The welfare impact of a managerial oligopoly with an altruistic firm

Abstract

I analyse the welfare impact of a mixed market with a private or public firm that is characterised by wider objectives or altruism, in the presence of an agency problem. Contrary to some earlier findings, the total surplus turns out to be increasing in the degree of altruism. This impact is stronger than without an agency problem, despite more stringent conditions for the market to remain mixed. The altruistic firm is more cost efficient, and viable if the market can remain mixed. A competition policy that encourages entry may increase welfare, but its scope is reduced by higher altruism.

Appendix A

Proof of Lemma 1

Use (3.1)–(3.3) to write the total surplus for the case $\alpha =0$ as

$$\begin{aligned} TS\vert _{\alpha =0} =\frac{(n^2+4n+3)(a-c)^2}{2(n+2)^2}. \end{aligned}$$

(7.1)

To get (3.6), calculate the percentage difference as defined by (3.4) and simplify. $\square$

Proof of Proposition 1

• (i) Output is higher in firm 1 because of (4.3) and (4.4). The expected marginal costs $c_{0}-e_{1}$ and $c_{0}-e_{p}$ are lower in firm 1 because of (4.6) and (4.7). The expected average costs $c_{0}-e_{1}+\phi e_1^2 /2x_1$ and $c_{0}-e_{p}+\phi e_p^2 /2x_p$ are also lower in firm 1, as follows from (4.3)–(4.4) and (4.6)–(4.7).

• (ii) Note that the weighted average $\bar{e}$ = ($x_{1}e_{1}+ x_{p}e_{p})$/$x$ is

  $$\begin{aligned} \bar{e}=\frac{A+B\alpha +C\alpha ^2}{(D+\alpha )E}, \end{aligned}$$

  (7.2)

  where

  $$\begin{aligned} A&= (n+1)(\phi -1)^2(a-c_0 )^2,\end{aligned}$$

  (7.3)
  $$\begin{aligned} B&= 2(\phi -1)^2(a-c_0 ),\end{aligned}$$

  (7.4)
  $$\begin{aligned} C&= \phi ^2(n^2+3n+1)-2(n+1)\phi +1,\end{aligned}$$

  (7.5)
  $$\begin{aligned} D&= (n+1)(a-c_0 ),\end{aligned}$$

  (7.6)
  $$\begin{aligned} E&= \left[ {(n+2)\phi -1} \right](\phi -1)^2. \end{aligned}$$

  (7.7)

  Differentiating with respect to $\alpha$ and rearranging shows that $\bar{e}$ is increasing in $\alpha$ if

  $$\begin{aligned} BD-A+2CD\alpha +C\alpha ^2>0 \end{aligned}$$

  (7.8)

  and vice versa. Note that $C>0$ because the roots of the equation $C=0$ are complex for $n>0$. It is also obvious that $BD>A$. The average marginal costs are therefore decreasing in $\alpha$, and it follows from the proof of part (i) that the weighted average of the unit costs is decreasing in $\alpha$. Part (ii) is thereby proved.

Proof of Proposition 2

• (i) Use (4.3)–(4.8) and add $CS=x^{2}/2$ to the profits, which are ($p-c_{0}+e_{1})x_{1}-\phi e_1^2 /2x_1$ and ($p-c_{0}+e_{p})x_{p}-\phi e_p^2 /2x_p$. Rearranging yields (4.9). It is obvious that its second term is positive for all $\alpha >$0. TS is trivially increasing in $\alpha$ if the third term is also positive; in the opposite case it is concave with a maximum for some value $\alpha ^{M}$ of $\alpha$:

  $$\begin{aligned} \alpha ^M=\frac{\phi (1-\phi )^2(a-c_0 )}{(\phi -1)^3-(3n+n^2)\phi ^2+2n\phi }. \end{aligned}$$

  (7.9)

  TS can be decreasing in the relevant interval only if $\alpha ^{M} <$ ($\alpha -1$)($a-c_{0})$/$\phi$ (see Proposition 1). Suppose as an antithesis that this is the case:

  $$\begin{aligned} \frac{\phi (1-\phi )^2(a-c_0 )}{(\phi -1)^3-(3n+n^2)\phi ^2+2n\phi }<\frac{(\phi -1)(a-c_0 )}{\phi }. \end{aligned}$$

  (7.10)

  This inequality can be satisfied only for the following values of $\phi$:

  $$\begin{aligned} \frac{2n+4}{12n+4n^2+8}>\phi >\frac{2n+2}{12n+4n^2+8}. \end{aligned}$$

  (7.11)

  It is obvious that this would require values of $\phi$ below unity, which is ruled out by the assumption $k>$1. TS($\alpha$) is therefore increasing in the relevant interval. Part (i) is thereby proved.

• (ii) Suppose as an antithesis that there exist strictly positive values of $n$ and values of $\phi$ such that $\phi > 1$ for which $m_{1} > m_{2}$. Use Lemma 1 and (4.10) to rearrange the condition $m_{1} > m_{2}$ to

$$\begin{aligned} (n^3+6n^2+10n+4)\phi ^2-(3+2n)(n+3)\phi +(n+3)<0. \end{aligned}$$

(7.12)

Combine this with the inequality

$$\begin{aligned} 1-\phi <0 \end{aligned}$$

(7.13)

so as to get

$$\begin{aligned} (n^3+6n^2+10n+4)\phi ^2-(2n^2+9n+8)\phi +(n+4)<0. \end{aligned}$$

(7.14)

Set the left-hand side of (7.14) equal to zero and solve the equation:

$$\begin{aligned} \phi _{1,2} =\frac{2n^2+9n+8}{2(n^3+6n^2+10n+4)}\pm \sqrt{\frac{-4n^3-23n^2-32n}{4(n^3+6n^2+10n+4)^2}}. \end{aligned}$$

(7.15)

These roots are complex unless the market is a monopoly ($n=0$). It follows that (7.14) is positive for all $\phi >1$ and $n>0$. Hence, the maximum percentage impact of altruism higher than in Sect. 3. $\square$

Proof of Lemma 3

• (i) Differentiate (4.9) with respect to $n$ when $\alpha =0$ and set the derivative equal to zero:

  $$\begin{aligned} \frac{-\phi n+2\phi ^2-4\phi +1}{(\phi n+2\phi -1)^3}=0. \end{aligned}$$

  (7.16)

  Solving for $n$ yields $\overset{\frown }{n} (0) \approx 2\phi -4 +1/\phi$. This is a maximum, because it is obvious that the derivative in question is decreasing in $n$. Part (i) is thereby proved.

• (ii) Differentiate $\overset{\frown }{n}$(0) with respect to $\phi$. It follows that $\overset{\frown }{n}$(0) is minimised if $\phi =\sqrt{2}$. However, as this would imply a negative $\overset{\frown }{n}$(0), it follows that $\overset{\frown }{n}$(0) is increasing in $\phi$ in the relevant area. Part (ii) is thereby proved.$\square$

Proof of Proposition 3

Differentiate (4.9) with respect to $n$, rearrange and solve for an extreme value that has to be a maximum:

$$\begin{aligned}&\overset{\frown }{n}(\alpha )\nonumber \\&\quad \approx \frac{(\phi -1)^2(2\phi ^2-4\phi +1)(a-c_0 )^2-4\phi ^2(\phi \!-\!1)^2(a\!-\!c_0 )\alpha \!+\!\phi ^2(2\phi ^2\!-\!1)\alpha ^2}{\phi (\phi \!-\!1)^2(a\!-\!c_0 )^2\!-\!\phi ^3\alpha ^2}.\nonumber \\ \end{aligned}$$

(7.17)

However, $\alpha = \overset{\frown }{a} = (\phi -1)(a-c_{0})/\phi$ satisfies the polynomials both in the numerator and denominator. Use this to simplify the expression to get (4.11). Note that its numerator is decreasing and its denominator is increasing in $\alpha$, so $\overset{\frown }{n} (\alpha )$ must be decreasing. $\square$