ECON2070 INTRODUCTION TO STRATEGIC THINKING
INTRODUCTION TO STRATEGIC THINKING
Hello, dear friend, you can consult us at any time if you have any questions, add WeChat: THEend8_
ECON2070 INTRODUCTION TO STRATEGIC THINKING
SUGGESTED ANSWERS TO TUTORIAL EXERCISE 1
1. (a) No. Univ. Quality is not dominant as it is strictly worse than the other two
choices when Alice is Street Smart. Real World U and Work are not dominant
as both are strictly worse than Univ. Quality when Alice is School Smart.
(b) No. Obviously Univ. Quality and Work cannot be dominated as they are
strictly the best when Alice is School Smart and Street Smart, respectively.
You may be tempted to think that Real World U may be dominated, but it
is not. There is no single choice that is at least as good as Real World U
in every state, so no other choice dominates it. (In fact, the answer to the
next part tells you that Real World U can be an expected utility maximizing
choice.)
(c) We will calculate Alice’s expected utility:
EU [Univ. Quality] = 0.4× 10 + 0.6× 2 = 5.2
EU [Real World U] = 0.4× 7 + 0.6× 5 = 5.8
EU [Work] = 0.4× 0 + 0.6× 8 = 4.8.
Alice will pick Real World U as it is giving the highest expected utility.
(d) The functions we are plotting are
EU [Univ. Quality] = 2p+ 10(1− p)
EU [Real World U] = 5p+ 7(1− p)
EU [Work] = 8p+ 0(1− p).
The plot is given in Figure 1. (Notice that the expected utility functions are
all linear. Thus as long as you know two points on the line, you can plot the
whole function. And you do know two points — the payoff at p = 0 (i.e.,
when Alice is School Smart) and the payoff at p = 1 (i.e., when Alice is Street
Smart). I draw the graph simply by joining the these points (for each choice)
with a straight line.
1
(e) How should we read Figure 1? Notice that every point on the x-axis represents
a probability on Alice being Street Smart. The y-axis gives Alice’s expected
utility. So take a p (a point on the x-axis) and move vertically up. The higher
we go (along this vertical line), the higher is Alice’s expected utility. Hence
the highest line (at this p) correspond to the expected utility maximizing
choice given this probability p.
Now it is clear. Going to Univ. Quality is expected utility maximizing if
p ≤ 1/2. If you would like to solve for this range algebraically, it is obtained
by:
EU [Univ. Quality] ≥ EU [Real World U]
2p+ 10(1− p) ≥ 5p+ 7(1− p)
3(1− p) ≥ 3p
p ≤ 1
2
.
(The graph is telling me that I don’t have to worry about Work in this range,
so I skip that inequality.)
Similarly, Going to Real World U is expected utility maximizing if 1/2 ≤ p ≤
7/10. The 7/10 threshold is obtained by
EU [Real World U] ≥ EU [Work]
5p+ 7(1− p) ≥ 8p
7(1− p) ≥ 3p
p ≤ 7
10
.
(Again, the graph is telling me that I don’t have to worry about Univ. Quality
in this range.)
Finally, Work is expected utility maximizing if p ≥ 7/10.
(f) The plots for Univ. Quality and Work are the same for Alice and Bob. The
only difference is the expression for Real World U, which is
EU [Real World U] = 3p+ 4(1− p)
for Bob. See Figure 2 for the plot for Bob.
From Figure 2 it is immediate that Real World U is never on the top of the
graph. At the range of p where it is better than Work, Univ. Quality is
strictly better than Real World U; at the range of p where it is better than
Univ. Quality, Work is strictly better. In other words, Real World U is never
an expected utility maximizer for Bob.
2
School
Smart
Street
Smart
p
10
UQ
2
7 RWU
5
0
Work
8
1
2
7
10
Figure 1. Expected Utility Plot for Alice
School
Smart
Street
Smart
p
10
UQ
2
4
RWU
3
0
Work
8
5
8
Figure 2. Expected Utility Plot for Bob
(g) Bob’s state-contingent payoff table (after adding the new choice) is given by:
School Smart Street Smart
Univ. Quality 10 2
Real World U 4 3
Work 0 8
Coin 5 5
3
Note: The numbers for Coin are obtained by:
Payoff at School Smart =
1
2
× 10 + 1
2
× 0 = 5
Payoff at Street Smart =
1
2
× 2 + 1
2
× 8 = 5.
(h) Ah! Now Real World U is strongly dominated by Coin! If Bob is School
Smart, Coin gives 5 while Real World U gives 4; If Bob is Street Smart, Coin
gives 6 while Real World U gives 3. In other words, Coin yields a strictly
higher payoff than Real World U in each and every state of nature. Therefore
Real World U is strongly dominated by Coin.
Does this question make you think that there is some relationship between
being dominated (by a randomization between other choices) and not being
an expected utility maximizer for any probability p? Yes your hinge is cor-
rect. There is a theorem saying that a choice is strongly dominated by a
randomization between other choices if and only if it is not an expected util-
ity maximizer for any probability distribution over the states of nature. That
theorem is beyond this course, but talk to me if you are interested.
2. (a) The expected utility of drilling is equal to the expected utility from the com-
pound lottery l defined below.
l′ = (0.5 ◦ 4, 0.5 ◦ 2) : 50% tax or no tax on operating profits
with equal probability
l = (0.6 ◦ l′, 0.4 ◦ 0) : oil with 0.6 probability and no oil otherwise
Note that there are three outcomes: {$0, $2 million, $4 million}. Reduction
of compound lottery property implies that the lottery l is indifferent to the
simple lottery l′′ = {0.4 ◦ 0, 0.3 ◦ 2, 0.3 ◦ 4}. Therefore the expected utility
from l is equal to expected utility from l′′ computed as
EU(l) = 0.4× 0 + 0.3× 2 + 0.3× 4 = 1.8.
Since 1.8 > 1, therefore the expected utility from drilling is higher than the
cost of drilling, hence the company should go ahead and drill.
(b) We re-write the lottery l as a function of p as
l = (p ◦ l′, (1− p) ◦ 0)
This implies that the induced simple lottery l′′ is
l′′ = {(1− p) ◦ 0, 0.5p ◦ 2, 0.5p ◦ 4}
4
Then the expected utility from drilling is
EU(l) = (1− p)× 0 + 0.5p× 2 + 0.5p× 4 = 3p.
Since EU(l) ≥ 1 if and only if p ≥ 1/3, the company should go ahead and
drill if p ≥ 1/3.
3. Assume X is the choice set and % be a complete and transitive relation on X which
satisfies the independence axiom. First, it follows from the independence axiom
that for all x, y, z ∈ X and all λ ∈ [0, 1], x y ⇐⇒ λx+ (1−λ)z λy+ (1−λ)z.
Define the following notation:
xλy := λx+ (1− λ)y for all x, y ∈ X and λ ∈ [0, 1].
Now pick λ, β ∈ [0, 1]. Assume λ > β. Then xλy = [y(1 − β)x]
(
1− λ−β
1−β
)
x =
xλ−β
1−β (xβy). Note that it follows from the independence axiom that if x y and
δ ∈ I\{0, 1}, then x xδy y. Then x xβy and hence xλ−β
1−β (xβy) xβy.
Therefore xλy xβy. Now assume λ ≤ β. If λ = β, then reflexivity of % implies
xλy ∼ xβy. If λ < β, then the above argument implies xβy xλy. Completeness
of % completes the proof. Q.E.D.