Tax incentives for inefficient executive pay and reward for luck

Abstract

I study the economic consequences of tax deductibility limits on salaries for the design of incentive contracts. The analysis is based on an agency model in which the firm’s cash flow is a function of the agent’s effort and an observable random factor beyond the agent’s control. According to my analysis, limiting the tax deductibility of fixed wages has two consequences. The principal rewards the agent on the basis of the observable random factor and adjusts the amount of performance-based pay in the optimal incentive contract. The new contract can have weaker or stronger work incentives than without the tax. The theoretical findings have implications for empirical compensation research. First, the analysis shows that reward for luck can be the optimal response to recent tax law changes, whereas earlier empirical literature has attributed this phenomenon to managerial entrenchment. Second, I demonstrate that a simple regression analysis that fails to control for separable measures of luck is likely to find an increased pay for performance sensitivity as a response to the introduction of tax deductibility limits on salaries even if the pay for performance sensitivity has actually declined.

Appendix

1. Derivation of the unconstrained incentive weight (Eq. 15):

Substituting the results for the optimal effort and $v_z^{\ast}$ from (9) and (12) into (10) yields

$$ \Uppi _u=(1-\tau)\cdot \left(b\cdot a+c\cdot \mu -C(a)-\frac{r}{2}\left(\frac{ C^{\prime }(a)}{b}\right) ^2\,\cdot\,\sigma _\varepsilon ^2-\underline {U}\right). $$

Maximizing this expression with respect to a yields the following first order condition

$$ \frac{\partial \Uppi _u}{\partial a}=(1-\tau)\cdot \left[ b-C^{\prime }(a)\left(1+r\cdot C^{\prime \prime }(a)\cdot \sigma _\varepsilon ^2/b^2\right) \right] =0. $$

Solving this condition for C′(a) yields (14), and using the fact that v _x  = C′(a)/b from (9) yields (15).$\square$

2. Proof of Lemma 1:

(a) Substituting the expression for the variance and the expectation of the agent’s pay from (7) and (8) into (17) yields the optimal salary as a function of v _x and v _z

$$ w(v_x,v_z)=\underline {U}+C(a)+\frac{r}{2}\cdot v_x^2\cdot \sigma _\varepsilon ^2+(v_x\cdot c+v_z)^2\cdot \sigma _z^2-v_x\cdot b\cdot a-(v_x\cdot c+v_z)\cdot \mu. $$

Differentiating this expression and evaluating the solution for the unconstrained optimal contract parameters yields

$$ \left.\frac{\partial w(v_x,v_z)}{\partial v_z}\right| _{v_x=v_x^{\ast},v_z=v_z^{\ast}} =-\mu < 0 $$

(43)

$$ \left. \frac{\partial w(v_x,v_z)}{\partial v_x}\right| _{v_x=v_x^{\ast},v_z=v_z^{\ast}} =r\cdot v_x^{\ast}\cdot \sigma _\varepsilon ^2-b\cdot a $$

(44)

Since (43) is strictly negative, but (44) is positive if $r\cdot v_x^{\ast}\cdot \sigma _\varepsilon ^2 > b\cdot a,$ condition (18) is always satisfied for $v_z^{\ast}$ but not for $v_x^{\ast}.$ $\square$

3. Proof of Proposition 2:

(a) Substituting the results for the optimal effort and $v_z^{\ast\ast}$ from (9) and (20) into (16) yields

$$ \Uppi _c=(1-\tau)(b\cdot a+c\cdot \mu)-C(a)-\frac{r}{2}\left(\frac{C^{\prime }(a)^2}{b^2}\cdot \sigma _\varepsilon ^2+\delta ^2\cdot \sigma _z^2\right) - \underline {U}+\tau \cdot (C^{\prime }(a)\cdot a+\delta \cdot \mu +\overline{w}). $$

Maximizing this expression with respect to a yields the following first-order condition

$$ \frac{\partial \Uppi _c}{\partial a}=(1-\tau)\cdot b-C^{\prime }(a)\left(1+r\cdot C^{\prime \prime }(a)\cdot \sigma _\varepsilon ^2/b^2\right) +\tau \cdot (C^{\prime \prime }(a)\cdot a+C^{\prime }(a))=0. $$

(45)

Rearranging terms and solving this condition for C′(a) yields the condition for the desired effort level, a**, in (22).

(b) After rearranging terms, (45) can be written as follows:

$$ \frac{\partial \Uppi _c}{\partial a}=\frac{\partial \Uppi _u} {\partial a}+\tau \cdot C^{\prime \prime }(a)\cdot \left[ a-C^{\prime }(a)\cdot r\cdot \sigma _\varepsilon ^2/b^2\right]. $$

(46)

Evaluating the expression in (46) at the unconstrained optimal effort level a* yields

$$ \left. \frac{\partial \Uppi _c}{\partial a}\right| _{a=a^{\ast}}=\tau \cdot C^{\prime \prime }(a^{\ast})\cdot \left[ a^{\ast}-C^{\prime }(a^{\ast})\cdot r\cdot \sigma _\varepsilon ^2/b^2\right]. $$

Evidently, a** > a* if the term in brackets is positive, or, because $C^{\prime }(a^{\ast})/b=v_x^{\ast}$ from (9), if

$$ b\cdot a^{\ast} > v_x^{\ast}\cdot r\cdot \sigma _\varepsilon ^2. $$

(47)

Otherwise, a** < a*. Evidently, (47) is equivalent to (18) from (44). $\square$

4. Proof of Proposition 1 for the binary agency model:

(a) The principal’s problem consists of maximizing (27) subject to (28) and (29), which is equivalent to maximizing the following Lagrangian:

$$ \begin{aligned} L\,=\,&(1-\tau)\cdot \left[ y_L+p_H\left(y_H-y_L-b_y\right) +E[z]-w-\pi \cdot b_z\right] -\tau \cdot (w-\overline{w})\\ &+\lambda \cdot \left[ E\left[ U_L\right] +p_H\cdot \left(E\left[ U_H\right] -E\left[ U_L\right] \right) -c_H-\underline {U}\right]\\ &+\mu \cdot \left[\left(E\left[ U_H\right] -E\left[ U_L\right] \right) - \frac{c_H-c_L}{p_H-p_L}\right]. \end{aligned} $$

For given values of λ, μ, and b _y assume that w > 0 and b _z  = 0, so that the Kuhn–Tucker-conditions are satisfied if

$$ \frac{\partial L}{\partial w} =-1+\lambda \cdot \frac{\partial E\left[ U_L\right]}{\partial w}+(\lambda \cdot p_H+\mu)\cdot \left(\frac{\partial E\left[ U_H\right]}{\partial w}-\frac{\partial E\left[ U_L\right]}{\partial w}\right) =0 $$

(48)

$$ \frac{\partial L}{\partial b_z} =-\pi \cdot (1-\tau)+\lambda \cdot \frac{\partial E\left[ U_L\right]}{\partial b_z}+(\lambda \cdot p_H+\mu)\cdot \left(\frac{\partial E\left[ U_H\right] }{\partial b_z}-\frac{ \partial E\left[ U_L\right]}{\partial b_z}\right) < 0, $$

(49)

where

$$ \begin{aligned} \frac{\partial E\left[ U_L\right] }{\partial w}\,=\,&\pi \cdot U^{\prime }(w+b_z)+(1-\pi)\cdot U^{\prime }(w)\\ \frac{\partial E\left[ U_H\right]}{\partial w}\,=\,&\pi \cdot U^{\prime }(w+b_y+b_z)+(1-\pi)\cdot U^{\prime }(w+b_y)\\ \frac{\partial E\left[ U_L\right]}{\partial b_z} \,=\,&\pi \cdot U^{\prime }(w+b_z) \\ \frac{\partial E\left[ U_H\right] } {\partial b_z}\,=\,&\pi \cdot U^{\prime }(w+b_y+b_z), \end{aligned} $$

so that for b _z  = 0

$$ \begin{aligned} \frac{\partial E\left[ U_L\right]}{\partial b_z}\,=\,&\pi \cdot \frac{ \partial E\left[ U_L\right] }{\partial w}\\ \frac{\partial E\left[ U_H\right]}{\partial b_z}\,=\,&\pi \cdot \frac{ \partial E\left[ U_H\right] }{\partial w}\\ \end{aligned} $$

substituting these expressions into (48) and (49) rearranging terms yields

$$ \begin{aligned} \lambda \cdot \frac{\partial E\left[ U_L\right]}{\partial w}+(\lambda \cdot p_H+\mu)\cdot \left(\frac{\partial E\left[ U_H\right] }{\partial w}- \frac{\partial E\left[ U_L\right] }{\partial w}\right) \,=\,&1\\ \lambda \cdot \frac{\partial E\left[ U_L\right]}{\partial w}+(\lambda \cdot p_H+\mu)\cdot \left(\frac{\partial E\left[ U_H\right]}{\partial w}- \frac{\partial E\left[ U_L\right] }{\partial w}\right) < &(1-\tau), \end{aligned} $$

a contradiction. I conclude that $b_z^{\ast} > 0$ $\square$