**Proof of Lemma 1:** In this simple setting involving a transfer, the Pareto-efficient {*t, π*} pair can be derived by solving the following problem:

$\underset{t,\text{\hspace{0.17em}}\pi}{\text{max}}\text{\hspace{0.17em}}W=[t\mathrm{(}1-\pi \mathrm{)}+\pi \mathrm{(}1-\gamma \mathrm{)}]x,\text{\hspace{0.17em}}\text{s}\text{.t}\text{.\hspace{0.17em}}x=\theta \mathrm{(}1-t\mathrm{)}\mathrm{(}1-\pi \mathrm{)}.\text{\hspace{1em}(A.1)}$

Substituting for *x* in (13) yields:

$W=\theta \mathrm{(}1-t\mathrm{)}\mathrm{(}1-\pi \mathrm{)}[t\mathrm{(}1-\pi \mathrm{)}+\pi \mathrm{(}1-\gamma \mathrm{)}].\text{\hspace{1em}(A.2)}$

Its derivatives with respect to *t* and *π* are:

$\frac{\partial W}{\partial t}=\theta \mathrm{(}1-\pi \mathrm{)}[\mathrm{(}1-2t\mathrm{)}\mathrm{(}1-\pi \mathrm{)}-\pi \mathrm{(}1-\gamma \mathrm{)}],\text{\hspace{1em}(A.3)}$

$\frac{\partial W}{\partial \pi}=\theta \mathrm{(}1-t\mathrm{)}[\mathrm{(}1-\gamma -2t\mathrm{)}-2\pi \mathrm{(}1-\gamma -t\mathrm{)}].\text{\hspace{1em}(A.4)}$

We use a technique similar to a phase diagram to prove that *t*=1/2 and *π*=0 is the unique optimum although *W* is not concave in the whole {*π, t*} space. Define the two loci *π*_{t} and *π*_{π} as the functions of *t* such that (A.3) and (A.4), respectively, are equal to zero:

$\frac{\partial W}{\partial t}=0\text{\hspace{0.17em}if\hspace{0.17em}}\pi ={\pi}_{t}\equiv \frac{1-2t}{2-\gamma -2t}\text{\hspace{0.17em}and\hspace{0.17em}}\frac{\partial W}{\partial \pi}=0\text{\hspace{0.17em}if\hspace{0.17em}}\pi ={\pi}_{\pi}\equiv \frac{1-2t-\gamma}{2\mathrm{(}1-\gamma -t\mathrm{)}}.\text{\hspace{1em}(A.5)}$

Both loci have strictly negative slopes in the relevant range and we can prove that *π*_{t}>*π*_{π}∀*t*∈[0, 1/2]. The loci intersect the axes as shown in Figure A.1, i.e., with *π*_{t} lying further from the origin than *π*_{π} Then the above inequality could only be wrong if the two loci intersected twice for 0<*t*≤(1–*γ*)/2. However, it can be checked that:

${\pi}_{t}-{\pi}_{\pi}=\frac{\gamma \mathrm{(}1-\gamma \mathrm{)}}{2\mathrm{(}2-\gamma -2t\mathrm{)}\mathrm{(}1-\gamma -t\mathrm{)}}>0\text{\hspace{0.17em}if\hspace{0.17em}}t<1-\gamma .\text{\hspace{1em}(A.6)}$

Hence, no intersection exists between *π*_{t} and *π*_{π} in the relevant range.

Then, these two loci divide the relevant range of the {*π, t*} space into three zones where the signs of ∂*W*/∂*t* and ∂*W*/∂*π* are liable to change. We depict the signs of these derivatives by arrows pointing in the direction of the changes in *π* or *t* that increase *W*.

Then, we identify easily two potential candidates for the optimum, namely *E*_{1}={*t*=1/2, *π*=0} and *E*_{2}={*t*=0, *π*=1/2}. However, *E*_{1} is a (one-sided) saddle point from which *W* increases monotonically as *t* increases monotonically and *π* decreases monotonically, i.e., as one moves to the south-east within the corridor between the two loci. This leaves *E*_{2} as the unique optimum.

**Proof of Proposition 2:** Assume that both players adopt the standard trigger strategy (see e.g., Gibbons 1992). If they choose to cooperate and refrain from raiding the traders in the hope of getting a transfer from the government *g*, the potential syndicated bandits get the following present value:

${V}^{C}=\frac{g}{1-{\delta}^{B}}.\text{\hspace{1em}(A.7)}$

If they choose instead to deviate and attack the traders, the syndicated bandits will be punished first within the same period, as the government withholds the transfer *g*, and the static Nash equilibrium outcome will then prevail ever after. Assuming that this deviation was not expected by the traders in the first period, this yields the following present value:

${V}^{D}=\frac{\mathrm{(}1-\gamma \mathrm{)}\theta}{2}+\frac{{\delta}^{B}\mathrm{(}1-\gamma \mathrm{)}\theta}{8\mathrm{(}1-{\delta}^{B}\mathrm{)}}.\text{\hspace{1em}(A.8)}$

The potential bandits will thus refrain from raiding the traders if *V*^{C}≥*V*^{D}, i.e., if:

$g\ge \mathrm{(}1-\gamma \mathrm{)}\theta \mathrm{(}\frac{4-3{\delta}^{B}}{8}\mathrm{)}.\text{\hspace{1em}(A.9)}$

On its part, if the government chooses to cooperate, i.e., to deliver the agreed amount *g* when observing *π*=0, it gets the following present value:

${W}^{C}=\frac{\theta -4g}{4\mathrm{(}1-{\delta}^{G}\mathrm{)}}.\text{\hspace{1em}(A.10)}$

If it chose instead to withhold the transfer in spite of the potential bandits’ compliance with the promised *π*=0, the government would first keep the whole fiscal revenue in the current period, and then get its Nash-equilibrium payoff *G*^{N} from then on. This yields the following present value:

${W}^{D}=\frac{\theta}{4}+\frac{{\delta}^{G}\theta}{16\mathrm{(}1-{\delta}^{G}\mathrm{)}}.\text{\hspace{1em}(A.11)}$

Then, the government will cooperate if *W*^{C}≥*W*^{D}, i.e., if:

$g\le \frac{3\theta {\delta}^{G}}{16}.\text{\hspace{1em}(A.12)}$

Hence, (A.9) and (A.12) give us the range of values of the transfer *g* that can sustain the efficient equilibrium and conditions (15) and (16) ensure that this range is not empty.

Figure A.1Determination of the pareto optimum.

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