Question

Alpha is deciding whether to invest $1 million in a project. There is a 70% chance that the project will be successful, yielding a return of 20% on investment. However, there is a 30% chance that the project will fail, in which case Alpha will only recover 80% of his investment. 1. What is the expected value of investing in the project? 2. Suppose Alpha evaluates the project in accordance with prospect theory. Specifically, v(x) = ( (x − r) 0.8 if x ≥ r −λ(r − x) 0.8 if r > x, where λ > 1, r = $1 million is the reference point, and x is the amount of money received by Alpha at the end of the project. The corresponding probability weighting functions are such that γ = δ = 0.6. (a) Argue (mathematically) that Alpha is loss averse. (b) What is Alpha’s value of investing in the project if λ = 2? (c) Find the value of λ such that Alpha is indifferent between investing and not investing in the project. Beta’s boss assigns him a task on Monday which must be completed before Wednesday. The task takes a total of 10 hours. If Beta works on the task for xt hours on day t, then he suffers a disutility of x 2 t on day t. Throughout the problem t ∈ {1, 2, 3}, where 1 stands for Monday, 2 for Tuesday, and 3 for Wednesday. Beta’s time preferences are given by exponential discounting with the discount factor of δ = 0.8 per day. 1. What is the present value (as measured on Monday) of Beta’s disutility if he works for 6 hours on Monday and 4 hours on Tuesday? 2. Beta’s problem is to complete the task before Wednesday in such a way to minimize the present value of his distuility (as measured on Monday). Write down the mathematical version of Beta’s problem (that is, minimize some function subject to some constraint) 3. How many hours does Beta choose to work the task on Monday? 4. Now suppose that the boss wants to assign a new task to Beta on Tuesday and would therefore like Beta to have more time on Tuesday. She incentivizes Beta by offering him a reward of r3 = 10x1 on Wednesday if Beta works on Monday for x1 hours on the first task. How many hours does Beta choose to work on Monday?

Answer 1

1. What is the expected value of investing in the project?

Answer: 0.7(1, 200, 000) + 0.3(800, 000) = 1, 080, 000. Give full credit in case a
student calculates expected value of Alpha’s gain/loss, which equals 0.7(200, 000)−
0.3(200, 000) = 80, 000.

2.

(a) Argue (mathematically) that Alpha is loss averse.

Answer: This follows from the fact that λ > 1. Gain of $y is worth y^0.8, which is less than not losing $y because the latter is worth λy^0.8.

(b) What is Alpha’s value of investing in the project if λ= 2?

Answer: Alpha’s value of investing in the project is π_g(0.7)(200, 000)^0.8 − 2π_l(0.3)(200, 000)^0.8 = −1859.03.

(c) Find the value of λ such that Alpha is indifferent between investing and not investing in the project.

Answer : Alpha will be indifferent if π_g(0.7) − λπ_l(0.3) = 0. Therefore, λ =1.66

1. What is the present value (as measured on Monday) of Beta’s disutility if he works for 6 hours on Monday and 4 hours on Tuesday?
Answer: Present value equals 36 + 0.8(16) = 48.8.

2. Beta’s problem is to complete the task before Wednesday in such a way to minimize the present value of his distuility (as measured on Monday). Write down the mathematical version of Beta’s problem (that is, minimize some function subject to
some constraint)

Answer: Beta wants to minimize( x₁)²+0.8(x₂)² subject to the constraint that x₁+x₂ = 10.

3. How many hours does Beta choose to work the task on Monday?

Answer: Substitute x₂ = 10−x₁ in the objective function to get ( x₁)²+0.8(10−x₁)² At the point of minimum, the derivative of the objective function must equal zero.
That is,

2x₁ − 1.6(10 − x₁) = 0 ==> x1 =40/9

4.

Answer: Beta now wants to maximize 0.8²(10x₁) − (x₁)² − 0.8(x₂)² subject to the constraint that x1 + x2 = 10. Substitute x2 = 10 − x1 in the objective function to get 0.8²(10x₁) − (x₁)² − 0.8(10 − x1)². At the point of maximum, the derivative of the
objective function must equal zero. That is

6.4 − 2x₁ + 1.6(10 − x1) = 0 ==>  x1 =56/9