Two-part tariff contracting with competing unreliable suppliers in a supply chain under asymmetric information

Abstract

We employ a two-stage game to study a two-part tariff contracting under asymmetric information in a supply chain, which consists of two unreliable suppliers and one retailer. The suppliers compete to sell their products, which are partial substitute, through a common retailer, who faces a stochastic demand and has superior information about the market. In the first stage, the suppliers simultaneously and independently announce the two-part tariff contract. The retailer, who is close to customers, decides whether to accept the two-part tariff contract. In the second stage, the uncertainty in market information, the supply information and the demand information are resolved. Then, the retailer determines the demand rates of products to optimize his profit. In this paper, we first derive the retailer’s optimal strategy and fully characterize the supplier’s optimal contract design. Subsequently, we study the impact of the degree of substitution on the equilibrium. We find that a higher degree of substitution implies a lower purchasing price but a higher fixed fee. We also evaluate the impact of supply uncertainty on the equilibriums. Finally, we conduct numerical experiments to show that the information rent is increasing with the degree of substitution. However, a larger intensity of competition is disadvantageous to the supplier.

Appendix

Proof of Lemma 1

Given the two-part tariff contract bundles $(B_{i\tau }, B_{j\eta })$, first of all, we substitute the payment function (2) into the retailer’s optimization problem (3). And then we claim that the retailer’s objective function $\pi _{\theta }(B_{i\tau }, B_{j\eta })$ is jointly concave in the pair of demand rates $q_\theta =(q_{i\theta },q_{j\theta })$.

Since

$$\begin{aligned} \frac{\partial \pi _{\theta }(B_{i\tau },B_{j\eta }) }{\partial q_{i\theta }}= & {} E(\epsilon _i)A_{i\theta }-2E(\epsilon _i)^{2}q_{i\theta }-2\gamma E(\epsilon _i\epsilon _j) q_{j\theta }-E(\epsilon _i)w_{i\tau }\\= & {} \mu _i A_{i\theta }-2(\mu ^{2}_{i}+\sigma ^{2}_{i})q_{i\theta }-2\gamma \mu _i\mu _j q_{j\theta }-\mu _iw_{i\tau }, \end{aligned}$$

and

$$\begin{aligned}&\frac{\partial ^{2}\pi _{\theta }(B_{i\tau },B_{j\eta }) }{\partial q^{2}_{i\theta }}=-2E(\epsilon _i)^{2}=-2(\mu ^{2}_{i}+\sigma ^{2}_{i}) <0;\\&\frac{\partial ^{2}\pi _{\theta }(B_{i\tau },B_{j\eta }) }{\partial q^{2}_{j\theta }}=-2E(\epsilon _j)^{2}=-2(\mu ^{2}_{j}+\sigma ^{2}_{j}) <0;\\&\frac{\partial ^{2}\pi _{\theta }(B_{i\tau },B_{j\eta }) }{\partial q_{i\theta }\partial q_{j\theta }}=-2\gamma E(\epsilon _i\epsilon _j)=-2\gamma \mu _i\mu _j. \end{aligned}$$

As $4(\mu ^{2}_{i}+\sigma ^{2}_{i})(\mu ^{2}_{j}+\sigma ^{2}_{j})-4\gamma ^{2}(\mu _i\mu _j)^{2}>0$, $\pi _{\theta }(B_{i\tau },B_{j\eta })$ is jointly concave in $(q_{i\theta }, q_{j\theta })$.

In this situation, the optimal solution is unique and can be derived by the first order condition:

$$\begin{aligned} \left\{ \begin{array}{ll} \mu _i A_{i\theta }-2(\mu ^{2}_{i}+\sigma ^{2}_{i})q_{i\theta }-2\gamma \mu _i\mu _j q_{j\theta }-\mu _iw_{i\tau }=0; \\ \mu _jA_{j\theta }-2(\mu ^{2}_{j}+\sigma ^{2}_{j})q^{*}_{j\theta }-2\gamma \mu _i\mu _jq^{*}_{i\theta }-\mu _jw_{j\eta }=0. \end{array} \right. \end{aligned}$$

Finally, we get $q^{*}_{i\theta }(w_{i\tau },w_{j\eta })=\frac{\mu _i(\mu ^{2}_{j}+\sigma ^{2}_{j})(A_{i\theta }-w_{i\tau })-\gamma \mu _i\mu ^{2}_j(A_{j\theta }-w_{j\eta })}{2(\mu ^{2}_{i}+\sigma ^{2}_{i})(\mu ^{2}_{j}+\sigma ^{2}_{j})-2\gamma ^{2}(\mu _i\mu _j)^{2}}.$ $\square$

Proof of Lemma 2

The proof of Lemma 2 follows from the proof of Lemma 3. $\square$

Proof of Theorem 1

The proof of Theorem 1 follows directly from the proof of Theorem 2. $\square$

Proof of Lemma 3

By substituting the payment function $s_{i\theta }$, supplier i’s profit and retailer’s profit from supplier i into Problem (4), we can simplify the contract design problem as follows

$$\begin{aligned} \Pi _i= & {} \max _{(T_{iH},w_{iH}),(T_{iL},w_{iL})}\{\rho [T_{iH}+(w_{iH}\mu _i-c_i)q^{*}_{iH}(w_{iH}, w_{jH})]\\&\qquad \qquad +(1-\rho )[T_{iL}+(w_{iL}\mu _i-c_i)q^{*}_{iL}( w_{iL}, w_{jL})]\}\\&s.t.\left\{ \begin{array}{ll} \frac{\mu _i^2(\mu ^{2}_{j}+\sigma ^{2}_{j})(A_{iL}- w_{iL})^{2}-\gamma (\mu _i\mu _j)^{2}(A_{iL}-w_{iL})(A_{jL}-w_{jL})}{ 4(\mu ^{2}_{i}+\sigma ^{2}_{i})(\mu ^{2}_{j}+\sigma ^{2}_{j})-4\gamma ^{2}(\mu _i\mu _j)^{2}}-T_{iL}\ge 0,&{}\quad (5) \\ \frac{\mu _i^2(\mu ^{2}_{j}+\sigma ^{2}_{j})(A_{iH}- w_{iH})^{2}-\gamma (\mu _i\mu _j)^{2}(A_{iH}-w_{iH})(A_{jH}-w_{jH})}{ 4(\mu ^{2}_{i}+\sigma ^{2}_{i})(\mu ^{2}_{j}+\sigma ^{2}_{j})-4\gamma ^{2}(\mu _i\mu _j)^{2}}-T_{iH}\ge 0,&{}\quad (6) \\ \frac{\mu _i^2(\mu ^{2}_{j}+\sigma ^{2}_{j})(A_{iL}- w_{iL})^{2}-\gamma (\mu _i\mu _j)^{2}(A_{iL}-w_{iL})(A_{jL}-w_{jL})}{ 4(\mu ^{2}_{i}+\sigma ^{2}_{i})(\mu ^{2}_{j}+\sigma ^{2}_{j})-4\gamma ^{2}(\mu _i\mu _j)^{2}}-T_{iL}\\ ~~~~\ge \frac{\mu _i^2(\mu ^{2}_{j}+\sigma ^{2}_{j})(A_{iL}- w_{iH})^{2}-\gamma (\mu _i\mu _j)^{2}(A_{iL}-w_{iH})(A_{jL}-w_{jL})}{ 4(\mu ^{2}_{i}+\sigma ^{2}_{i})(\mu ^{2}_{j}+\sigma ^{2}_{j})-4\gamma ^{2}(\mu _i\mu _j)^{2}}-T_{iH},&{}\quad (7)\\ \frac{\mu _i^2(\mu ^{2}_{j}+\sigma ^{2}_{j})(A_{iH}- w_{iH})^{2}-\gamma (\mu _i\mu _j)^{2}(A_{iH}-w_{iH})(A_{jH}-w_{jH})}{ 4(\mu ^{2}_{i}+\sigma ^{2}_{i})(\mu ^{2}_{j}+\sigma ^{2}_{j})-4\gamma ^{2}(\mu _i\mu _j)^{2}}-T_{iH}\\ ~~~~\ge \frac{\mu _i^2(\mu ^{2}_{j}+\sigma ^{2}_{j})(A_{iH}- w_{iL})^{2}-\gamma (\mu _i\mu _j)^{2}(A_{iH}-w_{iL})(A_{jH}-w_{jH})}{ 4(\mu ^{2}_{i}+\sigma ^{2}_{i})(\mu ^{2}_{j}+\sigma ^{2}_{j})-4\gamma ^{2}(\mu _i\mu _j)^{2}}-T_{iL}.&{}\quad (8) \end{array} \right. \end{aligned}$$

Under the condition $0\le [A_{jH}-w_{jH}-(A_{jL}-w_{jL})]\le \frac{(\mu ^{2}_{j}+\sigma ^{2}_{j})(A_{iH}-A_{iL})}{\gamma \mu _j^2}$, we claim that the first and the last constraints are binding.

First of all, we show that if the first and the last constraints hold, then the second constraint hold automatically. Notice that from the first constraint, we have:

$$\begin{aligned} -T_{iL}\ge -\frac{\mu _i^2(\mu ^{2}_{j}+\sigma ^{2}_{j})(A_{iL}- w_{iL})^{2}-\gamma (\mu _i\mu _j)^{2}(A_{iL}-w_{iL})(A_{jL}-w_{jL})}{ 4(\mu ^{2}_{i}+\sigma ^{2}_{i})(\mu ^{2}_{j}+\sigma ^{2}_{j})-4\gamma ^{2}(\mu _i\mu _j)^{2}}. \end{aligned}$$

(9)

Replace (9) into the last constraint, we get:

$$\begin{aligned}&\frac{\mu _i^2(\mu ^{2}_{j}+\sigma ^{2}_{j})(A_{iH}- w_{iH})^{2}-\gamma (\mu _i\mu _j)^{2}(A_{iH}-w_{iH})(A_{jH}-w_{jH})}{ 4(\mu ^{2}_{i}+\sigma ^{2}_{i})(\mu ^{2}_{j}+\sigma ^{2}_{j})-4\gamma ^{2}(\mu _i\mu _j)^{2}}-T_{iH}\\&\quad \ge \frac{\mu _i^2(\mu ^{2}_{j}+\sigma ^{2}_{j})(A_{iH}- w_{iL})^{2}-\gamma (\mu _i\mu _j)^{2}(A_{iH}-w_{iL})(A_{jH}-w_{jH})}{ 4(\mu ^{2}_{i}+\sigma ^{2}_{i})(\mu ^{2}_{j}+\sigma ^{2}_{j})-4\gamma ^{2}(\mu _i\mu _j)^{2}}\\&\qquad -\,\frac{\mu _i^2(\mu ^{2}_{j}+\sigma ^{2}_{j})(A_{iL}- w_{iL})^{2}-\gamma (\mu _i\mu _j)^{2}(A_{iL}-w_{iL})(A_{jL}-w_{jL})}{ 4(\mu ^{2}_{i}+\sigma ^{2}_{i})(\mu ^{2}_{j}+\sigma ^{2}_{j})-4\gamma ^{2}(\mu _i\mu _j)^{2}}\\&\quad =(A_{iH}- w_{iL})\frac{\mu _i^2(\mu ^{2}_{j}+\sigma ^{2}_{j})(A_{iH}- w_{iL})-\gamma (\mu _i\mu _j)^{2}(A_{jH}-w_{jH})}{ 4(\mu ^{2}_{i}+\sigma ^{2}_{i})(\mu ^{2}_{j}+\sigma ^{2}_{j})-4\gamma ^{2}(\mu _i\mu _j)^{2}}\\&\qquad -\,(A_{iL}- w_{iL})\frac{\mu _i^2(\mu ^{2}_{j}+\sigma ^{2}_{j})(A_{iL}- w_{iL})-\gamma (\mu _i\mu _j)^{2}(A_{jL}-w_{jL})}{ 4(\mu ^{2}_{i}+\sigma ^{2}_{i})(\mu ^{2}_{j}+\sigma ^{2}_{j})-4\gamma ^{2}(\mu _i\mu _j)^{2}}\\&\quad \ge (A_{iL}- w_{iL})\frac{\mu _i^2(\mu ^{2}_{j}+\sigma ^{2}_{j})(A_{iH}- w_{iL})-\gamma (\mu _i\mu _j)^{2}(A_{jH}-w_{jH})}{ 4(\mu ^{2}_{i}+\sigma ^{2}_{i})(\mu ^{2}_{j}+\sigma ^{2}_{j})-4\gamma ^{2}(\mu _i\mu _j)^{2}}\\&\qquad -\,(A_{iL}- w_{iL})\frac{\mu _i^2(\mu ^{2}_{j}+\sigma ^{2}_{j})(A_{iL}- w_{iL})-\gamma (\mu _i\mu _j)^{2}(A_{jL}-w_{jL})}{ 4(\mu ^{2}_{i}+\sigma ^{2}_{i})(\mu ^{2}_{j}+\sigma ^{2}_{j})-4\gamma ^{2}(\mu _i\mu _j)^{2}}\\&\quad =(A_{iL}- w_{iL})\frac{\mu _i^2(\mu ^{2}_{j}+\sigma ^{2}_{j})(A_{iH}- A_{iL})-\gamma (\mu _i\mu _j)^{2}(A_{jH}-w_{jH}-A_{jL}+w_{jL})}{ 4(\mu ^{2}_{i}+\sigma ^{2}_{i})(\mu ^{2}_{j}+\sigma ^{2}_{j})-4\gamma ^{2}(\mu _i\mu _j)^{2}}\\&\quad \ge 0. \end{aligned}$$

Secondly, we claim that the first constraint is binding. If the first constraint is not tight, we can increase $T_{iL}$ and $T_{iH}$ by the same amount $\varepsilon$ such that all the constraints still hold and the objective function will increase, which violates the optimality condition.

Then, we claim that the last constraint is also binding. If at the optimal solution, the last constraint is not binding, then we can increase $T_{iH}$ by $\varepsilon$ such that the last constraint is binding. In this case, all other constraints will hold and the objective function will increase, which violates the optimality condition.

Now we show as long as $w_{iL}> w_{iH}$, constraint (7) holds automatically. Since the first and the last constraints are binding, we get

$$\begin{aligned} \left\{ \begin{array}{ll} T_{L}=\frac{\mu _i^2(\mu ^{2}_{j}+\sigma ^{2}_{j})(A_{iL}- w_{iL})^{2}-\gamma (\mu _i\mu _j)^{2}(A_{iL}-w_{iL})(A_{jL}-w_{jL})}{ 4(\mu ^{2}_{i}+\sigma ^{2}_{i})(\mu ^{2}_{j}+\sigma ^{2}_{j})-4\gamma ^{2}(\mu _i\mu _j)^{2}},\\ T_{H}=T_{L}+\frac{(w_{iL}-w_{iH})[\mu _i^2(\mu ^{2}_{j}+\sigma ^{2}_{j})(2A_{iH}-w_{iH}-w_{iL})-\gamma (\mu _i\mu _j)^{2}(A_{jH}-w_{jH})]}{4(\mu ^{2}_{i}+\sigma ^{2}_{i})(\mu ^{2}_{j}+\sigma ^{2}_{j})-4\gamma ^{2}(\mu _i\mu _j)^{2}}. \end{array} \right. \end{aligned}$$

Substituting $T_{iL}$ and $T_{iH}$ into the third constraint, we get:

$$\begin{aligned}&\frac{\mu _i^2(\mu ^{2}_{j}+\sigma ^{2}_{j})(A_{iL}- w_{iL})^{2}-\gamma (\mu _i\mu _j)^{2}(A_{iL}-w_{iL})(A_{jL}-w_{jL})}{ 4(\mu ^{2}_{i}+\sigma ^{2}_{i})(\mu ^{2}_{j}+\sigma ^{2}_{j})-4\gamma ^{2}(\mu _i\mu _j)^{2}}-T_{iL}\\&\qquad -\,\frac{\mu _i^2(\mu ^{2}_{j}+\sigma ^{2}_{j})(A_{iL}- w_{iH})^{2}-\gamma (\mu _i\mu _j)^{2}(A_{iL}-w_{iH})(A_{jL}-w_{jL})}{ 4(\mu ^{2}_{i}+\sigma ^{2}_{i})(\mu ^{2}_{j}+\sigma ^{2}_{j})-4\gamma ^{2}(\mu _i\mu _j)^{2}}+T_{iH}\\&\quad =\frac{(w_{iH}-w_{iL})[\mu _i^2(\mu ^{2}_{j}+\sigma ^{2}_{j})(2A_{iL}-w_{iH}-w_{iL})- \gamma (\mu _i\mu _j)^{2}(A_{jL}-w_{jL})]}{4(\mu ^{2}_{i}+\sigma ^{2}_{i})(\mu ^{2}_{j}+\sigma ^{2}_{j})-4\gamma ^{2}(\mu _i\mu _j)^{2}}\\&\qquad +\,\frac{(w_{iL}-w_{iH})[\mu _i^2(\mu ^{2}_{j}+\sigma ^{2}_{j})(2A_{iH}-w_{iH}-w_{iL})-\gamma (\mu _i\mu _j)^{2}(A_{jH}-w_{jH})]}{4(\mu ^{2}_{i}+\sigma ^{2}_{i})(\mu ^{2}_{j}+\sigma ^{2}_{j})-4\gamma ^{2}(\mu _i\mu _j)^{2}}\\&\quad =\frac{(w_{iL}-w_{iH})[\mu _i^2(\mu ^{2}_{j}+\sigma ^{2}_{j})(2A_{iH}-2A_{iL})-\gamma (\mu _i\mu _j)^{2}(A_{jH}-w_{jH}\!-\!A_{jL}\!+\!w_{jL})]}{4(\mu ^{2}_{i}+\sigma ^{2}_{i})(\mu ^{2}_{j}+\sigma ^{2}_{j})-4\gamma ^{2}(\mu _i\mu _j)^{2}}\\&\quad \ge 0. \end{aligned}$$

Relaxing the third constraint, we can convert the original constrained optimization problem into an unconstrained optimization problem. And we can show that the supplier’s objective function is concave in $w_{iL}$ and $w_{iH}$ respectively. In this case, the optimal response function is determined by the first order condition:

$$\begin{aligned} \left\{ \!\begin{array}{ll} \frac{\partial \Pi _{i}}{\partial w^{*}_{iL}}=\frac{2\rho \mu _i^2(\mu ^{2}_{j}+\sigma ^{2}_{j})(A_{iH}-A_{iL})-\gamma (\mu _i\mu _j)^{2}\rho (A_{jH}-w_{jH})-\gamma (\mu _i\mu _j)^{2}(1-2\rho )(A_{jL}-w_{jL})-2(1-\rho )\mu _i^2(\mu ^{2}_{j}+\sigma ^{2}_{j})( w^{*}_{iL}-\frac{c_i}{\mu _i})}{4(\mu ^{2}_{i}+\sigma ^{2}_{i})(\mu ^{2}_{j}+\sigma ^{2}_{j})-4\gamma ^{2}(\mu _i\mu _j)^{2}}\\ \quad \qquad =0,\\ \frac{\partial \Pi _{i}}{\partial w^{*}_{iH}}=\frac{\rho [-\gamma (\mu _i\mu _j)^{2}(A_{jH}- w_{jH})-2\mu _i^2(\mu ^{2}_{j}+\sigma ^{2}_{j})(w^{*}_{iH}-\frac{c_i}{\mu _i})]}{4(\mu ^{2}_{i}+\sigma ^{2}_{i})(\mu ^{2}_{j}+\sigma ^{2}_{j})-4\gamma ^{2}(\mu _i\mu _j)^{2}}\\ \qquad \quad =0. \end{array} \right. \end{aligned}$$

Then we get:

$$\begin{aligned} \left\{ \begin{array}{ll} {w}^{*}_{iL}({w}_{jL},{w}_{jH})= \frac{c_i}{\mu _i}-\frac{\gamma \mu ^{2}_j}{2(\mu ^{2}_{j}+\sigma ^{2}_{j})}(A_{jH}-w_{jH})+\frac{\rho }{1-\rho }(A_{iH}-A_{iL})\\ ~~~~~~~~~~~~~~~~~~~~~\quad -\,\frac{\rho }{1-\rho }\frac{\gamma \mu ^{2}_j}{2(\mu ^{2}_{j}+\sigma ^{2}_{j})}[A_{jH}-w_{jH}-(A_{jL}-w_{jL})],\\ w^{*}_{iH}(w_{jL},w_{jH})=\frac{c_i}{\mu _i}-\frac{\gamma \mu ^{2}_j}{2(\mu ^{2}_{j}+\sigma ^{2}_{j})}(A_{jH}-w_{jH}). \end{array} \right. \end{aligned}$$

At last, we need to verify that $w^{*}_{iL}(w_{jL},w_{jH})>w^{*}_{iH}(w_{jL},w_{jH})$.

$$\begin{aligned}&w^{*}_{iL}(w_{jL},w_{jH})=\frac{c_i}{\mu _i}+\frac{\rho }{1-\rho }(A_{iH}-A_{iL})\\&\qquad -\,\frac{\rho }{1-\rho }\frac{\gamma \mu ^{2}_j}{2(\mu ^{2}_{j}+\sigma ^{2}_{j})}(A_{jH}-w_{jH})-(1-\frac{\rho }{1-\rho })\frac{\gamma \mu ^{2}_j}{2(\mu ^{2}_{j}+\sigma ^{2}_{j})}(A_{jL}-w_{jL})\\&\quad \ge \frac{c_i}{\mu _i}+\frac{\rho }{1-\rho }(A_{iH}-A_{iL})-\frac{\rho }{1-\rho }\frac{\gamma \mu ^{2}_j}{2(\mu ^{2}_{j}+\sigma ^{2}_{j})}(A_{jH}-w_{jH})\\&\qquad -\,\left( 1-\frac{\rho }{1-\rho }\right) \frac{\gamma \mu ^{2}_j}{2(\mu ^{2}_{j}+\sigma ^{2}_{j})}(A_{jH}-w_{jH})\\&\quad \ge \frac{c_i}{\mu _i}+\frac{\rho }{1-\rho }(A_{iH}-A_{iL})-\frac{\gamma \mu ^{2}_j}{2(\mu ^{2}_{j}+\sigma ^{2}_{j})}(A_{jH}-w_{jH})\\&\quad >\frac{c_i}{\mu _i}-\frac{\gamma \mu ^{2}_j}{2(\mu ^{2}_{j}+\sigma ^{2}_{j})}(A_{jH}-w_{jH})=w^{*}_{iH}(w_{jL},w_{jH}). \end{aligned}$$

Similarly, under the condition $[A_{jH}-w_{jH}-(A_{jL}-w_{jL})]> \frac{(\mu ^{2}_{j}+\sigma ^{2}_{j})(A_{iH}-A_{iL})}{\gamma \mu _j^2}$, we can solve for $w^{*}_{iL}(w_{jL},w_{jH})$ and $w^{*}_{iH}(w_{jL},w_{jH})$ in the same way. $\square$

Proof of Theorem 2

With a two-part tariff contract, from Lemma 3 we can get the supplier’s best response in a symmetric game, that is:

Under the condition $0\le \gamma \mu ^2[A_{H}-w_{H}-(A_{L}-w_{L})] \le (\mu ^2+\sigma ^2)(A_{H}-A_{L})$, we can easily get the optimal wholesale price in both the high and low demand state:

$$\begin{aligned} \left\{ \begin{array}{ll} w^{*}_{L}=A_L+\frac{2(\mu ^2+\sigma ^2)[(1-\rho )\frac{c}{\mu }+\rho A_H-A_L]+\frac{\rho 2(\mu ^2+\sigma ^2)\gamma \mu ^2}{2(\mu ^2+\sigma ^2)-\gamma \mu ^2}(\frac{c}{\mu }-A_H)}{2(1-\rho )(\mu ^2+\sigma ^2)-(1-2\rho )\gamma \mu ^{2}},\\ w^{*}_{H}=\frac{\frac{2c}{\mu }(\mu ^2+\sigma ^2)-\gamma \mu ^2 A_H}{2(\mu ^2+\sigma ^2)-\gamma \mu ^2}. \end{array} \right. \end{aligned}$$

Thus, we have $A_H-w_H=\frac{2(\mu ^{2}+\sigma ^{2})(A_H-\frac{c}{\mu })}{2(\mu ^{2}+\sigma ^{2})-\gamma \mu ^{2}}$, and also $A_L-w_L=-\frac{2(\mu ^2+\sigma ^2)[(1-\rho )\frac{c}{\mu }+\rho A_H-A_L]+\frac{\rho 2(\mu ^2+\sigma ^2)\gamma \mu ^2}{2(\mu ^2+\sigma ^2)-\gamma \mu ^2}(\frac{c}{\mu }-A_H)}{2(1-\rho )(\mu ^2+\sigma ^2)-(1-2\rho )\gamma \mu ^{2}}$.

In this scenario, for a symmetric game the condition $0\le \gamma \mu ^2[A_{H}-w_{H}-(A_{L}-w_{L})]\le (\mu ^2+\sigma ^2)(A_{H}-A_{L})$ is equivalent to $\frac{\gamma \mu ^2}{\mu ^2+\sigma ^2}\le \frac{2(1-\rho )}{3-2\rho }$.

Similarly, we can also derive the symmetric equilibrium under the condition that $\frac{\gamma \mu ^2}{\mu ^2+\sigma ^2}>\frac{2(1-\rho )}{3-2\rho }$. $\square$

Proof of Proposition 1

From Theorem 1, we can easily derive the first derivative of $w^{S}_{\theta }$ with respect to $\gamma$

$$\begin{aligned} \frac{\partial w^{S}_{\theta }}{\partial \gamma }= & {} \frac{-\mu ^2 A_\theta [2(\mu ^2+\sigma ^2)-\gamma \mu ^2]+\mu ^2[\frac{2c}{\mu }(\mu ^2+\sigma ^2)-\gamma \mu ^2 A_\theta ]}{[2(\mu ^2+\sigma ^2)-\gamma \mu ^2]^2}\\= & {} \frac{\mu ^2(\mu ^2+\sigma ^2)(\frac{2c}{\mu }-2A_\theta )}{[2(\mu ^2+\sigma ^2)-\gamma \mu ^2]^2}\\< & {} 0. \end{aligned}$$

As $A_\theta -w^{S}_{\theta }=\frac{2(\mu ^{2}+\sigma ^{2})(A_\theta -\frac{c}{\mu })}{2(\mu ^{2}+\sigma ^{2})-\gamma \mu ^{2}}$, $q^{S}_\theta =\frac{\mu (\mu ^{2}+\sigma ^{2})(A_\theta -\frac{c}{\mu })}{[2(\mu ^{2}+\sigma ^{2})-\gamma \mu ^{2}](\mu ^{2}+\sigma ^{2}+\gamma \mu ^{2})}$, and $T^{S}_{\theta }=\frac{\mu ^2(\mu ^{2}+\sigma ^{2})^2(A_\theta -\frac{c}{\mu })^{2}}{[2(\mu ^{2}+\sigma ^{2})-\gamma \mu ^{2}]^2(\mu ^{2}+\sigma ^{2}+\gamma \mu ^{2})}$, we have

$$\begin{aligned} \frac{\partial T^{S}_{\theta }}{\partial \gamma }= & {} -\frac{\mu ^4(\mu ^{2}+\sigma ^{2})^2(A_\theta -\frac{c}{\mu })^{2}}{[2(\mu ^{2}+\sigma ^{2})-\gamma \mu ^{2}]^2(\mu ^{2}+\sigma ^{2}+\gamma \mu ^{2})^2}\\&+\frac{2\mu ^4(\mu ^{2}+\sigma ^{2})^2(A_\theta -\frac{c}{\mu })^{2}}{[2(\mu ^{2}+\sigma ^{2})-\gamma \mu ^{2}]^3(\mu ^{2}+\sigma ^{2}+\gamma \mu ^{2})}\\= & {} \frac{3\gamma \mu ^6(\mu ^{2}+\sigma ^{2})^2(A_\theta -\frac{c}{\mu })^{2}}{[2(\mu ^{2}+\sigma ^{2})-\gamma \mu ^{2}]^3(\mu ^{2}+\sigma ^{2}+\gamma \mu ^{2})^2}\\> & {} 0. \end{aligned}$$

$\square$

Proof of Proposition 2

With a two-part tariff contract under symmetric information, the supplier’s optimal profit is given by

$$\begin{aligned} \Pi ^{S}_{\theta }= & {} T^{S}_{\theta }+(w^{S}_{\theta }\mu -c)q^{S}_{\theta }\\= & {} \frac{\mu ^2(\mu ^{2}+\sigma ^{2})^2(A_\theta -\frac{c}{\mu })^{2}}{[2(\mu ^{2}+\sigma ^{2})- \gamma \mu ^{2}]^2(\mu ^{2}+\sigma ^{2}+\gamma \mu ^{2})}\\&-\,\frac{\gamma \mu ^{2}(A_\theta -\frac{c}{\mu })}{2(\mu ^{2}+\sigma ^{2})-\gamma \mu ^{2}}\frac{\mu ^2(\mu ^{2}+\sigma ^{2})(A_\theta -\frac{c}{\mu })}{[2(\mu ^{2}+\sigma ^{2})-\gamma \mu ^{2}](\mu ^{2}+\sigma ^{2}+\gamma \mu ^{2})} \end{aligned}$$

Then we get

$$\begin{aligned} \frac{\partial \Pi ^{S}_{\theta }}{\partial \gamma }= & {} -\frac{\mu ^4(\mu ^{2}+\sigma ^{2})^2(A_\theta -\frac{c}{\mu })^{2}}{[2(\mu ^{2}+\sigma ^{2})-\gamma \mu ^{2}]^2(\mu ^{2}+\sigma ^{2}+\gamma \mu ^{2})}\\&-\,\frac{\mu ^4(\mu ^{2}+\sigma ^{2})^2(A_\theta -\frac{c}{\mu })^{2}(\mu ^{2}+\sigma ^{2}-\gamma \mu ^{2})}{[2(\mu ^{2}+\sigma ^{2})-\gamma \mu ^{2}]^2(\mu ^{2}+\sigma ^{2}+\gamma \mu ^{2})^2}\\&+\,\frac{2\mu ^4(\mu ^{2}+\sigma ^{2})^2(A_\theta -\frac{c}{\mu })^{2}(\mu ^{2}+\sigma ^{2}-\gamma \mu ^{2})}{[2(\mu ^{2}+\sigma ^{2})-\gamma \mu ^{2}]^3(\mu ^{2}+\sigma ^{2}+\gamma \mu ^{2})}\\= & {} -\frac{2\mu ^4(\mu ^{2}+\sigma ^{2})^2(A_\theta -\frac{c}{\mu })^{2}[\gamma ^2\mu ^{4}-\gamma \mu ^{2}(\mu ^{2}+\sigma ^{2})+(\mu ^{2}+\sigma ^{2})^2]}{[2(\mu ^{2}+\sigma ^{2})-\gamma \mu ^{2}]^3(\mu ^{2}+\sigma ^{2}+\gamma \mu ^{2})^2}\\< & {} 0. \end{aligned}$$

The inequality is due to the fact that the substitution coefficient $\gamma \in [0,1]$. $\square$

Proof of Proposition 3

Define $CV=\frac{\sigma ^2}{\mu ^2}$, and from Theorem 1 the optimal wholesale price can be written as $w^{S}_{\theta }=\frac{\frac{2c}{\mu }(1+CV)-\gamma A_\theta }{2(1+CV)-\gamma }$. The first derivative of $w^{S}_{\theta }$ with respect to CV will be

$$\begin{aligned} \frac{\partial w^{S}_{\theta }}{\partial CV}= & {} \frac{\frac{2c}{\mu }[2(1+CV)-\gamma ]-2[\frac{2c}{\mu }(1+CV)-\gamma A_\theta ]}{[2(1+CV)-\gamma ]^{2}}\\= & {} \frac{2\gamma (A_{\theta }-\frac{c}{\mu })}{[2(1+CV)-\gamma ]^{2}}\\> & {} 0. \end{aligned}$$

As $A_\theta -w^{S}_{\theta }=\frac{2(A_\theta -\frac{c}{\mu })(1+CV)}{2(1+CV)-\gamma }$, and $T^{S}_{\theta }=\frac{(1+CV)^2(A_\theta -\frac{c}{\mu })^{2}}{(1+CV+\gamma )[2(1+CV)-\gamma ]^{2}}$, we have

$$\begin{aligned} \frac{\partial T^{S}_{\theta }}{\partial CV}= & {} \frac{2(1+CV)(A_\theta - \frac{c}{\mu })^{2}}{(1+CV+\gamma )[2(1+CV)-\gamma ]^{2}}-\frac{(1+CV)^2(A_\theta -\frac{c}{\mu })^{2}}{(1+CV+\gamma )^2[2(1+CV)-\gamma ]^{2}}\\&-\,\frac{4(1+CV)^2(A_\theta -\frac{c}{\mu })^{2}}{(1+CV+\gamma )[2(1+CV)-\gamma ]^{3}}\\= & {} \frac{[-2\gamma ^2-\gamma (1+CV)-2(1+CV)^2](1+CV)(A_\theta -\frac{c}{\mu })^{2}}{(1+CV+\gamma )^2[2(1+CV)-\gamma ]^{3}} <0. \end{aligned}$$

$\square$

Proof of Proposition 4

With a two-part tariff contract under symmetric information, the supplier’s optimal profit is given by

$$\begin{aligned} \Pi ^{S}_{\theta }= & {} T^{S}_{\theta }+(w^{S}_{\theta }\mu -c)q^{S}_{\theta }\\ \!= & {} \!\frac{(1+CV)^2(A_\theta -\frac{c}{\mu })^{2}}{(1+CV+\gamma )[2(1+CV)-\gamma ]^{2}}-\frac{\gamma (A_\theta -\frac{c}{\mu })}{2(1\!+\!CV)\!-\!\gamma }\frac{(1+CV)(A_\theta -\frac{c}{\mu })}{[2(1+CV)-\gamma ](1\!+\!CV\!+\!\gamma )} \end{aligned}$$

Then we get

$$\begin{aligned} \frac{\partial \Pi ^{S}_{\theta }}{\partial CV}= & {} \frac{(2+2CV-\gamma )(A_\theta -\frac{c}{\mu })^{2}}{(1+CV+\gamma )[2(1+CV)-\gamma ]^{2}}-\frac{(1+CV)(1+CV-\gamma )(A_\theta -\frac{c}{\mu })^{2}}{(1+CV+\gamma )^2[2(1+CV)-\gamma ]^{2}}\\&-\,\frac{4(1+CV)(1+CV-\gamma )(A_\theta -\frac{c}{\mu })^{2}}{(1+CV+\gamma )[2(1+CV)-\gamma ]^{3}}\\= & {} \frac{[\gamma ^3-3\gamma (1+CV)^2-2(1+CV)^3](A_\theta -\frac{c}{\mu })^{2}}{(1+CV+\gamma )^2[2(1+CV)-\gamma ]^{3}}\\&<0. \end{aligned}$$

$\square$