− t̄)

)][(X̄ + β̂ (tk− t̄)

)− (α +β (tk− t̄))

]First note that

∑k(Xk− X̄) X̄ = ∑

k(Xk− X̄)α = ∑

kβ̂ (tk− t̄) X̄ = ∑

kβ̂ (tk− t̄)α = 0

Thus the mixed term is

∑k(Xk− X̄) β̂ (tk− t̄)−β ∑

k(Xk− X̄)(tk− t̄)

−β̂2∑k(tk− t̄)2 + β̂β ∑

k(tk− t̄)2

=(

β̂ −β

)∑k(Xk− X̄)(tk− t̄)+ β̂

(β − β̂

)∑k(tk− t̄)2

=(

β̂ −β

)β̂ ∑

j(t j− t̄)2 + β̂

(β − β̂

)∑k(tk− t̄)2 = 0