Income Distribution in Network Markets

Abstract

We enquiry about the effects of first and second order stochastic dominance shifts of the distribution of the consumers’ willingness to pay, within the standard model of a market with network externalities and hump-shaped demand curve. This issue is analyzed in the polar cases of perfect competition and monopoly. We find that, while under perfect competition both types of distributional changes result in higher output, provided marginal costs are low enough, in the monopoly case the final outcome depends on the way income distribution and the network externality interact in determining market demand elasticity.

Appendix

Proof Proof of Proposition 1

Let p(x,𝜃) be the implicit demand curve derived from (2). Proving the proposition amounts to proving that (a) there exists at least one value of x, $\overline {x}$ say, such that $\frac {\partial p(\overline {x} ,\theta )}{\partial \theta }=p_{\theta }(\overline {x},\theta )=0$; and (b) that $\frac {\partial ^{2}p(\overline {x},\theta )}{\partial x\partial \theta } =p_{\theta x}(\overline {x},\theta )>0$ if $\overline {x}$ is the highest value of x where a crossing takes place.

• (a) Observe that ${\int }_{y_{m}}^{y_{M}}F_{\theta }(y,\theta )dy\,=0$ implies that F _𝜃 (y,𝜃) takes on both negative and positive values, while ${\int }_{y_{m}}^{y}F_{\theta }(z,\theta )dz\leq 0$ for all y implies the existence of some $\overline {y}\in (y_{m},y_{M})$ such that F _𝜃 (y,𝜃)<0 for all $y< \overline {y}$, F _𝜃 (y,𝜃)>0 for any y greater than and close enough to $\overline {y}$, and $F_{\theta }(\overline {y},\theta )=0$. Take now any quantity x≥0 and let $y(p(x,\theta ),x)=\widehat {y}(x,\theta )$ be the income of the corresponding marginal consumers: using (2) and (3), it is straightforward that $\partial \widehat {y}(x,\theta )/\partial x=-1/f(\widehat {y}(x,\theta ),\theta )<0$, $\lim _{x\rightarrow 0}\widehat {y}(x,\theta )=y_{M}$, and $\lim _{x\rightarrow 1}\widehat {y}(x,\theta )=y_{m}$. Since this amounts to monotonicity, one can uniquely associate to $\overline {y}$ the quantity $\overline {x}$ such that $\widehat {y}(\overline {x},\theta )=\overline {y}$: there follows that, using (7), $p_{\theta }(\overline {x},\theta )=0$.

• (b) Let $x^{\prime }>\overline {x}$: the corresponding income value satisfies $\widehat {y}(x^{\prime },\theta )=y^{\prime }<\overline {y}$ so that the same argument as above can be invoked to yield $F_{\theta }(y^{\prime },\theta )<0$ and hence, by (7), $p_{\theta }(x^{\prime },\theta )>0$; on the other hand, the argument can be reversed for any $x^{\prime \prime }<\overline {x}$ close enough to $\overline {x}$, to the effect that for $y^{\prime \prime }> \overline {y}$ close enough to $\overline {y}$, $F_{\theta }(y^{\prime \prime },\theta )>0$ and hence $p_{\theta }(x^{\prime \prime },\theta )<0$. Since by construction $x^{\prime \prime }<\overline {x} <x^{\prime }$, while $p_{\theta }(x^{\prime \prime },\theta )<0= p_{\theta }(\overline {x},\theta )<p_{\theta }(x^{\prime },\theta )$, p _𝜃 (x,𝜃) is increasing in x around $\overline {x}$, and hence $p_{\theta x}(\overline {x},\theta )>0$. Finally, since $\overline {y}$ is the lowest value of y such that F _𝜃 (⋅,𝜃)=0, and $\widehat { y}(x,\theta )$ is monotonically decreasing in x, $\overline {x}$ is the highest (i.e. rightmost) value of x such that p _𝜃 (⋅,𝜃)=0.

□

Proof Proof of Proposition 3

Consider the marginal revenue defined in Eq. 13. Letting subscripts denote derivatives, differentiation wrt 𝜃 yields

$$\begin{array}{@{}rcl@{}} M_{\theta }(x,\theta ;\sigma ) &=&p_{\theta }(x,\theta )\left( 1-\frac{1}{ \eta (\widehat{y},\theta )}+\sigma \right) \\ &&+\frac{p(x,\theta )}{\eta^{2}(\widehat{y},\theta )}[\eta_{y}(\widehat{y} ,\theta )y_{p}(p(x,\theta ),x)p_{\theta }(x,\theta )+\eta_{\theta }(\widehat{y},\theta )] \end{array} $$

where $\widehat {y}=\widehat {y}(x,\theta )=y(p(x,\theta ),x)$. To ease notation, let λ(y,𝜃) = f _𝜃 (y,𝜃)/f(y,𝜃): we have $\eta _{\theta }(y,\theta )=\eta (y,\theta )\left (\lambda (y,\theta )+ \frac {F_{\theta }(y,\theta )}{1-F(y,\theta )}\right )$, and using (7) it is immediate to verify that $sign\left \{ M_{\theta }(x,\theta ;\sigma )\right \} =sign\left \{ L(x,\theta )\right \}$,with

$$L(x,\theta )=-\frac{F_{\theta }(\widehat{y},\theta )}{1-F(\widehat{y},\theta )}\left( 1-\frac{1}{\eta (\widehat{y},\theta )}+\sigma +\frac{\pi (\widehat{y },\theta )}{\eta (\widehat{y},\theta )}\right) +\lambda (\widehat{y},\theta ) $$

where π(y,𝜃) is Esteban’s income share elasticity. By differentiation it is easily seen that π _𝜃 (y,𝜃)>0 is equivalent to y λ _y (y,𝜃)>0, so that the class of fsd shocks identified by (15) delivers a monotonically increasing λ(y,𝜃).

• (a) Let $p_{\max }=p(x_{\max },\theta )$ be such that $p_{x}(x_{\max },\theta )=0$. Consider the income level y _σ , defined from (11) by the condition η(y _σ ,𝜃)=1/σ, and the associated price-quantity pair (p _σ ,x _σ ), such that p _σ = p(x _σ ,𝜃) and $y_{\sigma }=\widehat {y} (x_{\sigma },\theta )$. Clearly, $p_{\sigma }=p_{\max }$ and

  $$L(x_{\sigma },\theta )=-\frac{F_{\theta }(y_{\sigma },\theta )}{ 1-F(y_{\sigma },\theta )}\left( 1+\frac{\pi (y_{\sigma },\theta )}{\eta (y_{\sigma },\theta )}\right) +\lambda (y_{\sigma },\theta ) $$

  Now notice that

  • (a) F _𝜃 (y,𝜃)≤0 for all y, with at most equality at the endpoints of the support, by the definition of fsd;

  • (b) given (16), 1 + π(y,𝜃)/η(y,𝜃)≥0 for all y;

  • (c) given (15), λ(y,𝜃) is monotonically increasing: hence, $\lim _{y\rightarrow y_{m}}\lambda (y,\theta )<0$ and $\lim _{y\rightarrow y_{M}}\lambda (y,\theta )>0$, since by definition

    $$F_{\theta }(y_{M},\theta )={\int}_{y_{m}}^{y_{M}}f(y,\theta )\lambda (y,\theta )dy=0 $$

    and f(y,𝜃)>0 for $y\in \left (y_{m},y_{M}\right )$. So there is a unique value $y_{\theta }\in \left (y_{m},y_{M}\right )$ such that λ(y _𝜃 ,𝜃)=0, with $\lambda (y,\theta )\lessgtr 0$ as $y\lessgtr y_{\theta }$.

  From (a), (b) and (c) there follows that L(x _σ ,𝜃)>0 if y _σ >y _𝜃 , in which case there exists some $y^{\prime }\in \lbrack y_{\theta },y_{\sigma })$ such that L(x,𝜃)>0 for any x such that y ^∗(p(x,𝜃),x) lies in $[y^{\prime },y_{\sigma }]$. Since y _𝜃 depends only on the income distribution, while clearly y _σ is decreasing in σ and such that $\lim _{\sigma \rightarrow \infty }y_{\sigma }=y_{m}$ and $\lim _{\sigma \rightarrow 0}y_{\sigma }=y_{M}$, there exists a positive threshold value $\overline { \sigma }$ of σ (for which $y_{\overline {\sigma }}=y_{\theta }$), such that for all $\sigma \in (0,\overline {\sigma })$ it is indeed true that y _σ >y _𝜃 : hence, there is a corresponding value $y^{\prime }$ and an associated quantity $x^{\prime }=1-F(y^{\prime },\theta )$, such that L(x,𝜃)>0 for $x\in \lbrack x_{\sigma },x^{\prime }]$. There follows that M _𝜃 (x,𝜃)>0 for x in that interval, and hence output increases for all cost levels c such that $c>\overline {c}= M(x^{\prime },\theta )$ (obviously, it must be $c<M(x_{\sigma },\theta )=p_{\max }$).

• (b) Let y _{1 + σ} be defined by condition (12), i.e. η(y _{1 + σ} ,𝜃)=1/(1 + σ): letting the associated price-quantity pair be (p _{1 + σ} ,x _{1 + σ} ), such that y(p _{1 + σ} ,x _{1 + σ} ) = y _{1 + σ} and p _{1 + σ} = p(x _{1 + σ} ,𝜃), it must be M(x _{1 + σ} ,𝜃;σ)=0 , to which there corresponds the income y _{1 + σ} of the marginal consumer, such that $y_{1+\sigma }=\widehat {y}(x_{1+\sigma },\theta )$, $\widehat {y}(x,\theta )$ monotonically decreasing in x. As before, $sign\left \{ M_{\theta }(x,\theta ;\sigma )\right \}$ $=sign\left \{ L(x,\theta )\right \}$, while at x = x _{1 + σ} we have

  $$L(x_{1+\sigma },\theta )=-\frac{F_{\theta }(y_{1+\sigma },\theta )}{ 1-F(y_{1+\sigma },\theta )}\frac{\pi (y_{1+\sigma },\theta )}{\eta (y_{1+\sigma },\theta )}+\lambda (y_{1+\sigma },\theta ) $$

  To ease notation, let the RHS of the above be Λ(y _{1 + σ} ,𝜃), and observe that

  $$\lim_{y\rightarrow y_{m}}{\Lambda} (y,\theta )\leq 0 $$

  since by assumption η(y _m ,𝜃) = y _m f(y _m ,𝜃)>0 and by definition F _𝜃 (y _m ,𝜃) = F(y _m ,𝜃)=0, while π(y _m ,𝜃) is finite by assumption and $\lim _{y\rightarrow y_{m}}{\Lambda } (y,\theta )=\lim _{y\rightarrow y_{m}}\lambda (y,\theta )<0$.

  Now notice that, using Hopital’s rule,

  $$\lim_{y\rightarrow y_{M}}{\Lambda} (y,\theta )=\lim_{y\rightarrow y_{M}}\lambda (y,\theta )\left( 1+\frac{\pi (y,\theta )}{\eta (y,\theta )} \right) >0 $$

  as by (16) π(y,𝜃) + η(y,𝜃)>0 for all y∈[y _m ,y _M ], and $\lim _{y\rightarrow y_{M}}\lambda (y,\theta )>0$. Hence, by continuity there exists some $\widetilde {y}\in \left (y_{m},y_{M}\right )$ such that ${\Lambda } (\widetilde {y},\theta )=0$ and Λ(y,𝜃)<0 for all $y\in \lbrack y_{m},\widetilde {y})$. Let now $\widetilde {\sigma } \geq 0$ be uniquely defined by $\eta (\widetilde {y},\theta )=\min \left \{ 1,1/(1+\widetilde {\sigma })\right \} =\eta (y_{1+\widetilde {\sigma }},\theta )$: since η(y,𝜃) is monotonically increasing in y, we have $y_{1+\sigma }<\widetilde {y}$ for any $\sigma >\widetilde {\sigma }$ and, by the monotonicity of $\widehat {y}(x,\theta )$, we can associate to $\widetilde {y}$ the quantity $\widetilde {x}=x_{1+\widetilde {\sigma }}$ such that $\widetilde {y}=\widehat {y}(\widetilde {x},\theta )$: so we have that L(x _{1 + σ} ,𝜃)<0, and hence M _𝜃 (x _{1 + σ} ,𝜃;σ)<0, for any $\sigma >\widetilde {\sigma }$; there follows that, for any such σ, there exists an $x^{\prime }<x_{1+\sigma }$, such that M _𝜃 (x,𝜃;σ)<0 for all $x\in (x^{\prime },x_{1+\sigma }]$ : output then decreases for all cost levels c such that $M(x_{1+\sigma },\theta ;\sigma )=0\leq c<\widetilde {c}=M(x^{\prime },\theta ;\sigma )$.

□

Proof Proof of Proposition 4

We first need the following Lemma: □

Lemma 1

Assume 𝜃is a ssd parameter of the distribution F, and let λ(y,𝜃)=f _𝜃 (y,𝜃)/f(y,𝜃). Then $\lim _{y\rightarrow y_{m}}\lambda (y,\theta )\leq 0$.

Proof

By definition (9) of ssd, we have that ${\int }_{y_{m}}^{y}F_{\theta }(x,\theta )dx\leq 0$ for all y∈[y _m ,y _M ]. Since F _𝜃 (y _m ,𝜃)=0, F _𝜃 (⋅,𝜃) cannot be positive around y _m , which also holds for its derivative f _𝜃 (⋅,𝜃) as F _𝜃 (⋅,𝜃) cannot point upwards there. Recall now that f(y,𝜃)>0 for all y∈(y _m ,y _M ): the result then follows trivially, as $sign\left \{ \lambda (y,\theta )\right \} =sign\left \{ f_{\theta }(y,\theta )\right \}$. □

Assume now that σ>0, and recall from the proof of Proposition 3 above that $sign\left \{ M_{\theta }(x,\theta ;\sigma )\right \} =sign\left \{ L(x,\theta )\right \}$,with

$$L(x,\theta )=-\frac{F_{\theta }(\widehat{y},\theta )}{1-F(\widehat{y},\theta )}\left( 1-\frac{1}{\eta (\widehat{y},\theta )}+\sigma +\frac{\pi (\widehat{y },\theta )}{\eta (\widehat{y},\theta )}\right) +\lambda (\widehat{y},\theta ) $$

where as before $\widehat {y}=\widehat {y}(x,\theta )$, monotonically decreasing in x.

Let the RHS be denoted as $\mathcal {L}(\widehat {y},\theta )$. Clearly, $\lim _{x\rightarrow 1}L(x,\theta )=\lim _{y\rightarrow y_{m}} \mathcal {L}(y,\theta )$ $=\lim _{y\rightarrow y_{m}}\lambda (y,\theta )\leq 0$ by the Lemma above, since F _𝜃 (y _m ,𝜃)=0, and the term in brackets tends to a finite value as y tends to y _m , which follows from y _m f(y _m ,𝜃)>0 and π(y _m ,𝜃) being finite.

Moreover, by the definition of ssd, there exists an income level $\overline {y}$ such that F _𝜃 (y,𝜃)<0 for all $y\in \left (y_{m},\overline {y}\right )$ and $F_{\theta }(\overline {y},\theta )=0$. Indeed, from Proposition 2 we know that 𝜃 being ssd implies the existence of a quantity $\overline {x}\in (0,1)$ such that $p_{\theta }(\overline {x},\theta )=0$ and $p_{\theta x}(\overline {x},\theta )>0$: this is the rightmost point where the demand curves cross following a marginal increase in 𝜃, and it satisfies $\overline {y}=\widehat {y}(\overline {x },\theta )$. Using (7), it is then immediate to verify that it must be $f_{\theta }(\overline {y},\theta )>0$ and hence $\mathcal {L}(\overline {y} ,\theta )=\lambda (\overline {y},\theta )=f_{\theta }(\overline {y},\theta )/f(\overline {y},\theta )>0$. All of which implies the existence of some $\widetilde {y}\in \lbrack y_{m},\overline {y})$, such that $\mathcal {L}(\widetilde {y},\theta )=0$ and $\mathcal {L}(y,\theta )>0$ for all $y\in (\widetilde {y},\overline {y}]$, and hence M _𝜃 (x,𝜃;σ)>0 for all $x\in \lbrack \overline {x},\widetilde {x})$, where $\widetilde {x}$ is uniquely identified by the condition $\widetilde {y}=\widehat {y}(\widetilde {x} ,\theta )$.

Let now $\widetilde {\sigma }>0$ be identified by the condition $\widetilde { \sigma }=1/\eta (\widetilde {y},\theta )$. Clearly, this implies that for any σ such that $0<\sigma <\widetilde {\sigma }$ we have $y_{\sigma }> \widetilde {y}$ and hence $\widetilde {x}>x_{\max }$, as the latter satisfies $\sigma =1/\eta (\widehat {y}(x_{\max },\theta ),\theta )$. For any such σ, this means that M _𝜃 (x,𝜃;σ)>0 in the downward sloping region of the demand curve, the only viable for the monopolist problem. Hence, output increases for all costs levels $c\in (\widetilde {c}, \overline {c})$, where $\widetilde {c}=\max \left \{ M(\widetilde {x},\theta ;\sigma ),0\right \}$ and $\overline {c}=\min \left \{ M(x_{\max },\theta ;\sigma ),M(\overline {x},\theta ;\sigma )\right \}$. Take now some positive $\sigma <\widetilde {\sigma }$, and consider the income level y _{1 + σ} , defined by η(y _{1 + σ} ,𝜃)=1/(1 + σ), so that at the corresponding quantity x _{1 + σ} marginal revenue is zero. We then have two possibilities: either $y_{1+\sigma }\geq \widetilde {y}$, i.e. $x_{1+\sigma }\leq \widetilde {x}$, in which case M _𝜃 (x,𝜃;σ)>0 for all $x\in \lbrack \overline {x},x_{1+\sigma }]$, $\max \left \{ M(\widetilde {x},\theta ;\sigma ),0\right \} =0$, and output increases for all $c\in \lbrack 0,\overline {c}]$; or $y_{1+\sigma }<\widetilde {y}$, i.e. $x_{1+\sigma }>\widetilde {x}$, in which case $\max \left \{ M(\widetilde { x},\theta ;\sigma ),0\right \} =M(\widetilde {x},\theta ;\sigma )$, and output increases for all $c\in \lbrack \widetilde {c},\overline {c}]$, $\widetilde {c} =M(\widetilde {x},\theta ;\sigma )>0$. The latter cannot apply if $\sigma < \widetilde {\sigma }-1$, since in that case $\overline {\sigma }=\widetilde { \sigma }-1$ satisfies $\eta (\widetilde {y},\theta )=1/(1+\overline {\sigma })$ and we always have $[\overline {x},x_{1+\sigma }]\subset \lbrack \overline {x}, \widetilde {x})$: hence, output increases for any cost level low enough, if $\widetilde {\sigma }>1$ and $\sigma \in (0,\widetilde {\sigma }-1)$.