Roy is coming to plant flowers in Zoe’s garden. Zoe loves flowers, her utility for a garden with flowers is

Roy plants a unit mass of seeds and the fraction of these that will bloom into flowers depends on how attentive Roy is as a gardener. Roy’s attentiveness is his type . In particular when Roy’s type is , absent any sabotage by Zoe, there will be flowers in Zoe’s garden in Spring. Roy’s attentiveness is unknown to everyone and it is believed by all to be uniformly distributed on the unit interval.

Jane, Zoe’s neighbor, is looking for a gardener for the following Spring. Jane has high standards, she will hire Roy if and only if he is sufficiently attentive. In particular, Jane’s utility for hiring Roy when his true type is is given by

(Her utility is zero if she does not hire Roy.)

Roy tends to one and only one garden per year. Therefore Roy will continue to plant flowers in Zoe’s garden for a second year if and only if Jane does not hire him away from her.

Consequently, Zoe is contemplating sabotaging Roy’s flowers this year. If Zoe destroys a fraction of Roy’s seeds then the total number of flowers in Zoe’s garden when Spring arrives will be . Of course sabotage is costly for Zoe because she loves flowers.

There will be no sabotage in the second year because after two years of gardening Roy goes into retirement. Therefore, if Zoe destroys in the first year and Roy continues to work for Zoe in the second year, Zoe’s total payoff will be

whereas if Roy is hired away by Jane, then Zoe’s total payoff is just .

This is a two-player (Zoe and Jane) extensive-form game with incomplete information. The timing is as follows. First, Roy’s type is realized. Nobody observes Roy’s type. Zoe moves first and chooses . Then Spring arrives and the flowers bloom. Jane does not observe but does observe the number of flowers in Zoe’s garden. Then Jane chooses whether or not to hire Roy away from Zoe. Then the game ends.

Describe the set of all Perfect Bayesian Equilibria.

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## 6 comments

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September 9, 2013 at 12:57 am

afinetheoremIWithout giving the full result, here are some immediate properties:

1) alpha must be not be invertible (else Jane learns theta with certainty always, hence there is no gain to destroying crops in the first stage, hence since alpha is a scalar it must be random)

2) for any alpha such that alpha*theta>=2/3, Jane hires regardless of Zoe’s strategy.

3) the maximal deterministic strategy pays off 5/6 (all deterministic strategies except alpha=0 are invertible)

4) the optimal random strategy involves Jane hiring sometimes (alpha>2/3 cannot be in the support if theta=1 is to never be hired away, hence no strategy where Ray always stays can generate more than payoff 5/6)

5) Of course one can do better if alpha was known to Zoe: note that letting Ray go when theta=1 earns at most 1 (since Ray will be hired away), but he can be kept by mixing theta=1 and theta=1/3 in round 1 (letting alpha given theta=1 be equal to 1/3), earning 1.33>1. In general, if alpha is known to JZoe, she can earn (as a lower bound) an expectation of (1/6)*(1/3)+(1/2)*(1/3)+(1/2)*(1/3)+1/2=8/9>5/6.

6) With known alpha to Zoe, both the gain from strategic destruction, and the amount of crops destroyed, are increasing as theta increases above 2/3. The additional tradeoff with unknown alpha is that sometimes crops will be destroyed unnecessarily when theta is lower.

September 9, 2013 at 4:40 pm

enriqueBut why does Jane’s utility function decrease by 2/3 …?

September 10, 2013 at 10:18 am

tyroIs it the empty set?

September 10, 2013 at 10:23 am

EJMRAfter some clumsy crowdsourcing, 6493 provides a promising answer

http://www.econjobrumors.com/topic/great-prelim-question-cheap-talk

September 10, 2013 at 12:10 pm

jeffEJMR you are my favorite blog follower so I will give you a hint. A necessary condition for a prelim question to be considered good is that the solution must be elegant.

A further necessary condition for the question to be considered not just good but great is that the solution must have a touch of tragedy.

September 13, 2013 at 5:44 pm

José Antonio Espín SánchezTo the comment above:

– Jane’s Utility decreasing by 2/3 implies that she only hires Roy if she thinks that thetha>2/3. Remember that not hiring gives Jane zero utility and hiring Roy with thetha2/3, then Roy will be hired in the second period, Zoe gets 5/6

if thetha1/2

So destroying the flowers is a dominated strategy… so much for a deterministic solution.

Zoe will randomize using f(y)=P(alpha=y) over (0,1] so that it will make Jane more uninformed, that is, x is still a signal of thetha, but is less precise, since now it is made of the product of 2 random variables, thetha and p.

The cheapest way to do so is to find a function f(y) such that the signal x is uninformative, and the posterior is such that Jane is indifferent between hiring Roy or not, that is E(thetha-2/3/x,f)=E(thetha-2/3/f)=0.