TA: Yankun Wang
Economics 617
Problem Set 9 Solution Key
1. [On Slater’s Condition]
Let
X
=
R
2
+
,
and
f, g
1
and
g
2
be functions from
X
to
R
,
de
fi
ned by:
f
(
x
1
, x
2
) = 2
x
1
+
x
2
g
1
(
x
1
, x
2
) = 1
−
(
x
1
+
x
2
)
g
2
(
x
1
, x
2
) = (
x
1
+
x
2
)
−
1
⎫
⎬
⎭
Consider the following optimization problem:
Max
f
(
x
1
, x
2
)
subject to
g
j
(
x
1
, x
2
)
≥
0
for
j
∈
{
1
,
2
}
and
x
∈
X
⎫
⎬
⎭
(
P
)
(a) Show that
ˆ
x
= (1
,
0)
solves problem
(
P
)
.
Solution: In order to have
g
j
(
x
1
, x
2
)
≥
0
for
j
= 1
,
2
at the same time, we
must have
(
x
1
+
x
2
)
−
1 = 0
.
Thus
x
1
= 1
−
x
2
;
f
(
x
1
, x
2
) = 2
x
1
+
x
2
= 2
−
x
2
.
To maximize
f
, we want
x
2
to be as small as possible, and the minimum of
x
2
is
0
.
When
x
2
= 0
, x
1
= 1
−
x
2
= 1
.
Remark: The problem can be view at a utility maximization problem with
linear utility function.
Think about the economic explanation as well as the
graphical illustration for this problem. Just notice that neither the graph nor
the intuition serves as the formal proof.
(b) Note that
X
is a convex set in
R
2
,
and that
f, g
1
and
g
2
are concave
functions from
X
to
R
.
Show that Slater’s Condition is not satis
fi
ed.
Solution: If we know what the slater’s condition is, this is kind of obvious:
we can never have a point
x
∈
X
such that
1
−
(
x
1
+
x
2
)
>
0
as well as
(
x
1
+
x
2
)
−
1
>
0
, i.e., x
1
+
x
2
<
1
and
x
1
+
x
2
>
1
.
Remark: Make sure you know how to use the de
fi
nition to prove that linear
functions are both concave and convex.
(c) Does there exist
ˆ
λ
∈
R
2
+
such that
(ˆ
x,
ˆ
λ
)
is a saddle point? Explain.
Solution: We want to know if we can
fi
nd
ˆ
λ
∈
R
2
+
such that:
2
x
1
+
x
2
+ (
ˆ
λ
1
−
ˆ
λ
2
)(1
−
(
x
1
+
x
2
))
≤
2
≤
2
,
for all
x
∈
X, X
=
R
2
+
.
where the de
fi
nition of saddle point has been applied.
If we let
ˆ
λ
1
−
ˆ
λ
2
= 2
,
the above inequality becomes:
2
−
x
2
≤
2
,
and this is
always true for
x
2
≥
0
.
Thus by de
fi
nition we have
fi
nd that
(ˆ
x,
ˆ
λ
)
is a saddle point as long as:
ˆ
x
= (1
,
0)
,
ˆ
λ
1
−
ˆ
λ
2
= 2
,
and
ˆ
λ
∈
R
2
+
.
Remark: The Lecture Notes has an example (right after Theorem 45 ) in
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
which Slater’s Condition is violated, and we can’t
fi
nd a saddle point. Try to
understand the di
ff
erence between that example and this problem.
2. [Application of the theorem: “GM
⇒
SP”]
Let
a
∈
(0
,
1)
and
b
∈
(0
,
1)
be given parameters. Let
A
be the set de
fi
ned
by:
A
=
{
(
x, y, z
)
∈
R
3
+
:
y
≤
x, z
≥
bx
and
[
y
+
a
(
z
−
bx
)]
≤
1
}
Consider the following constrained optimization problem:
Maximize
y
subject to
(
x, y, z
)
∈
A
and
z
≥
x
⎫
⎬
⎭
(
B
)
Problem
(
B
)
has a solution (you do not have to show this); denote an arbitrary
solution of problem
(
B
)
by
(¯
x,
¯
y,
¯
z
)
.
Show that there is
¯
p >
0
such that:
y
+ ¯
pz
−
¯
px
≤
¯
y
for all
(
x, y, z
)
∈
A
Proof:
The title of the problem serves as a hint: use Theorem 45 in the Lecture
Notes. It is given that
(¯
x,
¯
y,
¯
z
)
is a constrained global maximum. Let
g
(
x, y, z
) =
z
−
x, f
(
x, y, z
) =
y, X
=
A,
we need to check
fi
rst that
A
is convex,
f, g
are
concave. All of these can be done by checking the de
fi
nitions of convex sets and
concave functions. Is the Slater’s Condition satis
fi
ed? Yes, since
(0
,
0
,
1)
in an
element in
A,
and
g
(0
,
0
,
1) = 1
>
0
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 STAFF
 Economics, Derivative, Optimization, Convex set, Convex function

Click to edit the document details