Group buying and consumer referral on a social network

Abstract

This paper examines a unique selling strategy, group buying on a social E-commerce platform. We proposed a framework on how to design a group buying strategy in the presence of different social network attributes (different structures, different referral costs, and different network externalities) to examine the following questions: (1) How will the group buying threshold depend on consumer referrals under social networks attributes? (2) What determines the optimal group buying threshold and the group buying price in different social networks attributes? (3) What is the efficiency of group buying with threshold compared to the traditional group buying without referrals and referral reward program? We show that the threshold causes an induction effect, but this induction effect is meaningless when the externality is considerably greater than the referral cost. Moreover, we also show that higher referral costs and prices result in a larger threshold when referral cost and externality have approximate impact on the utility of the focal consumer. We also find that the firm cannot set a low price of group buying even when the referral cost is much higher than the network externality. Finally, we present a sufficient condition that a group buying strategy is better than a referral reward strategy.

Appendix

The proof of Proposition 2.

When $\;p_{1} < \frac{c - \delta }{c}b$, note $f\left( {\underline{x} } \right) = \frac{{b\left( {b - p_{1} } \right)}}{{\left( {b - \delta \underline{x} } \right)^{2} }}\int\limits_{{\underline{x} }}^{m} {g\left( n \right)dn} - \left( {\frac{{b - p_{1} }}{{b - \delta \underline{x} }}\underline{x} + 1} \right)g\left( {\underline{x} } \right)$. $f\left( 0 \right) = \frac{{b - p_{1} }}{b} > 0$,$f\left( m \right) = - \left( {\frac{{b - p_{1} }}{b - \delta m}m + 1} \right)g\left( m \right) < 0$. Thus, So the function in Eq. (5) has the increase and decrease in $\left[ { 0 ,m} \right]$, and function in Eq. (5) is increasing at first. Suppose $f\left( {x^{ * } } \right) = 0$. If $0\le x^{ * } \le \frac{{bc - \sqrt {bc\left( {bc - 4b\delta + 4bp_{1} } \right)} }}{2\delta c}$, the maximum point is $x^{ * }$. But if $x^{ * } > \frac{{bc - \sqrt {bc\left( {bc - 4b\delta + 4bp_{1} } \right)} }}{2\delta c}$, the maximum point must be $\frac{{bc - \sqrt {bc\left( {bc - 4b\delta + 4bp_{1} } \right)} }}{2\delta c}$. This is an extreme case, and it is impossible to react much to the association between the various external factors, and we are temporarily ignoring this situation. Therefore, when $p_{1} < \frac{c - \delta }{c}b$, the optimal threshold is in $\left( {1,\frac{{\left( {b - p} \right)\left( {bc - \sqrt {bc\left( {bc - 4b\delta + 4bp_{1} } \right)} } \right)}}{{\delta \left( {bc + \sqrt {bc\left( {bc - 4b\delta + 4bp_{1} } \right)} } \right)}}} \right)$ should satisfy Eq. (8).

We can proof the conclusion when $p_{1} \ge \hbox{max} \left( {\frac{c - \delta }{c}b,\frac{\delta - c}{\delta }b} \right)$ in same way.

The proof of Proposition 6.

First, we consider the first subsection of Eq. (10), we note it as Eqs. (20) and (21):

$$\left[ {\int\limits_{{\underline{x} }}^{m} {\frac{b - p}{{b - \delta \underline{x} }}\underline{x} g\left( n \right)dn} + 1- G\left( {\underline{x} } \right)} \right]\;\;p$$

(20)

$$\left[ {\int\limits_{{\underline{x} }}^{{\overline{x} }} {\frac{b - p}{{b - \delta \underline{x} }}\underline{x} g\left( n \right)dn} + \int\limits_{{\overline{x} }}^{m} {\frac{b - p}{b - \delta n}ng\left( n \right)dn} + 1- G\left( {\underline{x} } \right)} \right]\;p$$

(21)

Then, according to the first derivative, we can easily know that both the functions in two subsections are increased first and then reduced in $\left( { 0 ,\frac{p}{\delta }} \right)$. Therefore, it is obvious that the maximum point of Eq. (10) can only be taken at the maximum point of Eq. (20), the maximum point of Eq. (21), and the point of discontinuity $\frac{c - \delta }{c}b$.

According to the first derivative of Eq. (11), we got the maximum point of the Eq. (20) is $\frac{b}{ 2} + \frac{{b - \delta \underline{x} }}{{2\underline{x} }}$, which is bigger than $\frac{b}{2}$

When $c > 2\delta$, $\frac{c - \delta }{c}b > \frac{b}{2}$, and the maximum point of Eq. (21) is bigger than $\frac{c - \delta }{c}b$,

So, the Proposition 6 can be proved.

The proof of Proposition 7.

• (1) When $\delta < c$.

If $p_{1} < \frac{c - \delta }{c}b$, compare Eq. (11) to (5). Note $F\left( {\underline{x} } \right) = Q\left( {\underline{x} } \right) - Q_{T} = \int\limits_{{\underline{x} }}^{m} {\left( {\frac{{b - p_{1} }}{{b - \delta \underline{x} }}\underline{x} + 1} \right)g\left( n \right)dn} - 1$. Clearly, $F\left( 0 \right) = 0$. $\frac{{\partial F\left( {\underline{x} } \right)}}{{\partial \underline{x} }} = \frac{{b\left( {b - p_{1} } \right)}}{{\left( {b - \delta \underline{x} } \right)^{2} }}\int\limits_{{\underline{x} }}^{m} {g\left( n \right)dn} - \left( {\frac{{b - p_{1} }}{{b - \delta \underline{x} }}\underline{x} + 1} \right)g\left( {\underline{x} } \right)$, $\left. {\frac{{\partial F\left( {\underline{x} } \right)}}{{\partial \underline{x} }}} \right|_{{\underline{x} = 0}} = \frac{{b - p_{1} }}{b} > 0$, thus, the function in Eq. (5) is increasing at first. So, we can always find a $\underline{x}$ makes $Q\left( {\underline{x} } \right) > Q_{T}$, even if the $\underline{x}$ is not optimal.

If $p_{1} > \frac{c - \delta }{c}b$, we should compare Eq. (11) to (6). From the present of Eqs. (5) and (6). We can easily see that $\left( {M + 1} \right)\int\limits_{{\underline{x} }}^{{\overline{x} }} {g\left( n \right)} dn + \int\limits_{{\overline{x} }}^{m} {\left( {\frac{{b - p_{1} }}{b - \delta n}n + 1} \right)} g\left( n \right)dn \ge \int\limits_{{\underline{x} }}^{m} {\left( {M + 1} \right)g\left( n \right)dn}$ for any $\underline{x}$. Thus, the above conclusion is also established when $p_{1} > \frac{c - \delta }{c}b$.

When $\delta \ge c$ and $p_{1} > \frac{c - \delta }{c}b$, compare Eq. (12) to (6). Note $F\left( {\underline{x} } \right) = Q\left( {\underline{x} } \right) - Q_{T}$

$$= \left( {\frac{{b - p_{1} }}{{b - \delta \underline{x} }}\underline{x} + 1} \right)\int\limits_{{\underline{x} }}^{{\overline{x} }} {g(n)dn} + \int\limits_{{\overline{x} }}^{m} {\left( {\frac{{b - p_{1} }}{b - \delta n}n + 1} \right)g(n)dn} - \int\limits_{0}^{{x^{\prime}}} {g(n)dn} - \int\limits_{{x^{\prime}}}^{m} {\frac{{b - p_{1} }}{b - \delta n}} ng\left( n \right)dn$$

Clearly, $F\left( 0 \right) = 0$, $\left. {\frac{{\partial F\left( {\underline{x} } \right)}}{{\partial \underline{x} }}} \right|_{{\underline{x} = 0}} = \frac{{b - p_{1} }}{b} > 0$, thus, the function in Eq. (6) is increasing at first. So, we can always find a $\underline{x}$ makes $Q\left( {\underline{x} } \right) > Q_{T}$, even if the $\underline{x}$ is not optimal.