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 < 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, burgers, and salads. 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) > ρA (x) implies ρB (x) ρB (x) + ρB (y) ρB (y) ⇒ ρA (y) > ρB (x) + ρB (y) ρB (x) + ρB (y) 1 ⇒ > ρB (y) ρA (y) 1 − ρA (x) > 1 − Now observe that ρB (x) ρB (x) + ρB (y) ρB (x) = · ρB (y) ρB (x) + ρB (y) ρB (y) 1 ρB (x) ρB (x) + ρB (y) 1 > ρA (x) · since > ρA (x) and > 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 · 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 · 0.25 = 0.125, ρD (y) = 0.5 · 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