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note Now, consider that <math>A>C \iff N> \frac{(1-\exp[-\rho(\sum_{i\neq j} z(e_{j}N-1)])]}{1-\exp[-\rho(\frac{1}{N}\sum_{i\neq j} z(e_{j})+\frac{1}{N}z(e_{i})-1)]}</math>.
BPP Field Exam 2010 Answers (view source)
Revision as of 00:08, 30 March 2011
, 00:08, 30 March 2011→Question B1.3
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 if all other workers are working, then <math>C=\frac{1}{N}(1-\exp[-\rho z (N-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.
