Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

9.6. A Regression Problem 543

Since ˆαis a linear function of independent and normally distributed random
variables, ˆαhas a normal distribution with mean

E(ˆα)=E

(

1

n

∑n

i=1

Yi

)

=

1

n

∑n

i=1

E(Yi)=

1

n

∑n

i=1

[α+β(xi−x)] =α

and variance

var( ˆα)=

∑n

i=1

(

1

n

) 2

var(Yi)=

σ^2

n

.

The estimatorβˆis also a linear function ofY 1 ,Y 2 ,...,Ynand hence has a normal
distribution with mean

E(βˆ)=

∑n

i=1(x∑i−x)[α+β(xi−x)]

n

i=1(xi−x)

2

=

α

∑n

i=1(xi−x)+β

∑n

i=1(xi−x)

2

∑n

i=1(xi−x)

2 =β

and variance

var(βˆ)=

∑n

i=1

[

xi−x

∑n

i=1(xi−x)

2

] 2

var(Yi)

=

∑n

i=1(xi−x)

2

[

∑n

i=1(xi−x)

(^2) ]^2
σ^2 =
σ^2
∑n
i=1(xi−x)
2.
In summary, the estimators ˆαandβˆare linear functions of the independent
normal random variablesY 1 ,...,Yn. In Exercise 9.6.4 it is further shown that the
covariance between ˆαandβˆis zero. It follows that ˆαandβˆare independent random
variables with a bivariate normal distribution; that is,
(
αˆ
βˆ
)
has aN 2
((
α
β
)
,σ^2
[ 1
n^0
0 Pn^1
i=1(xi−x)^2
])
distribution. (9.6.9)
Next, we consider the estimator ofσ^2. It can be shown (Exercise 9.6.9) that
∑n
i=1
[Yi−α−β(xi−x)]^2 =
∑n
i=1
{(ˆα−α)+(βˆ−β)(xi−x)
+[Yi−αˆ−βˆ(xi−x)]}^2
= n(ˆα−α)^2 +(βˆ−β)^2
∑n
i=1
(xi−x)^2 +nσˆ^2 ,
or for brevity,
Q=Q 1 +Q 2 +Q 3.
HereQ, Q 1 ,Q 2 ,andQ 3 are real quadratic forms in the variables
Yi−α−β(xi−x),i=1, 2 ,...,n.