Measuring the Power of Regulatory Regimes

Abstract

The power of a regulatory regime (PRR) measures the strength of the incentives for cost-reducing innovation. A static PRR metric is derived that is a function of: the regulated firm’s share of profits; the ratchet effect; and the Arrow effect. The analysis reveals that the dynamic PRR is significantly lower than the static PRR even for long-lived regulatory regimes. A key finding is that the literature may significantly overstate the PRR. This may help explain why the empirical findings with regard to the efficiency gains from price cap regulation do not offer unequivocal validation of the prevailing theory.

Appendix: Proofs of Select Propositions

Proof of Proposition 3(i)

$$\frac{{\partial \rho^{D} }}{\partial n} \approx \rho^{D} (n,l) - \rho^{D} (n - 1,l) = \frac{{\rho}}{{\sum\limits_{j = 1}^{l } {\left( {\frac{1}{1 + r}} \right)}^{j - 1} }} \times \left[ {\frac{1}{n}\sum\limits_{i = 1}^{n} {(n + 1 - i)\frac{1}{{(1 + r)^{i - 1} }}} - \frac{1}{n - 1}\sum\limits_{i = 1}^{n - 1} {(n - 1 + 1 - i)\frac{1}{{(1 + r)^{i - 1} }}} } \right].$$

(11)

The first term on the right-hand side of (11) is positive. It suffices to show that the second term in brackets is positive. The term is rewritten as follows:

$$\left[ {\left( {\frac{n}{n} - \frac{n - 1}{{n - 1}}} \right) + \left( {\frac{n - 1}{n} - \frac{n - 2}{{n - 1}}} \right)\left( {\frac{1}{1 + r}} \right) + \left( {\frac{n - 2}{n} - \frac{n - 3}{{n - 1}}} \right)\left( {\frac{1}{1 + r}} \right)^{2} + \cdot \cdot \cdot + \left( {\frac{2}{n} - \frac{1}{n - 1}} \right)\left( {\frac{1}{1 + r}} \right)^{n - 2} + {\kern 1pt} \,\,\frac{1}{n}\left( {\frac{1}{1 + r}} \right)^{n - 1} } \right].$$

(12)

The first and last terms in (12) are zero and positive, respectively. Each of the remaining terms has the general form $\left( {\frac{n - j}{n} - \frac{n - j - 1}{{n - 1}}} \right){\kern 1pt} {\kern 1pt} {\kern 1pt} {\text{for}}{\kern 1pt} \,n > j > 0.{\kern 1pt}$ Hence, it is sufficient to show that $\frac{n - j}{n} > \frac{n - j - 1}{{n - 1}}{\kern 1pt} {\kern 1pt} \Rightarrow n^{2} - nj - n + j > n^{2} - nj - n{\kern 1pt} .{\kern 1pt} {\kern 1pt}$ This inequality is satisfied $\forall j > 0.$□

Proof of Proposition 3(ii)

$$\frac{{\partial \rho^{D} }}{\partial l} \approx \rho^{D} (n,l) - \rho^{D} (n,l - 1) = \left[ {\frac{\rho }{n} \times \sum\limits_{i = 1}^{n} {(n + 1 - i)\frac{1}{{(1 + r)^{i - 1} }}} } \right]\left[ {\frac{1}{{\sum\nolimits_{j = 1}^{l} {\left( {\frac{1}{1 + r}} \right)}^{j - 1} }} - \frac{1}{{\sum\nolimits_{j = 1}^{l - 1} {\left( {\frac{1}{1 + r}} \right)}^{j - 1} }}} \right].$$

(13)

The first term on the right-hand side of (13) is positive. It is therefore sufficient to show that the second term in large brackets is negative. This is the case when

$$\sum\limits_{j = 1}^{l} {\left( {\frac{1}{1 + r}} \right)}^{j - 1} > \sum\limits_{j = 1}^{l - 1} {\left( {\frac{1}{1 + r}} \right)}^{j - 1} \Rightarrow \left( {\frac{1}{1 + r}} \right)^{l - 1} > 0,$$

(14)

which is satisfied since $r \ge 0.$□

Proof of Proposition 4(i)

Let $x = \frac{1}{1 + r}.$ We show that $\frac{{\partial \rho^{D} }}{\partial x} < 0 \Rightarrow \frac{{\partial \rho^{D} }}{\partial r} = \frac{{\partial \rho^{D} }}{\partial x}\frac{dx}{{dr}} > 0\;{\text{since}}\,\frac{dx}{{dr}} = - \frac{1}{{(1 + r)^{2} }} < 0.$ Without loss of generality, set $\rho = 1.$ The proof is established for ${\kern 1pt} n = l{\kern 1pt}$, which implies that it must hold ${\kern 1pt} \forall n < l.$

$$\rho^{D} = \frac{{1 + \left( {\frac{n - 1}{n}} \right)x + \left( {\frac{n - 2}{n}} \right)x^{2} + \cdots + \left( \frac{1}{n} \right)x^{n - 1} }}{{1 + x + x^{2} + \cdots + x^{l - 1} }}.$$

(15)

$${\text{sgn}} \left\{ {\frac{{\partial \rho^{D} }}{\partial x}} \right\} = {\text{sgn}} \left\{ \begin{gathered} \left[ {\left( {\frac{n - 1}{n}} \right) + 2\left( {\frac{n - 2}{n}} \right)x + 3\left( {\frac{n - 3}{n}} \right)x^{2} + \cdots + (n - 1)\left( \frac{1}{n} \right)x^{n - 2} } \right]\left[ {1 + x + x^{2} + \cdots + x^{l - 1} } \right] \hfill \\ \, - \left[ {1 + 2x + \cdots + (l - 1)x^{l - 2} } \right]\left[ {1 + \left( {\frac{n - 1}{n}} \right)x + \left( {\frac{n - 2}{n}} \right)x^{2} + \cdots + \left( \frac{1}{n} \right)x^{n - 1} } \right] \hfill \\ \end{gathered} \right\}.$$

(16)

Reorder the terms on the right-hand side of (16) by the power of x and set $l = n.$

$$\begin{aligned} & x^{0} \left[ {\frac{n - 1}{n} - 1} \right] + x\left[ {\frac{n - 1}{n} + 2\left( {\frac{n - 2}{n}} \right) - \left( {\frac{n - 1}{n}} \right) - 2} \right] + x^{2} \left[ {\frac{n - 1}{n} + 2\left( {\frac{n - 2}{n}} \right) + 3\left( {\frac{n - 3}{n}} \right) - 2\left( {\frac{n - 1}{n}} \right) - \left( {\frac{n - 2}{n}} \right) - 3} \right]{\kern 1pt} \\ & \quad + x^{3} \left[ {\frac{n - 1}{n} + 2\left( {\frac{n - 2}{n}} \right) + 3\left( {\frac{n - 3}{n}} \right) + 4\left( {\frac{n - 4}{n}} \right) - 3\left( {\frac{n - 1}{n}} \right) - 2\left( {\frac{n - 2}{n}} \right) - \left( {\frac{n - 3}{n}} \right) - 4} \right] + \cdots + x^{2n - 3} \left[ {\left( {\frac{n - 1}{n}} \right) - \left( {\frac{n - 1}{n}} \right)} \right]. \\ \end{aligned}$$

(17)

The last term in (17) is equal to zero, and all of the other terms are strictly negative.□

Proof of Proposition 4(ii)

It is sufficient to show there exists an $n > l$ such that $\frac{{\partial \rho^{D} }}{\partial r} < 0.$ The result is obvious $\forall n > l = 1.$ We show that it holds for $n=3 \,{\text{and}}\,l=2,$ which implies that it holds $\forall n > l = 2.$ Let $x = \frac{1}{1 + r}\,{\text{and}}\,\rho = 1.$

$$\rho^{D} = \frac{{1 + \frac{2}{3}x + \frac{1}{3}x^{2} }}{1 + x} = \frac{{3 + 2x + x^{2} }}{3 + 3x}.$$

(18)

$${\text{sgn}} \left\{ {\frac{{\partial \rho^{D} }}{\partial x}} \right\} = {\text{sgn}} \left\{ {\left[ {2 + 2x} \right]\left[ {3 + 3x} \right] - 3[3 + 2x + x^{2} ]} \right\} = {\text{sgn}} \left\{ { - 3 + 6x + 3x^{2} } \right\} = {\text{sgn}} \left\{ { - 1 + 2x + x^{2} } \right\}.$$

(19)

$$\frac{{\partial \rho^{D} }}{\partial x} > 0 \Rightarrow x > 0.4142 \Rightarrow r < 1.4143.$$

(20)

The result follows since $\frac{{\partial \rho^{D} }}{\partial r} = \frac{{\partial \rho^{D} }}{\partial x}\frac{dx}{{dr}} < 0$ when $\frac{{\partial \rho^{D} }}{\partial x} > 0{\kern 1pt}$ given that $\frac{dx}{{dr}} = - \frac{1}{{(1 + r)^{2} }} < 0.$□