2
LECTURE #26: THU, NOV 29, 2007
PRINT DATE: AUGUST 21, 2007
(COMPSTAT3)
Proof again is a trivial exercise in differentiation: since
f
(
α
, g
(
α
))
≡
f
(¯
α
,
¯
x
)
we can take the partial derivative of
f
w.r.t.
α
j
:
d
f
(
α
, g
(
α
))
d
α
j
= 0
=
∂f
(
α
, g
(
α
))
∂α
j
+
∂f
(
α
, g
(
α
))
∂x
∂g
(
α
))
∂α
j
rearranging yields:
∂g
(
α
))
∂α
j
=

∂f
(
α
,g
(
α
))
∂α
j
∂f
(
α
,g
(
α
))
∂x
An important feature to note is that the domain of
f
has one more dimension than the domain of
g
. Reason is that it is the
graph
of
g
, i.e.,
{
α
, g
(
α
) :
α
∈
etc
}
that recovers the level set. That is,
the graph of a real valued function is a set that lives in a Euclidean space one dimension higher
than the dimension of the domain of the function. In this case, a point (
α
, g
(
α
)) is a point on the
level set of
f
.
Example: argued that the solution to
any
economic system can be represented as the level set of
some function. Here’s a simple economic model:
S
=
S
(
t, p
),
D
=
D
(
y, p
),
S
=
D
, where
p
denotes
market price,
t
denotes a tax rate paid by the producer and
y
denotes consumer income level. The
solution to this model can be represented as the level set
f
(
α
, x
)
≡
0, where
f
=
S

D
,
α
=
t, y
and
x
=
p
. The level set of
f
corresponding to zero is the set of all (
price , tax , income
) triples
such that the price clears the market for the corresponding values of the exogenous variables.