On the social (sub)optimality of divisionalization under product differentiation

Abstract

We revisit the interplay between differentiation and divisionalization in a duopoly version of Ziss (Econ Lett 59:133–138, 1998). We model divisionalization as a discrete problem to prove that (i) firms may choose not to become multidivisional; and (ii) there may arise asymmetric outcomes in mixed strategies, due to the existence of multiple symmetric equilibria. If industry-wide divisionalization is the unique equilibrium, it can be socially efficient provided goods are almost perfect substitutes. Even small degrees of differentiation may suffice to make industry-wide divisionalization socially desirable because of the prevalence of consumers’ taste for variety over the replication of fixed costs.

Appendix

In order to characterise the Nash equilibrium in mixed strategies relevant for region II, we may assume that firm 1 (resp., 2) attaches probability $\mathfrak {p}\in \left[ 0,1\right]$ (resp., $\mathfrak {q}\in \left[ 0,1 \right]$) to strategy S. Firm 1 must choose $\mathfrak {p}$ so as to make firm 2 indifferent between S and M, and the problem of firm 2 in choosing $\mathfrak {q}$ is analogous. The relevant expected profits for the two firms, calculated along columns or rows, are the following:

$$\begin{aligned} E\pi _{2}\left( S\right)= & {} \mathfrak {p}\pi _{SS}^{{ CN}}+\left( 1-\mathfrak {p} \right) \pi _{SM}^{{ CN}} \nonumber \\ E\pi _{2}\left( M\right)= & {} \mathfrak {p}\pi _{MS}^{{ CN}}+\left( 1-\mathfrak {p} \right) \pi _{{ MM}}^{{ CN}} \end{aligned}$$

(A1)

$$\begin{aligned} E\pi _{1}\left( S\right)= & {} \mathfrak {q}\pi _{SS}^{{ CN}}+\left( 1-\mathfrak {q}\right) \pi _{SM}^{{ CN}} \nonumber \\ E\pi _{1}\left( M\right)= & {} \mathfrak {q}\pi _{MS}^{{ CN}}+\left( 1-\mathfrak {q} \right) \pi _{{ MM}}^{{ CN}} \end{aligned}$$

(A2)

Solving the system $E\pi _{i}\left( S\right) =E\pi _{i}\left( M\right) ,$$i=1,2,$ we obtain

$$\begin{aligned} \mathfrak {p}^{*}=\mathfrak {q}^{*}=\frac{\pi _{{ MM}}^{{ CN}}-\pi _{SM}^{{ CN}} }{\pi _{{ MM}}^{{ CN}}-\pi _{SM}^{{ CN}}+\pi _{SS}^{{ CN}}-\pi _{MS}^{{ CN}}} \end{aligned}$$

(A3)

which is positive and lower than one in region II.

The probabilities of observing each of the four possible outcomes are

$$\begin{aligned} P\left( S,S\right)= & {} \mathfrak {p}^{*}\mathfrak {q}^{*}=\left( \mathfrak {p}^{*}\right) ^{2}=\left( \mathfrak {q}^{*}\right) ^{2} \nonumber \\ P\left( M,M\right)= & {} \left( 1-\mathfrak {p}^{*}\right) \left( 1-\mathfrak {q }^{*}\right) =\left( 1-\mathfrak {p}^{*}\right) ^{2}=\left( 1-\mathfrak {q}^{*}\right) ^{2} \nonumber \\ P\left( M,S\right)= & {} P\left( S,M\right) =\left( 1-\mathfrak {p}^{*}\right) \mathfrak {q}^{*}=\left( 1-\mathfrak {q}^{*}\right) ^{2}\mathfrak {p} ^{*} \end{aligned}$$

(A4)

The last step consists in assessing the probability for firms to play one of the two Nash equilibria in pure strategies, $P\left( \mathcal {N}\right) =P\left( S,S\right) +P\left( M,M\right)$ against the probability of making a mistake, $P\left( \mathcal {M}\right) =P\left( M,S\right) +P\left( S,M\right)$, with

$$\begin{aligned} P\left( \mathcal {N}\right) -P\left( \mathcal {M}\right) =\frac{\left( \pi _{{ MM}}^{{ CN}}-\pi _{SM}^{{ CN}}-\pi _{SS}^{{ CN}}+\pi _{MS}^{{ CN}}\right) ^{2}}{\left( \pi _{{ MM}}^{{ CN}}-\pi _{SM}^{{ CN}}+\pi _{SS}^{{ CN}}-\pi _{MS}^{{ CN}}\right) ^{2}}>0. \end{aligned}$$

(A5)