Gift policy, bribes and corruption

Abstract

A ban on gifts in public offices partially deters bribery but may induce undeterred bribers to switch from gifts to money. The rise in the use of a more liquid instrument (money) increases the bribe acceptance rate and, possibly, the measure of unqualified applications approved by the office. A gift ban may thus amplify the social costs of corruption by allocating public resources to the wrong people. It is shown that the optimal limited gift ban with a value cap binding on “expensive” gifts dominates the free gifts policy. The limited ban has weaker deterrence than the complete ban but it may reduce the social costs of corruption by minimizing the switch from gifts to money.

Appendix

Proof of Proposition 2

Comparing (6) and (3), it is easy to verify that $B_\pi < B_M^0$. Thus, there exist client types who prefer bribing with the $\pi$-gift to offering the money bribe $m_B$ defined in (1).

Because the agent’s final utility depends on the gift or cash he receives and the expected fine, his sequential rational strategy is to accept any bribe that yields a non-negative expected utility. This fact is used in defining the critical money bribe $m_B$ in (1) and the values of the $\pi$-gift bribe and the 1-gift bribe. Given the cost and acceptance probabilities of these bribes, each client type’s strategy is reduced to chosing the bribe instrument that maximizes his individual expected utility.

(a) Assume (10), thus, $B_C \ge B_M$. For $B > B_M$, we know by (7) that the best instrument of bribery is the 1-gift.

Consider $B \in (B_M, B_C )$. By definition, $U_\pi > U_1$ if $B < B_C$. Thus, for B in the range $(B_M, B_C )$, we have $U_M< U_1 < U_\pi$. These B types prefer the $\pi$-gift to the 1-gift, and the 1-gift to the money bribe $m_B$.

Consider now $B \le B_M$ and recall that $B_M> B_M^0 > B_\pi$. For client types $B \le B_M$, the expected utilities are ranked as $U_\pi > U_1$ and $U_M \ge U_1$, with $U_M = U_1 < U_\pi$ at $B = B_M$. Note that the expressions of $U_M$ and $U_\pi$ are both linear and increasing in B. Since $B_\pi < B_M^0$ and $U_\pi > U_M$ at $B = B_M$, it follows that $U_\pi > U_M$ for all $B \in [B_\pi , B_M ]$. Finally, for $B < B_\pi$, all bribe options yield a negative expected utility by definition, so the optimal strategy of these client types is not to apply. The equilibrium strategies are therefore as stated in Lemma 1.

(b) Assume $B_C < B_M$. As in part (a), consider first large enough B types, such that $B > B_M$. In this range, the expected utilities rank as $U_1> U_M > U_\pi$, so these client types will offer the 1-gift bribe. Consider $B \le B_M$. Decreasing B gradually in the left neighborhood of $B_M$ (recall that $U_M = U_1 > U_\pi$ at $B_M$), the ranking of expected utilities becomes $U_M> U_1 > U_\pi$. This ranking holds as long as $B > B_C$. Thus all client types $B \in (B_C, B_M)$ offer the money bribe. At $B=B_C$ both gift options yield the same expected utility, $U_1 = U_\pi$, yet this client’s best option remains the money bribe because $B_C < B_M$ and so, $U_M > U_1 = U_\pi$.

Moving to the left neighborhood of $B_C$, observe that by continuity of the client’s expected utility we have $U_M> U_\pi > U_1$ for B sufficiently close to $B_C$. In this benefit range the money bribe remains the first best option but the $\pi$-gift becomes the second-best option, better than the 1-gift bribe.

Decreasing B further, the benefit at which $U_M =0$ is reached, $B=B_M^0$($> B_\pi$) where $U_\pi > U_M =0$. Given continuity of $U_M$ and $U_\pi$ and the fact that $U_M > U_\pi$ at $B_C$, there must exist a unique benefit $\underline{B}_M \in (B_M^0 , B_C)$ at which $U_\pi = U_M$, such that $U_M > U_\pi$ for $B > \underline{B}_M$ and $U_M < U_\pi$ for $B < \underline{B}_M$. Therefore the money bribe is optimal in the right neighborhood of $\underline{B}_M$ whereas the $\pi$-gift bribe is optimal for $B \in (B_\pi , \underline{B}_M]$. $\square$

Proof of Proposition 5

The proof proceeds by evaluation of the derivatives of the right and left hand sides of (13) in the case $B_C < B_M$, to verify if (13) is reinforced or not by marginal changes in the variables listed in the proposition.

Condition (13) is more likely to hold when $f_A$ increases, if

$$\begin{aligned} \frac{ \partial \Delta NA}{\partial f_A} \lesseqqgtr&\,\,0 \text{ if } (1-\mu )g( B_M^0 ) \frac{\mu +(1- \mu )r}{(1- \mu )(1-r)}\nonumber \\ \gtreqqless&\,\,\pi g(B_\pi ) \frac{r}{1-r} + (1-\mu -\pi ) g( \underline{B}_M ) \frac{\mu +(1- \mu - \pi )r}{(1- \mu - \pi )(1-r)}. \end{aligned}$$

(15)

If the densities are equal, $g(B_\pi ) = g( B_M^0 ) = g( \underline{B}_M ) =g(B_M)$, by simplifying and rearranging terms reveals that (15) holds with equality. Therefore the sign of the impact on $\Delta NA$ depends on the densities. The proof for case of a larger $f_C$ is identical.

Consider a marginal increase in r. The impact on $\Delta NA$ is

$$\begin{aligned}&\frac{ \partial \Delta NA}{\partial f_A} \lesseqqgtr 0 \,\,\text{ if } \,\, (1- \mu )g( B_M^0 ) \frac{\partial B_M^0 }{\partial r} \gtreqqless \pi g(B_\pi ) \frac{\partial B_\pi }{\partial r} + (1-\mu - \pi ) g( \underline{B}_M ) \frac{\partial \underline{B}_M }{\partial r},\\&\quad \text{ where } \,\, \frac{\partial B_M^0 }{\partial r} = \frac{f_A+f_C}{1-r}[1 + \frac{\mu + (1-\mu )r}{(1-\mu )(1-r)}, \quad \quad \frac{\partial B_\pi }{\partial r} = \frac{f_A+f_C}{1-r}\left[ 1 + \frac{r}{1-r} \right] ,\\&\quad \text{ and } \quad \frac{\partial \underline{B}_M }{\partial r} = \frac{f_A+f_C}{1-r}[1 + \frac{\mu + (1-\mu - \pi )r}{(1-\mu - \pi )(1-r)}. \end{aligned}$$

Under equal densities $g(B_\pi ) = g( B_M^0 ) = g( \underline{B}_M ) =g(B_M)$, we get $\partial \Delta NA / \partial r =0$ after simplifying the terms. For a larger r to have an impact on $\Delta NA$, the densities around these critical B levels must be different.

First-stage detection probability $\mu$ does not affect $B_\pi$. Thus

$$\begin{aligned} \frac{ \partial \Delta NA}{\partial f_A} \lesseqqgtr 0 \,\,\text{ if } \,\, - G(B_M^0 ) + (1- \mu )g( B_M^0 ) \frac{\partial B_M^0 }{\partial \mu } \gtreqqless - G( \underline{B}_M ) + (1-\mu - \pi ) g( \underline{B}_M ) \frac{\partial \underline{B}_M }{\partial \mu } , \end{aligned}$$

that is,

$$\begin{aligned}&- G(B_M^0 ) + (1- \mu )g( B_M^0 ) \left[ \frac{(f_A+f_C)(1-r)}{(1- \mu )(1-r)} \right. \nonumber \\&\qquad \left. + \frac{(\mu +(1- \mu )r)(f_A+f_C)(1-r)}{(1- \mu )^2(1-r)^2} \right] \\&\quad \gtreqqless - G( \underline{B}_M ) + (1-\mu - \pi ) g( \underline{B}_M ) \left[ \frac{(f_A+f_C)(1-r)}{(1- \mu -\pi )(1-r)}\right. \nonumber \\&\qquad \left. + \frac{(\mu +(1- \mu -\pi )r)(f_A+f_C)(1-r)}{(1- \mu -\pi )^2(1-r)^2}\right] . \end{aligned}$$

These terms can be arranged as

$$\begin{aligned}&G( \underline{B}_M )- G(B_M^0 ) \lesseqgtr (f_A+f_C) g( \underline{B}_M ) \left[ 1 + \frac{(\mu +(1- \mu -\pi )r)}{(1- \mu -\pi )(1-r)}\right] \\&\quad - g( B_M^0 )(f_A+f_C) \left[ 1 + \frac{(\mu +(1- \mu )r)}{(1- \mu )(1-r)}\right] . \end{aligned}$$

The left hand side is positive because $\underline{B}_M > B_M^0$. The sign of the right hand side is positive under constant densities at $\underline{B}_M$ and $B_M^0$, ambiguous otherwise. However, unless the density $g( \underline{B}_M )$ is too large (and, larger than $g( B_M^0 )$), the left hand side will exceed the right hand side, hence, $\mu$ and $\Delta NA$ will move in the same direction.

Finally, since $B_\pi$ and $B_M^0$ do not depend on $\pi$, the impact of an increase in $\pi$ is

$$\begin{aligned} \frac{ \partial \Delta NA}{\partial f_A} \lesseqqgtr 0 \,\,\text{ if } \,\, G(B_\pi ) - G( \underline{B}_M ) + g( \underline{B}_M ) \frac{(f_A+f_C)}{(1-r)} \left[ \frac{\mu }{1- \mu -\pi } \right] \lesseqqgtr 0. \end{aligned}$$

The last term is positive whereas $G(B_\pi ) - G( \underline{B}_M )$, the direct impact at constant bribe strategies, is negative. Therefore the overall impact inclusive of the change induced in $\underline{B}_M$ rests ambiguous. $\square$