Seller competition on two-sided platforms

Abstract

Two-sided platforms connect two or more distinct user groups. Agents on such a platform not only value the participation of users from a different group but are also affected by the same-side network effects that arise from the participation of agents in their own group. We study how negative same-side network effects among sellers affect the participation levels and profit of a monopoly platform. We use a novel specification of the CES utility function to model our consumer preferences, where taste for variety and substitutability are not interrelated. We find that when the platform implements subscription pricing on both sides, an increase in the intensity of competition (higher negative same-side network effects) amongst sellers leads to more participation from both buyers and sellers and there is an increase in the profit of the platform. On the other hand, when the platform can only charge a fee from the seller, participation on both sides first rises and then falls. The platform’s profit also follows the same trend. We also briefly discuss how prices of competing platforms change when there is an increase in the intensity of competition amongst sellers.

Appendices

Appendices

1.1 Subscription pricing

1.1.1 Derivation of demand

Each consumer maximises 3 subject to the following budget constraint.

$$\begin{aligned} \int _{0}^{n_s}{p_i q_i di}=y. \end{aligned}$$

(41)

Given an expenditure y on differentiated good, the first order condition of the maximisation problem of the consumer for good i is

$$\begin{aligned} {n_s}^\theta \frac{1}{\rho }\left[ \int _{0}^{n_s}{q_i}^\rho di\right] ^{\dfrac{1}{\rho }-1} \rho {q_i}^{\rho -1}-\lambda p_i=0. \end{aligned}$$

(42)

Similarly for another good j,

$$\begin{aligned} {n_s}^\theta \frac{1}{\rho }\left[ \int _{0}^{n_s}{q_i}^\rho di\right] ^{\dfrac{1}{\rho }-1} \rho {q_j}^{\rho -1}-\lambda p_j=0. \end{aligned}$$

(43)

The above first two order conditions imply

$$\begin{aligned} \frac{q_i}{q_j}={\left[ \frac{p_j}{p_i}\right] }^{\frac{1}{1-\rho }}. \end{aligned}$$

(44)

Each seller has an identical marginal cost of production. We consider a symmetric equilibrium amongst sellers where goods are equally priced and therefore, equally demanded.

$$\begin{aligned} p_i=p_j \rightarrow \frac{p_i}{p_j}=1\quad \rightarrow \frac{q_i}{q_j}=1 \quad \rightarrow q_i=q_j. \end{aligned}$$

(45)

Substituting the value of $q_i$ from 44 in the budget constraint we get

$$\begin{aligned} y=\int _{o}^{n_s} q_j {\left( \frac{p_j}{p_i}\right) }^{\frac{1}{1-\rho }} p_i di={p_j}^{\frac{1}{1-\rho }}q_j \int _{0}^{n_s}{p_i}^{\frac{-\rho }{\rho -1}}di. \end{aligned}$$

(46)

This implies

$$\begin{aligned} q_j=y{p_j}^{\frac{-1}{1-\rho }}{{\textbf {P}}}^{-1}, \quad \text {where}\quad {\textbf {P}}=\int _{0}^{n_s}{p_i}^{-\frac{\rho }{1-\rho }} di. \end{aligned}$$

(47)

The expression for $q_j$ is the demand function and can be generalised for good i.

1.1.2 Optimal price and quantity of each individual seller

Each seller chooses a price $p_i$ to maximise the following profit function taking P as given.

$$\begin{aligned} \pi _i=(1-t)p_i q_i n_b -c q_i n_b, \quad \text {where} \quad q_i=y{p_i}^{\frac{-1}{1-\rho }}{{\textbf {P}}}^{-1} \end{aligned}$$

(48)

The first order condition is

$$\begin{aligned} p\left( 1-\frac{1}{\frac{1}{(1-\rho )(1-t)}}\right) =c \quad \rightarrow \quad p=\frac{c}{\rho (1-t)}. \end{aligned}$$

(49)

Additionally, the quantity sold by each seller is given by

$$\begin{aligned} q=\frac{\rho (1-t) y }{c n_s}. \end{aligned}$$

(50)

Proof of Proposition 1

We derive $\dfrac{d{n_s}^*}{d\rho }$ and $\dfrac{d{n_b}^*}{d\rho }$ by totally differentiating $\dfrac{\partial \Pi }{\partial n_s}$ and $\dfrac{\partial \Pi }{\partial n_b}$ against $n_s$, $n_b$ and $\rho$ and re-arranging the resulting equations.

$$\begin{aligned} \frac{d{n_s}^*}{d\rho }&= \frac{1}{K}\left( -\frac{\partial ^2 \Pi }{\partial n_s \partial \rho }\frac{\partial ^2\Pi }{{n_b}^2} + \frac{\partial ^2 \Pi }{\partial n_b \partial \rho }\frac{\partial ^2 \Pi }{\partial n_b \partial n_s}\right) \end{aligned}$$

(51)

$$\begin{aligned} \frac{d{n_b}^*}{d\rho }&= \frac{1}{K}\left( -\frac{\partial ^2 \Pi }{\partial n_b \partial \rho }\frac{\partial ^2\Pi }{{n_s}^2} + \frac{\partial ^2 \Pi }{\partial n_s \partial \rho }\frac{\partial ^2 \Pi }{\partial n_b \partial n_s}\right) \end{aligned}$$

(52)

Here, $K= \dfrac{\partial ^2\Pi }{\partial {n_b}^2}\dfrac{\partial ^2\Pi }{\partial {n_s}^2}-{\left( \dfrac{\partial ^2 \Pi }{\partial n_b \partial n_s}\right) }^2 > 0$ due to second-order conditions.

We also show the intermediate calculations to calculate the above mathematical expressions.

$$\begin{aligned} \frac{{\partial }^2\Pi }{\partial {n_s}^2}&=\frac{(k-1) k n_b \rho y {n_s}^{k-2}}{c}-2,\\ \frac{{\partial }^2\Pi }{\partial {n_b}^2}&=-2,\\ \frac{\partial ^2 \Pi }{\partial n_s \partial \rho }&=\frac{kn_b y{n_s}^{k-1}}{c},\\ \frac{\partial ^2 \Pi }{\partial n_s \partial n_b}&=\frac{k \rho y {n_s}^{k-1}}{c},\\ \frac{\partial ^2 \Pi }{\partial n_b \partial \rho }&=\frac{y{n_s}^k}{c}-y. \end{aligned}$$

Let $\eta _s=-\dfrac{\partial ^2 \Pi }{\partial n_s \partial \rho }\dfrac{\partial ^2\Pi }{{n_b}^2} + \dfrac{\partial ^2 \Pi }{\partial n_b \partial \rho }\dfrac{\partial ^2 \Pi }{\partial n_b \partial n_s}$ and $\eta _b= -\dfrac{\partial ^2 \Pi }{\partial n_b \partial \rho }\dfrac{\partial ^2\Pi }{{n_s}^2} + \dfrac{\partial ^2 \Pi }{\partial n_s \partial \rho }\dfrac{\partial ^2 \Pi }{\partial n_b \partial n_s}$.

To understand how $\dfrac{d{n_s}^*}{d\rho }$ and $\dfrac{d{n_b}^*}{d\rho }$ behave, we need to determine the signs of $\eta _s$ and $\eta _b$.

$$\begin{aligned} \eta _b&= \frac{k^2n_b{n_s}^{2k-2}\rho y^2}{c^2}+\left( y-\frac{{n_s}^ky}{c}\right) \left( -2+\frac{(k-1)kn_b{n_s}^{k-2}\rho y}{c}\right) , \end{aligned}$$

(53)

$$\begin{aligned} \eta _s&=\frac{2kn_b{n_s}^{k-1}y}{c}+\frac{k{n_s}^{k-1}\rho y^2 \left( \dfrac{{n_s}^k}{c}-1\right) }{c}. \end{aligned}$$

(54)

We find that $\eta _s ,\eta _b>0$, for $(n_s)^k>c$. $\square$

Table 1 Optimal values- ${\Pi }^*,{m_s}^*, {m_b}^*, {n_s}^*$ and ${n_b}^*$ for different values of $\rho$

1.2 Buyer price restricted to zero

We consider that the platform keeps the entry free for consumers and only charges a subscription fee to the seller. Popular listing platforms such as Yellow Pages, Yelp , Google My Business, and Angie’s List charge businesses either for basic listing or for a premium spot. Typically consumers can access these directories free of charge if they are connected to the internet. In this case, the profit and utility function will be identical to (12) and (13). We assume that $r_b=0$ (standalone utility of the consumer is zero) for easier calculations. Putting $m_b=0$, the platform will maximise its profit function w.r.t. $n_s$. Subsequently, the FOC will be given by

$$\begin{aligned} \frac{\partial \Pi }{\partial n_s}&=r_s+\pi (n_b,n_s)+\frac{\partial \pi (n_b,n_s)}{\partial n_s }n_s-2n_s-f_s=0. \end{aligned}$$

(55)

We summarise our results as follows.

When buyer prices are restricted to zero,

• (i) $\dfrac{d {\Pi }^*}{d \rho } > 0$ for $\rho <\dfrac{1}{2}$ and $\dfrac{d{\Pi }^*}{d\rho }< 0$ for $\rho >\dfrac{1}{2}$.

• (ii) $\dfrac{d {n_s}^*}{d \rho }>0$ for $\rho <\dfrac{1}{2}$ and $\dfrac{d {n_s}^*}{d \rho }<0$ for $\rho >\dfrac{1}{2}$

• (iii) $\dfrac{d {n_b}^*}{d \rho }>0$ for $\rho <\dfrac{1}{1+k}$ and $\dfrac{d {n_b}^*}{d \rho }<0$ for $\rho >\dfrac{1}{1+k}$

Proof

• (i) To determine the sign of $\dfrac{d {\Pi }^*}{d\rho }$ we write the first order condition given in 55 as follows

  $$\begin{aligned} \dfrac{\partial {\Pi }}{\partial n_s}= (r_s-n_s-f_s)n_s +n_b{\tilde{\pi }}, \end{aligned}$$

  (56)

  where ${\tilde{\pi }}$ is the profit per buyer for each seller. We know that $n_b=\dfrac{(n_s)^k \rho y}{c}$. Using the envelope theorem we have

  $$\begin{aligned} \dfrac{\partial {\Pi }^*}{\partial \rho }={n_s}^{k-1}\frac{y^2}{c^2}(1-2\rho ). \end{aligned}$$

  (57)

  This expression is positive for $\rho <\dfrac{1}{2}$ and negative for $\rho >\dfrac{1}{2}$.

• (ii) By Totally differentiating $\dfrac{\partial \Pi }{\partial n_s}$, we compute $\dfrac{d{n_s}^*}{d\rho }=-\dfrac{{\partial }^2\Pi }{\partial {n_s}\partial \rho } \Big / \dfrac{{\partial }^2\Pi }{\partial {n_s}^2}$ which is equal to

  $$\begin{aligned} \dfrac{d{n_s}^*}{d\rho }=-\frac{k (2 \rho -1) y^2 {n_s}^{k+1}}{2 c^2 {n_s}^2+(k-1) k (\rho -1) \rho y^2 {n_s}^k}. \end{aligned}$$

  (58)

  The above expression is positive for $\rho <\frac{1}{2}$ and negative for $\rho >\frac{1}{2}$

• (iii) We know that $n_b=\dfrac{(n_s)^k \rho y}{c}$, therefore

  $$\begin{aligned} \dfrac{d{n_b}^*}{d\rho ^*}=\dfrac{k {n_s}^{k-1}.\frac{d{n_s}^*}{d\rho }\rho y}{c} + \dfrac{{n_s}^{k} y}{c}. \end{aligned}$$

  (59)

  Putting the value of $\dfrac{d{n_s}^*}{d\rho }$ in the above equation we have

  $$\begin{aligned} \dfrac{d{n_b}^*}{d\rho ^*}=\dfrac{y {ns}^k \left( 2 c^2 {ns}^2-k \rho y^2 {ns}^k (k \rho +\rho -1)\right) }{2 c^3 {ns}^2+c (1-k) k (1-\rho ) \rho y^2 {ns}^k}. \end{aligned}$$

  (60)

  In the above expression the denominator is positive. For all values of y such that $c^2 {n_s}^2 < {n_s}^k y^2$, the numerator is positive for $\rho <\frac{1}{1+k}$ and negative for $\rho >\frac{1}{1+k}$.

$\square$

We substitute the value of $n_b$ in the profit of seller (12) to get the following expression, $\pi (n_s)=\dfrac{{n_s}^{k-1}\rho y^2(1-\rho )n_b}{c}$. It is straightforward to calculate that $\dfrac{\partial \pi }{\partial \rho }>0$ for $\rho <\dfrac{1}{2}$. On the other hand, $\dfrac{\partial \pi }{\partial \rho }<0$ for $\rho >\dfrac{1}{2}$.

Higher substitutability between the goods reduces the sellers’ profit for $\rho >\frac{1}{2}$. To incentivize more sellers to join the platform, the platform must charge a subscription fee less than zero while charging positive fees on the buyer side. However, since the platform cannot do so in this case, it cannot attract more sellers to the platform for $\rho >\frac{1}{2}$. Sellers exit the platform as the effect of competition intensifies for $\rho >\frac{1}{2}$. Therefore, the platform profit increases for $\rho <\frac{1}{2}$ and then falls. Consequently, the number of buyers go up till the effect of decrease in price compensate the negative cross-side network effect caused by the exit of sellers and then falls when $\rho >\dfrac{1}{1+k}$.

Since the optimal subscription fee for seller (${m_s}^*$) cannot be characterised, we simulate the value for optimal variables for a given value of parameters in Table 2. We also plot ${m_s}^*$ for different values of $\rho$ in Fig. 3 . We see that as $\rho$ increases ${m_s}^*$ increases for $\rho <\frac{1}{2}$ and then falls. This is because till $\rho <\frac{1}{2}$, as the extent of competition intensifies, more buyers join the platform. The cross-side network effects exerted by buyers overpowers the increase in competition effect and the profit of sellers go up. Therefore, sellers derive more value from the platform and therefore platform charges them a higher subscription fee. For $\rho >\frac{1}{2}$, the competition effect dominates and seller profit goes down forcing the platform to charge sellers a lower subscription fee.

Fig. 3

Optimal Seller subscription fee for different values of $\rho$ Notes: We plot the values of ${m_s}^*$ for $0.05\le \rho \le 0.95$ at an increment of 0.05. We consider $k=0.7$, $c=2$, $r_s=100$ and $y=100$

Table 2 Optimal values- ${\Pi }^*,{m_s}^*, {n_s}^*$ and ${n_b}^*$ for different values of $\rho$

1.3 Usage pricing

1.3.1 When $\theta =k-\dfrac{1-\rho }{\rho }$ in the consumer utility function

We calculate optimal quantity q sold by each seller and the price p charged for it as $q=\dfrac{y (\rho -\rho t)}{c{n_s}}$ and $p=\dfrac{c}{\rho (1-t)}$. Profit for each seller is given by $\pi (n_s,n_b)=\dfrac{n_b(1-\rho )(1-t)y}{n_s}$ and the utility for each buyer is $u(n_s)=\dfrac{\rho (1-t) y {n_s}^k}{c}$. Substituting the value of $u(n_s)$ in Eq. 10 with stand-alone utilities equal to zero and no subscription fees) gives us the number of buyers $n_b=\dfrac{\rho (1-t) y {n_s}^k}{c}$. We substitute this value of $n_b$ in $\pi (n_s,n_b)$, which gives us the profit of each sellers as $\pi (n_s)=\dfrac{(1-\rho ) \rho (t-1)^2 y^2{n_s}^{k-1}}{c}$. We then substitute $\pi (n_s)$ in 9 to get the number of sellers as follows.

$$\begin{aligned} n_s= \left( \frac{(1-\rho ) \rho (t-1)^2 y^2}{c}\right) ^{\frac{1}{2-k}}. \end{aligned}$$

(61)

Substituting the value of $n_s$ back in $n_b$ we get, $n_b=\dfrac{\rho (1-t) y \left( \left( \frac{(1-\rho ) \rho (t-1)^2 y^2}{c}\right) ^{\frac{1}{2-k}}\right) ^k}{c}$. The quantity sold by each sellers is $q=\dfrac{\rho (1-t) y \left( \dfrac{(1-\rho ) \rho (t-1)^2 y^2}{c}\right) ^{\frac{1}{k-2}}}{c}$. Hence, the profit of the monopoly platform will be given by 22. Furthermore, we can also calculate ${n_s}^*$ and ${n_b}^*$ by substituting $t^*$. These expressions are given in 23 and 24.

1.3.2 When $\theta =0$ in consumer utility function

We substitute the values of p and q in utility and profit function of the buyer and seller. Subsequently we have $\pi (n_s,n_b)= \dfrac{n_b (1-\rho ) (1-t) y}{n_s}$ and $u(n_s)= \dfrac{\rho (1-t) y {n_s}^{\frac{1-\rho }{\rho }}}{c}$. Using the profit and utilty expressions we calculate $n_b=\dfrac{\rho (1-t) y {n_s}^{\frac{1-\rho }{\rho }}}{c}$. By substituting the value of $n_b$ in the profit of sellers, we have $\pi (n_s)=\frac{(1-\rho ) \rho (t-1)^2 y^2 {n_s}^{\frac{1}{\rho }-2}}{c}$. We calculate $n_s$ using equation 9 , where stand-alone utilities are zero and there are no subscription charges. We have

$$\begin{aligned} n_s= \left( \frac{(1-\rho ) \rho (t-1)^2 y^2}{c}\right) ^{\frac{\rho }{3 \rho -1}}. \end{aligned}$$

(62)

We can substitute $n_s$ from above to get $q= \frac{\rho (1-t) y \left( \frac{(1-\rho ) \rho (t-1)^2 y^2}{c}\right) ^{\frac{\rho }{1-3 \rho }}}{c}$. Substituting the final values of p, q and $n_s$ in 21, we get the profit of the monopoly platform as given in 28. Furthermore, we use the value of $t^*$ to calculate the optimal number of buyers (${n_b}^*$) and sellers (${n_s}^*$).

Proof of Proposition 4

We use ${n_b}^*$ and ${n_s}^*$ in 29 and 30 to calculate

$$\begin{aligned} \dfrac{d{n_b}^*}{d\rho } =&\frac{16^{\frac{\rho }{1-3 \rho }} \left( \frac{(1-\rho ) (\rho +1)^2 y^2}{c \rho }\right) ^{\frac{\rho }{3 \rho -1}} \left( \left( 1-\rho ^2\right) \log \left[ \frac{(1-\rho ) (\rho +1)^2 y^2}{16 c \rho }\right] +\rho (\rho (6 \rho -5)+4)-1\right) }{(1-3 \rho )^2 \left( \rho ^2-1\right) }, \end{aligned}$$

(63)

$$\begin{aligned} \dfrac{d{n_s}^*}{d\rho }=&\phi \left( (\rho -1) \rho \left( \log \left[ \frac{(1-\rho )(\rho +1)^2 y^2}{16 c \rho }\right] +3 \rho -1\right) \right) \nonumber \\&-\phi \left( (1-3 \rho )^2 \log \left[ \left( \frac{(1-\rho ) \rho \left( \frac{3 \rho -1}{4 \rho }-1\right) ^2 y^2}{c}\right) ^{\frac{\rho }{3 \rho -1}}\right] \right) ,\nonumber \\&\text {where} \quad \phi =\frac{(\rho +1) y \left( \left( \frac{(1-\rho ) \rho \left( \frac{3 \rho -1}{4 \rho }-1\right) ^2 y^2}{c}\right) ^{\frac{\rho }{3 \rho -1}}\right) ^{\frac{1}{\rho }-1}}{4 c (1-3 \rho )^2 \rho ^2}. \end{aligned}$$

(64)

Careful examinations of both these equations reveal that they are negative for $\rho >\frac{1}{3}$. $\square$

1.4 Extensions

1.4.1 Alternative functional form of $\theta$

1.4.2 Proof of Proposition 9

• (1) Following identical steps as we did in Lemma 1, we find utility per seller ${{{\widetilde{u}}}(n_s)}$ and profit per buyer ${{{{\widetilde{\pi }}}}(n_s)}$ as follows

  $$\begin{aligned} {\widetilde{\pi }}(n_s)= \dfrac{(1-\rho )y}{n_s}, \qquad \widetilde{u}(n_s)= \dfrac{{n_s}^{k\rho -\frac{\rho }{2}-1}\rho y}{c}. \end{aligned}$$

  (65)

  The profit function of the platform then becomes

  $$\begin{aligned} \Pi = n_b (r_b-n_b-f_b)+ n_s(r_s-n_s-f_s)+ (\widetilde{u}(n_s)+ {\widetilde{\pi }}(n_s))n_s n_b. \end{aligned}$$

  (66)

  We calculate

  $$\begin{aligned} \frac{\partial \widetilde{u}}{\partial \rho }&=\frac{y {n_s}^{k \rho -\frac{\rho }{2}-1}}{c}+\frac{\left( k-\frac{1}{2}\right) \rho y \log ({n_s}){n_s}^{k \rho -\frac{\rho }{2}-1}}{c}, \end{aligned}$$

  (67)
  $$\begin{aligned} \frac{\partial {\widetilde{\pi }}}{\partial \rho }&=\dfrac{-y}{n_s} \end{aligned}$$

  (68)

  Using the envelope theorem, we can find how profit changes w.r.t. $\rho$.

  $$\begin{aligned} \frac{d{\Pi }^*}{d\rho } = \Big (\frac{\partial \widetilde{u}(n_s)}{\partial \rho } + \frac{\partial {\widetilde{\pi }}(n_s)}{\partial \rho }\Big ) n_s n_b \end{aligned}$$

  (69)

  For $n_s>>0$ and $k>\dfrac{1}{2}$, the expression above is always positive.

• (2) Using similar expression as specified in 51, we calculate $\dfrac{d{n_s}^*}{d\rho }$. To determine the sign of this expression we need to establish the sign of $\eta _s$ as mentioned in proposition 1. Therefore,

  $$\begin{aligned} \eta _s=\frac{(2 k-1) \rho y {n_s}^{(k-1) \rho -1} \left( \rho y {n_s}^{k \rho } ((2 k-1) \rho \log ({n_s})+2)-2 c {n_s}^{\rho /2} ((1-2 k) {n_b} \rho \log ({n_s})-4 {n_b}+\rho y)\right) }{4 c^2} \end{aligned}$$

• (3) We calculate $\dfrac{d{n_b}^*}{d\rho }$ using 52 and find the expression for $\eta _b$ as follows.

  $$\begin{aligned} \eta _b=&\dfrac{y {n_s}^{-\rho -2}}{4 c^2} (-8 c^2 {n_s}^{\rho +2}+(2 k-1) \rho \log ({n_s}) \left( 4 c {n_s}^{\left( k+\frac{1}{2}\right) \rho +2}+(2 k-1) {n_b} \rho ^2 y{n_s}^{2 k \rho }\right) \\&+c {n_s}^{\left( k+\frac{1}{2}\right) \rho } \left( (2 k-1) {n_b} \rho ^2 y (2 k \rho -\rho -2)+8 {n_s}^2\right) +(2 k-1) {n_b} \rho ^2 y (2 k \rho -\rho +2){n_s}^{2 k \rho }) \end{aligned}$$

  The above expression is always positive for $k>\dfrac{1}{2}$ and sufficiently large values of $n_s$ and $n_b$.

1.4.3 Competing platforms

The profit for platform i is given by

$$\begin{aligned} {\Pi }^i= ({m_s}^i-f_s){n_s}^i+({m_b}^i-f_b){n_b}^i. \end{aligned}$$

The equations for indifferent buyer and seller between the two platforms can be written as

$$\begin{aligned} n_b&=\frac{1}{2}+\frac{1}{2t_b}(\Delta u({n_s}^i,{n_b}^i))-\frac{1}{2t_b}({m_b}^i-{m_b}^j),\\ n_s&=\frac{1}{2}+\frac{1}{2t_s}(\Delta \pi ({n_s}^i,{n_b}^i))-\frac{1}{2t_s}({m_s}^i-{m_s}^j), \end{aligned}$$

where

$$\begin{aligned} \Delta u({n_s}^i,{n_b}^i)&=u({n_s}^i,{n_b}^i)-u({1-n_s}^i,{1-n_b}^i),\\ \Delta \pi ({n_s}^i,{n_b}^i)&=\pi ({n_s}^i,{n_b}^i)-\pi (1-{n_s}^i,1-{n_b}^i). \end{aligned}$$

Substituting the value of indifferent buyer and indifferent seller in the profit equation and then differentiating w.r.t. ${m_s}^i$ and ${m_b}^i$ gives us the following first order conditions.

$$\begin{aligned} \frac{\partial {\Pi }^i}{\partial {m_s}^i}&= \left( \Delta \pi _s \frac{d{n_s}^i}{d{m_s}^i} {m_s}^i+\Delta \pi \right) \frac{1}{2t_s}-\frac{{m_s}^i}{t_s}+\frac{{m_s}^j}{2t_s}+\frac{1}{2}\\&\quad -\Delta \pi _s\frac{d{n_s}^i}{d{m_s}^i} f_s+\Delta u_s \frac{dn_s}{d{m_s}^i}({m_b}^i-f_b)+ \frac{f_s}{2t_s},\\ \frac{\partial {\Pi }^i}{\partial {m_b}^i}&= \left( \Delta u_b\frac{d{n_b}^i}{d{m_b}^i}{m_b}^i+\Delta u\right) \frac{1}{2t_b} -\frac{{m_b}^i}{t_b}+\frac{{m_b}^j}{2t_b}+\frac{1}{2}\\&\quad -\Delta u_b\frac{d{n_b}^i}{d{m_b}^i}f_b+\Delta \pi _b \frac{d{n_b}^i}{d{m_b}^i}({m_s}^i-f_s) + \frac{f_b}{2t_b}, \end{aligned}$$

where $\Delta u_s({n_s}^i,{n_b}^i) \equiv \dfrac{\partial \Delta u({n_s}^i,{n_b}^i)}{\partial n_s}$, $\Delta \pi _s({n_s}^i,{n_b}^i) \equiv \dfrac{\partial \Delta \pi ({n_s}^i,{n_b}^i)}{\partial n_s}$, $\Delta u_b({n_s}^i,{n_b}^i) \equiv \dfrac{\partial \Delta u({n_s}^i,{n_b}^i)}{\partial n_b}$ and $\Delta \pi _b({n_s}^i,{n_b}^i) \equiv \dfrac{\partial \Delta \pi ({n_s}^i,{n_b}^i)}{\partial n_b}$. Symmetric equations can be written down for platform j.

We can use the equation for indifferent buyer and seller to find $\dfrac{dn_s}{d{m_s}^i}$ and $\dfrac{dn_b}{d{m_b}^i}$ which can be written as

$$\begin{aligned} \frac{dn_b}{d{m_b}^i}= & {} \frac{\frac{-1}{2t_b}}{(1-\frac{\Delta u_b}{2t_b})},\\ \frac{dn_s}{d{m_s}^i}= & {} \frac{\frac{-1}{2t_s}}{(1-\frac{\Delta \pi _s}{2t_s})}. \end{aligned}$$

1.4.4 Proof of Proposition 10

Since it is symmetric equilibrium, all the expressions can be calculated using the fact that ${n_b}^1={n_s}^1={n_b}^2={n_s}^2=\frac{1}{2}$. This gives us the two equilibrium values of ${m_s}^*$ and ${m_b}^*$ as follows.

$$\begin{aligned} {m_s}^*= & {} f_s+t_s-\frac{1}{2}\left( \Delta u_s\left( \frac{1}{2},\frac{1}{2}\right) +\Delta \pi _s\left( \frac{1}{2},\frac{1}{2}\right) \right) ,\\ {m_b}^*= & {} f_b+t_b-\frac{1}{2}\left( \Delta u_b\left( \frac{1}{2},\frac{1}{2}\right) +\Delta \pi _b\left( \frac{1}{2},\frac{1}{2}\right) \right) . \end{aligned}$$

In order to get the final expressions, we calculate the values of $\Delta u_s(\frac{1}{2},\frac{1}{2})$, $\Delta \pi _s(\frac{1}{2},\frac{1}{2})$, $\Delta u_b(\frac{1}{2},\frac{1}{2})$ and $\Delta \pi _b(\frac{1}{2},\frac{1}{2})$. We substitute the utility of the buyer and profit of the seller from equation 2.

$$\begin{aligned} \Delta u({n_s}^i)&= \frac{\rho y}{c}({{n_s}^i}^k-{(1-{n_s}^i)}^k),\\ \Delta \pi ({n_s}^i,{n_b}^i)&= (1-\rho )y\left( \frac{{n_b}^i}{{n_s}^i}-\frac{1-{n_b}^i}{1-{n_s}^i}\right) . \end{aligned}$$

Differentiating the above expressions with respect to ${n_s}^i$ and ${n_b}^i$, we get

$$\begin{aligned} \Delta u_s({n_s}^i)&=\frac{\rho y}{c}k{({n_s}^i)}^{k-1}+{(1-{n_s}^i)}^{k-1},\\ \Delta \pi _s({n_s}^i,{n_b}^i)&=(1-\rho )y\left( \frac{-{n_b}^i}{{({n_s}^i)}^2}-\frac{1-{n_b}^i}{{(1-{n_s}^i)}^2}\right) ,\\ \Delta \pi _b({n_s}^i,{n_b}^i)&=(1-\rho )y\left( \frac{1}{{n_s}^i}+\frac{1}{1-{n_s}^i}\right) . \end{aligned}$$

In the above equations, we substitute ${n_s}^i, {n_b}^i=\frac{1}{2}$. Since, $\Delta u({n_s}^i,{n_b}^i )$ is not a function of ${n_b}^i$ , $\Delta u_b(\frac{1}{2},\frac{1}{2})=0$. This gives us

$$\begin{aligned} \Delta u_s\left( \frac{1}{2}\right)&=\frac{\rho y}{c}2k{\frac{1}{2}}^{k-1},\\ \Delta \pi _s\left( \frac{1}{2},\frac{1}{2}\right)&=-2(1-\rho )y,\\ \Delta \pi _b\left( \frac{1}{2},\frac{1}{2}\right)&=4(1-\rho )y. \end{aligned}$$