ECON30020 Mathematical Economics Tutorial 8. Monopoly and Matrices
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ECON30020 Mathematical Economics
Tutorial 8. Monopoly and Matrices
Question 1. Monopoly Pricing with Non-Concave Revenue
Consider a monopoly that faces the inverse demand function
P (Q) = (|4(1) ↓ Q,
| 9 (1 ↓ Q)
if
if
Q ↓ [0, 1/4)
Q ↓ [1/4, 1]
and has the cost function C(Q) = Q2 /10.
(a) Assuming first that the monopoly is restricted to setting a uniform (or market clearing)
price, show that there are two solutions to the first-order condition R\ (Q) ↓ C\ (Q) = 0. Denote
these by QL and QH , say what they are, explain why each characterises a local maximum, and determine which one corresponds to the global maximum.
Solution. Denoting by R1(Q) and R2(Q) the revenue to the left and right of the Q = 1/4, we have (as noted in class)
R1(Q) = ╱ 1 ↓ Q、Q =e MR1(Q) = R(Q) = 1 ↓ Q, Q < 1/4 R2(Q) = (1 ↓ Q)Q =e MR2(Q) = R(Q) = ↓ Q, Q > 1/4
To maximise profit, we solve the first order condition (FOC): MR(Q) = MC(Q). Marginal cost is given by MC(Q) = C\ (Q) = Q/5.
Solving MR1(QL ) = MC(QL ), Q < 1/4:
1 ↓ QL = QL =e QL ╱ + 、 = 1 =e QL = = 类 0.1807.
Solving MR2(QH ) = MC(QH ), Q > 1/4:
↓ QH = QH =e QH ╱ + 、 = =e QH = = 类 0.4082.
Therefore, there are two solutions to FOC: QL = 类 0.1807 and QH = 类 0.4082.
Both characterise local maxima because MR(Q) < 0, MR(Q) < 0, and MC\ (Q) = C\\ (Q) > 0, so the second-order condition for a maximum is satisfied:
π \\ (QL ) = R(QL ) ↓ C\\ (QL ) = MR(QL ) ↓ MC\ (QL ) = ↓ 16/3 ↓ 1/5 < 0 π \\ (QH ) = R(QH ) ↓ C\\ (QH ) = MR(QH ) ↓ MC\ (QH ) = ↓8/9 ↓ 1/5 < 0.
To find the global maximum, we need to compare profits at both critical points:
π(QL ) = R1(QL ) ↓ C(QL ) = ╱ 1 ↓ QL 、QL ↓ = 类 0.0904 π(QH ) = R2(QH ) ↓ C(QH ) = ╱ ↓ QH 、QH ↓ = 类 0.0907,
so, by a whisker, QH is the global maximiser. The optimal profit is π(QH ) = 0.0907. The graph below illustrates the two solutions to the first-order conditions.
From the profit function we can see that both outputs give us almost the same profit:
(b) Now allow the monopoly to use a two-price mechanism as derived in lectures. Show that the monopoly will indeed achieve higher profit than that under market clearing pricing.
Solution.
First, we need to concavify the revenue. That was done in lectures. We need to find Q1 and Q2 such that
MR1(Q1 ) = MR2(Q2 ) =
The solution was given in lectures:
Q1 = 类 0.17603 and
R2(Q2 ) ↓ R1(Q1 )
Q2 ↓ Q1 .
Q2 = 类 0.43118.
To concavify the revenue function, we need to replace the part of the revenue between outputs Q1 and Q2 with a straight line tangent to it, as in the graph below:
R(Q) = R1(Q1 ) + MR1(Q1 )(Q ↓ Q1 ) = 0.09340 + 0.06117(Q ↓ 0.17603), Q1 < Q < Q2 .
Therefore, the new (concavified) revenue R(Q) is given by:
╱ 1 ↓ Q、Q, 0 ≤ Q ≤ 0.176
R(Q) = ( 0.0934 + 0.06117 (Q ↓ 0.176) , 0.176 < Q < 0.431
ì(╱ 9(4) ↓ Q、Q, 0.431 ≤ Q < 1.
The new marginal revenue is given by
1 ↓ Q,
MR(Q) = ( 0.06117,
ì
ì
ì(9(4) ↓ Q,
0 ≤ Q ≤ 0.176
0.176 < Q < 0.431
0.431 ≤ Q < 1.
The graph below illustrates the concavification of the revenue function R (the new part is in red) and the new MR function (old MR is in blue, new MR is on top in red).
Revenues R(Q) and R(Q)
R
0.11
0.10
0.09
0.08
Marginal revenues MR(Q) and MR(Q)
MR
0.4
0.2
Second, we need to find the optimal output.
Notice that 0.176 = Q1 < QL < QH < Q2 = 0.431. Because MC is increasing, this implies that MC(Q) intersects MR(Q) at the output level Q where the marginal revenue function is horizontal, MR = 0.06117, see the graph below.
MR
0.3
0.2
0.1
Solving the first order condition:
MR(Q) = MC(Q) =e MC(Q) = 0.06117 =e Q = 0.06117 =e Q* = 0.3058.
The corresponding profit is π(Q* ) = R(Q* ) ↓ C(Q* ) = 0.0920, which is indeed larger than the
maximum profit under market clearing pricing, π(QH ) = 0.0907.
Please note that your answers will crucially depend on the precision you kept during all calcu- lations!
The last step is to determine the two prices, remember, it is the two-price mechanism! So far, we found Q1 = 0.176, Q2 = 0.431 and the optimal output Q* = 0.3058. This is the output the monopoly will produce and sell.
The monopoly will set two prices, p1 and p2 , with p1 > p2 . Each price is targeted at a specific group of consumers.
The consumers ”located between Q = 0 and Q1 ”, i.e. the consumers who value the good at least at P (Q1 ), will be buying the good at high price p1 (for sure, with probability 1), so the monopoly will sell the amount Q1 at price p1 .
The consumers ”located between Q1 and Q2 ”, i.e. the consumers with values of the good be- tween P (Q2 ) and P (Q1 ), will be trying to buy the good at the lower price p2 , and will succeed
with probability α . The monopoly will sell amount Q* ↓ Q1 at price p2 .
The price p2 is given by the inverse demand at Q2 = 0.431:
p2 = P (Q2 ) = (1 ↓ Q2 ) 类 0.253.
Thus, those consumers who will be trying to buy at price p2 , will be able to buy with probability
α = quantity available at price p2 Q* ↓ Q1 0.3058 ↓ 0.1760
quantity demanded at price p2 Q2 ↓ Q1 0.4312 ↓ 0.1760
To find price p1 , we need to consider a person ”located at Q1 ”, i.e. the person who has value
v (willingness to pay) = P (Q1 ) = P (0.176) = 1 ↓ Q1 = 1 ↓ * 0.176 = 0.531. This person
must be indifferent between buying at price p1 for sure and buying at price p2 = 0.253 with probability α = 0.51 (lecture 6, slide 39); that allows us to calculate price p1 :
p1 = (1 ↓ α)P (Q1 ) + αP (Q2 ) = 0.491 * 0.531 + 0.509 * 0.253 = 0.39.
To summarise:
The monopoly produces Q* = 0.305 and offers two prices: p1 = 0.39 and p2 = 0.25.
❼ Some consumers will prefer to pay p1 = 0.39 to get the good for sure. These are the
consumers with value (willingness to pay) for the good satisfying v 女 0.53. The monopoly sells Q1 = 0.176 to them.
❼ Some consumers will prefer to take a risk and attempt to buy at p2 = 0.25. They will
be rationed and have probability α = 0.51 of getting the good. These are the consumers
with values satisfying 0.25 ≤ v < 0.53. The monopoly sells Q* ↓ Q1 = 0.129 to them.
Question 2. Evolution of Income Distributions
Consider a society with 18 million individuals who are in one of three income categories (low,
middle, high). At date 0, the distribution of individuals is described by the row vector x0 = (10, 6, 2),
meaning that initially 10 million individuals have low incomes, 6 middle incomes and 2 million high incomes.
a) Assume first that the transition probability from one period to the next is given by the matrix
╱ 0.80 P = 0.15
(
0.1
0.70
0.05
0.1 、
0.15 ,
|
where pij is the probability that an individual who in a given period is in category i will be in
category j in the next period. Compute next period’s distribution x1 = x0 少 P and plot it on
a graph. If you have a computer program that allows you to compute it easily, derive then the
distributions for the subsequent periods, using the facts that xt = x0 少 Pt . Discuss what you
observe.
Solution. Computing yields (rounding to two decimal places)
x0 少 P = (9, 5.3, 3.7)
x0 少 P2 = (8.18, 4.795, 5.025) 类 (8.18, 4.80, 5.03)
x0 少 P3 = (7.5145, 4.42575, 6.05975) 类 (7.51, 4.43, 6.06)
This is plotted in the figure:
Over time, the distribution moves to the right in the sense that the lower and medium income categories have fewer people in period t+1 than in period t. In the limit, when t goes to infinity, the distribution converges to the stationary distribution (4 .91, 3.27, 9.82), see the graph above.
b) Consider now a different pattern of income dynamics, one in which individuals are much less likely to stay in the middle income category. Specifically, let
╱ 0.80 0.1 0.1 、
= 0.45 0.10 0.45
( 0.05 0.05 0.90 |
be the new transition probability matrix. (Notice that first and last rows are the same as for P but the probability of staying in the middle income category is decreased from 0.7 to 0.1, with the 0.6 probability mass added equally to the extremes, that is, 21 = p21 + 0.3 and pˆ23 = p23+0.3.) Do the same as in part a) and then discuss what you observe and the differences to a).
Solution. Computing yields (rounding to two decimal places)
x0 少 = (10.8, 1.7, 5.5)
x0 少 2 = (9.68, 1.525, 6.795) 类 (9.68, 1.53, 6.80)
x0 少 3 = (8.77, 1.46025, 7.76975) 类 (8.77, 1.46, 7.77)
This is plotted in the figure:
There are “centrifugal” forces at work for the middle income category, which becomes smaller over time. Initially, some of that “flight” goes to the low income category, but that then is reversed and mass seems to shift from the low income category to the high income category as well. In the limit, when t goes to infinity, the distribution converges to the stationary distribution (5.59, 1.24, 11.2), see the graph above. Compared to (a), more people end up in the low and high income categories, and fewer in the middle income one.
Question 3. Matrix multiplication
Obtain for the row vector a and the column vector b, the products ab and ba:
┌↓1┐
| 1 |
Solution. We have
┌↓1┐
ab = ì 1 2 0í 0 = 1( ↓ 1) + 2(0) + 0(1) = ↓ 1
1
and
ba = 0 ì 1 | | |
2 0í = 0 | |
↓2 0 2 |
0┐ 0 | |
Notice that the two are not equal, particularly, the first product is a scalar (number), but the second one is a 3 by 3 matrix. This result holds generally.
Question 4. More matrix multiplication
Perform the following matrix multiplications to obtain AB and BA where possible:
A = ┐ , B =
Solution. Multiplying A and B gives us
AB = ┐ = ┌0(4) 2(3)┐
Multiplying B and A gives us a completely different matrix, even demension is different!
BA = ┐ =
Do not treat matrices as mere numbers! They get offended.
Question 5: Even more matrix multiplication
Let
A = ┌3(1) 6(2)┐ , B = ┌2(3) 8┐ , C = ┌ ┐ .
Verify that AB = AC even though B C .
Solution. To verify the statement in the question, simply calculate AB and AC:
AB = ┌3(1) 6(2)┐ ┌2(3) 8┐ = AC = ┌3(1) 6(2)┐ ┌ ┐ =
So in this particular case, AB = AC .
┐ = ┌ 21(7) 6(2)┐
62 12(4)┐ = ┌ 21(7) 6(2)┐
2022-06-20