The next several questions use the following public good setting.
There are two individuals, $i=1,2$. Individual $i$ has income $w_i$, contributes $g_i$ to a public good, and consumes the rest privately:
$$\begin{align*}x_i=w_i-g_i,\qquad G=g_1+g_2.\end{align*}$$Individual $i$’s preferences are given by
$$\begin{align*}u_i(x_i,G)=x_i+G\left(4-\frac{G}{2}\right).\end{align*}$$Substituting $x_i$ and $G$, we can write each individual’s utility as a function of the two contributions:
$$\begin{align*}u_i(g_1,g_2)=w_i-g_i+(g_1+g_2)\left(4-\frac{g_1+g_2}{2}\right).\end{align*}$$Expanded:
$$\begin{align*} u_1(g_1,g_2)&=w_1+3g_1+4g_2-\frac{g_1^2}{2}-g_1g_2-\frac{g_2^2}{2}\\ u_2(g_1,g_2)&=w_2+4g_1+3g_2-\frac{g_1^2}{2}-g_1g_2-\frac{g_2^2}{2} \end{align*}$$Note: a superscript is a power exponent and a subscript identifies the individual. In $g_1^2$, the subscript 1 means individual 1 and the superscript 2 means squared.
Using Mathematica syntax, which of the following correctly defines the payoff function $u_1(x_1,G)=x_1+G\cdot\left(4-\frac{G}{2}\right)$?
Answer. u1[x1_, G_] := x1 + G*(4 - G/2)
The arguments of a Mathematica function go in square brackets, and on the left-hand side each argument carries an underscore (x1_, G_) so that it stands for any value. The right-hand side uses the plain names, without underscores. Terms are grouped with parentheses, never square brackets, and G*(4 - G/2) keeps the division by 2 inside the parentheses, as in $G\left(4-\frac{G}{2}\right)$. Each of the other options breaks at least one of these rules.
Ignoring the constraints $0\le g_1\le w_1$, use Recipe #0 and write down player 1’s best response to $g_2$. Please write only player 1’s best response; do not include your reasoning or work.
Answer. $BR_1(g_2)=3-g_2$.
Recipe #0 sets the derivative of $u_1$ with respect to player 1’s choice equal to zero:
$$\begin{align*} \frac{\partial u_1}{\partial g_1}&=3-g_1-g_2=0,\\ g_1&=3-g_2. \end{align*}$$Assume wealth levels are $w_1=5$ and $w_2=7$. In any Nash equilibrium, what is the total level of public good?
Hint: unlike the classroom examples, we may have many Nash equilibria here.
Answer. $G=3$.
Player 2’s best response comes from the same recipe:
$$\begin{align*} \frac{\partial u_2}{\partial g_2}&=3-g_1-g_2=0,\\ BR_2(g_1)&=3-g_1. \end{align*}$$Both best responses describe the same line,
$$\begin{align*}g_1+g_2=3,\end{align*}$$so every pair $(g_1,g_2)$ on this line with $0\le g_1\le 3$ is a Nash equilibrium, and in each of them $G=g_1+g_2=3$. The wealth bounds never bind, because no contribution exceeds $3$, which is less than both $w_1=5$ and $w_2=7$.
Now assume there is only one individual, so $g_2=0$, but there are two periods, $t=0$ and $t=1$. Let $(g_0,g_1)$ be the individual’s contributions in periods 0 and 1.
The public good is a flow: it does not carry over to the next period, so $G_t=g_t$. In each period $t$, the individual consumes $x_t=w_t-g_t$ privately and has utility $x_t+G_t\left(4-\frac{G_t}{2}\right)$.
Wealth can be transferred costlessly over time, and total wealth is 9:
$$\begin{align*}w_0+w_1=9.\end{align*}$$The individual chooses $(g_0,g_1,w_0,w_1)$ to maximize discounted utility subject to the total wealth constraint.
Write explicitly the agent’s discounted utility for this problem as a function of the choice variables and the discount factor. Use $d$ to denote the discount factor.
Notice: “explicitly” means the expression may contain only $g_0$, $g_1$, $w_0$, $w_1$, $d$, and numbers. Do not leave any function or shorthand symbols such as $u(\cdot)$, $x_t$, or $G_t$ in your answer.
Answer.
$$\begin{align*} U&=w_0-g_0+g_0\left(4-\frac{g_0}{2}\right)+d\left(w_1-g_1+g_1\left(4-\frac{g_1}{2}\right)\right). \end{align*}$$Consider the Lagrangian for the above problem. Obtain the first-order condition of the Lagrangian with respect to $\lambda$.
Hint: remember that a first-order condition entails a bit more than just computing a derivative.
Answer. With $U$ the discounted utility of Question 6 and the constraint $w_0+w_1=9$, the Lagrangian is
$$\begin{align*}\mathcal{L}=U(g_0,g_1,w_0,w_1)-\lambda\,(w_0+w_1-9),\end{align*}$$and its first-order condition with respect to $\lambda$ sets the derivative equal to zero:
$$\begin{align*} \frac{\partial\mathcal{L}}{\partial\lambda}&=-(w_0+w_1-9)=0,\\ \text{that is,}\quad w_0+w_1&=9. \end{align*}$$The first-order condition with respect to the multiplier recovers the constraint.
We are back to the static setting of Question 1, with two players. Player 1 chooses $g_1$ to maximize $u_1(g_1,g_2)$ subject to two constraints:
$$\begin{align*} \max_{g_1}\;&w_1+3g_1+4g_2-\frac{g_1^2}{2}-g_1g_2-\frac{g_2^2}{2}\\ \text{subject to}\;&0\le g_1\quad\text{(first constraint)}\\ &g_1\le w_1\quad\text{(second constraint)} \end{align*}$$Which constraint should we rewrite in order to apply Recipe #2?
Answer. The first constraint.
Recipe #2 needs every constraint written as a function of the choice variables less than or equal to a constant. The second constraint, $g_1\le w_1$, already has that form. The first, $0\le g_1$, does not, and is rewritten as
$$\begin{align*}-g_1\le 0.\end{align*}$$In the case $\lambda_1=0$ and $\lambda_2\gt 0$, write down the corresponding first-order conditions that we must consider.
Answer. With the constraints written as $-g_1\le 0$ and $g_1\le w_1$, the Lagrangian is
$$\begin{align*}\mathcal{L}=u_1(g_1,g_2)-\lambda_1\,(-g_1)-\lambda_2\,(g_1-w_1).\end{align*}$$In this case $\lambda_1=0$ drops out, and Recipe #1 is applied with the variables $g_1$ and $\lambda_2$:
$$\begin{align*} \frac{\partial\mathcal{L}}{\partial g_1}&=3-g_1-g_2-\lambda_2=0,\\ \frac{\partial\mathcal{L}}{\partial\lambda_2}&=-(g_1-w_1)=0,\quad\text{that is,}\quad g_1=w_1. \end{align*}$$A solution in this case must also satisfy $\lambda_2\gt 0$ and the first constraint, $-g_1\le 0$.
If $g_2=2$ and $w_1=2$, does the case $\lambda_1=0$ and $\lambda_2\gt 0$ correspond to a solution? Why or why not?
Answer. No. The first-order conditions of Question 10 give
$$\begin{align*} g_1&=w_1=2,\\ \lambda_2&=3-g_1-g_2=3-2-2=-1\lt 0, \end{align*}$$which contradicts $\lambda_2\gt 0$. The solution is interior instead: $g_1=3-g_2=1$, which lies between $0$ and $w_1=2$.
Does Ehrenberg & Bognanno’s article have a theoretical model?
Answer. No. The paper has only an empirical model, their equation (1):
$$s_{ji}=a_0+a_1\,\text{TPRIZE}_i+a_2\,\mathbf{x}_i+a_3\,\mathbf{y}_j+a_4\,\mathbf{z}_i+v_{ij}$$where $s_{ji}$ is the final score of player $j$ in tournament $i$, $\text{TPRIZE}_i$ is the total prize money awarded in the tournament, $\mathbf{x}_i$ controls for the difficulty of the course and the weather, $\mathbf{y}_j$ proxies for player $j$’s ability, $\mathbf{z}_i$ controls for the quality of the other players in the field, and $v_{ij}$ is a random error term.
What is the main hypothesis that Ehrenberg & Bognanno’s article is trying to test?
Answer. Does the incentive structure (prizes) affect individual output (players’ scores)?
Does Dobson, Goddard & Stähler’s article have a theoretical model?
Answer. Yes. Dobson et al. use a contest model similar to the one in the lecture notes. The home team (team 1) and the away team (team 2) choose efforts $e_1$ and $e_2$:
$$r=p(0,0)=\frac{\beta_1+h}{\beta_1+\beta_2+h}\tag{1}$$ $$p(e_1,e_2)=\frac{e_1\,r}{e_1\,r+e_2\,(1-r)}\tag{2}$$ $$\pi_1=p\,U_1-c_1e_1,\qquad \pi_2=(1-p)\,U_2-c_2e_2\tag{3}$$where $\beta_i$ is the quality of team $i$, $h$ is the home-field advantage, $r$ is the home team’s success probability when neither team exerts effort, $\pi_i$ is team $i$’s expected payoff, $U_i$ is team $i$’s payoff from winning, and $c_i$ is team $i$’s marginal cost of effort.
The main differences: the lecture notes assume $r=\tfrac12$ (no home-team advantage and teams of equal quality) and $c_1=c_2=1$ (the cost of effort is the same for both teams). Also, the lecture notes assume a quadratic cost of effort, while the article, like some other examples in the course, assumes a linear cost.
What is the main prediction that Dobson, Goddard & Stähler’s article is trying to test?
Answer. The main prediction is the paper’s Result 2: when the two teams have the same cost of effort and the same payoff from a win, both exert the most effort in an evenly balanced game, that is, when $r=\tfrac12$ after allowing for home-field advantage. Asymmetries between the teams lower effort. The equilibrium efforts are
$$\begin{aligned} e_1^*&=\frac{r(1-r)\,c_2U_1^2U_2}{\left(r\,c_2U_1+(1-r)\,c_1U_2\right)^2}\\[4pt] e_2^*&=\frac{r(1-r)\,c_1U_1U_2^2}{\left(r\,c_2U_1+(1-r)\,c_1U_2\right)^2} \end{aligned}\tag{4}$$With $c_1=c_2=c$ and $U_1=U_2=U$, both reduce to $e^*=r(1-r)\,U/c$, which is largest at $r=\tfrac12$. This is why the empirical model includes $\text{UNCERT}=\text{HPROB}\cdot(1-\text{HPROB})$, where HPROB is their measure of $r$: the prediction is a positive coefficient on UNCERT.
The paper also tests Result 1: a team exerts more effort when its own payoff from a win, $U_i$, is larger. A game that still matters for a team’s championship, European qualification, promotion or relegation chances has a larger $U_i$. The empirical model captures this with the dummies HSIG (home team) and ASIG (away team).
The coefficients on UNCERT are positive in every equation, as Result 2 predicts. Result 1 holds for away teams, whose fouls and cards rise when the game matters to them. The home team’s effort does not respond significantly to its own stakes.
How is “effort” measured in Dobson, Goddard & Stähler’s article?
Answer. By fouls and cards. In the authors’ words (Section 5, p. 75): “In the empirical model the effort of the teams is reflected in the numbers of fouls committed and the numbers of yellow and red cards awarded.”
Do you think this is a good measure of effort? Why or why not?
Answer. There is no single correct answer, but any answer must acknowledge that red cards are a problematic measure of effort: a red card leaves the team with one player fewer, which most likely lowers its probability of winning.