118
+ 4A ^ A 2^ov{
2
+ 4A j A 2 Cov{
2
+ 4A^A2Cov{
, 2 ,
+ 4A ^ A 2 Cov{
2
+ 4A ^A2Cov{
Recalling that,
6i=6k=1>6je
(2
,3)}
(6,6j)e(2,
3)
6k =
U
(5i,6k)e(2,
3)
6i =
1}
6k=1>(6j6k
) e
(2,3
>}
61e(2,3) ,6j
= 6
k = 1l
(4.3.1)
n
1
n
= (proportion of sample which are type l's)
P
+ A^ = (probability of being a type 1),
and, thus, the A coefficients in front of each covariance
term are the probabilities necessary to uncondition each
covariance term. For example,
3
X1Ccv(aljb1J aikbikK-j-k-l>
pUi-ij-^-nc^Uijbi.
a ., b .,
lk lk
6i=6j=6k=1)
Cov(aijbij ailrbn. ^,6^-6^- 6^-1) .
ik ik
Now note that the eight covariance terms correspond to the
eight possibilities for the subscripts i, j and k (i.e., 3
subscripts with 2 possibilities for each, that is, each
3
subscript is either a 1 or (2,3) yields 2 =8 combinations)
and thus
'CD
(3,3) = Cov(aij b^j = 4y
The last step of the proof, (i.e. showing Wa N(0,j: )
W3
follows from observing that