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BPP Field Exam 2010 Answers (view source)
Revision as of 20:22, 29 March 2011
, 20:22, 29 March 2011→Question B1.3
The CEO can design a scheme that exploits the risk aversion of the agents using chance. The contract would work like this: If all employees exert work, each worker will get an equal share <math>1/N</math> of the effort. However, if any single worker does NOT work, then the payoffs will be determined by a lottery in which each employee gets a <math>\frac{1}{N}</math> chance of getting 100% of the combined output and a <math>1-\frac{1}{N}</math> chance of getting zero. I will now show that irrespective of what other players are doing, the dominant strategy is to work.
Note that CARA utility is <math>u(c)=1-e^{-\rho c}</math>. An employee i's utility from working (if all others work) is <math>A=1-\exp[-\rho(\frac{1}{N}\sum_{i\neq j} z(e_{j})+\frac{1}{N}z(e_{i})-1)]</math>.
If employee works but others aren't, the lottery is triggered and employee i's utility is <math>B=\frac{1}{N}(1-\exp[-\rho(\sum_{i\neq j} z(e_{j})+z(e_{i})-1)])</math>.
First, note that <math>B>C</math>. Within the algebra, note that the utilities are identical except for the exponents. Note that <math>\sum_{i\neq j} z(e_{j})+z-1>\sum_{i\neq j} z(e_{j})</math> because <math>z-1>0</math>. As such, we know that the inequality always holds. As for the intuition: Note that the lotteries are identical except for the payoff in <math>\frac{1}{N}</math> of the time. If he works, this value is higher, so he prefers to work.
With regards to <math>A>C</math>,note that <math>A>C \iff N> \frac{1-\exp[-\rho(\sum_{i\neq j} z(e_{j}))]}{1-\exp[-\rho(\frac{1}{N}\sum_{i\neq j} z(e_{j})+\frac{1}{N}z(e_{i})-1)]}</math>. In the above RHS expression, we know that the numerator is smaller than the denominator, so the fraction is less than 1. We know that <math>N>1</math>, so the inequality always holds.
===Question C1: Agenda Control and Status Quo===
