Optimal performance reward, tax compliance and enforcement

Abstract

This paper incorporates the incentives of tax inspectors into an equilibrium model of tax compliance and enforcement when the taxpayers’ true income is private information (‘adverse selection’) and the effort of tax inspectors to verify reported income is unobservable (‘moral hazard’). It characterizes the optimal remuneration for tax inspectors, which is a function of discovered tax evasion, paying particular attention to the determinants of the power of incentives and the curvature of the optimal reward scheme. It is shown that the structure of the optimal reward is increasing, and in general non-linear, in the magnitude of discovered tax evasion. The equilibrium characterized has the features that: taxpayers with higher true income underreport less and tax inspectors’ auditing effort, and hence the probability of detecting tax non-compliance, decreases with reported income.

Appendices

Appendix 1

Proof of Proposition 1

This appendix establishes that the tax inspector’s optimal award follows (16) and the taxpayer’s tax evasion (17).

Making use of (6) and (7), and of the fact that $y(x)=I-x=r^{-1}(x)-x$, into (10) and rearranging gives

$$\begin{aligned} (1+\pi )y(x)=-\frac{\big (1-p(b)(1+\pi )\big )}{p^{\prime }(b)b^{\prime }(y)y^{\prime }(x)}, \end{aligned}$$

(26)

which upon substituting (26) into (13) gives

$$\begin{aligned} b^{\prime }(y)y^{\prime }(x)=-\frac{t(1-p(b)(1+\pi ))}{p^{\prime }(b)b(y)+p(b)}. \end{aligned}$$

(27)

Substituting (27) into (26) gives

$$\begin{aligned} y(x)=\frac{1}{t(1+\pi )}(b(y(x))+\psi (b(y(x)))), \end{aligned}$$

(28)

where $\psi (b)\equiv p(b)/p^{\prime }(b)$. Since (28) holds for any admissible x, differentiating both sides with respect to x one obtains

$$\begin{aligned} y^{\prime }(x)=\frac{1}{t(1+\pi )}( b^{\prime }(y(x))y^{\prime }(x)+\psi ^{\prime }(b(y(x)))b^{\prime }(y(x))y^{\prime }(x)) . \end{aligned}$$

(29)

The second-order condition given by (11), making use of (8), becomes

$$\begin{aligned} A\equiv & {} -\{p^{\prime }(e)f^{\prime \prime }\left( b\right) (b^{\prime }(y)(y^{\prime }(x)))^{2}+p^{\prime }(e)f^{\prime }(b)b^{\prime \prime }(y)(y^{\prime }(x))^{2}\nonumber \\&+p^{\prime }(e)f^{\prime }(b)b^{\prime }(y)y^{\prime \prime }(x)+p^{\prime \prime }(e)(f^{\prime }(b)b^{\prime }(y)y^{\prime }(x))^2\} y(x)\nonumber \\&+2p^{\prime }(e)f^{\prime }(b)b^{\prime }(y)y^{\prime }(x)<0. \end{aligned}$$

(30)

Equation (29) points to two cases: (i) $y^{\prime }(x)=0$ and (ii) $y^{\prime }(x)\ne 0$.

Case (i): If $y^{\prime }(x)=0$, then from (27) follows that $p(b)=1/(1+\pi )$, which implies that $b(y)=f^{-1}(p^{-1}(1/(1+\pi )))=c'(p^{-1}(1/(1+\pi )))/p'(p^{-1}(1/(1+\pi )))$ (with the second equality following from the fact that $f(\cdot )\equiv z^{-1}(\cdot )$, where $z(\cdot )\equiv c^{\prime }(\cdot )/p^{\prime }(\cdot )$). Differentiating (10) with respect to x, allowing also I to change according to $r^{-1}(x)$ gives²⁵

$$\begin{aligned} A-p^{\prime }(b)b^{\prime }(y)y^{\prime }(x)r^{-1\prime }(x)=0, \end{aligned}$$

(31)

where $A<0$ is the second-order condition of the taxpayer in (30). With $y^{\prime }(x)=0$, $b^{\prime }(y)y^{\prime }(x)=0$, (31) is not satisfied. Clearly, for the constant reward and verification case to be an equilibrium, the taxpayer must be indifferent among all reports, in which case the second-order condition is satisfied with equality. Section 3 examines this case and shows that such an equilibrium does not exist.

Case (ii): With $y^{\prime }\ne 0$, then (29) is satisfied if (omitting the x for ease of exposition)

$$\begin{aligned} b^{\prime }(y)=\frac{t(1+\pi )}{1+\psi ^{^{\prime }}(b(y))}, \end{aligned}$$

(32)

which is (16). Consistency between (32) and (27) implies that $y'(x)$ takes the form of (17). $\square$

Appendix 2

Proof of Lemma 1

Since, following from Proposition 1, $b'(y)>0$, and, following from (30), $A<0$ then from (31) it is the case that

$$\begin{aligned} b^{\prime }(y)y^{\prime }(x)r^{-1\prime }(x)<0. \end{aligned}$$

(33)

It is thus the case that either (i) $b^{\prime }(y)>0$, $y^{\prime }(x)<0$, and $r^{-1\prime }(x)>0$, or (ii) $b^{\prime }(y)>0$, $y^{\prime }(x)>0$, and $r^{-1\prime }(x)<0$.

Case (ii) is ruled out since $y(x)=I-x=r^{-1}(x)-x$ and so $y^{\prime }(x)=r^{-1\prime }(x)-1>0$ is inconsistent with $r^{-1\prime }(x)<0$. It follows that only case (i) holds. $\square$

Appendix 3

Derivation of $\psi ''(b)$ and dependence of MCII on p(e) and c(e). Differentiating $\psi (b_0)$ twice gives

$$\begin{aligned} \psi ^{\prime \prime }(b_0)=-\frac{p^{\prime \prime }(b_0)}{p^{\prime }(b_0)}+\frac{ 2p(b_0)(p^{\prime \prime }(b_0))^2}{(p^{\prime }(b_0))^3}-\frac{p(b_0)p^{\prime \prime \prime }(b_0)}{ (p^{\prime }(b_0))^2}, \end{aligned}$$

(34)

where, recalling that $p(b)\equiv p(f(b))$, with $f(\cdot )\equiv z^{-1}(\cdot )$, $z(\cdot )\equiv c^{\prime }(\cdot )/p^{\prime }(\cdot )$,

$$\begin{aligned} p^{\prime }(b)= & {} p^{\prime }(e)f^{\prime }(b),\end{aligned}$$

(35)

$$\begin{aligned} p^{\prime \prime }(b)= & {} p^{\prime \prime }(e)f^{\prime }(b)+p^{\prime }(e)f^{\prime \prime }(b),\end{aligned}$$

(36)

$$\begin{aligned} p'''(b)= & {} p'''(e)\left( f'(b)\right) ^2+p''(e)[f''(b)(1+f'(b))+f'''(b)], \end{aligned}$$

(37)

where $f^{\prime }(b)=1/(z^{\prime }(z^{-1}(b)))>0$ and $z^{\prime }(\cdot )=(c^{\prime \prime }( \cdot )p^{\prime }(\cdot )-c^{\prime }(\cdot )p^{\prime \prime }(\cdot ))/(p^{\prime }(\cdot ))^2$.

We turn now to MCII. Since MCII is $b+\psi (b)$, with $\psi (b)\equiv p(b)/p'(b)$ that MCII depends on p(e) and c(e) follows from (35). $\square$

Appendix 4

Proof of Theorem 1

Let $\overline{x}$ be the highest reported income associated with the highest true income $\overline{I}$ and let $\overline{y}\equiv y(\overline{x} )\equiv \overline{I}-\overline{x}$ be the amount of tax evasion at the highest true income. The boundary condition of the ordinary differential equation (16) is the b that solves, following from (13),

$$\begin{aligned} b=t(1+\pi )\overline{y}-\psi (b). \end{aligned}$$

(38)

The claim now is that if (19) is satisfied the right-hand side of (38) must be strictly positive that is, $t(1+\pi )\overline{y} -\psi (0)>0$ which implies that $\overline{y}>\psi (0)/(t(1+\pi ))$. To see that this is the case suppose $\overline{y}\le \psi (0)/(t(1+\pi ))$. Then, the right-hand side of (38) is nonpositive at $b(\bar{y})=0$ (and for any positive b) and $b(\overline{y})=0$, since only that value satisfies (38) (with an inequality in the case of $b(\bar{y})=0$, assuming that the reward is not allowed to take negative values). But this boundary would imply zero effort on part of the tax inspector, that is, $f(0)=0$ and hence $p(0)=0$. This implies, following from (19), that $b_1(\overline{y})>0$, which in turn implies that $b(\overline{y} )=0$ violates the condition $b_0\ge b_1$. As it has just been established, $t(1+\pi ) \overline{y}-\psi (0)>0$ and, it will be recalled, that $\psi ^{\prime }(b)>0$. This implies that there exists a boundary condition for $b^{\prime }$, $b(\overline{y})>0$, that uniquely satisfies Eq. (38).

Next, we choose the boundary for the differential equation (17). Since $\overline{y}$ is the lowest tax evasion, and $b^{\prime }(y)>0$, $b( \overline{y})$ is also the lowest. Recall that $b_0\ge b_1$ must be satisfied. We choose as a boundary the $\overline{y}$ that solves $b_1( \overline{y})=b_0(\overline{y})$. This amounts to solving $y^{\prime }(b( \overline{y}))=-1$. If this solution does not exist, it means that either $y^{\prime }<-1$ or $y^{\prime }>-1$ for all possible boundary conditions. In the former case, a separating equilibrium does not exist because $r^{-1}(x)$ is decreasing in x. In the latter case, we solve $y^{\prime }(b(\overline{y }))=-k$ and we choose the $\overline{y}(k)$ that corresponds to the maximum $k\in (0,1)$.²⁶ Once the boundary condition $\overline{y}$ has been chosen, the $\overline{x}$ is given by $\overline{y}=\overline{I}-\overline{x}$. The lower bound $\underline{x}$ is the solution to $y_0(\underline{x})=\underline{I}- \underline{x}$. Such an $\underline{x}$ clearly exists since $y_0(x)$ increases as x decreases with slope less than one (in absolute) and x can approach $-\infty$.

In addition, it is assumed that

$$\begin{aligned}&b_0^{\prime }\left[ \frac{t(1+\pi )}{1+\psi ^{\prime }(b_0)}(\overline{I} -x)\left\{ 1-\frac{\psi ^{\prime \prime }(b_0)}{1+\psi ^{\prime }(b_0)}\right\} +2 \right] \ge 0\,\,\quad \text{ for } x>\overline{x} \nonumber \\&b_0^{\prime }\left[ \frac{t(1+\pi )}{1+\psi ^{\prime }(b_0)}(\underline{I} -x)\left\{ 1-\frac{\psi ^{\prime \prime }(b_0)}{1+\psi ^{\prime }(b_0)}\right\} +2 \right] \ge 0\,\,\quad \text{ for } x<\underline{x}. \end{aligned}$$

(39)

The above condition always holds if $\psi ^{\prime \prime }(b_0)\le 0$, that is if the reward is either linear or convex. It may also hold if $\psi ^{\prime \prime }(b_0)\ge 0$, and so the reward is concave.

Given the assumptions outlined in text, the solutions to the differential equations, (16) and (17), that govern the evolution of the reward and the taxpayer’s reporting, respectively, exist and are unique. Further, we assume that this is true throughout $[\underline{x}, \overline{x}]$.

We next show that the taxpayer has no incentive to report an x outside the $[\underline{x},\overline{x}]$ interval. Let $x>\overline{x}$. The belief of the inspector is $\tau (x)=\overline{I}$. So, $y(x)=\overline{I}-x$ and $y^{\prime }(x)=-1$. The optimal reward for any $x>\overline{x}$ is now $b_0(y)$, with $y=\overline{I}-x$, so $b_0(\overline{I}-x)$. Fix the type of the taxpayer at $\overline{I}$ and consider the first-order condition of the taxpayer given by (10), re-written here for convenience,

$$\begin{aligned} -p'(b)b'(y)y'(x)(1+\pi )y(x)+p(b)(1+\pi )-1=0. \end{aligned}$$

(40)

At $x=\overline{x}$, the first-order condition is satisfied with equality. Differentiating (40) with respect to x one obtains

$$\begin{aligned} -[p^{\prime }(b)(b^{\prime }(y))^2+p^{\prime }(b)b^{\prime \prime }(y)]y(x)-2p^{\prime }(b)b^{\prime }(y). \end{aligned}$$

(41)

The above expression is negative if and only if

$$\begin{aligned}{}[(b^{\prime }(y))^2+b^{\prime \prime }(y)]y(x)+2b^{\prime }(y)\ge 0, \end{aligned}$$

(42)

which, after using the expressions for $b^{\prime }$ and $b^{\prime \prime }$, becomes

$$\begin{aligned} b^{\prime }\left[ \frac{t(1+\pi )}{1+\psi ^{\prime }(b)}y(x)\left\{ 1-\frac{ \psi ^{\prime \prime }(b)}{1+\psi ^{\prime }(b)}\right\} +2\right] \ge 0, \end{aligned}$$

(43)

which is satisfied given (39).

Therefore, the slope of the taxpayer’s utility function decreases as x increases beyond $\overline{x}$. Given that the slope is zero at $x= \overline{x}$, it follows that it becomes negative when $x>\overline{x}$. Thus, the taxpayer has no incentive to report an income higher than $\overline{x}$. While so far we have set $I=\overline{I}$, the same steps can be used to show that even if $I<\overline{I}$ the taxpayer has no incentive to report an x higher than $\overline{x}$. Suppose $I<\overline{I}$. The slope of the taxpayer’s objective function is zero if he reports $x=r(I)< \overline{x}$ and the objective function is strictly concave, implying that the slope of the objective function is negative at $x=\overline{x}$. After this point, as we showed above, the slope becomes even more negative, making a deviation in that region unprofitable.

The same steps can be used to prove that any $x<\underline{x}$ is a dominated strategy, using in this case $b_0(\underline{I}-x)$ for any $x< \underline{x}$. $\square$

Appendix 5

Proof of Proposition 2

Following the proof of Proposition 1, $y^{\prime }=0$ (and so tax evasion is the same across all taxpayer types), implies $b=f^{-1}(p^{-1}(1/(1+\pi ))).$ This, however, cannot be an equilibrium. The reason is as follows. The constant reward gives rise to a constant probability of detection $p(e(b))=1/(1+\pi ),$ which implies that the taxpayer is indifferent among all income reports x for any I. So, he will choose some constant tax evasion consistent with this equilibrium (the constant y, denoted by $\hat{y}$, can be determined from the first-order condition of the revenue authority (13)).

The income reports are in the interval $[\underline{I}-\hat{y}, \overline{I}- \hat{y}]$. Note, using (19), that the reward associated with the constant solution is higher than the reward from the separating equilibrium.

First, assume that $\hat{y}<\tilde{y}(\underline{x})$. This implies that $\underline{I}-\hat{y}>\underline{I}-\tilde{y}(\underline{x})=\underline{x}$. Consider now the taxpayer with $I=\underline{I}$ who, according to the constant reward equilibrium, should report $\underline{I}-\hat{y}$. Suppose this taxpayer deviates to $x\in (\underline{x},\underline{I}-\hat{y})$. The beliefs are $\tau (x)=\underline{I}$ and the optimal reward is now the $b_0(y)$ which is discretely lower than the constant reward. The taxpayer becomes better off because tax evasion increased, while the probability of detection decreased. This is true for a positive measure of taxpayers.

Assume now that $\hat{y}>\tilde{y}(\overline{x})$. This implies that $\overline{I}-\hat{y}<\overline{I}-\tilde{y}(\overline{x})=\overline{x}$. Consider now the taxpayer with $I=\overline{I}$ who, according to the constant reward equilibrium, should report $\overline{I}-\hat{y}$. Suppose this taxpayer deviates to $x\in (\overline{I}-\hat{y}, \overline{x})$. The beliefs are $\tau (x)=\overline{I}$ and the optimal reward is now the $b_0(y)$ which is discretely lower than the constant reward. The taxpayer becomes better off because tax evasion has decreased marginally, while the probability of detection decreased discontinuously. This is true for a positive measure of taxpayers.

Finally, because $\tilde{y}(\overline{x})<\tilde{y}(\underline{x})$ the above two cases exhaust all the possibilities. Therefore, the constant reward cannot be an equilibrium, provided that the separating equilibrium satisfies (19). $\square$