9.3Distribution of the Estimators 359

or

E

[

SSR

n− 2

]

=σ^2

ThusSSR/(n−2) is an unbiased estimator ofσ^2. In addition, it can be shown thatSSR
is independent of the pairAandB.

REMARKS

A plausibility argument as to whySSR/σ^2 might have a chi-square distribution withn− 2
degrees of freedom and be independent ofAandBruns as follows. Because theYiare
independent normal random variables, it follows that (Yi−E[Yi])/

√
Var(Yi),i=1,...,n
are independent standard normals and so

∑n

i= 1

(Yi−E[Yi])^2

Var(Yi)

=

∑n

i= 1

(Yi−α−βxi)^2

σ^2

∼χn^2

Now if we substitute the estimatorsAandBforαandβ, then 2 degrees of freedom
are lost, and so it is not an altogether surprising result thatSSR/σ^2 has a chi-square
distribution withn−2 degrees of freedom.
The fact thatSSRis independent ofAandBis quite similar to the fundamental result
that in normal samplingXandS^2 are independent. Indeed this latter result states that
ifY 1 ,...,Ynis a normal sample with population meanμand varianceσ 2 , then if in the
sum of squares

∑n

i= 1 (Yi−μ)

(^2) /σ (^2) , which has a chi-square distribution withndegrees
of freedom, one substitutes the estimator∑ Y forμto obtain the new sum of squares
i(Yi−Y)
(^2) /σ (^2) , then this quantity [equal to (n−1)S (^2) /σ (^2) ]will be independent of
Y and will have a chi-square distribution withn−1 degrees of freedom. SinceSSR/σ^2
is obtained by substituting the estimators∑ AandBforαandβin the sum of squares
n
i= 1 (Yi−α−βxi)^2 /σ^2 , it is not unreasonable to expect that this quantity might be
independent ofAandB.
When theYiare normal random variables, the least square estimators are also the
maximum likelihood estimators. To verify this remark, note that the joint density of
Y 1 ,...,Ynis given by
fY 1 ,...,Yn(y 1 ,...,yn)=
∏n
i= 1
fYi(yi)

∏n
i= 1
1
√
2 πσ
e−(yi−α−βxi)
(^2) /2σ 2

1
(2π)n/2σn
e−
∑n
i= 1 (yi−α−βxi)^2 /2σ^2