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An If employee works but others aren't, the lottery is triggered and employee i's utility from working <i>if others are not working</i> is: <math>B=\frac{1}{N}(1-\exp[-\rho(\sum_{i\neq j} z(e_{j})+z(e_{i})])</math>.
An If employee i's does NOT work, the lottery is triggered and his utility from NOT working if others are not working is: <math>1-\exp[C=\frac{1}{N}(1-\exp[-\rho(\sum_{i\neq j} z(e_{j})+\frac{1}{N}z(e_{i}])-1]</math>
If employee i does NOT work, the lottery is triggered I will now show that <math>A>C</math> and his utility is: <math>1-\exp[sum_{i\neq j} z(e_{j})]B>C</math>. An employee i's utility from NOT working guarantees that the lottery will be triggered.
BPP Field Exam 2010 Answers (view source)
Revision as of 19:41, 29 March 2011
, 19:41, 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. 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 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>.
===Question C1: Agenda Control and Status Quo===
