Tag Archives: stochastic

Modeling Dynamic Incentives: Application to Basketball

I will give a talk on “Modeling Dynamic Incentives: Application to Basketball” at the GERAD (Groupe d’études et de recherche en analyse des décisions) on June, 10th June, 6th. This is some joint work with Nathalie Colombier and Romuald Elie

An important aspect of the strategy of most organizations is the provision of incentives to the employees to meet the organization’s objectives. Typically this implies tying pay to performance (see Prendergast, 1999). In order to reward employees for their effort, firms spend considerable resources on performance evaluations. In many cases, evaluation consists of comparing actual performance to a pre-defined individual target. Another frequently used format is relative performance evaluation. Relative performance evaluation may motivate employees to work harder.But it may also be demoralizing and create an excessively competitive workplace, which may hinder overall performance; see Lazear (1989). Determining the overall impact of relative performance evaluation is crucial for companies. Economic research on relative performance evaluation has mainly focused on the comparison of final performances between competitors,like in tournament theory, and on quantitative and subjective performance ratings (Lazear and Gibbs, 2009). In contrast, what happens during a competition and the impact of feedback frequency on effort have so far received little attention. Following Berger and Pope (2011), we decided to use a basketball application to get a better understanding of the role of the feedback information. Sports datasets allow to observe score and team behavior continuously (during a game but also during the season) which can be use as a proxy of the effort. Berger an Pope (2010) asked ”can loosing lead to winning ?” looking at the impact of the halftime score difference on winning probability in NCAA (college) and NBA (pro) games. More precisely, they studied whether a team loosing at halftime is more likely to win than expected using a logit model. They find that usually the higher the score difference the more likely the are to win. But if the halftime score difference is around 0 they observe a discontinuity: loosing with a small difference (e.g. down by 1 point) can lead to increase the effort and win the game. In this paper we try answer the question ”when loosing lead to winning ?”.

When should I optimally shoot at my son ? (part 2)

Following my previous post of yesterday, online here, assume now that I do not know if my son came when I turned my back at time, and missed me… Then the payoff function is the one propose by Vincent, i.e.


In that particular case,






If we draw those functions, on [0,1], the optimal value is solution of https://perso.univ-rennes1.fr/arthur.charpentier/latex/duel103.png

i.e. https://perso.univ-rennes1.fr/arthur.charpentier/latex/duel104.png (we focus only on solutions in [0,1]). Because here the game is symmetric, my son should also shoot at time https://perso.univ-rennes1.fr/arthur.charpentier/latex/duel105.png
Thus, the payoff is then


Since we consider here a  zero-sum game, this cannot be a solution of the game. So the game does not have pure strategy solution.

Assume that now I have a mixed strategy, i.e. a distribution of the optimal time to shot. My strategy has distribution https://perso.univ-rennes1.fr/arthur.charpentier/latex/duel110.png, with density https://perso.univ-rennes1.fr/arthur.charpentier/latex/duel111.png (we assume here that the density exists, or we seek only solution that are differentiable). Assume further that there exists https://perso.univ-rennes1.fr/arthur.charpentier/latex/duel112.png>0 such that the support of my optimal shooting time (the time to shoot is now a random variable) is (https://perso.univ-rennes1.fr/arthur.charpentier/latex/duel112.png,1] (or [https://perso.univ-rennes1.fr/arthur.charpentier/latex/duel112.png,1] since we assume that https://perso.univ-rennes1.fr/arthur.charpentier/latex/duel110.png is differentiable). There is a discussion at the end of Vincent’s post where he needs that assumption, at the end. Actually, I think we can make it now, since we can rationally assume that
I will not shot at time 0 (even on a neighborhood of 0 since I have zero chance to hit my son).
The expected payoff function, assuming that my son shoots at time y is


Since the zero-sum game is symmetric, again, the expected payoff should be zero. It comes that necessarily,


if https://perso.univ-rennes1.fr/arthur.charpentier/latex/duel121.png. Hence, if we differentiate (with respect to y), we have


and if we differentiate one more time, it comes


i.e. a general solution should be of the form https://perso.univ-rennes1.fr/arthur.charpentier/latex/duel130.png.
Here, we have the same solution as the one considered in Vincent’s blog. His solution is obtained as follows (with slightly different expressions) conditional to https://perso.univ-rennes1.fr/arthur.charpentier/latex/duel03.png, my expected payoff is




With a simple integration by parts,


where https://perso.univ-rennes1.fr/arthur.charpentier/latex/duel134.png, i.e.




So, if we want to be indifferent to https://perso.univ-rennes1.fr/arthur.charpentier/latex/duel03.png‘s strategy, https://perso.univ-rennes1.fr/arthur.charpentier/latex/duel141.png, where


with https://perso.univ-rennes1.fr/arthur.charpentier/latex/duel143.png,


Consider solutions https://perso.univ-rennes1.fr/arthur.charpentier/latex/duel150.png, then https://perso.univ-rennes1.fr/arthur.charpentier/latex/duel151.png=1, i.e. either https://perso.univ-rennes1.fr/arthur.charpentier/latex/duel152.png=1 and then https://perso.univ-rennes1.fr/arthur.charpentier/latex/duel153.png is constant, or https://perso.univ-rennes1.fr/arthur.charpentier/latex/duel152.png=-1. This means that https://perso.univ-rennes1.fr/arthur.charpentier/latex/duel154.png is in proportional to https://perso.univ-rennes1.fr/arthur.charpentier/latex/duel155.png.
If we substitute in the equation we had, initially, it comes that




If we consider https://perso.univ-rennes1.fr/arthur.charpentier/latex/duel03.png=a and https://perso.univ-rennes1.fr/arthur.charpentier/latex/duel03.png=1, it comes that https://perso.univ-rennes1.fr/arthur.charpentier/latex/duel112.png=1/3 while https://perso.univ-rennes1.fr/arthur.charpentier/latex/duel162.png=1/4 (but we don’t really care about that normalizing constant).
It means that we should not start shooting before 1/3 of the tank is fulled. Actually, it makes sense, since


if https://perso.univ-rennes1.fr/arthur.charpentier/latex/duel03.png<1/3 (while https://perso.univ-rennes1.fr/arthur.charpentier/latex/duel164.png=0 if  https://perso.univ-rennes1.fr/arthur.charpentier/latex/duel03.png>1/3).