# 481Assignment1 Solutions

```Assignment 1: Solutions
ECON481, Winter 2021
Question 2
(a) This is agent is rational: choices are governed by the utility function u(x) = −x.
(b) This agent is not rational. Consider B = {1, 2, 3} and B 0 = {2, 3, 4}. Then c(B) = {2}
and c(B 0 ) = {3}. Since 2 and 3 each belong to B and B 0 , this is a violation of WARP.
(c) This agent is rational: choices are governed by the utility function

x
if x &lt; 1000
u(x) =
1000 if x ≥ 1000
(d) This agent is not rational. Once again, let B = {1, 2, 3} and B 0 = {2, 3, 4}. Then
Avg(B) = 2 and Avg(B 0 ) = 3, so c(B) = {2, 3} and c(B 0 ) = {3, 4}. Since 2 and 3 each
belong to B and B 0 , this is a violation of WARP.
Question 3
(a) Lots of possible answers. For example, suppose customers at local pubs choose x (pizza)
twice as often as they choose y (burgers), and that they are five times more likely to
choose y than z (salad). Pub A only serves pizza and burgers, while pub B serves pizza,
To represent choices as a Luce rule, we just need to choose Luce values vx , vy , vz such
that vx = 2vy and vy = 5vz . So, vx = 10, vy = 5, and vz = 1 will work. Using these
10
5
5
1
values, we get ρA (x) = 15
, ρA (y) = 15
, ρB (x) = 10
, ρB (y) = 16
, ρB (z) = 16
. Notice that
16
ρA (x)
ρB (x)
= 2 = ρB (y) , consistent with the Luce axiom (and our story that customers are
ρA (y)
twice as likely to choose pizza over burgers at any pub).
(b) A good approach is to give a story about attraction or compromise effects, as the given
inequality is the one from that part of the lecture slides.
1
A common mistake in these types of exercises is to confuse probability ratios with probability levels. The Luce IIA axiom does not require ρA (x) = ρB (x) and ρA (y) = ρB (y); it
only requires the ratio r = ρρAA(x)
to be the same in menus A and B. To see the difference,
(y)
suppose that in menu A we have ρA (x) = 2/3 and ρA (y) = 1/3. Then the ratio is r = 2.
Now suppose that in menu B, we have ρB (x) = 0.02, ρB (y) = 0.01, and ρB (z) = 0.97.
We still get the same ratio of 2 even though x and y are chosen with tiny probabilities,
so this is consistent with the Luce axiom.
(c) First, note that ρA (x) + ρA (y) = 1. Therefore,
ρB (x)
ρB (x)+ρB (y)
&gt; ρA (x) implies
ρB (x)
ρB (x) + ρB (y)
ρB (y)
⇒ ρA (y) &gt;
ρB (x) + ρB (y)
ρB (x) + ρB (y)
1
⇒
&gt;
ρB (y)
ρA (y)
1 − ρA (x) &gt; 1 −
Now observe that
ρB (x)
ρB (x) + ρB (y)
ρB (x)
=
&middot;
ρB (y)
ρB (x) + ρB (y)
ρB (y)
1
ρB (x)
ρB (x) + ρB (y)
1
&gt; ρA (x) &middot;
since
&gt; ρA (x) and
&gt;
ρA (y)
ρB (x) + ρB (y)
ρB (y)
ρA (y)
ρA (x)
=
ρA (y)
and this contradicts the Luce IIA axiom.
Question 4
(a) In menu A = {x, y}, where x = (1, 2) and y = (2, 1) we have ρA (x) = 0.25 and
ρA (y) = 0.75 because the agent chooses to maximize flavor (the first coordinate) with
probability 0.75 and nutrition (second coordinate) with probability 0.25. Since y and x
are the unique maximizers of the first and second coordinates, respectively, we end up
with the given probabilities.
In menu B, outcome z = (2, 2) dominates outcome w = (1, 1) in both coordinates, so
ρB (z) = 1 and ρB (w) = 0.
In menu C, x is chosen with probability 0.5 &middot; 0.25 = 0.125 because z is chosen whenever
the agent maximizes flavor and x and z are tied when the agent maximizes nutrition.
2
So, ρC (x) = 0.125 and ρC (z) = 0.875.
In menu D, w is never chosen because it is dominated in both dimensions by z. Option
z is chosen with probability 0.5 because it maximizes both flavor and nutrition but is
tied with y when maximizing flavor and tied with x when maximizing nutrition. Thus,
ρD (w) = 0, ρD (x) = 0.5 &middot; 0.25 = 0.125, ρD (y) = 0.5 &middot; 0.75 = 0.375, and ρD (z) = 0.5.
(b) To be represented by a Luce rule, the choice frequencies must satisfy the Luce IIA
axiom. Using the results from A, we have ρρDD (x)
= 0.125
= 41 and ρρCC (x)
= 0.125
= 71 . This
(z)
0.5
(z)
0.875
contradicts the axiom, so choices can’t be represented by a Luce rule.
3
```