Paper, Order, or Assignment Requirements
Take Home Assignment for Eco460
Due at 6:10pm on November 25, 2014
No late work will be accepted
Note: Please hand in your assignment in class.
(40 points) Consider a two-period extension of the model in Chapter 4 of
the Lecture Notes. The two periods are indexed by t = 0 and 1. There is no
uncertainty in period 0 and let c(0) be the household’s consumption in period
There is uncertainly in the second period and the household’s consumption
plan is denoted by c(1) = (c1, c2, …, cN ), where N is the number of states. The
household’s preferences for consumption in two periods are summarized by the
following utility function:
U(c(0), c(1)) = u(c(0)) + β
Here u(c) = 1 −
1−γ − 1) and β is the time discount factor, 0 < β < 1. That is, household value the utility from future consumption at a discount. The household is endowed with a certain income y(0) in period 0, which the household can be used for (1) consumption in period 0, (2) invest in a risk-free bond that yields a risk-free return of e rf in period 1 or (3) invest in a risky asset that yields a risky return of e er = (e r1 , …, erN ) in period 1. So the household’s budget constraint in period 0 is: c(0) + b + s = y0 where b is the amount invested in the risk-free bond and s is the amount invested in the risky asset (stock). There is no endowment income in period 1. Since this is a two-period economy, there would be no investment in period 1 either. So the household’s budget constraint in period 1 is ci = berf + seri , i = 1, …, N (10 points) The household’s optimal investment problem is to choose c(0), b and s to maximize U(c(0), c(1)) subject to the budget constraints in both periods. Use the Lagrangian method to derive the first-order conditions for optimal choices of c(0), b and s. [Hint: you can substitute the budget constraints for ci into the expected utility function.] (10 points) Let gi = ln [ci/c(0)] be the consumption growth rate in state i, and g = (g1, …, gN ). Use the first-order conditions to prove that 1 = βE e rf −γg , and 1 = βE e r˜−γg . 1
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