Sorting the good guys from bad: on the optimal audit structure with ex-ante information acquisition

Abstract

In a costly state verification model under commitment, the principal may acquire a costly public and imperfectly revealing signal before or after contracting. If the project remains profitable after all signal realizations, optimally the signal is collected, if at all, only after contracting, and it may be acquired randomly or deterministically. Moreover, audit is deterministic and targeted on some signal-state combinations. The paper provides a detailed characterization of the optimal contract and performs a comparative static analysis of signal acquisition strategy and payoffs with respect to enforcement costs and informativeness of the signal. We explore the robustness of the results, including commitment issues. The results are interpreted in light of the observed features of financial contracts, showing that the optimality of the standard debt contract with deterministic audit extends to a setting with random audit.

Appendix

1.1 Signal after the contract

Proof of Proposition 1

The proof proceeds in steps. We first show that the incentive constraints (20) and (21) bind. Then, that $m_{M}^{N}=m_{M}^{\sigma }=0$. Further that also the incentive constraints ( 24) and (25) bind. Last, we prove that (22) and (23) are slack. This allows us to set the contract problem as one in two states, where the top two, $H$ and $M$, are pooled into a single state.

• 1. The incentive constraints (20) and (21) bind. It follows from Border and Sobel (1987) that the bottom non-monitored state has a binding incentive constraint with the top monitored state. So if $m_{M}^{N},m_{M}^{\sigma }>0,$ (20) and (21) must bind. Conversely, if $m_{M}^{N}=m_{M}^{\sigma }=0,$ repayments must be pooled in the top two states and $R_{\hat{H}}^{N}=R_{\hat{ M}}^{N},R_{\hat{H}}^{\sigma }=R_{\hat{M}}^{\sigma }$.

• 2. $m_{M}^{N}=m_{M}^{\sigma }=0$ (this is proved below for the no-signal case, but the proof is analogous for the with-signal case).

  The rationale is that when $m_{M}^{N}>0$ and $R_{\hat{M}}^{N}<f_{M}$, there are savings to be made on audit cost by reducing $m_{M}^{N}$ and raising $R_{ \hat{M}}^{N}$ in a way which has a neutral effect on the participation constraint (keeping $R_{\hat{M}}^{N}\left( 1-m_{M}^{N}\right)$ constant), but slackens (24), reducing the enforcement cost. This goes on until we hit one of the two corners, i.e., either $m_{M}^{N}=0,$ or $R_{ \hat{M}}^{N}=f_{M}$ or both.

  Suppose we first reach $m_{M}^{N}=0$ and $R_{\hat{M}}^{N}<f_{M}.$ Then, $R_{ \hat{M}M}^{N}$ is irrelevant and from (20) binding $R_{\hat{M} }^{N}=R_{\hat{H}}^{N}.$ If in constraint (24) we replace $R_{\hat{H} }^{N}$ with $R_{\hat{M}}^{N}$ and rearrange, this becomes $R_{\hat{M} }^{N}\le \left( 1-m_{L}^{N}\right) f_{L}+m_{L}^{N}f_{M}.$ If we hit $R_{ \hat{M}}^{N}=f_{M}$ with $m_{M}^{N}>0,$ we can reduce both $R_{\hat{H}}^{N}$ and $m_{M}^{N},$ preserving the participation constraint. Doing this, we can attain $m_{M}^{N}=0.$ To see this, replace $R_{\hat{H}}^{N}$ by $f_{M},$ $R_{ \hat{M}}^{N}$ by $f_{M}$ and $m_{M}^{N}$ by zero in the participation constraint (19), to get

  $$\begin{aligned} \left( \pi _{H}+\pi _{M}\right) f_{M}+\pi _{L}f_{L}-I-\left( \alpha \Sigma _{\sigma }\pi _{L\sigma }m_{L}^{\sigma }+\left( 1-\alpha \right) \pi _{L}m_{L}^{N}\right) c_{m}-\alpha c_{a}\ge EP. \end{aligned}$$

  This is positive by assumption 1 and shows that there exists a value of $R_{\hat{H}}^{N}\ge f_{M}$ with $m_{M}^{N}=0,R_{\hat{M}}^{N}=f_{M}$, which satisfies the participation constraint.

• 3. The incentive constraints (24) and (25) bind (this is proved below for the no-signal case, but the proof is analogous for the with-signal case).

  Consider constraints (22) and (24). Suppose we have an optimum with (24) slack and, as we know from Border and Sobel (1987), since the lowest state always has positive monitoring, $m_{L}^{N}>0.$ The optimum must satisfy both the above inequalities and the participation constraint.

  Keep all the $R_{\hat{H}}^{N},R_{\hat{M}}^{N}$ fixed, so the right-hand sides of the incentive constraints are all fixed. Reducing $m_{L}^{N}$ will keep the second inequality valid for small changes but could violate the first inequality (if it was binding initially). Suppose it was binding initially. Then, it is possible to reduce $R_{\hat{H}}^{N}$ too to keep (22) binding. This will also slacken (20). The effect, if possible, is to reduce agency cost. Reducing $R_{\hat{H}}^{N}$ and $m_{L}^{N}$ also has implications for the participation constraint. Can you keep doing this until (24) binds at the given $R_{\hat{M} }^{N}$? If so to keep (22) binding, there will be finite changes $\Delta R_{\hat{H}}^{N},\Delta m_{L}^{N}$ such that

  $$\begin{aligned} \left( f_{H}-f_{L}\right) \left( 1-m_{L}^{N}-\Delta m_{L}^{N}\right)&= f_{H}-R_{\hat{H}}^{N}-\Delta R_{\hat{H}}^{N}, \end{aligned}$$

  (51)
  $$\begin{aligned} \left( f_{M}-f_{L}\right) \left( 1-m_{L}^{N}-\Delta m_{L}^{N}\right)&= f_{M}-R_{\hat{M}}^{N}. \end{aligned}$$

  (52)

  Equality (51) will certainly hold if $\left( f_{H}-f_{L}\right) \Delta m_{L}^{N}\ge \Delta R_{\hat{H}}^{N}$ or $\Delta R_{\hat{H} }^{N}/\Delta m_{L}^{N}\le f_{H}-f_{L}.$ Within a given signal situation, the participation constraint is $\pi _{H}R_{\hat{H}}^{N}+\pi _{M}R_{\hat{M} }^{N}+\pi _{L}\left( f_{L}-m_{L}^{N}c_{m}\right) .$ To prevent this falling with simultaneous reductions in $R_{\hat{H}}^{N}$ and $m_{L}^{N}$ needs $\pi _{H}\Delta R_{\hat{H}}^{N}\ge \pi _{L}c_{m}\Delta m_{L}^{N}$ or $\Delta R_{\hat{H}}^{N}/\Delta m_{L}^{N}\ge \pi _{L}c_{m}/\pi _{H}.$ So we need both $\Delta R_{\hat{H}}^{N}/\Delta m_{L}^{N}\le f_{H}-f_{L}$ and $\Delta R_{\hat{H}}^{N}/\Delta m_{L}^{N}\ge \pi _{L}c_{m}/\pi _{H}.$ We can find such simultaneous changes $\Delta R_{\hat{H}}^{N},\Delta m_{L}^{N}$ if $\pi _{H}(f_{H}-f_{L})\ge \pi _{L}c_{m}.$ We can reduce $m_{L}^{N}$ and $R_{ \hat{H}}^{N}$ to reach a solution where (16) binds if $\pi _{H}(f_{H}-f_{L})\ge \pi _{L}c_{m}$ so long as there are $0\le m_{L}^{N^{*}}\le 1,R_{\hat{H}}^{N^{*}}<f_{H}$ such that

  $$\begin{aligned} \left( f_{H}-f_{L}\right) \left( 1-m_{L}^{N^{*}}\right)&\le f_{H}-R_{ \hat{H}}^{N^{*}} \\ \left( f_{M}-f_{L}\right) \left( 1-m_{L}^{N^{*}}\right)&= f_{M}-R_{\hat{ M}}^{N} \end{aligned}$$

  which would imply

  $$\begin{aligned} \left( f_{M}-f_{L}\right) \left( 1-m_{L}^{N^{*}}\right)&= f_{M}-R_{\hat{ M}}^{N} \\&\le f_{M}-f_{L}\le f_{H}-f_{L} \end{aligned}$$

  But this is always true. So it is always possible to reduce $m_{L}^{N}$ sufficiently to make (24) bind.

• 4. The incentive constraints (22) and (23) are slack. Setting $m_{M}^{N},m_{M}^{\sigma }=0$, $R_{\hat{H}}^{N}=R_{\hat{M}}^{N}$ and (24) binding, the remaining incentive constraints (22) and (24) become, respectively,

  $$\begin{aligned} R_{\hat{M}}^{N}-f_{L}&\le m_{L}^{N}\left( f_{H}-f_{L}\right) ,\quad R_{\hat{M} }^{\sigma }-f_{L}\le m_{L}^{\sigma }\left( f_{H}-f_{L}\right) \\ R_{\hat{M}}^{N}-f_{L}&= m_{L}^{N}\left( f_{M}-f_{L}\right) ,\quad R_{\hat{M} }^{\sigma }-f_{L}=m_{L}^{\sigma }\left( f_{M}-f_{L}\right) \end{aligned}$$

  whence we deduce that if the last set is binding, the first must be slack.

$\square$

The rest of the proof proceeds as follows. We have four variables to choose $m_{L}^{G},m_{L}^{B},m_{L}^{N},\alpha .$ We first show that one set of optimal solution values for $m_{L}^{G},m_{L}^{B}$ is $m_{L}^{G}=1,m_{L}^{B}=0$ for any admissible values of $0\le \alpha ,m_{L}^{N}\le 1.$ In fact if $c_{a}>0$, these are the only optimal values when $\alpha >0$ (Proposition 2). When $\alpha =0$, they are admissible but so are any others (in fact, $m_{L}^{G},m_{L}^{B}$ are then irrelevant). Using $m_{L}^{G}=1,m_{L}^{B}=0$ then gives information about the optimal values of $\alpha ,m_{L}^{N}$ (through the trade-off between $\alpha$ and $m_{L}^{N}$ in the objective and constraint). Proposition 3 shows that if $NC_{a}>0$ and $c_{a}>0$, then optimally $\alpha =0$ and $m_{L}^{N}$ assumes the value of the no-signal contract. Proposition 4 shows that if $NC_{a}<0$ and $c_{a}>0,$ then optimally $\alpha >0$, and depending on the probability distribution of revenues after signal collection, either $m_{L}^{N}=0$ or $m_{L}^{N}=1,$ or $\alpha =1$ and $m_{L}^{N}$ is irrelevant.

Proof of Proposition 2

Suppose that $0<\alpha <1$ and $c_{a}>0.$

• 1. An optimum cannot have all $m_{L}^{\sigma }=m_{L}^{N}=1,\sigma =G,B.$ If it did, the participation constraint (35) would be slack $(\left( \pi _{H}+\pi _{M}\right) f_{M}+\pi _{L}f_{L}-I-\pi _{L}c_{m}-\alpha c_{a}>0$ by assumption 1) and a rent would be needlessly left to the principal. Similarly, it cannot have all $m_{L}^{\sigma }=m_{L}^{N}=0,\sigma =G,B,$ since then the expected repayments would not meet the reservation level of the principal $(f_{L}-I-\alpha c_{a}<0)$. Thus, at least one audit probability must be positive.

• 2. Suppose that there are $m_{L}^{\sigma },m_{L}^{N}>0,\sigma =G,B,$ and $m_{L}^{G}<1$ such that the participation constraint (35) holds. This is not an optimum because it is possible to lower $m_{L}^{B}$ (i.e., lower $R_{\hat{M}}^{B}$) and increase $m_{L}^{G}$ (i.e., increase $R_{ \hat{M}}^{G}$) reducing the expected cost (34) and slackening the participation constraint (35). Indeed, because $\pi _{HG}+\pi _{MG}>\pi _{HB}+\pi _{MB}$, the increase in $m_{L}^{G}\, (R_{\hat{M}}^{G})$ and reduction in $m_{L}^{B}\, (R_{\hat{M}}^{B})$ allow an increase in the expected revenues after a good signal larger than the fall in revenues after the bad signal. Similarly, this shift reduces the with-signal expected audit cost since $\pi _{LG}<\pi _{LB}.$ This can be seen in the following expression:

  $$\begin{aligned}&\frac{1}{f_{H}-f_{L}}\left\{ \left[ \left( \pi _{HG}+\pi _{MG}\right) \left( f_{M}-f_{L}\right) -\pi _{LG}c_{m}\right] dR_{\hat{M}}^{G}\right. \\&\left. +\left( \left( \pi _{HB}+\pi _{MB}\right) \left( f_{M}-f_{L}\right) -\pi _{LB}c_{m}\right) dR_{\hat{M}}^{B}.\right\} \end{aligned}$$

  Since $\left[ \left( \pi _{HG}\!+\!\pi _{MG}\right) \left( f_{H}\!-\!f_{L}\right) \!-\!\pi _{LG}c_{m}\right] \!-\!\left[ \left( \pi _{HB}\!+\!\pi _{MB}\right) \left( f_{H}-f_{L}\right) -\pi _{LB}c_{m}\right]$ $>0,$ the participation constraint is slackened by an increase in $m_{L}^{G}$ matched by an equal reduction in $m_{L}^{B}.$ This allows a further reduction in $m_{L}^{B}$ and thus a reduction in the objective function. So an optimum cannot have $m_{L}^{G}<1$ and $m_{L}^{B}>0$ and must at least entail one of $m_{L}^{G}=1$ or $m_{L}^{B}=0.$

• 3. Suppose that $m_{L}^{G}=1$ but $1>m_{L}^{B}>0.$ Then, if $m_{L}^{N}<1,$ it is possible to vary $m_{L}^{N}$ and $m_{L}^{B}$ keeping the participation constraint constant

  $$\begin{aligned} \mathrm{d}m_{L}^{N}=-\frac{\alpha \left[ \left( \pi _{HB}+\pi _{MB}\right) \left( f_{M}-f_{L}\right) -\pi _{LB}c_{m}\right] }{\left( 1-\alpha \right) \left[ \left( \pi _{H}+\pi _{M}\right) \left( f_{M}-f_{L}\right) -\pi _{L}c_{m} \right] }\mathrm{d}m_{L}^{B}<0 \end{aligned}$$

  (53)

  Thus, $m_{L}^{B}$ and $m_{L}^{N}$ must vary in opposite directions to satisfy the participation constraint. The effect of such variations in the objective function (34) is $dEC=\alpha c_{m}\left[ -\pi _{LG}\mathrm{d}m_{L}^{N}+\pi _{LB}(\mathrm{d}m_{L}^{B}-\mathrm{d}m_{L}^{N})\right] +\pi _{L}c_{m}\mathrm{d}m_{L}^{N},$ which, using (53), becomes $dEC=\alpha c_{m}\frac{\rho (f_{M}-f_{L})}{\left( \pi _{H}+\pi _{M}\right) \left( f_{M}-f_{L}\right) -\pi _{L}c_{m}}\mathrm{d}m_{L}^{B},$ where $\rho$ is the correlation index (4). Since the coefficient of $\mathrm{d}m_{L}^{B}$ is positive, it is best to reduce $m_{L}^{B}$ and raise $m_{L}^{N}$ as far as possible.

  • (a) One possibility is to hit $m_{L}^{B}=0$ with $m_{L}^{N}>0.$ This proves the result.

  • (b) Alternatively, $m_{L}^{N}=1$ with $m_{L}^{B}>0$ still. In this last case, the optimization problem becomes

    $$\begin{aligned}&\min \alpha \left[ c_{a}-\pi _{LB}c_{m}\left( 1-m_{L}^{B}\right) \right] +\pi _{L}c_{m} \\&\text {s.t. }-\alpha \left\{ \left( 1-m_{L}^{B}\right) \left[ \left( \pi _{HB}+\pi _{MB}\right) (f_{M}-f_{L})-\pi _{LB}c_{m}\right] +c_{a}\right\} \\&\quad +\left( \left( \pi _{H}+\pi _{M}\right) f_{M}+\pi _{L}f_{L}-\pi _{L}c_{m}\right) =I \end{aligned}$$

    It is then possible to vary $\alpha$ and $m_{L}^{B}$ keeping the participation constraint constant

    $$\begin{aligned} \mathrm{d}\alpha =\frac{\alpha \left\{ \left[ \left( \pi _{HB}+\pi _{MB}\right) (f_{M}-f_{L})-\pi _{LB}c_{m}\right] \right\} }{\left( 1-m_{L}^{B}\right) \left[ \left( \pi _{HB}+\pi _{MB}\right) (f_{M}-f_{L})-\pi _{LB}c_{m}\right] +c_{a}}\mathrm{d}m_{L}^{B}>0 \qquad \end{aligned}$$

    (54)

    The effect on the objective function is $dEC=\left[ c_{a}-\pi _{LB}c_{m}\left( 1-m_{L}^{B}\right) \right] \mathrm{d}\alpha +\alpha \pi _{LB}c_{m}\mathrm{d}m_{L}^{B},$ whence using (54) gives

    $$\begin{aligned} dEC=\frac{\alpha c_{a}\left( \pi _{HB}+\pi _{MB}\right) \left( f_{M}-f_{L}\right) }{\left( 1-m_{L}^{B}\right) \left[ \left( \pi _{HB}+\pi _{MB}\right) (f_{M}-f_{L})-\pi _{LB}c_{m}\right] +c_{a}}\mathrm{d}m_{L}^{B}>0. \end{aligned}$$

    It is possible to reduce $m_{L}^{B}$ to zero while keeping $\alpha >0$ since, with $m_{L}^{G}=m_{L}^{N}=1,\,m_{L}^{B}=0,$ the participation constraint becomes $\alpha EPC_{\sigma }+\left( 1-\alpha \right) \left[ \left( \pi _{H}+\pi _{M}\right) f_{M}+\pi _{L}f_{L}-I-\pi _{L}c_{m}\right] \ge 0.$ Since $\left( \pi _{H}+\pi _{M}\right) f_{M}+\pi _{L}f_{L}-I-\pi _{L}c_{m}>0,$ this can be satisfied with $\alpha >0.$

• 4. Alternatively, suppose the optimum has $m_{L}^{B}=0$ but $0<m_{L}^{G}<1$ and $m_{L}^{N}>0.$ It is then possible to reduce $m_{L}^{N},$ raise $m_{L}^{G}$ so as to keep the participation constraint constant: $\mathrm{d}m_{L}^{N}=- \frac{\alpha \left\{ \left( \pi _{HG}+\pi _{HB}\right) (f_{M}-f_{L})-\pi _{LG}c_{m}\right\} }{(1-\alpha )\left\{ \left( \pi _{H}+\pi _{M}\right) (f_{M}-f_{L})-\pi _{L}c_{m}\right\} }\mathrm{d}m_{L}^{G}.$ The change in objective ( 34) is $dEC=\alpha c_{m}\frac{-\rho (f_{H}-f_{L})}{\left( \pi _{H}+\pi _{M}\right) (f_{M}-f_{L})-\pi _{L}c_{m}}\mathrm{d}m_{L}^{G}<0.$ So it is best to raise $m_{L}^{G}$ to maximum and reduce $m_{L}^{N}.$

  • (a) One possibility is that we hit $m_{L}^{G}=1$ with $m_{L}^{N}>0,$ which proves the result.

  • (b) The other possible case has $m_{L}^{N}=0$ with $m_{L}^{G}<1$ still. In this last case, the optimization problem becomes

    $$\begin{aligned}&\displaystyle \min \alpha \left( c_{a}+\pi _{LG}m_{L}^{G}c_{m}\right) \\&\displaystyle \text {s.t. }\alpha \left\{ \left[ \left( \pi _{HG}+\pi _{HB}\right) \left( f_{M}-f_{L}\right) -\pi _{LG}c_{m}\right] m_{L}^{G}-c_{a}\right\} =I-f_{L}. \end{aligned}$$

    Varying $m_{L}^{G}$ and $\alpha$ so as to keep the participation constraint constant gives $\mathrm{d}m_{L}^{G}=-\frac{\left( \left( \pi _{HG}+\pi _{HB}\right) \left( f_{M}-f_{L}\right) -\pi _{LG}c_{m}\right) m_{L}^{G}-c_{a}}{\alpha \left[ \left( \pi _{HG}+\pi _{HB}\right) \left( f_{M}-f_{L}\right) -\pi _{LG}c_{m}\right] }\mathrm{d}\alpha .$ Using the resulting $\mathrm{d}m_{L}^{G}$ in the objective function $dEC=\left( c_{a}+c_{m}\pi _{LG}m_{L}^{G}\right) \mathrm{d}\alpha +\alpha c_{m}\pi _{LG}\mathrm{d}m_{L}^{G}$ gives $dEC=\frac{\left( \pi _{HG}+\pi _{HB}\right) \left( f_{M}-f_{L}\right) c_{a}}{\left( \pi _{HG}+\pi _{HB}\right) \left( f_{M}-f_{L}\right) -\pi _{LG}c_{m}}\mathrm{d}\alpha >0.$ So it is best to reduce $\alpha$ and raise $m_{L}^{G}$ as far as possible. In fact, we can attain $m_{L}^{G}=1$ with $\alpha >0.$

Hence, the optimum must involve $m_{L}^{G}=1,m_{L}^{B}=0$.$\square$

Proposition 2 is valid if $\alpha >0.$ If $\alpha =0,m_{L}^{\sigma }$ have no impact on participation constraint or objective and so without loss of generality, we can also set them to these values if $\alpha =0.$

To prove Propositions (3) and (4), we use the following lemma.

Lemma 1

When $0\le \alpha <1$ and $0<m_{L}^{N}<1$ are optimal, if $EPC_{\sigma }\ne 0,$ a simultaneous variation in $\alpha$ and $m_{L}^{N}$ keeping the participation constraint constant changes expected enforcement cost according to

$$\begin{aligned} dEC\!=\!\frac{\left\{ \left( \pi _{H}\!+\!\pi _{M}\right) c_{a}\!-\!\left[ \pi _{LB}\left( \pi _{HG}\!+\!\pi _{MG}\right) \!-\pi _{LG}\left( \pi _{HB}\!+\!\pi _{MB}\right) \right] c_{m}\right\} \left( f_{M}\!-\!f_{L}\right) }{\left( \pi _{H}\!+\!\pi _{M}\right) \left( f_{M}\!-\!f_{L}\right) \!-\!\pi _{L}c_{m}}\mathrm{d}\alpha , \end{aligned}$$

whose sign depends on the sign of $\left( \pi _{H}+\pi _{M}\right) c_{a}-\rho c_{m}\equiv NC_{a}.$

Proof

Consider Program $\mathcal {P}_{EC-\mathrm{after}}^{\prime }.$ The participation constraint (35) is

$$\begin{aligned}&\alpha \left\{ \left( \pi _{HG}+\pi _{MG}\right) f_{M}+\left( 1-\pi _{HG}-\pi _{MG}\right) f_{L}-I-\pi _{LG}c_{m}-c_{a}\right\} \\&\quad +\left( 1-\alpha \right) \left\{ \left( \pi _{H}+\pi _{M}\right) R_{\hat{M} }^{N}+\pi _{L}\left( f_{L}-m_{L}^{N}c_{m}\right) -I\right\} =0 \end{aligned}$$

which can also be written as:

$$\begin{aligned}&\alpha \left\{ \left( \pi _{HG}+\pi _{MG}\right) f_{M}+\left( 1-\pi _{HG}-\pi _{MG}\right) f_{L}-I-\pi _{LG}c_{m}-c_{a}\right\} +\left( 1-\alpha \right) \\&+\left\{ \left( \pi _{H}+\pi _{M}\right) \left( f_{M}-f_{L}\right) m_{L}^{N}+\left( \pi _{H}+\pi _{M}\right) f_{L}+\pi _{L}\left( f_{L}-m_{L}^{N}c_{m}\right) -I\right\} =0 \end{aligned}$$

Suppose $0\le \alpha <1$ and $0<m_{L}^{N}<1$ are optimal. When $EPC_{\sigma }\ne 0$ simultaneous variation in $\alpha ,m_{L}^{N}$ to keep the participation constraint constant must satisfy

$$\begin{aligned} \left[ EPC_{\sigma }-\left( PC_{N}-I\right) \right] \mathrm{d}\alpha +\left( 1-\alpha \right) \left[ \left( \pi _{H}+\pi _{M}\right) \left( f_{M}-f_{L}\right) -\pi _{L}c_{m}\right] \mathrm{d}m_{L}^{N}=0 \end{aligned}$$

$$\begin{aligned} \frac{\mathrm{d}m_{L}^{N}}{\mathrm{d}\alpha }=-\frac{EPC_{\sigma }-\left( PC_{N}-I\right) }{ \left( 1-\alpha \right) \left[ \left( \pi _{H}+\pi _{M}\right) \left( f_{M}-f_{L}\right) -\pi _{L}c_{m}\right] } \end{aligned}$$

(55)

While the sign of the denominator is positive by Assumption 5, the sign of the numerator is ambiguous. However, it can be inferred from the sign of $EPC_{\sigma }.$ For $\alpha \in \left( 0,1\right) ,$ to satisfy the participation constraint with equality, $PC_{N}-I\lesseqgtr 0$ if $EPC_{\sigma }\gtreqless 0.$ Thus, if

• $EPC_{\sigma }>0,$ $\frac{\mathrm{d}m_{L}^{N}}{\mathrm{d}\alpha }|_{PC}<0;$

• $EPC_{\sigma }<0,$ $\frac{\mathrm{d}m_{L}^{N}}{\mathrm{d}\alpha }|_{PC}>0.$

The simultaneous variation in $m_{L}^{N},\alpha$ changes the expected enforcement cost by $\left( c_{a}+\pi _{LG}c_{m}-\pi _{L}m_{L}^{N}c_{m}\right) \mathrm{d}\alpha +\left( 1-\alpha \right) \pi _{L}c_{m}\mathrm{d}m_{L}^{N},$ i.e., using 55,

$$\begin{aligned} dEC=\frac{\left\{ \left( \pi _{H}\!+\!\pi _{M}\right) c_{a}\!-\!\left[ \pi _{LB}\left( \pi _{HG}\!+\!\pi _{MG}\right) \!-\!\pi _{LG}\left( \pi _{HB}\!+\!\pi _{MB}\right) \right] c_{m}\right\} \left( f_{M}\!-\!f_{L}\right) }{\left( \pi _{H}\!+\!\pi _{M}\right) \left( f_{M}\!-\!f_{L}\right) \!-\!\pi _{L}c_{m}}\mathrm{d}\alpha .\nonumber \\ \end{aligned}$$

(56)

• If $EPC_{\sigma }=0,\,PC=(1-\alpha )\left\{ m_{L}^{N}\left[ \left( \pi _{H}+\pi _{M}\right) \left( f_{M}-f_{L}\right) -\pi _{L}c_{m}\right] -I+f_{L}\right\}$ and given $\alpha <1$ we must have $m_{L}^{N}\left[ \left( \pi _{H}+\pi _{M}\right) \left( f_{M}-f_{L}\right) -\pi _{L}c_{m} \right] -I+f_{L}=0.$ In this case, no marginal compensation between $m_{L}^{N},\alpha$ is possible.

$\square$

Proof of Proposition 3

If $NC_{a}>0,$ from (56), enforcement costs are minimized by reducing $\alpha .$ However, the change in $m_{L}^{N}$ following a change in $\alpha$ necessary to preserve the participation constraint depends on the sign of $EPC_{\sigma }.$ Three cases can arise:

• 1. If $EPC_{\sigma }>0,$ $\frac{\mathrm{d}m_{L}^{N}}{\mathrm{d}\alpha }|_{PC}<0.$ Thus, $m_{L}^{N}$ and $\alpha$ vary in opposite directions to satisfy the participation constraint. So, to minimize the enforcement cost, it is best to raise $m_{L}^{N}$ and decrease $\alpha$. However, an optimum cannot have $m_{L}^{N}=1$ and $\alpha >0$, because the participation constraint would be slack ($\alpha EPC_{\sigma }+(1-\alpha )\left( PC_{N}-I\right) =0.$ If $m_{L}^{N}=1,\,PC_{N}-I=\left( \pi _{H}+\pi _{M}\right) f_{M}+\pi _{L}f_{L}-I-\pi _{L}c_{m},$ which is positive by Assumption 1. Given $EPC_{\sigma }>0,\,PC>0$). The firm can then reduce $m_{L}^{N}$ until the participation constraint binds, giving $PC_{N}-I<0$. But because $\frac{ dEC}{\mathrm{d}m_{L}^{N}}=-\frac{\left( 1-\alpha \right) \left( f_{M}-f_{L}\right) \left( \left( \pi _{H}+\pi _{M}\right) c_{a}-\rho c_{m}\right) }{EPC_{\sigma }-\left( PC_{N}-I\right) }<0,$ it is best to raise $m_{L}^{N}$ and lower $\alpha .$ However, $m_{L}^{N}$ only has to rise to balance $PC$ and we know that the no-signal solution with $\alpha =0$ is achievable with $m_{L}^{N}<1.$ If $\alpha$ were positive but small, there would be further gains to be realized. Thus, optimally $\alpha =0,$ with $R_{\hat{M}}^{N}=R_{ \hat{M}}^{N}|_{\alpha =0}$ (40).

• 2. If $EPC_{\sigma }<0,\,\frac{\mathrm{d}m_{L}^{N}}{\mathrm{d}\alpha }|_{PC}>0.$ Now, $\alpha ,m_{L}^{N}$ must both fall to balance $PC$. Again, we know that we can achieve $\alpha =0$ with $PC=0$ at the no-signal solution. And if we left $\alpha >0$, there would still be gains to be realized from reducing both $\alpha ,m_{L}^{N}$. Thus, the solution must have $m_{L}^{N}>0,\,R_{ \hat{M}}^{N}=R_{\hat{M}}^{N}|_{\alpha =0}$ (40), and $\alpha =0.$

• 3. If $EPC_{\sigma }=0,\,PC=(1-\alpha )(PC_{N}-I).$ One possibility is $\alpha =1,$ in which case $m_{L}^{N}$ is immaterial, the other is $PC_{N}=I,$ in which case, since $c_{a}>0,$ it is best to set $\alpha =0$ and $R_{\hat{M}}^{N}=R_{\hat{M}}^{N}|_{\alpha =0}$ (40). Of these two possibilities, if $NC_{a}>0$, it is best to set $\alpha =0.$

In either case, enforcement costs are $EC_{\alpha =0}=\pi _{L}m_{L}^{N}c_{m}$ (41).$\square$

Proof of Proposition 4

If $NC_{a}<0,$ from (56) $\frac{dEC}{\mathrm{d}\alpha }<0$, enforcement costs are minimized by raising $\alpha .$ However, the change in $m_{L}^{N}$ following a change in $\alpha$ necessary to preserve the participation constraint depends on the sign of $EPC_{\sigma }.$ Three cases can arise:

• 1. If $EPC_{\sigma }>0,$ from (55) $\frac{\mathrm{d}m_{L}^{N}}{\mathrm{d}\alpha } |_{PC}<0.$ Thus, $m_{L}^{N}$ and $\alpha$ vary in opposite directions to satisfy the participation constraint. To minimize the enforcement cost, it is best to reduce $m_{L}^{N}$ as far as possible resulting in $m_{L}^{N}=0$ ( $R_{\hat{M}}^{N}=f_{L}$) and raise $\alpha$ just far enough to satisfy the participation constraint ($\alpha =\underline{\alpha }$). Setting $\alpha =1$ (and $m_{L}^{N}$ interior) is not optimal as it would give a slack participation constraint. The enforcement costs are $\underline{\alpha } \left( c_{a}+\pi _{LG}c_{m}\right)$.

• 2. If $EPC_{\sigma }=0,$ since $NC_{a}<0$, now it is best to set $\alpha =1$ and then $m_{L}^{N}$ is immaterial. Enforcement costs are $c_{a}+\pi _{LG}c_{m}$.

• 3. If $EPC_{\sigma }<0,\,\frac{\mathrm{d}m_{L}^{N}}{\mathrm{d}\alpha }|_{PC}>0.$ Thus, $m_{L}^{N}$ and $\alpha$ vary in the same direction to satisfy the participation constraint and $\alpha$ has to be increased to minimize the enforcement cost. Setting $\alpha =1$ is not optimal as it would violate participation constraint. Thus, $m_{L}^{N}$ must be raised as far as possible, giving $m_{L}^{N}=1$ and $\alpha$ must be raised sufficiently to satisfy the participation constraint $(\alpha =\bar{\alpha })$. The enforcement costs are $\bar{\alpha }\left( c_{a}+\pi _{LG}c_{m}\right) +\left( 1-\bar{\alpha }\right) \pi _{L}c_{m}$.

To see why policing involves concentrated audit with $m_{L}^{G}=1$ and $\alpha$ interior, consider the enforcement costs when $m_{L}^{G}$ is interior: $EC=\alpha \left( c_{a}+\pi _{LG}m_{L}^{G}c_{m}\right) .$ Proportional and opposite changes in $m_{L}^{G}$ or $\alpha$ have identical expected audit cost effects, but in addition the expected signal acquisition cost is $\alpha c_{a}$. Thus, the extra cost of a proportional increase in the audit probability $m_{L}^{G}$ is more than offset by the saving induced by a reduction in signal acquisition frequency $\alpha$. It is then optimal to increase the audit probability all the way until $m_{L}^{G}=1$ and reduce the chance of signal acquisition until the participation constraint binds. Thus, it is the direct cost of collecting the signal that ultimately makes the signal acquisition strategy more expensive than auditing and leads to deterministic audits.

If the project is not ex-ante profitable after a signal $(EPC_{\sigma }<0)$, then high state repayments in no-signal states must be increased to satisfy the participation constraint, thus necessitating audit of no-signal states too. Signal contingent deterministic monitoring is still optimal $( m_{L}^{G}=1,m_{L}^{B}=0),$ but now signal acquisition has effects on the relative expected audit costs of good signal and no-signal low reports. The optimal combination $\alpha ,m_{L}^{N}$ minimizes $EC=\alpha \left( c_{a}+\pi _{LG}c_{m}\right) +\left( 1-\alpha \right) \pi _{L}m_{L}^{N}c_{m}.$ In trading off $\alpha$ and $m_{L}^{N},$ each has a specific cost $(c_{a}+\pi _{LG}c_{m})$ for $\alpha$ and $\pi _{L}m_{L}^{N}c_{m}$ for $m_{L}^{N}$, as well as a common saving of $\alpha \pi _{L}m_{L}^{N}c_{m},$ since, when there is a signal, $m_{L}^{N}$ is not applied. From the fact that $NC_{a}<0$ and the signal is relatively cheap, it turns out that the cost saving from reducing $\alpha$ exceeds that from reducing $m_{L}^{N}$ and it is optimal to increase the audit probability all the way until $m_{L}^{N}=1$ and use $\alpha$ just to satisfy the participation constraint.$\square$

Proof of Proposition 5

Let $c_{m}^{*},c_{a}^{*}$ (44) solve $EP=EPC_{\sigma }$. Similarly, let $\widehat{c}_{m},\widehat{c}_{a}$ (45) solve $EPC_{\sigma }=NC_{a}.$ Feasibility requires $c_{a},c_{m}$ are below the $EP$ line The difference $c_{a}^{*}-\widehat{c}_{a}$ (which has the opposite sign of $c_{m}^{*}-\widehat{c}_{m}$) is equal to $\rho \left( \pi _{HG}+\pi _{MG}\right) \left( f_{M}-f_{L}\right) -\left( \pi _{H}+\pi _{M}\right) \pi _{LB}(I-f_{L})$. It is increasing in $\rho$ and zero when $\rho =\rho ^{*}\equiv \frac{\pi _{LB}\left( \pi _{H}+\pi _{M}\right) \left( I-f_{L}\right) }{\left( \pi _{HG}+\pi _{MG}\right) \left( f_{M}-f_{L}\right) }.$ Hence, if $\rho \le \rho ^{*}$, we have $c_{a}^{*}\le \widehat{c}_{a}$ and all three parts of Proposition 4 apply. Conversely, if $\rho >\rho ^{*},c_{a}^{*}> \widehat{c}_{a}$ and $c_{m}^{*}<\widehat{c}_{m}.$ In this case, the only feasible points $c_{a},c_{m}$ below the $EP$ line and with $NC_{a}\le 0$ must entail $EPC_{\sigma }>0$ and hence, whenever $NC_{a}\le 0,$ the signal strategy can only be that described by $MS_{a},$ in Proposition 4. $\square$

Comparative statics: Signal strategy as a function of acquisition and audit cost. We first plot $\alpha$ against $c_{a}$ and $c_{m}.$ When $NC_{a}<0$ and $EPC_{\sigma }\ge 0,\, \tfrac{\partial \underline{\alpha }}{\partial c_{m}}= \tfrac{\pi _{LG}\left( I-f_{L}\right) }{\left( \left( 1-\pi _{LG}\right) \left( f_{M}-f_{L}\right) -\pi _{LG}c_{m}-c_{a}\right) ^{2}}>0,\,\tfrac{ \partial ^{2}\underline{\alpha }}{\partial c_{m}^{2}}=\tfrac{2\pi _{LG}^{2}\alpha }{\left( \left( 1-\pi _{LG}\right) \left( f_{M}-f_{L}\right) -\pi _{LG}c_{m}-c_{a}\right) ^{2}}>0.$

When $EPC_{\sigma }<0,\,\tfrac{\partial \bar{\alpha }}{\partial c_{m}}=- \tfrac{\pi _{L}-\alpha \pi _{LB}}{\left( \left( 1-\pi _{LB}\right) \left( f_{M}-f_{L}\right) -\pi _{LB}c_{m}+c_{a}\right) ^{2}}<0,\,\tfrac{\partial ^{2}\bar{\alpha }}{\partial c_{m}^{2}}=-\tfrac{2\pi _{LB}\left( \pi _{L}-\alpha \pi _{LB}\right) }{\left( \left( 1-\pi _{LB}\right) \left( f_{M}-f_{L}\right) -\pi _{LB}c_{m}+c_{a}\right) ^{2}}<0.$ At its maximum, the slope of $\alpha \left( c_{m}\right)$ changes from $\frac{\partial \alpha }{\partial c_{m}}=\frac{\pi _{LG}}{I-f_{L}}$ to $\frac{\partial \alpha }{\partial c_{m}}=\frac{\pi _{LG}^{2}}{\rho \left( f_{M}-f_{L}\right) -\pi _{LB}\left( I-f_{L}\right) -\pi _{L}c_{a}}<0$.²¹

When $EPC_{\sigma }\ge 0,\,\tfrac{\partial \underline{\alpha }}{\partial c_{a}}=\tfrac{I-f_{L}}{\left( \left( \pi _{HG}+\pi _{MG}\right) \left( f_{M}-f_{L}\right) -\pi _{LG}c_{m}-c_{a}\right) ^{2}}>0,$ $\tfrac{\partial ^{2}\underline{\alpha }}{\partial c_{a}^{2}}=\tfrac{2\left( I-f_{L}\right) }{ \left( \left( \pi _{HG}+\pi _{MG}\right) \left( f_{M}-f_{L}\right) -\pi _{LG}c_{m}-c_{a}\right) ^{3}}>0.$

When $EPC_{\sigma }<0,\, \tfrac{\partial \bar{\alpha }}{\partial c_{a}}=- \tfrac{Ef-I-\pi _{L}c_{m}}{\left( \left( \pi _{HB}+\pi _{MB}\right) \left( f_{M}-f_{L}\right) -\pi _{LB}c_{m}+c_{a}\right) ^{2}}<0,\,\tfrac{\partial ^{2}\bar{\alpha }}{\partial c_{a}^{2}}=\tfrac{2\left( Ef-I-\pi _{L}c_{m}\right) }{\left( \left( \pi _{HB}+\pi _{MB}\right) \left( f_{M}-f_{L}\right) -\pi _{LB}c_{m}+c_{a}\right) ^{3}}>0.$

Since utility is given by $U=Ef-I-EC=\pi _{H}f_{H}+\pi _{M}f_{M}+\pi _{L}f_{L}-I-EC,$ where $EC$ is the cost function (34), we can easily plot $U$ against $c_{a}$ and $c_{m}.$ For $c_{a}>0$ and increasing $c_{m},$ optimally $\alpha =0$ and utility is given by $U_{\alpha =0}=\frac{\left( \pi _{H}+\pi _{M}\right) \left( f_{M}-f_{L}\right) \left( Ef-I-\pi _{L}c_{m}\right) }{\left( \pi _{H}+\pi _{M}\right) \left( f_{M}-f_{L}\right) -\pi _{L}c_{m}}$. Then, $\tfrac{ \partial U_{\alpha =0}}{\partial c_{m}}=-\tfrac{\left( \pi _{H}+\pi _{M}\right) \pi _{L}\left( f_{M}-f_{L}\right) \left( I-f_{L}\right) }{\left( \left( \pi _{H}+\pi _{M}\right) \left( f_{M}-f_{L}\right) -\pi _{L}c_{m}\right) ^{2}}<0;$ $\tfrac{\partial U_{\alpha =0}^{2}}{\partial c_{m}^{2}}=-\tfrac{2\left( \pi _{H}+\pi _{M}\right) \pi _{L}^{2}\left( f_{M}-f_{L}\right) \left( I-f_{L}\right) }{\left( \left( \pi _{H}+\pi _{M}\right) \left( f_{M}-f_{L}\right) -\pi _{L}c_{m}\right) ^{3}}<0.$

For higher $c_{m},$ optimally $\alpha >0$ and utility is given by $U_{MS_{a}}\,=\frac{\left( f_{M}-f_{L}\right) \left[ \left( \pi _{HB}+\pi _{MB}\right) \left( I-f_{L}\right) +\left( \pi _{H}+\pi _{M}\right) EPC_{\sigma }\right] }{\left( \pi _{HG}+\pi _{MG}\right) \left( f_{M}-f_{L}\right) -\pi _{LG}c_{m}-c_{a}}$ and $\tfrac{\partial U_{MS_{a}}}{ \partial c_{m}}=-\tfrac{\pi _{LG}\left( \pi _{HG}+\pi _{MG}\right) \left( f_{M}-f_{L}\right) \left( I-f_{L}\right) }{\left( \left( \pi _{HG}+\pi _{MG}\right) \left( f_{M}-f_{L}\right) -\pi _{LG}c_{m}-c_{a}\right) ^{2}}<0;$ $\tfrac{\partial U_{MS_{a}}^{2}}{\partial c_{m}^{2}}=-\tfrac{2\pi _{LG}^{2}\left( \pi _{HG}+\pi _{MG}\right) \left( f_{M}-f_{L}\right) \left( I-f_{L}\right) }{\left( \left( \pi _{HG}+\pi _{MG}\right) \left( f_{M}-f_{L}\right) -\pi _{LG}c_{m}-c_{a}\right) ^{2}}<0.$

Last, for sufficiently high $c_{m},$ optimally still $\alpha >0,$ but utility is given by $U_{MS_{b}}=\frac{\left( Ef-I-\pi _{L}c_{m}\right) \left( \pi _{HB}+\pi _{MB}\right) \left( f_{M}-f_{L}\right) }{\left( \pi _{HB}+\pi _{MB}\right) \left( f_{M}-f_{L}\right) -\pi _{LB}c_{m}+c_{a}}\,$ and $\tfrac{\partial U_{MS_{b}}}{\partial c_{m}}=\tfrac{\left( \pi _{HB}+\pi _{MB}\right) \left( f_{M}-f_{L}\right) \left\{ \pi _{LB}EPC_{\sigma }-\pi _{LG}\left[ \left( \pi _{HB}+\pi _{MB}\right) \left( f_{M}-f_{L}\right) -\pi _{LB}c_{m}+c_{a}\right] \right\} }{\left( \left( \pi _{HB}+\pi _{MB}\right) \left( f_{M}-f_{L}\right) -\pi _{LB}c_{m}+c_{a}\right) ^{2}}<0;\,\tfrac{ \partial U_{MS_{b}}^{^{2}}}{\partial c_{m}^{2}}=-\tfrac{2\pi _{LB}\left( \pi _{HB}+\pi _{MB}\right) \left( f_{M}-f_{L}\right) \left\{ \pi _{LB}EPC_{\sigma }-\pi _{LG}\left[ \left( \pi _{HB}+\pi _{MB}\right) \left( f_{M}-f_{L}\right) -\pi _{LB}c_{m}+c_{a}\right] \right\} }{\left( \left( \pi _{HB}+\pi _{MB}\right) \left( f_{M}-f_{L}\right) -\pi _{LB}c_{m}+c_{a}\right) ^{3}}<0.$

The pattern of utility for varying $c_{a}$ and $c_{m}$ is depicted in Figs. 8 and 9.

Fig. 8

$U\left( c_{m}\right)$ for $c_{a}>0$

Fig. 9

$U\left( c_{a}\right)$ for $c_{m}>0$

1.2 Signal before the contract

Proof of Proposition 6

When $U^{B}>0,$ using program $\mathcal {P}_{b4}^{^{\prime }}$, solving the participation constraint for $R_{\hat{M}}^{\sigma }$ (48) and substituting out both in $m_{L}^{\sigma }$ and in the utility function $U^{\sigma }$ gives $m_{L|b4}^{\sigma }$ and $EU_{b4}$ reported in Proposition 6 (47 and 49), respectively). Notice that $m_{L|b4}^{G}-m_{L|b4}^{B}=\tfrac{-\left( I-f_{L}\right) \rho \left( f_{H}-f_{L}+c_{m}\right) }{\left( \pi _{HG}\left( f_{H}-f_{L}\right) -\pi _{LG}c_{m}\right) \left( \pi _{HB}\left( f_{H}-f_{L}\right) -\pi _{LB}c_{m}\right) }<0.$ When $U^{B}\le 0,$ the principal knows he will make zero profit at best from proceeding after a bad signal and generally will abandon the project. The outcomes after a good signal are not affected, so the expected return to the agent arises only from the good signal outcome ( 50).   $\square$

1.3 Globally optimal signal strategy

Proof of Proposition 7

Let $R_{b4}$ be the optimal repayments in the left-hand (LH) branch (signal before the contract), $\alpha ,R_\mathrm{after}$ the optimal signal strategy and repayments in the right-hand (RH) branch (signal after the contract).

$R_{b4}$ satisfies the truthtelling constraint (TT), the participation constraint (PC) in each signal state $E_{s|\sigma }PC^{\sigma }=0,\sigma =G,B,$ and the relevant limited liability (LL) constraints (18).

The maximum payoff in the LH branch is

$$\begin{aligned} EU_{b4}(R_{b4})=\sum _{\sigma }\pi _{\sigma }E_{s|\sigma }U^{\sigma }(R_{b4}^{\sigma }). \end{aligned}$$

The RH contract has the same TT constraints and LL constraints, and the principal participation constraint is $\alpha \sum _{\sigma }E_{s|\sigma }PC^{\sigma }+(1-\alpha )E_{s|N}PC^{N}.$ The maximum payoff on RH branch is

$$\begin{aligned} EU_\mathrm{after}(\alpha ,R_\mathrm{after})=\alpha \sum _{\sigma }\pi _{\sigma }E_{s|\sigma }U^{\sigma }\left( R_\mathrm{after}\right) +(1-\alpha )EU^{N}(R_\mathrm{after}). \end{aligned}$$

In more detail,

$$\begin{aligned} EU_{b4}(R_{b4})&= \pi _{G}E_{s|G}U^{G}+\pi _{B}E_{s|B}U^{B} \\&= \pi _{G}\max _{R^{G}}\left[ E_{s|G}U^{G}|E_{s|G}PC^{G},TT^{G},LL^{G}\right] \\&\quad +\pi _{B}\max _{R^{B}}\left[ E_{s|B}U^{B}|E_{s|B}PC^{B},TT^{B},LL^{B}\right] \\&= \pi _{G}E_{s|G}U^{G}(R_{b4}^{G})+\pi _{B}E_{s|B}U^{B}(R_{b4}^{B}) \\&\le \max _{R^{G},R^{B}}\Bigg [\left( \pi _{G}E_{s|G}U^{G}+\pi _{B}E_{s|B}U^{B}\right) |\left( \pi _{G}E_{s|G}PC^{G}\right. \nonumber \\&\quad \left. +\pi _{B}E_{s|B}PC^{B}\right) ,TT^{G},LL^{G},TT^{B},LL^{B}\Bigg ] \\&\equiv EU_\mathrm{after}(1,R^{G}\left( 1\right) ,R^{B}\left( 1\right) ) \\&\le \underset{\alpha }{\max }\left[ EU_\mathrm{after}(\alpha ,R^{G}\left( \alpha \right) ,R^{B}\left( \alpha \right) )|0\le \alpha \le 1\right] \end{aligned}$$

where $R^{\sigma }\left( \alpha \right)$ is the optimal RH branch repayment at the fixed value of $\alpha$. In fact, the first inequality is strict since for $\alpha =1$, there is cross-subsidization in the participation constraint with the principal making a loss in the bad signal state and a gain in the good signal state. $\alpha =1$ is optimal if $EPC_{\sigma }=0.$ In other cases, $(EPC_{\sigma }\lessgtr 0)$ optimally $\alpha <1$ in which case the second inequality is then also strict. Incidentally, in these cases, there is also cross-subsidization in the different parts of the ex-ante participation constraint but now between with-signal and no-signal states. $\square$