Suppose a student must finish a problem set over the course of six days, t =0, 1, 2, 3, 4, 5. Unlike in the example in class, the student needs to work on the problem set for two consecutive days to finish it. If she works on it one day and then does not finish it on the next day, her previous work is completely lost (for example, she forgets what she was doing). If the student starts the problem set in period t = 0, it takes 2 units in period 0 and 6 units in period 1 to finish it. If she starts in period t = 1 or t = 2, it takes 2 units in period t and 9 units in period t + 1 to finish it. And if she starts in period t = 3 or t = 4, it takes 2 units in period t and 15 units in period t + 1 to finish it. That is, the problem set is easy to start, but harder and harder to finish. The student has no access to a commitment technology and decides in each period whether to start (if she has not done so), and whether to finish (if she has started the last period). She is a hyperbolic discounter with β =1/2 and δ = 1. (Note: We can certainly assume she would never start the problem set unless she plans to finish it. In addition to the usual choices in a given period - to not do the PS (here: not start) or to do the PS (and plan to finish it) - she will have an additional choice if she started the PS last period: to finish a started PS.)

(a) Start by constructing the table of discounted utilities as part of your answer (see Lecture 8, which will help you see what her tradeoffs are from each period's perspective).

(b) What does the student do if she is naive? Show your workings. Explain the intuition.

(c) What does the student do if she is sophisticated? Show your workings. Explain the intuition. 

Answer rating: 100% (QA)

a The table of discounted utilities for the student is as follows Period t 0t 1t 2t 3t 4t 5 Utility2... View the full answer