Architecture of Decision Support

In Model C1:

• The per-feature likelihood ratios are $\frac{5}{2},\frac{5}{8}$ for $T$ present or absent, $2,\frac{2}{3}$ for $R$ and $4,\frac{2}{3}$ for $D$. The naive-Bayes score is not the true ratio — for the cell with all three features it is $20$ against $30$ — and it orders two pairs of cells differently from the true ratios (Table 11.1 (p. 454)).

• Scorecard A's three classes hold $524$, $2444$ and $7032$ consignments and $140$, $140$ and $120$ violators, with violation densities $\frac{35}{131}$, $\frac{35}{611}$ and $\frac{5}{293}$; they are H $=\{TRD,T\bar{R}D,\bar{T}RD\}$, M $=\{TR\bar{D},\bar{T}\bar{R}D,T\bar{R}\bar{D}\}$ and L $=\{\bar{T}R\bar{D},\bar{T}\bar{R}\bar{D}\}$. Cutting the true ratio order at the same class sizes gives the same three classes: at this resolution the misspecification of the score is invisible.

• Scorecards A and B are not comparable in Blackwell's order: their optimal operating characteristics cross (B is above at false-alarm rate $\frac{1}{40}$, $\frac{3}{10}$ against $\frac{7}{32}$; A is above at $\frac{7}{25}$, $\frac{7}{10}$ against $\frac{14}{25}$). Accordingly, in Model O1's decision with $c_I=1$, the value of information per consignment is $0.0176$ for A and $0.0240$ for B when $c_M=5$, and $0.2632$ for A and $0.2080$ for B when $c_M=20$. Both are garblings of the full eight-cell table, which is worth $0.2828$ at $c_M=20$.

• With a budget of $1500$ inspections spent cell by cell in score order (the last cell partly, at random), the naive-Bayes order detects $196.91$ violators and the true order $209.30$: the misordering costs $12.39$ detections once a budget splits a class.

If no violator adapts ($\lambda =0$), the static rule is optimal: it minimises the loss, which is then the backtest, over all feasible rates. In Model C2 it inspects H at rate $1$, M at $\frac{244}{611}=0.3993$ and L not at all, with backtest loss $302.05$.

At base values, the leader-optimal announcement under leader-favourable ties is

which deters every adaptive violator, at expected loss $\frac{64065}{262}\approx 244.52$. Class $k$ is deterred iff $(g+p)c_k\ge g-e_k$, that is iff $c_k\ge \frac{1}{5},\frac{3}{20},\frac{1}{10}$ for H, M, L; $c^{\star }$ holds M and L exactly at their thresholds and spends the rest of the budget on H. The best announcements that instead accept the adaptive violators into H, M or L lose $358.65$, $329.09$ and $326.53$; the last is the static rule. At $c^{\star }$ the adaptive violators are indifferent between abstaining and presenting in M or L; if they resolve the tie against the authority (presenting in L), the loss is $\frac{88955}{262}\approx 339.52$, exactly $95$ more.

Write ${\theta }_k=(g-e_k)/(g+p)$ for the deterrence threshold of class $k$ and $D={\sum }_k{n}_k{\theta }_k$ for the inspections needed to hold every class at its threshold. If $D\le B$ and deterring all adaptive violators is optimal, the certified rates are $c_k={\theta }_k+y_k$, where $y$ fills the residual budget $B-D$ by the static rule: densest class first, each up to rate $1$. At base values $\theta =(\frac{1}{5},\frac{3}{20},\frac{1}{10})$, $D=1174.6$, and the residual $325.4$ inspections all go to H.

For every follower option the polytope $P_j$ does not depend on $\lambda$, and the loss is affine in $\lambda$ at each vertex. Hence the certified loss is a piecewise-linear function of $\lambda$ with finitely many exactly computable breakpoints. Deterring all adaptive violators is optimal iff $\lambda \ge {\lambda }^{\ast }$, where

where $\Delta$ measures, in loss units, the detections that the deterrent allocation gives up on history. Below ${\lambda }^{\ast }$ the certified allocation is the static rule and the adaptive violators present in L. The static rule's loss grows linearly with $\lambda$, while its backtest does not move (Figure 11.3 (p. 436)).

Suppose the authority measures each class's violation density from one period's inspections, which are drawn at random within each class and so estimate without bias the density of the violators actually present. Under the certified allocation the measured densities are $\frac{105}{524}$, $\frac{105}{2444}$ and $\frac{15}{1172}$ ($0.2004$, $0.0430$, $0.0128$), against historical densities $0.2672$, $0.0573$ and $0.0171$: the deterred violators appear in no class, and L looks safer than history said. Under the static rule L is never inspected, and its density is not measured at all. A static rule retrained on the certified period's measurements orders the classes as before, reproduces the static allocation exactly, and readmits the adaptive violators into L (loss $326.53$). The certified allocation recomputed with the measured counts in place of the historical ones is unchanged.

Split the $100$ adaptive violators into $100-k$ with Model C2's disguise costs and $k$ with cheaper ones, $(0,\frac{1}{5},\frac{2}{5})$, each type choosing its class separately; the authority knows the masses and commits to rates, as in a Bayesian Stackelberg game. For $k\le 15$ the certified allocation is $c^{\star }$: it deters the ordinary type and concedes L to the cheap type, at loss $244.52$ plus $\frac{19}{20}$ per cheap violator ($249.27$ at $k=5$, $258.77$ at $k=15$). For $k\ge 16$ it deters both, with rates $(0.5060,0.16,0.12)$ and loss $259.64$, whatever the split. The switch is at $k^{\ast }=\frac{39600}{2489}\approx 15.91$.

In Model C3 at base values (Table 11.2 (p. 463)):

• The backtest ranks the static rule first ($302.05$) and the certified allocation third ($326.03$), behind the imitation policy; against best-responding violators with leader-favourable ties the certified allocation loses least ($244.52$) and the static rule falls to third ($326.53$), with gaps $82.01$ for the static rule and $125.48$ for the uniform rate.

• The imitation policy keeps the M rate of $c^{\star }$, falls $\frac{1}{200}$ short of it in L and spends the released inspections in H, whose rate rises by $\frac{879}{13100}$ (about $6.7$ points); it has a better backtest ($321.63$ against $326.03$); against best-responding violators its loss is $336.48$ and its gap $91.95$, because the shortfall in L moves the adaptive violators there.

• The gap is not continuous in the implemented rates: as $q_{\mathrm{L}}\uparrow \frac{1}{10}$ with the rest of $c^{\star }$ fixed, the loss tends to $339.52$, while at $q_{\mathrm{L}}=\frac{1}{10}$ it is $244.52$; the jump, $95$, equals the cost of the worst tie in Proposition 11.5 (p. 457) (Figure 11.4 (p. 439)a).

• No margin, no certificate. For any audit that observes $m$ independent decisions on class L — any test, not only threshold tests — the pass probability of a policy with rate $q$ is a polynomial in $q$. Hence for every $\epsilon >0$ there are policies with $q_{\mathrm{L}}<\frac{1}{10}$, each exploitable by Proposition 11.14 (p. 462)(c), that pass with probability within $\epsilon$ of the certified allocation's. At $m=2800$ with the threshold test, rates $\frac{1}{10}-10^{-3}$, $\frac{1}{10}-10^{-5}$, $\frac{1}{10}-10^{-7}$ pass with probability $0.033292$, $0.048772$, $0.048951$, approaching the knife edge's $0.048953$.

• With a margin, a computable certificate. Against the margin allocation ($q_{\mathrm{L}}=\frac{3}{25}$), with $t$ fixed so that a policy at the knife edge passes with probability at most $\frac{1}{20}$ (and every policy with $q_{\mathrm{L}}<\frac{1}{10}$ with less): at $m=400$ the threshold is $t=51$ and the margin policy passes with probability $0.3441$; at $m=1600$, $t=181$ and $0.8112$; at $m=2800$, $t=307$ and $0.9582$. Probabilities are exact rationals, truncated.

• Unlogged release. At the certified allocation every $\omega >0$ moves the adaptive violators into L and the loss jumps from $244.52$ towards $339.52$. At the margin allocation deterrence survives for $\omega \le \frac{1}{6}$ under leader-favourable ties, and for $\omega <\frac{1}{6}$ under pessimistic ones — $\frac{1}{6}$ is the share at which $(1-\omega )\frac{3}{25}=\frac{1}{10}$, L's threshold — while the loss rises linearly on the way, from $261.56$ to $262.46$, through missed detections of the non-adaptive violators in L (Figure 11.5 (p. 441)a).

• Hunches. Inspecting all of H takes $283.32$ inspections from L, whose rate falls to $0.0797$, below its threshold: the adaptive violators move into L and the loss rises from $261.56$ to $331.00$.

• Release on a clear check. Unanticipated, the filter lowers the effective rate for violators in M to $\frac{4}{5}\cdot \frac{17}{100}$, below M's threshold; they move into M and the loss rises to $362.45$, with the saved inspections unspent. Designed in, the recomputed selection rates are $(1,1,\frac{923}{1848})$ and the loss falls to $198.02$, with every adaptive violator strictly deterred.