Midterm Practice

The Public Good Game

The setting for every problem

All problems below use the public good game of the lecture notes, with one change: the utility function. Two individuals $i=1,2$ have incomes $w_1,w_2\gt 0$. Individual $i$ contributes $g_i$ to a public good and consumes the rest privately: $$x_i=w_i-g_i,\qquad G=g_1+g_2.$$ Each individual's utility is $$u_i(x_i,G)=\sqrt{x_i}+\sqrt{G}.$$

This is not the utility of the lecture-notes example, where $u_i(x_i,G)=x_i\cdot G$. We write $G_{-i}$ for the total contribution of everyone other than $i$; with two individuals, $G_{-1}=g_2$ and $G_{-2}=g_1$.

Basic Mathematica Commands

Problem 1

Substituting the budget, individual 1's utility as a function of the two contributions is $$u_1(g_1,g_2)=\sqrt{w_1-g_1}+\sqrt{g_1+g_2}.$$ Answer in Mathematica syntax, using the commands on the cheat sheet.

Define the function u1[g1_, g2_], keeping w1 as a symbol.
Use Grad to write individual 1's first-order condition with respect to g1, and store it in foc.
Solve foc for g1.
Use a replacement rule to evaluate your solution at $w_1=13$ and $g_2=5$.

Unconstrained Optimization

Problem 2

Take the total contribution of the others, $G_{-i}$, as given, and ignore the constraint $0\le g_i\le w_i$. Use Recipe #0.

Write individual $i$'s utility as a function of $g_i$ alone.
Write the first-order condition.
Find individual $i$'s best response $\mathrm{BR}_i(G_{-i})$.
Compare $\mathrm{BR}_i$ with the best response in the lecture-notes example, where $u_i=x_i\cdot G$. What do the two utilities have in common at the optimum?

Unconstrained Optimization

Problem 2 — Answers

$u_i(g_i)=\sqrt{w_i-g_i}+\sqrt{g_i+G_{-i}}$.
First-order condition: $$\begin{aligned}\frac{d u_i}{d g_i}&=-\frac{1}{2\sqrt{w_i-g_i}}+\frac{1}{2\sqrt{g_i+G_{-i}}}\\ &=0.\end{aligned}$$
Move the negative term to the right-hand side and flip both fractions, then square both sides and solve for $g_i$: $$\begin{aligned}\sqrt{w_i-g_i}&=\sqrt{g_i+G_{-i}}\\ w_i-g_i&=g_i+G_{-i}\\ g_i&=\frac{w_i-G_{-i}}{2}.\end{aligned}$$ So $\mathrm{BR}_i(G_{-i})=\dfrac{w_i-G_{-i}}{2}$.
It is the same best response as in the lecture-notes example, $g_1=\frac{w_1-g_2}{2}$. In both cases the individual stops where the marginal utility of the private good equals the marginal utility of the public good: the marginal loss from less private consumption is perfectly offset by the marginal gain from more public good. This happens when $x_i=G$ for both of these utility functions.

Optimization with Equality Constraints

Problem 3

Now the individuals contribute sequentially: individual 1 chooses $g_1$ first; individual 2 observes $g_1$ and then chooses $g_2$. The public good is not a flow—individual 1's contribution is still there when individual 2 contributes—so $G=g_1+g_2$. As in Problem 2, write payoffs as functions of the contributions only, $$u_i(g_1,g_2)=\sqrt{w_i-g_i}+\sqrt{g_1+g_2},$$ and ignore $0\le g_i\le w_i$.

Individual 2 moves last, so it best-responds to $g_1$. Use Problem 2 to write $g_2=\mathrm{BR}_2(g_1)$.
Individual 1 anticipates this. Write individual 1's problem, $$\max_{g_1,g_2}\;u_1(g_1,g_2)\quad\st\quad g_2=\mathrm{BR}_2(g_1),$$ and its Lagrangian.
Use Recipe #1 to find $g_1$, $g_2$, and the Lagrange multiplier $\lambda$. These are the actions played in the subgame perfect Nash equilibrium.
Let $w_1=13$ and $w_2=5$. Compute $g_1$, $g_2$, $G$, and $\lambda$.

Optimization with Equality Constraints

Problem 3 — Answers

$g_2=\mathrm{BR}_2(g_1)=\dfrac{w_2-g_1}{2}$.
Individual 1's problem and its Lagrangian: $$\begin{aligned}&\max_{g_1,g_2}\;\sqrt{w_1-g_1}+\sqrt{g_1+g_2}\\ &\st\quad g_2=\frac{w_2-g_1}{2},\end{aligned}$$ $$\begin{aligned}\mathcal{L}&=\sqrt{w_1-g_1}+\sqrt{g_1+g_2}\\ &\quad-\lambda\paren{g_2-\frac{w_2-g_1}{2}}.\end{aligned}$$
Writing $G=g_1+g_2$, the first-order conditions are $$\begin{aligned} \diff{\mathcal{L}}{g_1}&=-\frac{1}{2\sqrt{w_1-g_1}}+\frac{1}{2\sqrt{G}}-\frac{\lambda}{2}=0,\\ \diff{\mathcal{L}}{g_2}&=\frac{1}{2\sqrt{G}}-\lambda=0,\\ \diff{\mathcal{L}}{\lambda}&=-\paren{g_2-\frac{w_2-g_1}{2}}=0. \end{aligned}$$ The second gives $\lambda=\frac{1}{2\sqrt{G}}$; substituted into the first, $w_1-g_1=4G$. The constraint gives $G=\frac{g_1+w_2}{2}$, so $$g_1=\frac{w_1-2w_2}{3},\qquad g_2=\frac{5w_2-w_1}{6},$$ $$G=\frac{w_1+w_2}{6},\qquad \lambda=\frac{1}{2\sqrt{G}}.$$
$g_1=1$, $g_2=2$, $G=3$, and $\lambda=\frac{1}{2\sqrt3}\approx 0.289$.

Nash Equilibrium of Static Games

Problem 4

Back to simultaneous contributions: both individuals choose at the same time. Use the best responses from Problem 2, and ignore $0\le g_i\le w_i$.

Write the system of best responses and solve it for the Nash equilibrium contributions $(g_1^e,g_2^e)$ and the public good $G^e$.
Let $w_1=w_2=w$. Find $g_1^e$, $g_2^e$, and $G^e$.
Let $w_1=13$ and $w_2=5$. Compute $g_2^e$. Why is this not an acceptable answer? (Problem 7 fixes it.)

Nash Equilibrium of Static Games

Problem 4 — Answers

The best responses are $$g_1=\frac{w_1-g_2}{2},\qquad g_2=\frac{w_2-g_1}{2}.$$ Substitute $g_2$ into individual 1's best response, multiply both sides by $4$, and collect the $g_1$ terms: $$\begin{aligned}g_1&=\frac{1}{2}\left(w_1-\frac{w_2-g_1}{2}\right)\\ 4g_1&=2w_1-w_2+g_1\\ 3g_1&=2w_1-w_2\\ g_1^e&=\frac{2w_1-w_2}{3}.\end{aligned}$$ Plug $g_1^e$ into individual 2's best response: $$\begin{aligned}g_2^e&=\frac{1}{2}\left(w_2-\frac{2w_1-w_2}{3}\right)\\ &=\frac{3w_2-2w_1+w_2}{6}\\ &=\frac{2w_2-w_1}{3}.\end{aligned}$$ There is no system to solve for $G^e$: just add the two equilibrium contributions. $$\begin{aligned}G^e&=g_1^e+g_2^e\\ &=\frac{(2w_1-w_2)+(2w_2-w_1)}{3}\\ &=\frac{w_1+w_2}{3}.\end{aligned}$$
$g_1^e=g_2^e=\frac{w}{3}$ and $G^e=\frac{2w}{3}$.
$g_2^e=\frac{10-13}{3}=-1$. A negative contribution would mean individual 2 takes resources out of the public good, which the model rules out ($g_2\ge 0$). The formula assumed an interior solution; when $w_1$ is large relative to $w_2$, the constraint $g_2\ge 0$ binds.

Subgame Perfect Nash Equilibrium

Problem 5

Return to the sequential game of Problem 3.

Write the subgame perfect Nash equilibrium as a strategy profile. Why is individual 2's strategy a function rather than a number?
Let $u_1$ be any differentiable payoff, and assume only that individual 2's best response $\mathrm{BR}_2(g_1)$ is decreasing in $g_1$. Use the chain rule to compare $$\frac{\partial}{\partial g_1}\,u_1(g_1,g_2)$$ with $$\frac{d}{d g_1}\,u_1\big(g_1,\mathrm{BR}_2(g_1)\big).$$ When is the second smaller than the first? What does that mean for how much a first mover contributes?
With $w_1=13$ and $w_2=5$, individual 1 is richer but contributes less than individual 2. Explain why.

Subgame Perfect Nash Equilibrium

Problem 5 — Answers

The SPNE is the profile $\paren{g_1^*,\ \mathrm{BR}_2(\cdot)}$ with $g_1^*=\frac{w_1-2w_2}{3}$ and $\mathrm{BR}_2(g_1)=\frac{w_2-g_1}{2}$ for every $g_1$. Individual 2's strategy must say what it does after every possible $g_1$, including those never played, because subgame perfection requires optimal play in every subgame. The number $\frac{5w_2-w_1}{6}$ is only the action on the equilibrium path.
By the chain rule, $$\begin{aligned}\frac{d}{d g_1}\,u_1\big(g_1,\mathrm{BR}_2(g_1)\big)&=\frac{\partial u_1}{\partial g_1}\\ &\quad+\frac{\partial u_1}{\partial g_2}\cdot\mathrm{BR}_2'(g_1).\end{aligned}$$ Since $\mathrm{BR}_2'\lt 0$, the derivative that lets $g_2$ respond is smaller than the one that holds $g_2$ fixed exactly when $\frac{\partial u_1}{\partial g_2}\gt 0$—individual 1 gains from individual 2's contribution, as with any public good. Each unit the first mover contributes crowds out part of the follower's contribution, so the first mover gains less from contributing and gives less than it would if it took $g_2$ as fixed.
In our particular example we can see the same thing without the chain rule for $\mathrm{BR}_2$. There, $$u_1(g_1,g_2)=\sqrt{w_1-g_1}+\sqrt{g_1+g_2},$$ $$\mathrm{BR}_2(g_1)=\frac{w_2-g_1}{2}.$$

Holding $g_2$ fixed. Differentiate with respect to $g_1$: $$\frac{\partial u_1}{\partial g_1}=-\frac{1}{2\sqrt{w_1-g_1}}+\frac{1}{2\sqrt{g_1+g_2}}.$$

Letting $g_2$ respond. Replace $g_2$ by $\mathrm{BR}_2(g_1)$ first. The public good becomes $$\begin{aligned}g_1+\mathrm{BR}_2(g_1)&=g_1+\frac{w_2-g_1}{2}\\ &=\frac{g_1+w_2}{2},\end{aligned}$$ so individual 1's payoff depends on $g_1$ alone: $$\begin{aligned}u_1\big(g_1,\mathrm{BR}_2(g_1)\big)&=\sqrt{w_1-g_1}\\ &\quad+\sqrt{\frac{g_1+w_2}{2}}.\end{aligned}$$ Differentiate. The expression inside the second square root has slope $\frac12$: $$\begin{aligned}\frac{d}{dg_1}\,u_1\big(g_1,\mathrm{BR}_2(g_1)\big)&=-\frac{1}{2\sqrt{w_1-g_1}}\\ &\quad+\frac12\cdot\frac{1}{2\sqrt{\frac{g_1+w_2}{2}}}.\end{aligned}$$

The difference. Evaluate both derivatives at the same point, $g_2=\mathrm{BR}_2(g_1)$, and write $G=g_1+g_2=\frac{g_1+w_2}{2}$. The first terms cancel: $$\begin{aligned}&\frac{\partial u_1}{\partial g_1}-\frac{d}{dg_1}\,u_1\big(g_1,\mathrm{BR}_2(g_1)\big)\\ &\qquad=\frac{1}{2\sqrt G}-\frac12\cdot\frac{1}{2\sqrt G}\\ &\qquad=\frac{1}{4\sqrt G}\gt 0.\end{aligned}$$

With $g_2$ fixed, one more unit of $g_1$ adds a full unit to $G$. When individual 2 responds, it cuts its contribution by half a unit, so $G$ rises by only $\frac12$ and individual 1's marginal gain from the public good is cut in half. The missing half is exactly the chain-rule term: $$\begin{aligned}\frac{\partial u_1}{\partial g_2}\cdot\mathrm{BR}_2'(g_1)&=\frac{1}{2\sqrt G}\cdot\paren{-\frac12}\\ &=-\frac{1}{4\sqrt G}.\end{aligned}$$
Here $\mathrm{BR}_2'=-\frac12$: each unit individual 1 gives lowers individual 2's contribution by half a unit, so it raises $G$ by only $\frac12$ while costing a full unit of private consumption. Individual 1 commits to a small contribution ($g_1=1$) and lets individual 2 make up the difference ($g_2=2$)—the first mover free-rides on the follower.

Discount Factors

Problem 6

Now the agent receives the same income $w$ in every period and must choose a donation $g_t$ in every period. Assume there is only one agent and two periods, $t=0,1$. The agent cares about the stock of the public good: $G_0=g_0$ in period $0$ and $G_1=g_0+g_1$ in period $1$. The agent's utility in period $t$ is $$u_t=\sqrt{G_t}+\sqrt{x_t},\qquad x_t=w-g_t.$$ This is a different utility from the lecture-notes example, where $u_i=x_i\cdot G$. The agent discounts period-1 utility by the discount factor $\delta\in(0,1)$.

Write the agent's discounted utility as a function of $(g_0,g_1)$. Notice that $0$ and $1$ are time indexes, not individuals.
Use the first-order condition for $g_1$ to write $g_1$ as a function of $g_0$. Does $\delta$ appear?
Substitute your answer and write the first-order condition for $g_0$. Use it to show that $g_0\gt\tfrac{w}{2}$.
How does $\delta$ affect $g_0$? What happens to $g_0$ and $g_1$ as $\delta\to 0$? Explain.

Discount Factors

Problem 6 — Answers

The discounted utility is $$\begin{aligned}U(g_0,g_1)&=\sqrt{g_0}+\sqrt{w-g_0}\\ &\quad+\delta\left[\sqrt{g_0+g_1}+\sqrt{w-g_1}\right].\end{aligned}$$
The first-order condition for $g_1$ is $$\begin{aligned}\diff{U}{g_1}&=\delta\left[\frac{1}{2\sqrt{g_0+g_1}}-\frac{1}{2\sqrt{w-g_1}}\right]=0\\ &\iff g_1=\frac{w-g_0}{2}.\end{aligned}$$ $\delta$ multiplies the whole condition, so it cancels: the period-1 donation does not depend on $\delta$.
Substituting, $G_1=x_1=\frac{w+g_0}{2}$, so $$U=\sqrt{g_0}+\sqrt{w-g_0}+\delta\sqrt{2\,(w+g_0)},$$ $$\frac{1}{2\sqrt{g_0}}-\frac{1}{2\sqrt{w-g_0}}+\frac{\delta}{\sqrt{2\,(w+g_0)}}=0.$$ The last term is positive, so $\frac{1}{\sqrt{g_0}}\lt\frac{1}{\sqrt{w-g_0}}$, that is, $g_0\gt w-g_0$, or $g_0\gt\frac{w}{2}$.
The left side of the first-order condition is decreasing in $g_0$ and increasing in $\delta$, so a higher $\delta$ raises $g_0$ (and lowers $g_1=\frac{w-g_0}{2}$): a donation made at $t=0$ is still in the stock at $t=1$, so a more patient agent gives earlier. With $w=1$, for example, $g_0\approx 0.68$ at $\delta=0.5$ and $g_0\approx 0.75$ at $\delta=0.8$. As $\delta\to 0$ the future stops mattering and $g_0\to\frac{w}{2}$, $g_1\to\frac{w}{4}$.

Optimization with Inequality Constraints

Problem 7

Return to the simultaneous game of Problem 4, now with the constraints $0\le g_i\le w_i$. Use Recipe #2.

Write each constraint in the form $h_j(g_i)\le c_j$, and write individual $i$'s Lagrangian with multipliers $\lambda_1$ (for $g_i\ge 0$) and $\lambda_2$ (for $g_i\le w_i$).
Find $\mathrm{BR}_i(G_{-i})$. Hint: write $\mathrm{BR}_i$ as a piecewise function.
Explain why the constraint $g_i\le w_i$ never binds.
Let $w_1=13$ and $w_2=5$. Find the Nash equilibrium and the value of the multiplier on individual 2's constraint $g_2\ge 0$.

Optimization with Inequality Constraints

Problem 7 — Answers

The constraints are $-g_i\le 0$ and $g_i\le w_i$: $$\begin{aligned}\mathcal{L}&=\sqrt{w_i-g_i}+\sqrt{g_i+G_{-i}}\\ &\quad-\lambda_1\,(-g_i-0)-\lambda_2\,(g_i-w_i).\end{aligned}$$
The first-order condition is $$-\frac{1}{2\sqrt{w_i-g_i}}+\frac{1}{2\sqrt{g_i+G_{-i}}}+\lambda_1-\lambda_2=0.$$ With $\lambda_1=\lambda_2=0$: $g_i=\frac{w_i-G_{-i}}{2}$, valid when $G_{-i}\le w_i$. With $\lambda_1\gt 0$, $\lambda_2=0$: $g_i=0$ and $\lambda_1=\frac{1}{2\sqrt{w_i}}-\frac{1}{2\sqrt{G_{-i}}}$, positive when $G_{-i}\gt w_i$. The cases with $\lambda_2\gt 0$ have no solution (see 3). So $$\mathrm{BR}_i(G_{-i})=\begin{cases}\dfrac{w_i-G_{-i}}{2}&\text{if } G_{-i}\le w_i,\\[6pt] 0&\text{if } G_{-i}\gt w_i.\end{cases}$$
At $g_i=w_i$ private consumption is zero and its marginal utility $\frac{1}{2\sqrt{w_i-g_i}}$ is infinite, so giving everything away is never optimal. Indeed, the interior best response is at most $\frac{w_i}{2}\lt w_i$.
Try $g_2=0$: then $g_1=\mathrm{BR}_1(0)=\frac{13}{2}$, and since $\frac{13}{2}\gt w_2=5$, $\mathrm{BR}_2\big(\tfrac{13}{2}\big)=0$. The Nash equilibrium is $g_1^e=\frac{13}{2}$, $g_2^e=0$, $G^e=\frac{13}{2}$, and the multiplier on $g_2\ge 0$ is $$\lambda_1=\frac{1}{2\sqrt5}-\frac{1}{\sqrt{26}}\approx 0.027\gt 0.$$

Click on the subtitle to reveal answers.

The Public Good Game

Basic Mathematica Commands

Unconstrained Optimization

Optimization with Equality Constraints

Nash Equilibrium of Static Games

Subgame Perfect Nash Equilibrium

Discount Factors

Optimization with Inequality Constraints