Leadership and Rules: Inspection

For $1\le m<n$, $V(n,m)$ is the value of the stage matrix

rows inspect now, wait; columns violate now, stay legal now. For $n=2$, $m=1$: $G=\left(\begin{array}{cc}1& -1\\ -1& 0\end{array}\right)$, $p=q=(\frac{1}{3},\frac{2}{3})$, $L=U=-\frac{1}{3}$, so $V(2,1)=-\frac{1}{3}$. For $m=1$ the recursion gives $V(n,1)=-(n-1)/(n+1)$ (Dresher 1962, eq. (8), p. 10); this is checked here for $n\le 8$ (Figure 6.1 (p. 228)).

For $n=2$, $m=1$, let the inspector's plans be inspect event 1, wait and inspect event 2, never inspect, and the inspectee's plans be violate in event 1 and the four plans legal in event 1, then (action if inspected, action if not inspected) in event 2. The $3\times 5$ matrix has value $-\frac{1}{3}$, certified by $p=(\frac{1}{3},\frac{2}{3},0)$ and $q$ putting $\frac{1}{3}$ on violate in event 1 and $\frac{2}{3}$ on legal, then violate if inspected and stay legal if not.

If in every event the realised inspection decision is observed before the violation decision, then $V(n,m)=-1$ for all $1\le m<n$.

The first-event game has no pure equilibrium and a unique mixed one: the inspector inspects with probability $p^{\circ }=d/(c+2d)$ and the inspectee violates with probability $q^{\circ }=a/(2a-b)$. The inspector's payoff is $-a^2/(2a-b)$; the inspectee's is $d^2/(c+2d)$. At base parameters: $p^{\circ }=\frac{1}{3}$, $q^{\circ }=\frac{2}{3}$, payoffs $-\frac{2}{3}$ and $\frac{1}{3}$.

Suppose the inspector announces its first-event inspection probability $p$, the inspectee observes $p$ but not the draw, and responds. Legal is a best reply iff $p\ge p^{\circ }$; which reply is selected at equality depends on the tie convention.

• Leader-favourable ties (the inspectee, when indifferent, stays legal): the optimal announcement is $p=p^{\circ }$ and the inspector obtains $-a{p}^{\circ }=-ad/(c+2d)$; at base parameters $-\frac{1}{3}$, against $-\frac{2}{3}$ without commitment.

• Pessimistic ties (when indifferent, the inspectee violates): the supremum is the same, $-ad/(c+2d)$, but it is not attained; every $p>p^{\circ }$ is feasible and $p\downarrow p^{\circ }$ approaches it.

Under leader-favourable ties the optimal rule deters, inspects after a flag before it inspects after a clear signal, and gives the inspector, if $\beta (c+d)+\alpha d\ge d$,

otherwise $r_f=1$ and

Under pessimistic ties the supremum is the same; it is not attained when every optimal rule leaves the inspectee indifferent, as at base parameters. With $\alpha =0$ and $\theta =d/(c+d)$ the answer depends on $\beta$: if $\beta >\theta$ the value is $0$, every $r_f\ge \theta /\beta$ with $r_c=0$ is optimal, and a strictly deterring rule attains it under both conventions; if $\beta =\theta$ the value $0$ is attained under $(\mathrm{L})$ only; if $\beta <\theta$ then $r_f=1$, $r_c>0$, the value is negative, and under $(\mathrm{P})$ it is not attained. At $a=1$, $b=\frac{1}{2}$, $c=d=1$, $\beta =\frac{1}{4}$: $r_c=\frac{1}{5}$ and $V_2^{\mathrm{L}}=-\frac{1}{5}$; only $\beta =1$ is the case of Remark 6.13 (p. 260). At base parameters with $\beta =\frac{4}{5}$, $\alpha =\frac{1}{5}$: $r_f=\frac{5}{9}$, $r_c=0$, $V_2^{\mathrm{L}}=-\frac{1}{9}$.