1. No category

Solutions for HW 1, Econ 902, FALL 2014
1. Reed Chapter 1.
• Reed 1.1.8. Prove that if x and y are real numbers, then 2xy ≤ x2 + y 2 .
Since x and y are real numbers, we know that the quantity (x − y)(x − y) ≥ 0. Thus,
(x − y)(x − y) ≥ 0
x2 − 2xy + y 2 ≥ 0
x2 + y 2 ≥ 2xy
Q.E.D.
• Reed 1.4.1: Many examples are possible.
• Reed 1.4.3: Suppose that a, b, c, and d are postive numbers such that a/b < c/d. Prove that:
a
a+c
c
<
<
b
b+d
d
Proof: The proof requires two steps proving each inequality separately. First, ad < bc since
they are all positive real numbers. Then,
ad + ab < bc + ab
a(b + d) < b(a + c)
a(b + d)
<a+c
b
a+c
a
<
b
b+d
Similarly,
ad + cd < bc + cd
d(a + c) < c(b + d)
c(b + d)
a+c<
d
a+c
c
<
b+d
d
Q.E.D.
2. Dadkhah: Ch. 2: 11, 19
• E. 2.11.
1
1
1
+ (1+r)
– The present value of the lottery ticket is given by 200, 000 1 + 1+r
.
2 + ··· +
20
(1+r)
For r = 0.12, the NPV is 1,693,900; for r = 0.15, NPV = 1,451,900; and for r = 0.20,
NPV = 1,173,900.
P∞
1
1
– The PV of a one dollar a year payment in perpetuity is P V = t=0 (1+r)
t = 1 + r.
• E. 2.19. Proposition if n is a positive integer, then
12 + 22 + 32 + · · · + n2 =
n (n + 1) (2n + 1)
.
6
The above statement if true for the case when n = 1: 12 =
the function is true for some k. That is,
12 + 22 + 32 + · · · + k 2 =
1
(1+1)(2+1)
.
6
k (k + 1) (2k + 1)
.
6
Now, suppose that
Then, by the induction hypothesis,
12 + 22 + 32 + · · · + k 2 + (k + 1)
2
2
k(k+1)(2k+1)
+ (k + 1)
6
(k+1)[k(2k+1)+6(k+1)]
6
(k+1)(2k+2)(k+3)
6
(k+1)((k+1)+1)(2(k+1)+1)
6
=
=
=
=
We have shown that if the function is true for k, then k + 1 is also true. Therefore, by
induction, the function is true for all n.
3. Dadkhah: Ch. 6: 2, 6 (i, ii, iii, iv, v, vii, viii), 7
• E. 6.2.
R (Q + ∆Q) − C (Q + ∆Q)
Π + ∆Π
=
therefore,
∆Π
=
∆Q
=
R(Q+∆Q)R(Q)
∆Q
∆R
∆C
∆Q − ∆Q
−
C(Q+∆Q)C(Q)
∆Q
The marginal profit is the difference between the marginal revenue and the marginal cost.
The marginal profit is equal to zero when the marginal revenue equals the marginal cost.
• E. 6.6.
i.
ii.
iii.
iv.
v.
vii.
viii.
∂y
∂x
∂y
∂x
∂y
∂x
∂y
∂x
∂y
∂x
∂y
∂x
∂y
∂x
= − x23
−1/2
= (2x)
= x−3/2 x + 21
= 23 x−1/2 + 13 x−2/3 − x−2
1−x2
= (1+x
2 )2
2
+2
= x33x
+2x−5
= e−x (1 − x)
• E. 6.7. (i) this function is monotonically increasing for x > 0; (ii) this function is monotonically increasing; (iii) this function is monotonically decreasing for x > 0; and (iv) is
monotonically decreasing.
4. Dadkhah: Ch. 7: 1, 2, 4
• E. 7.1. See attached.
• E. 7.2. See attached.
• E. 7.4.
– M PK = α if αK < βL, 0 otherwise ; M PL = β if βL < αK, 0 otherwise .
– M PK = αAK α−1 Lβ > 0; M PL = βAK α Lβ−1 > 0.
– M PK = αK ρ−1 (αK ρ + βLρ )
1/ρ−1
> 0; M PL = βLρ−1 (αK ρ + βLρ )
5. Solution:
(a)
1 + β + β2 + β3
1−β
1 + β + β2 + β3
1−β
1−β+β(1−β)+β 2 (1−β)+β 3 (1−β)
1−β
1−β+β−β 2 +β 2 −β 3 +β 3 −β 4
1−β
1−β 4
1−β
2
=
1−β 4
1−β
=
=
=
=
1−β 4
1−β
1/ρ−1
> 0.
(b) Proposition if n is a positive integer and 0 < β < 1, then
1 + β + β2 + · · · + βn =
1 − β n+1
1−β
Proof. If Q (n) is the above statement, then Q (1) is true. The left side would simply be
2
(1−β)(1+β)
1 + β and the right side would be the same: 1−β
= 1 + β. Now, suppose that
1−β =
1−β
Q (k) is true for some k. That is,
1 + β + β2 + · · · + βk =
1 − β k+1
1−β
Then, by this induction hypothesis,
1 + β + β 2 + · · · + β k + β k+1
=
=
=
=
=
1−β k+1
k+1
1−β + β
k+1
1−β
k+1 1−β
+
β
1−β
1−β
1−β k+1 +β k+1 (1−β)
1−β
1−β k+1 +β k+1 −β (k+1)+1
1−β
1−β (k+1)+1
1−β
We have shown that if Q (k) is true, then Q (k + 1) is true. Therefore, by induction, Q (n)
is true for all n.
6. The following are common utility function over consumption (represent by c in each function
below) that you will see in your courses during the first year. Find dU/dc and d2 U/dc2 for
each and draw graphs of consumption against utility, consumption against marginal utility,
and consumption against the second derivative of utility w.r.t. consumption. Use Excel or some
other program of your choice to construct the graphs - no hand drawn graphs, please. For the
parameters simply choose one that makes sense to you. You should experiment with how the
parameters change the functions.
Solution: Notice that a always positive first derivative indicates that greater levels of consumption lead to higher utility levels. A always negative second derivative implies the marginal utility
is decreasing as consumption rises.
c1−σ −1
1−σ
(a) U = Ac
(b) U = cα
(c) U = ln c
(d) U = −e−αc
dU
dc
A
αcα−1
1
c
αe−αc
c−σ
d2 U
dc2
0
−α (1 − α) cα−2
− c12
−α2 e−αc
−σc−σ−1
Graphs are in the a,b,c,d,e order as givn in the exercise.
3
(e) U =
25
20
15
10
5
0
1
2
3
4
5
6
7
8
9
10
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
0
2
4
6
8
10
2.5
2
1.5
1
0.5
0
−0.5
−1
−1.5
−2
−2.5
0
2
4
6
8
10
0
−0.1
−0.2
−0.3
−0.4
−0.5
−0.6
−0.7
−0.8
−0.9
−1
0
2
4
6
8
10
4
3
2
1
0
−1
−2
−3
0
2
4
6
8
10
7. Find dU/dc and d2 U/dc2 and draw graphs of consumption against utility, consumption against
marginal utility, and consumption against the second derivative of utility w.r.t. consumption for
1−σ
U=
(c − k)
−1
, where k > 0
1−σ
Provide an economic interpretation of k. Defend your reasoning.
Solution:
U
=
(c−k)1−σ −1
,
1−σ
dU
dc
=
(c − k)
d2 U
dc2
=
−σ (c − k)
where k > 0
−σ
−σ−1
One possible interpretation of k is that it denotes the subsitence level of per capita consumption.
This subsistence consumption could be thought of as a consumption-based poverty, or survival,
line.
The following graphs illustrate consumption against marginal utility and consumption against the
second derivative of the marginal utility respectively.
The graph of consumption versus utility is same as part e of problem 6, shifted k units to the right.
9
5
4
3
2
1
0
−1
−2
−3
0
2
4
6
8
10
0
−5
−10
−15
−20
−25
−30
0
2
4
6
8
10
8. Latex: From the Mas-Collel handout: 1.B.1, 1.B.2. These two problems must be typed up in
Latex.
Solutions:
(a) 1.B.1 Since y z implies y z, the transitivity implies that x z. Suppose that z x.
Since y z, the transitivity then implies that y x. But this contradicts x y. Thus we
cannot have z x. Hence, x z.
(b) 1.B.2 By the completeness, x x for every x ∈ X. Hence there is no x ∈ X such that
x x. Suppose that x y and y z, then x y z. By (iii) of Proposition 1.B.1, which
was proved in the above exercise, we have x z. Hence is transitive. Property (i) is now
proved.
As for (ii), since x x for every x ∈ X, x ∼ x for every x ∈ X as well. Thus ∼ is reflexive.
Suppose that x ∼ y and y ∼ z. Then x y, y z, y x, and z y. By the transitivity, this
implies that x z and z x. Thus x ∼ z. Hence ∼ is transitive. Suppose that x ∼ y. Then
x y and y x. Thus y x and x y. Hence y ∼ x. Thus ∼ is symmetric. Property (ii)
is now proved.
12