Midterm Examination: Economics 210A
October 2011
The exam has 6 questions. Answer as many as you can. Good luck.
1) A) Must every quasiconcave function must be concave? If so, prove it. If not,
provide a counterexample. (In all answers where you provide a counterexample,
you must show that your example is really a counterexample.)
Answer:
Not every quasiconcave function is concave.
Here is
a counterexample.
Define the function
F
with domain
+
(the
positive real numbers.) such that
F
(
x
) =
x
2
. We show two things:
1) This function is quasiconcave.
1
To see this, note that
F
is a
strictly increasing function on
+
. Therefore if
F
(
y
)
≥
F
(
x
), it must
be that
y
≥
x
and hence for any
t
∈
[0
,
1],
ty
+ (1
−
t
)
x
≥
x
. Since
F
is an increasing function, it follows that
F
(
ty
+ (1
−
t
)
x
)
≥
F
(
x
)
.
Therefore
F
is quasiconcave. 2) The function
F
is not concave. To
see this, note that
F
(2) = 4 and
F
(0) = 0, but
F
(
1
2
2 +
1
2
0) =
F
(1) = 1
≤
1
2
F
(2) +
1
2
F
(0) = 2
.
This cannot be the case if
F
is a concave function.
B) Must every concave function be quasiconcave? If so, prove it. If not, provide
a counterexample.
Answer:
Every concave function is quasiconcave. Proof. If
f
is
concave, its domain is a convex set
A
. For all
x
and
y
in
A
, and
t
between 0 and 1, if
f
(
tx
+ (1
−
t
)
y
)
≥
tf
(
x
) + (1
−
t
)
f
(
y
)
.
(1)
From the Expression
??
it follows that
f
(
tx
+ (1
−
t
)
y
)
≥
f
(
y
) +
t
(
f
(
x
)
−
f
(
y
))
.
(2)
If
f
(
x
)
≥
f
(
y
), then it follows from Expression
??
that
f
(
tx
+ (1
−
t
)
y
)
≥
f
(
y
)
.
(3)
which means that
f
is quasiconcave.
1
By the way,
F
(
x
) =
x
2
defined on the entire real line would not be quasiconcave. Can
you show this?
1
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2) Let
F
and
G
be realvalued concave functions with the same domain,
A
.
Define the function
H
so that for all
x
∈
A
,
H
(
x
) =
F
(
x
) +
G
(
x
).
Is
H
a
concave function? If so, prove it. If not, provide a counterexample.
Answer:
Since
F
and
G
are concave functions with domain
A
, it
must be that if
x
∈
A
and
y
∈
A
, then for all
t
between 0 and 1,
F
(
tx
+ (1
−
t
)
y
)
≥
tF
(
x
) + (1
−
t
)
F
(
y
)
and
G
(
tx
+ (1
−
t
)
y
)
≥
tG
(
x
) + (1
−
t
)
G
(
y
)
Given these two inequalities, we see that
H
(
tx
+ (1
−
t
)
y
)
=
F
(
tx
+ (1
−
t
)
y
) +
G
(
tx
+ (1
−
t
)
y
)
≥
tF
(
x
) + (1
−
t
)
F
(
y
) +
tG
(
x
) + (1
−
t
)
G
(
y
)
=
t
(
F
(
x
) +
G
(
x
)) + (1
−
t
) (
F
(
y
) +
G
(
y
))
=
tH
(
x
) + (1
−
t
)
H
(
y
)
By definition, the left side of Expression
??
equals
H
(
tx
+ (1
−
t
)
y
).
The right side of Expression
??
equals
tF
(
x
)+
tG
(
x
)+(1
−
t
)
F
(
y
)+
(1
−
t
)
G
(
y
) =
tH
(
x
) + (1
−
t
)
HY
). Therefore
H
(
tx
+ (1
−
t
)
y
≥
tH
(
x
) + (1
−
t
)
H
(
y
)
for all
t
between 0 and 1, which means that
H
is a concave function.
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 Fall '09
 Bergstrom
 Economics, Derivative, Convex function, Shephard, Marshallian, Irving Fernandez

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