Where the Mathematics Stops

(a) For site value $w>0$ the game has a unique equilibrium: the defender defends site 1 with probability $1/(1+w)$, the attacker attacks site 1 with probability $w/(1+w)$, and the value is $w/(1+w)$. At $v=2$: $\frac{1}{3}$, $\frac{2}{3}$ and $\frac{2}{3}$. (b) Against an attacker playing the true equilibrium mixture, the misspecified mixture $\hat{x}=1/(1+v+\epsilon )$ loses nothing, for every $\epsilon >-v$. (c) Against an attacker who best-responds to $\hat{x}$, the loss relative to the value is

first order in $\epsilon$: near $0$ it grows at the rate $1/(1+v{)}^2$ per unit of $|\epsilon |$ on the right and $v/(1+v{)}^2$ on the left (the one-sided derivatives in $\epsilon$ are $1/(1+v{)}^2$ and $-v/(1+v{)}^2$). At $v=2$ the rates are $\frac{1}{9}$ and $\frac{2}{9}$; believing $\hat{v}=3$ costs $\frac{1}{12}$, believing $\hat{v}=1$ costs $\frac{1}{3}$. (d) The defender's model forecasts damage $\hat{v}/(1+\hat{v})$. For $\epsilon \ge 0$ the realised damage against a best-responding attacker equals this forecast exactly; for $\epsilon <0$ it exceeds the forecast by $-\epsilon /(1+v+\epsilon )$. At $\hat{v}=3$ forecast and outcome are both $\frac{3}{4}$; at $\hat{v}=1$ the forecast is $\frac{1}{2}$ and the outcome $1$.

Let $G$ be a finite zero-sum payoff matrix (row player maximises), $E$ a matrix of the same size with ${\max }_{ij}|E_{ij}|\le 1$, and $\hat{x}$ an optimal row mixture of $G+\epsilon E$. Then the exploitability of $\hat{x}$ in the true game satisfies $v(G)-{\min }_j({\hat{x}}^{\top }G{)}_j\le 2|\epsilon |$.

(a) At every admissible parameter point the equilibrium violation rate $q^{\circ }=a/(2a-b)$ and the inspector's payoff $-a^2/(2a-b)$ do not depend on $c$, while the inspection rate $p^{\circ }=d/(c+2d)$ and the inspectee's payoff $d^2/(c+2d)$ strictly decrease in $c$. At base parameters, raising the penalty from $1$ to $3$ moves $p^{\circ }$ from $\frac{1}{3}$ to $\frac{1}{5}$ and the inspectee's payoff from $\frac{1}{3}$ to $\frac{1}{5}$, and leaves $q^{\circ }=\frac{2}{3}$ and the inspector's payoff $-\frac{2}{3}$ unchanged. (b) The decision-theoretic forecast with $p_0=\frac{1}{3}$ predicts violation with probability $1$ for $c<1$, $0$ for $c>1$, and indifference at $c=1$: at $c=3$ it predicts $0$ where equilibrium play gives $\frac{2}{3}$. (c) The equilibrium violation rate responds to the inspector's payoffs instead: with $b=\frac{1}{2}$, $0$, $-\frac{1}{2}$ (a detected violation costing the inspector less, nothing, or rewarding it) $q^{\circ }$ is $\frac{2}{3}$, $\frac{1}{2}$, $\frac{2}{5}$, and $p^{\circ }$ stays $\frac{1}{3}$. (d) Under inspector leadership (${\Gamma }_1$, leader-favourable ties; Proposition 6.9 (p. 257)) the inspectee stays legal at every $c$, and the penalty lowers the threshold the inspector must announce: its value rises from $-\frac{1}{3}$ at $c=1$ to $-\frac{1}{5}$ at $c=3$. (e) In the cross-section the equilibrium violation rates are $\frac{2}{3},\frac{4}{7},\frac{6}{11},\frac{8}{15}$ and the inspection rates $\frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{1}{6}$. The least-squares slope of violation on penalty is $-\frac{82}{1925}\approx -0.0426$; the slope of violation on inspection is $\frac{14192}{17479}\approx 0.8119$, positive. The effect on violation of raising any one jurisdiction's penalty, all else fixed, is exactly $0$.

(a) For every $A>3$ both $(\mathrm{C},\mathrm{C})$ and $(\mathrm{H},\mathrm{H})$ are strict equilibria, and the mixed equilibrium plays $\mathrm{C}$ with probability $p^{\ast }=2/(A-1)$. $(\mathrm{C},\mathrm{C})$ is payoff dominant for every $A>3$; it is risk dominant iff $A>5$ (deviation losses $A-3$ against $2$), with a tie at $A=5$. At $A=4$: $p^{\ast }=\frac{2}{3}$; payoff dominance selects $(\mathrm{C},\mathrm{C})$ and risk dominance $(\mathrm{H},\mathrm{H})$. (b) The shares of initial beliefs leading to each outcome are: under D1, $1-p^{\ast }=(A-3)/(A-1)$ to $(\mathrm{C},\mathrm{C})$ and $p^{\ast }$ to $(\mathrm{H},\mathrm{H})$; under D2, $(1-p^{\ast }{)}^2$ to $(\mathrm{C},\mathrm{C})$, $(p^{\ast }{)}^2$ to $(\mathrm{H},\mathrm{H})$, and $2p^{\ast }(1-p^{\ast })$ to a perpetual alternation between $(\mathrm{C},\mathrm{H})$ and $(\mathrm{H},\mathrm{C})$; under D3, $1-p^{\ast }$ and $p^{\ast }$. At $A=4$: $\frac{1}{3}$ and $\frac{2}{3}$ under D1 and D3; $\frac{1}{9}$, $\frac{4}{9}$ and $\frac{4}{9}$ under D2. (c) Under D1 the risk-dominant equilibrium is the one with the larger basin, and the basin of $(\mathrm{C},\mathrm{C})$ is exactly $\frac{1}{2}$ at the switch $A=5$.

(a) The majority relation cycles, $\mathrm{A}\succ \mathrm{B}\succ \mathrm{C}\succ \mathrm{A}$, each by two factions to one; no transitive “state preference” exists and no policy is a Condorcet winner. (b) The three agendas produce three different sincere outcomes: $\mathrm{C}$ (first $\mathrm{A}$ against $\mathrm{B}$), $\mathrm{B}$ (first $\mathrm{A}$ against $\mathrm{C}$), $\mathrm{A}$ (first $\mathrm{B}$ against $\mathrm{C}$). Sophisticated voting also produces all three, assigned differently: $\mathrm{B}$, $\mathrm{A}$, $\mathrm{C}$. Whoever sets the agenda chooses the policy. (c) Under the impartial culture the probability of a majority cycle is $\frac{1}{18}\approx 0.0556$ for $n=3$, $\frac{5}{72}\approx 0.0694$ for $n=5$, $\frac{875}{11664}\approx 0.0750$ for $n=7$, and $\frac{65485}{839808}\approx 0.0780$ for $n=9$. (d) With a known threshold, government A obtains $x=\max \{\frac{1}{2},r\}$ for $r\le 1$, and there is no ratifiable agreement for $r>1$: a constraint that binds moves the split towards the constrained side. (e) With an unknown threshold, government B offers $x^{\ast }=\min \{\frac{1}{2},\bar{r}\}$, and ratification fails with probability $1-x^{\ast }/\bar{r}$: $0$ at $\bar{r}=\frac{2}{5}$, $\frac{1}{3}$ at $\bar{r}=\frac{3}{4}$, $\frac{1}{2}$ at $\bar{r}=1$. At $\bar{r}=1$ each government's expected payoff is $\frac{1}{4}$.

(a) Structures S and D induce the same joint law of (inspected, violated): $\frac{2}{9}$, $\frac{1}{9}$, $\frac{4}{9}$, $\frac{2}{9}$ on (I, V), (I, L), (N, V), (N, L) — inspection rate $\frac{1}{3}$, violation rate $\frac{2}{3}$, independent. No sample of any size from the observed regime distinguishes them. (b) Under the penalty $c=2$, structure S predicts $p=\frac{1}{4}$, $q=\frac{2}{3}$ and structure D predicts $p=\frac{1}{3}$, $q=\frac{1}{3}$; the total-variation distance between the predicted joint laws is $\frac{1}{3}$. (c) Under the detection reward $b=-\frac{1}{2}$, structure S predicts $q=\frac{2}{5}$ with $p=\frac{1}{3}$ and structure D predicts no change; the distance is $\frac{4}{15}$. (d) Within structure S, $(1,\frac{1}{2},1,1)$ and $(2,1,2,2)$ generate the same data. Raising the penalty by $1$ predicts the inspection rates $\frac{1}{4}$ and $\frac{2}{7}$ respectively; doubling it predicts $\frac{1}{4}$ for both.