
Sec. 4.2 Optimization Derivation 69
Cross multiplying gives
t
m
α
0
+ t
2
m
α
1
− t
m
r
m
= β
0
+ β
2
t
2
m
.
Simplifying and solving for t
m
now g ives
t
m
=
β
0
α
0
− r
m
,
which is the identical result obtained in Theorem 4.1 (b).
We can perform a similar analysis beginning with Theorem 4.2 (b).
Set t
m
= t
1
and x
m
= x
1
in Theorem 4.2 (b). Because x
m
is optimal
for (2.3), µ
m
= α
0
+ t
m
α
1
and σ
2
m
= β
0
+ t
2
m
β
2
. Then Theorem 4.2(b)
implies
r
m
= µ
′
x
m
−
x
′
m
Σx
m
t
m
(4.14)
= α
0
+ t
m
α
1
−
β
0
+ t
2
m
β
2
t
m
.
Multiplying through by t
m
, using α
1
= β
2
and simplifying gives
t
m
=
β
0
α
0
− r
m
and µ
m
= α
0
+
β
0
α
0
− r
m
α
1
, (4.15)
in agreement with Theorem 4.1(b).
There are other somewhat simpler methods to derive the results of
(4.15) (see ...