The power of regulatory regimes reexamined

Abstract

The literature finds that pure price cap regulation (PCR) is a high-powered regulatory regime because it elicits the first-best level of cost-reducing effort. Conversely, the regulated firm underinvests in cost-reducing effort under PCR with earnings sharing. These findings, while standard in the literature, do not consider the strategic behavior of the regulator ex post. When the regulator controls competitive entry and the Hope standard applies the regulated firm is no longer the residual claimant for its efficiency gains even under pure PCR. The findings of the literature are reversed in this setting. The power of the regulatory regime may be higher under PCR with earnings sharing than under pure PCR. In fact, over a wide range of welfare weights, equilibrium investment in cost-reducing effort is zero under pure PCR. This may explain why the empirical evidence on the efficiency gains from PCR falls short of an unequivocal validation of the theory.

Appendix: Proofs of Lemmas and Propositions

Proof of Proposition 1

Setting $\phi = \nu = 0\,$ in (3) and solving for $p(a*){\kern 1pt} {\kern 1pt} - c\,$ yields

$$ p(a*){\kern 1pt} \, - c = \frac{Q}{{Q_{p} }}\left[ {\frac{{\omega_{1} }}{{w_{2} s + \omega_{2} (1 - s)}} - 1} \right]. $$

(A1)

$p(a*){\kern 1pt} - c > 0 \Rightarrow \Uppi > 0 \Rightarrow$$\frac{{\omega_{1} }}{{w_{2} s + \omega_{2} (1 - s)}} < 1$ in (A1) since $Q_{p} < 0.$ Solving for $s$ yields$\,s > \frac{{\omega_{1} - \omega_{3} }}{{\omega_{2} - \omega_{3} }} > 0\,$ when $\omega_{2} > \omega_{1} > \omega_{3} \ge 0.$ □

Proof of Lemma 1

Totally differentiating (3) with respect to $s{\kern 1pt}$ with $\phi = \nu = 0$ and rearranging terms yields

$$ \frac{da*}{ds} = \frac{{ - [\omega_{2} - \omega_{3} ][(R_{p} - C_{Q} Q_{p} )p'(a)]}}{\Delta } < 0\; $$

(A2)

since $\;\Delta = \frac{{d^{2} W}}{{da^{2} }} < 0{\kern 1pt}$ by Assumption 1. □

Proof of Lemma 2

For part (i), totally differentiate (3) with respect to e with $\phi = \nu = 0.{\kern 1pt}$ Rearranging terms yields

$$ \frac{da*}{de} = \frac{{Q_{p} c_{e} p'(a){\kern 1pt} [s\omega_{2} + (1 - s)\omega_{3} ]}}{\Delta } > 0 $$

(A3)

since $Q_{p} c_{e} p'(a) < 0$ and $\Delta = \frac{{d^{2} W}}{{da^{2} }} < 0$ by Assumption 1.

For part (ii), totally differentiate (A3) with respect to s to obtain

$$ \frac{{d^{2} a^{*} }}{deds} = \frac{{ - [p'(a)]^{3} [Q_{p} ]^{2} [\omega_{2} - \omega_{3} ][\omega_{1} ]c_{e} }}{{\Delta^{2} }} < 0\, $$

(A4)

when ${\kern 1pt} \omega_{2} > \omega_{3} \; \Rightarrow$$\frac{{d^{2} p^{*} }}{deds} > 0.\;$ □

Proof of Lemma 3

$\,s < \frac{{\omega_{1} - \omega_{3} }}{{\omega_{2} - \omega_{3} }} \Rightarrow \phi > 0{\kern 1pt} {\kern 1pt}$ by Proposition 1. The first-order necessary condition for e in [FP-EX-1] is given by

$$ e:\; - \psi '(e) \le 0;\;{\text{and}}\;e[ - \psi '(e)] = 0 \Rightarrow e_{3}^{*} = 0\, $$

by complementary slackness since $\psi '(e) > 0\;\forall e > 0.$ □

Proof of Lemma 4

The Lagrangian for [FP-EX-2] is given by

$$ L = [1 - s][Q(p)(p - c)] - \psi (e) + \lambda \left[ {\frac{dW}{da}} \right]. $$

(A5)

Differentiating (A5) with respect to a and assuming an interior solution yields

$$ [1 - s][(R_{p} - C_{Q} Q_{p} )p'(a)] + \lambda \left[ {\frac{{d^{2} W}}{{da^{2} }}} \right] = 0. $$

(A6)

The first term in (A6) is negative $\forall s < 1.$ This implies that $\lambda < 0$ since $\frac{{d^{2} W}}{{da^{2} }} < 0$ by Assumption 1. The necessary first-order condition for e in (A5) is given by

$$ L{}_{e} = \;[1 - s][ - C_{e} ] - \psi '(e) + \lambda \left[ {\frac{da*}{de}} \right] \le 0;\;{\text{and}}\;e\left[ {L{}_{e}} \right] = 0. $$

(A7)

Since $\lambda < 0$ and $\frac{da*}{de} > 0{\kern 1pt}$ by Lemma 2(i), ${\kern 1pt} \lambda \left[ {\frac{da*}{de}} \right] < 0.$ It follows that

$$ \;[1 - s][ - C_{e} ] - \psi '(e) + \lambda \left[ {\frac{da*}{de}} \right]\;\; < \;{\kern 1pt} \;[1 - s][ - C_{e} ] - \psi '(e) $$

(A8)

for any given ${\kern 1pt} Q.$ Hence, $e_{3}^{*} < e_{2}^{*}$ when $\varepsilon$ is “small.” □

Proof of Proposition 2

The profit-maximizing level of effort $(e_{3}^{*} )$ in [FP-EX-2] is defined implicitly by

$$ [1 - s][ - C_{e} ] - \psi '(e) + \lambda \left[ {\frac{da*}{de}} \right]\;\; = \;{\kern 1pt} 0. $$

(A9)

Substituting $a*{\kern 1pt}$ into the regulated firm’s profit function, we obtain

$$ \hat{\Uppi } = [1 - s][Q(p(a*))][p(a*) - c] - \psi (e). $$

(A10)

The necessary first-order condition for e is given by

$$ \begin{aligned} {[1 - s] \underbrace {{[Q_{p} p'(a*)\frac{da*}{de}]}}_{ + }[p(a*) - c] + [1 - s][Q(p(a*))]p'(a*)\frac{da*}{de}]} \hfill \\ + \;[1 - s][Q(p(a*))][ - c_{e} ] - \psi '(e) \le 0\;{\text{and}}\;e\left[ {\hat{\Uppi }_{e} } \right] = 0. \hfill \\ \end{aligned} $$

(A11)

When $p(a*) \ge c,\,$ it is sufficient for an interior solution that

$$ [1 - s][Q(p(a*))][ - c_{e} + p'(a*)\frac{da*}{de}] - \psi '(e) = 0. $$

(A12)

Substituting for $\frac{da*}{de}$ upon appeal to Assumption 1 and Lemma 2(i) and simplifying yields

$$ [1 - s][Q(p(a*))][ - c_{e} ]\left[ {1 - \frac{{[s\omega_{2} + (1 - s)\omega_{3} ]}}{{2[s\omega_{2} + (1 - s)\omega_{3} ] - \omega_{1} }}} \right] - \psi '(e) = 0. $$

(A13)

Given that $c_{e} < 0$ and $\psi '(0) \approx 0,$ it suffices to show that

$$ \frac{{[s\omega_{2} + (1 - s)\omega_{3} ]}}{{2[s\omega_{2} + (1 - s)\omega_{3} ] - \omega_{1} }} < 1. $$

(A14)

The inequality in (A14) is satisfied when $\omega_{1} < s\omega_{2} + (1 - s)\omega_{3} \Rightarrow \,s > \frac{{\omega_{1} - \omega_{3} }}{{\omega_{2} - \omega_{3} }} \Rightarrow p(a*)\,\, > c\,$ from (A1) and Proposition 1. □

Proof of Proposition 3

$$ \hat{\Uppi }(p,e) = \;[1 - s][p - c(e)]Q - \psi (e ). $$

(A15)

Assuming an interior solution, the necessary first-order condition for $e_{3}^{*}$ is given by

$$ \frac{{\partial \hat{\Uppi }}}{\partial e} = [1 - s]\left[ {\frac{dp}{da}\frac{da}{de} - c_{e} } \right]Q + [1 - s][p - c(e)]\left[ {\frac{dQ}{dp}\frac{dp}{da}\frac{da}{de}} \right] - \psi '(e) = 0. $$

(A16)

Applying the Implicit Function Theorem to (A16),

$$ \frac{{de_{3}^{*} }}{ds} = - \frac{{F_{s} }}{{F_{e} }}, $$

(A17)

where $F_{s} = \frac{{\partial^{2} \hat{\Uppi }}}{\partial e\partial s}\,$ and $F_{e} = \frac{{\partial^{2} \hat{\Uppi }}}{{\partial e^{2} }} < 0\,$ by strict concavity. Rearranging terms and simplifying yields

$$ \begin{aligned} \frac{{de_{3}^{*} }}{ds} = \;\frac{{\left[ {p'(a)\frac{da}{de} - c_{e} } \right]\left[ { - Q + (1 - s)\frac{dQ}{dp}\frac{da}{ds}p'(a)} \right]}}{{ - F_{e} }} \hfill \\ \quad \;\; + \;\frac{{\left[ {\frac{dQ}{dp}\frac{da}{de}p'(a)} \right]\left[ { - \left( {p - c(e)} \right) + (1 - s)\frac{da}{ds}p'(a)} \right]}}{{ - F_{e} }} \hfill \\ \;\;\;\;\;\; + \;\frac{{[1 - s]\left[ {p'(a)\frac{{d^{2} a}}{deds}} \right]\left[ {Q + (p - c(e))\frac{dQ}{dp}} \right]}}{{ - F_{e} }}. \hfill \\ \end{aligned} $$

(A18)

Letting $s \to 1{\kern 1pt}$ in (A18) yields

$$ \frac{{de_{3}^{*} }}{ds} \approx \frac{{ - \left[ {p'(a)\frac{da}{de} - c_{e} } \right]\left[ Q \right] - \left[ {\frac{dQ}{dp}\frac{da}{de}p'(a)} \right]\left[ {p - c(e)} \right]}}{{ - F_{e} }} < 0 $$

(A19)

since $p'(a)\frac{da}{de} - c_{e} = c_{e} \left[ {\frac{{s\omega_{2} + (1 - s)\omega_{3} }}{{2s\omega_{2} + 2(1 - s)\omega_{3} - \omega_{1} }} - 1} \right]\,\; > 0\,$ when $\omega_{1} < s\omega_{2} + (1 - s)\omega_{3} \,$ and $\frac{dQ}{dp}\frac{da}{de}p'(a) > 0\,$ upon appeal to Lemma 2(i). This completes the proof of part (i).

To prove part (ii), let $\frac{dQ}{dP} \to 0\,$ in (A18) and recall that $- F_{e} > 0{\kern 1pt}$ which implies that

$$ \text{sgn} \left\{ {\frac{{de_{3}^{*} }}{ds}} \right\} = \text{sgn} \left\{ { - \left[ {p'(a)\frac{da}{de} - c_{e} } \right] + [1 - s]\left[ {p'(a)\frac{{d^{2} a}}{deds}} \right]} \right\}. $$

(A20)

Substituting $\frac{da}{de}\;$ and $\frac{{d^{2} a}}{deds}\;$ from Lemma 2(i) and 2(ii), respectively, (A20) is positive when

$$ \frac{{ - s\omega_{2} - (1 - s)\omega_{3} + \omega_{1} }}{{2s\omega_{2} + 2(1 - s)\omega_{3} - \omega_{1} }} + \frac{{[1 - s][\omega_{2} - \omega_{3} ]\omega_{1} }}{{[2s\omega_{2} + 2(1 - s)\omega_{3} - \omega_{1} ]^{2} }} > 0. $$

(A21)

Multiply (A21) through by $[2s\omega_{2} + 2(1 - s)\omega_{3} - \omega_{1} ]^{2}$ and rearranging terms yields

$$ [1 - s][\omega_{2} - \omega_{3} ]\omega_{1} - [s\omega_{2} + (1 - s)\omega_{3} - \omega_{1} ][2s\omega_{2} + 2(1 - s)\omega_{3} - \omega_{1} ] > 0. $$

(A22)

Simplifying (A22) and setting the resulting expression equal to zero yields the quadratic function

$$ [ - 2\omega_{2}^{2} + 4\omega_{2} \omega_{3} - 2\omega_{3}^{2} ]s^{2} + [ - 4\omega_{2} \omega_{3} + 4\omega_{3}^{2} + 2\omega_{1} \omega_{2} - 2\omega_{1} \omega_{3} ]s + [\omega_{1} \omega_{2} - 2\omega_{3}^{2} + 2\omega_{1} \omega_{3} - \omega_{1}^{2} ] = 0. $$

(A23)

Applying the quadratic formula to (A23) and simplifying yields the upper bound for s

$$ \hat{s} = \frac{{(2\omega_{3} {\kern 1pt} - \omega_{1} )(\omega_{2} {\kern 1pt} - \omega_{3} ) \pm \sqrt {[(\omega_{1} - 2\omega_{3} )(\omega_{2} {\kern 1pt} - \omega_{3} )]^{2} {\kern 1pt} + {\kern 1pt} \;\,[2(\omega_{2} {\kern 1pt} - \omega_{3} )^{2} ][\omega_{1} (\omega_{2} {\kern 1pt} - \omega_{1} ){\kern 1pt} - 2\omega_{3} (\omega_{3} {\kern 1pt} - \omega_{1} )]} }}{{ - 2(\omega_{2} {\kern 1pt} - \omega_{3} )^{2} }}. $$

(A24)

Factoring out the term $(\omega_{2} \, - \omega_{3} )\,$ in (A24) and simplifying yields (4) in Proposition 3(ii). □

Proof of Lemma 5

The Lagrangian expression associated with [FP-EN-1] is given by

$$ L = \;[1 - s][R(p) - C(Q(p),e)] - \psi (e) + \lambda_{1} \left[ {\frac{dW}{da}} \right] + \lambda_{2} [1 - s] + \lambda_{3} s. $$

(A25)

Differentiating (A25) with respect to a and assuming an interior solution yields

$$ [1 - s][(R_{p} - C_{Q} Q_{p} )p'(a)] + \lambda_{1} \left[ {\frac{{d^{2} W}}{{da^{2} }}} \right] = 0. $$

(A26)

The first term in (A26) is negative $\forall s < 1.$$\frac{{d^{2} W}}{{da^{2} }} < 0$ by Assumption 1. This implies that $\lambda_{1} < 0.$ Differentiating (A25) with respect to s yields

$$ - [R(p) - C(Q(p),e)] + \lambda_{1} \left[ {\frac{da*}{ds}} \right] - \lambda_{2} + \lambda_{3} = 0. $$

(A27)

Suppose now that $\lambda_{2} > 0 \Rightarrow s* = 1\,$ and $\lambda_{3} = 0.$ This implies from (A26) that

$\lambda_{1} \left[ {\frac{da*}{ds}} \right] = 0 \Rightarrow \lambda_{1} = 0\,$ since $\left[ {\frac{da*}{ds}} \right] < 0\,$ by Lemma 1. But if $\lambda_{1} = 0$, then from (A27) we have that $- [R(p) - C(Q(p),e)] - \lambda_{2} < 0$ which implies that $s* = 0$ by complementary slackness, which contradicts $\lambda_{2} > 0$ and $s* = 1 \Rightarrow s* < 1.$ Combining this result with Proposition 1 implies that

$$ \frac{{\omega_{1} - \omega_{3} }}{{\omega_{2} - \omega_{3} }} < s{\kern 1pt} * < 1.{\kern 1pt} $$

(A28)

□

Proof of Lemma 6

Lemma 5 implies that $\phi = 0{\kern 1pt} {\kern 1pt}$ at ${\kern 1pt} s{\kern 1pt} *.{\kern 1pt}$ The necessary first-order condition for e in (A25) is given by

$$ L{}_{e} = \;[1 - s*][ - C_{e} ] - \psi '(e) + \lambda_{1} \left[ {\frac{da*}{de}} \right] \le 0;\;{\text{and}}\;e\left[ {L{}_{e}} \right] = 0. $$

(A29)

$\lambda_{1} < 0$ by Lemma 5 and $\frac{da*}{de} > 0$ by Lemma 2(i) $\Rightarrow \lambda_{1} \left[ {\frac{da*}{de}} \right] < 0.$ It follows that

$$ \;[1 - s*][ - C_{e} ] - \psi '(e) + \lambda_{1} \left[ {\frac{da*}{de}} \right]\;\; < \;{\kern 1pt} \;[1 - s*][ - C_{e} ] - \psi '(e) $$

(A30)

when $\varepsilon \approx 0{\kern 1pt} \,$ which implies that $e_{3}^{*} < e_{2}^{*}$. □

Proof of Proposition 4

The proof of part (i) is virtually identical to the proof of Proposition 2 with s* replacing s upon appeal to Lemma 5. The proof of part (ii) follows directly from Proposition 1 and Lemma 5. □