Appendix
Proof of Proposition 1
We divide this proof into two parts.Footnote 33 In part 1, we derive the optimal prices and couponing strategies. This is the equilibrium candidate. Then in part 2, we show that neither firm has an incentive to deviate unilaterally.
Part 1: equilibrium candidate
Firms’ profit functions are
$$\begin{aligned} \pi _{1}= & {} p_{1}(1-\alpha )(1-\lambda _{2})L+p_{1}(1-\alpha )(1-\lambda _{1})(p_{2}-p_{1})+p_{1}(1-\alpha )\lambda _{2}(L\nonumber \\&+p_{2}-r_{2}-p_{1})+(p_{1}-r_{1})(1-\alpha )\lambda _{1}(p_{2}-p_{1}+r_{1})+p_{1}\alpha (L+p_{2}-p_{1})\nonumber \\&-r_{1}\alpha \lambda _{1}L-k(\lambda _{1}L)^{2}, \end{aligned}$$
(5)
$$\begin{aligned} \pi _{2}= & {} p_{2}(1-\alpha )(1-\lambda _{1})(L-p_{2}+p_{1})+p_{2}(1-\alpha )\lambda _{1}(L-p_{2}+p_{1}-r_{1}) \nonumber \\&+(p_{2}-r_{2})(1-\alpha )\lambda _{2}(p_{1}-p_{2}+r_{2})+p_{2}\alpha (L-p_{2}+p_{1})\nonumber \\&-r_{2}\alpha \lambda _{2}L-k(\lambda _{2}L)^{2}. \end{aligned}$$
(6)
Taking derivative of \(\pi _{2}\) with respect to \(p_{2}\), \(r_{2}\) and \(\lambda _{2}\) respectively, then imposing the symmetry conditions (\(p_{1}=p_{2},r_{1}=r_{2}\) and \(\lambda _{1}=\lambda _{2}\)), we can obtain
$$\begin{aligned}&\displaystyle \frac{\partial \pi _{2}}{\partial r_{2}}=-\lambda _{2}(\alpha L-p_{2}-2r_{2}\alpha +2r_{2}+\alpha p_{2})=0,\end{aligned}$$
(7)
$$\begin{aligned}&\displaystyle \frac{\partial \pi _{2}}{\partial \lambda _{2}}=-2k\lambda _{2}L^{2}-r_{2}\alpha L-\alpha p_{2}r_{2}+r_{2}^{2}\alpha +p_{2}r_{2}-r_{2}^{2}=0,\end{aligned}$$
(8)
$$\begin{aligned}&\displaystyle \frac{\partial \pi _2}{\partial p_2}=L-p_2-\lambda _{2}r_{2}\alpha +\lambda _{2} \alpha p_2+r_{2} \lambda _{2}-\lambda _{2} p_2=0. \end{aligned}$$
(9)
Since the cost of coupon distribution is quadratic in \(\lambda \), and the rest is roughly linear in \(\lambda \), it must be that the optimal \( \lambda _{2}>0\). Then, Eq. (7) implies,
$$\begin{aligned} r_{2}=\frac{\alpha L-p_2+\alpha p_2}{2(\alpha -1)}=\frac{1}{2}\left( p_2- \frac{\alpha L}{1-\alpha }\right) . \end{aligned}$$
(10)
From this expression, we can see that \(p_2>r_{2}\).
Next, we substitute the expression for \(r_{2}\) into Eq. (8) and solve for \(\lambda _{2}\). We obtain
$$\begin{aligned} \lambda _{2}=\frac{(\alpha L-p_{2}+\alpha p_{2})^{2}}{8(1-\alpha )kL^{2}}. \end{aligned}$$
(11)
Using \(r_{2}\) and \(\lambda _{2}\) in Eq. (9), we can solve for the equilibrium priceFootnote 34
$$\begin{aligned} p_{2}=\frac{\left( \frac{2}{3}A-\frac{3}{2}\frac{-\frac{4}{9}\alpha ^{2}+ \frac{16}{3}k}{A}-\frac{1}{3}\alpha \right) L}{-1+\alpha }, \end{aligned}$$
where
$$\begin{aligned} A=\left( \alpha ^{3}{+}36k\alpha {-}27k{+}3\sqrt{12\alpha ^{4}k{+}96\alpha ^{2}k^{2}{+}192k^{3}-6k\alpha ^{3}-216k^{2}\alpha +81k^{2}}\right) ^{\frac{1}{3 }}. \end{aligned}$$
We can then substitute \(p_{2}\) back into the expressions for \(r_{2}\) and \( \lambda _{2}\). The final expressions are too lengthy to report.
So far, we have used first-order conditions (FOCs) to solve for the optimal choices of prices and promotion intensities. However, FOCs are necessary but not sufficient. We need to make sure that the solution we obtained indeed constitutes an equilibrium. Instead of checking whether the Hessian matrix is negative semidefinite (which is quite messy), we show that neither firm has an incentive to unilaterally deviate from this pair of strategies (Bester and Petrakis use a similar method). Without loss of generality, we fix firm 1’s price and promotion strategies as given in Proposition 1, and show that firm 2 to has no incentive to deviate.
Part 2: firm 2 has no incentive to deviate
Note that, the demand/profit functions depend on the locations of marginal consumers and there are two cases. In the first case, \(p_{2}\ge p_{1}\) still holds and thus \(l_{a}\ge 0\). In the second case, \(p_{2}<p_{1}\). In both cases, we assume that \(l_{b1}<0\) and \(l_{b2}>0\).Footnote 35
Start with case 1 where \(p_{2}\ge p_{1}\). Firm 2’s deviation profit is given by equation (6), with \(p_{1}=p^{*}\), \( r_{1}=r^{*}\) and \(\lambda _{1}=\lambda ^{*}\). We normalize \(L=1\). The optimal choice requires that
$$\begin{aligned} \frac{\partial \pi _{2}^{dev}}{\partial r_{2}}=\frac{\partial \pi _{2}^{dev} }{\partial \lambda _{2}}=0. \end{aligned}$$
Solving the first order conditions, we obtain
$$\begin{aligned} r_{2}^{dev}=\frac{2(1-\alpha )p_2 - (1-\alpha ) p^* -\alpha }{2(1-\alpha )}, \end{aligned}$$
$$\begin{aligned} \lambda _{2}^{dev}= \frac{\alpha ^2+\alpha ^2 (p^*)^2+4\alpha ^2 p_2-2 p^* \alpha ^2+2\alpha p^*-4\alpha p_2-2\alpha (p^*)^2+(p^*)^2}{(1-\alpha )k}. \end{aligned}$$
The first order conditions are necessary and sufficient. Note that, we do not substitute \(p^{*}\) in these expressions as they would be too lengthy to report. Notice that, firm 2’s deviation profit depends only on \( p_{2}^{dev}\), \(\alpha \) and k. We want to check whether firm 2 can increase its profit by choosing \(p_{2}^{dev}\ne p^{*}\), i.e., to have
$$\begin{aligned} \pi _{2}^{dev}(p_{2}^{dev})>\pi ^{*},\quad \forall \alpha ,k. \end{aligned}$$
We tried various combinations of \(\alpha \) and k, and we found that firm 2 can never increase its profit by choosing a price different than \(p^{*}\) . Therefore, firm 2 has no incentive to deviate. We then proceeded to the case of \(l_{a}<0\) (i.e. \( p_{2}<p_{1}\)). The steps are similar and we found that firm 2 has no incentive to deviate if k is not too small relative to L.Footnote 36
\(\square \)