Manufacturer-To-Retailer versus Manufacturer-To-Consumer Rebates in a Supply Chain

Abstract

Starting with a newsvendor model (single-product, single-period, stochastic demand), we build a single-retailer, single-manufacturer supply chain with endogenous manufacturer rebates and retail pricing. The demand uncertainty is multiplicative, and the expected demand depends on the effective (retail) price of the product. A retailer rebate goes from the manufacturer to the retailer for each unit it sells. A consumer rebate goes from the manufacturer to the consumers for each unit they buy. Each consumer’s response to consumer rebates is characterized by two exogenous parameters: α, the effective fraction of the consumer rebate that the consumer values, leading to the lower effective retail price perceived by the consumer, and β, the probability that a consumer rebate will be redeemed. The type(s) of rebate(s) allowed and the unit wholesale price are given exogenously. Simultaneously, the manufacturer sets the size of the rebate(s) and the retailer sets the retail price. The retailer then decides how many units of the product to stock and the manufacturer delivers that amount by the beginning of the selling season. Compared to no rebates, an equilibrium retailer rebate leads to a lower effective price (hence, higher sales volume) and higher profits for both the supply chain and the retailer. An equilibrium consumer rebate also leads to a lower effective price and higher profits for the retailer, but not necessarily for the chain. Under our assumptions, such a consumer rebate (with or without a retailer rebate) allocates a fixed fraction of the (expected) supply chain profits to each player: The retailer gets $\alpha/(\alpha+\beta)$ and the manufacturer gets the rest, leading to interesting consequences. However, both firms prefer that α be higher and β lower: Even though the manufacturer gets a smaller share of the chain profits, the total amount received is higher. Neither the retailer nor the manufacturer always prefers one particular kind of rebate to the other. In addition, contrary to popular belief, it is possible for both firms to prefer consumer rebates even when all such rebates are redeemed.

APPENDIX

For the purposes of the appendix, let $h(\cdot)=\frac{\phi_\epsilon(\cdot)}{1 - \Phi_\epsilon(\cdot)}$ denote the failure rate of $\Phi_\epsilon$. Throughout the appendix, we will use the following short-hand notation by dropping the functional arguments: $f=f(p-\alpha r_{{C}})$, $z=z(p, r_{{C}}, \xi)$, $y^*=y^*(p, r_{{C}}, r_{{R}})$ and $h = h\left(\frac{y^*(p, r_{{C}}, r_{{R}})}{f(p - \alpha r_{{C}})}\right)$. In addition, define $\gamma:= -\frac{f'}{f}$ and $\theta:=\frac{f"}{f}$. Hence, by (A3), $\gamma>0,$ and, by (A4), $\gamma' =\gamma^2-\theta \geq 0.$ We first state and prove some lemmas that will be useful in the proofs of the propositions.

Lemma 1 Suppose (A1) holds. for $z(p, r_{{C}}, \xi)$ implicitly defined by ( 4 ), we have:

• (a) $\nabla_1z \hskip5.5pt= - \gamma z$,

• (b) $\nabla_2z \hskip5pt= \alpha \gamma z$,

• (c) $\nabla_{11}^2z \hskip2.5pt= \theta z$,

• (d) $\nabla_{22}^2z \hskip1.5pt= \alpha^2 \theta z$,

• (e) $\nabla_{12}^2z \hskip2.5pt= - \alpha \theta z$.

Proof of Lemma 1 By virtue of (A1), we can rewrite (4) as $\Phi_\epsilon\left(\frac{z}{f}\right) = \xi$. Now, implicit differentiation of this identity with respect to p yields the following:

$$\nabla_1z f - z f^{\,\prime} = 0.$$

The first part of the lemma follows from the above equality recalling the definition of $\gamma:= - \frac{f\,'}{f}$. The proof of the second part follows the same logic. The third part can be obtained directly by partial differentiation of the expression for $\nabla_1z$. Likewise, the fourth and fifth parts are obtained by partial differentiation of the expression for $\nabla_2z$.

Lemma 2 Suppose (A1) through (A4) hold. For $y^*(p, r_{{C}}, r_{{R}})$ implicitly defined by (2), we have:

• (a) $\nabla_1y^* = - \gamma y^* + \frac{f}{(p+r_{{R}}-v)h},$

• (b) $\nabla_2y^* = \alpha \gamma y^* > 0,$

• (c) $\nabla_3y^* = \frac{f}{(p+r_{{R}}-v)h} > 0,$

• (d) $\nabla_1y^* = - \frac{1}{\alpha} \nabla_2y^* + \nabla_3y^*$,

• (e) $\nabla_{22}^2y^* = \alpha^2 \theta y^*$,

• (f) $\nabla_{33}^2y^* = - \frac{f}{(p+r_{{R}}-v)^2 h} - \frac{f h^\prime}{(p+r_{{R}}-v)^2 h^3} < 0$,

• (g) $\nabla_{23}^2y^* = \alpha \gamma \frac{f} {(p+r_{{R}}-v) h} > 0$,

• (h) $\nabla_{13}^2y^* = - \frac{1}{\alpha} \nabla_{23}^2y^* + \nabla_{33}^2y^* < 0$.

Proof of Lemma 2

Proofs of (a) through (d) Due to (A1), we can rewrite (2) as

$$\begin{aligned} \Phi_\epsilon\left(\frac{y^*}{f}\right) = \frac{p+r_{{R}} - w}{p+r_{ {R}}-v}\end{aligned}$$

((9))

Now, implicit differentiation of (9) with respect to p yields

$$\frac{\nabla_1y^* f - f^\prime y^*} {f^2} \phi_\epsilon\left(\frac{y^*}{f}\right)= \frac{w-v}{(p+r_{{R}}-v)^2}.$$

Recalling the definition of $h(\cdot)=\frac{\phi_\epsilon(\cdot)}{1 - \Phi_\epsilon(\cdot)}$ and noting that $1 - \Phi_\epsilon\left(\frac{y^*}{f}\right)=\frac{w-v}{p+r_{{R}}-v}$ (this follows from (9)), we can leave $\nabla_1y^*$ alone in the above expression to obtain part (a) of the lemma. The proofs of parts (b) and (c) follow the same line of argument. Part (d) of the lemma follows directly from parts (a) through (c).

Proofs of (e) through (g) These follow from partial differentiation of the expressions obtained in parts (a) through (c). To see why $\nabla_{33}^2 y^* < 0$, recall that $h(\cdot)$ is the failure rate and it is an increasing function by (A2). To see why $\nabla_{23}^2 y^* > 0$, recall that $\gamma>0$ by (A3).

Proof of (h) This follows from part (d) of the lemma.

Lemma  3   Given $r_{{C}}$ and $r_{{R}}$, if $\tilde{p}$ satisfies $\nabla_1\Pi_{\rm R}(\tilde{p}, r_{{C}}, r_{{R}}) = 0$, then $\nabla_1y^*$ $\nabla_1y^*(\tilde{p}, r_{{C}}, r_{{R}}) < 0$.

Proof of Lemma 3 Omitted. See Aydin and Porteus (2006) for the proof of the same result under more general conditions.

Lemma 4 Let $f(x)$ be a twice-continuously-differentiable function of a single real variable defined on $[a,\infty).$ Suppose that $f^{\prime \prime}(x) < 0$ at any $x \geq a$ that satisfies $f^{\prime}(x)=0$. Then:

• (a) (Petruzzi and Dada, 1999) If $f'(a)>0$ and $f(x)$ is strictly decreasing in x as x tends to infinity, then there exists a unique $x^*>a$ that satisfies $f^{\prime}(x)=0$, and $x^*$ maximizes $f(x)$.

• (b) If $f'(a)\le 0$ then $f(x)$ is non-increasing for all $x \geq a$ , and $x^{*}=a$ maximizes $f(x)$.

Proof of Lemma 4 Omitted. Lemma 4 (a) is due to Petruzzi and Dada (1999). See Aydin and Porteus (2006) for a detailed proof. The proof of part (b) is very similar.

Lemma 5 In a two-player game, let $g_i(x_1,x_2)$ be the payoff function of player $i=1,2$ when the strategies chosen by players 1 and 2 are $x_1$ and $x_2$, respectively. The strategy space for player i is $X_i:=\{x: \underline{x}_i \leq x \leq \overline{x}_i\}$. Suppose that $g_i$ is continuous and quasi-concave with respect to $x_i$, $i=1,2$. Let $x_i^*(x_j)$ be the best response of player i when player j chooses strategy $x_j$; i.e., $x_i^*(x_j) = \arg \max_{x_i}(g_i(x_1, x_2))$. Then:

(a) There exists at least one pure strategy Nash equilibrium.

(b) If $\frac{dx_1^*(x_2)}{d x_2} \frac{dx_2^*(x_1)}{d x_1} < 1$, then there exists a unique pure strategy Nash equilibrium.

Proof of Lemma 5 Omitted. See Cachon and Netessine (2003) for a summary of standard results in game theory.

Lemma 6 Suppose (A1) through (A4) hold. Let $\Pi_{\rm R}(p, r_{{C}}, r_{{R}})$ and $\Pi_{\rm M}(p, r_{{C}}, r_{{R}})$ be as defined by (5) and (6), respectively. Then: (a)

$$\begin{aligned} \nabla_1\Pi_{\rm R}(p, r_{{C}}, r_{{R}}) &= \int_0^{\frac{p + r_{{R}} - w}{p+r_{{R}}-v}} z d \xi + \frac{w-v}{p+r_{{R}}-v} y^* - \gamma (p+r_{{R}}-v) \int_0^{\frac{p + r_{{R}} - w}{p+r_{{R}}-v}} z d \xi \end{aligned}$$

(b)

$$\begin{aligned} \nabla_2\Pi_{\rm M}(p, r_{{C}}, r_{{R}}) &=& (w-c) \alpha \gamma y^* -& \left[\beta + \alpha \gamma (r_{{R}} + \beta r_{{C}})\right] \left( \int_0^{\frac{p + r_{{R}} - w}{p+r_{ {R}}-v}} z d \xi + \frac{w-v}{p+r_{{R}}-v} y^* \right), \end{aligned}$$

(c)

$$\begin{aligned} \nabla_3\Pi_{\rm M}(p, r_{{C}}, r_{{R}}) = &\left[(w-c) - (r_{{R}} + \beta r_{{C}}) \frac{w-v}{p+r_{{R}}-v}\right]\frac{f }{(p+r_{{R}}-v)h}\\ &- \int_0^{\frac{p + r_{{R}} - w}{p+r_{ {R}}-v}} z d \xi - \frac{w-v}{p+r_{{R}}-v} y^* \end{aligned}$$

(d) For $p + r_{{R}} > v$:

$$\begin{aligned} \left.\nabla^2_{11}\Pi_{\rm R}(p, r_{ {C}}, r_{{R}})\right|_{\nabla_1\Pi_{\rm R}=0} &=& - \gamma \int_0^{\frac{p + r_{{R}} - w}{p+r_{{R}}-v}} z d \xi + \frac{w-v}{p+r_{{R}}-v} \nabla_1y^* - (p+r_{{R}}-v) \gamma' \int_0^{\frac{p + r_{{R}} - w}{p+r_{ {R}}-v}}z < 0\ \end{aligned}$$

(e) For $p + r_{{R}} > v$:

$$\begin{aligned} \left.\nabla^2_{22}\Pi_{\rm M}(p, r_{{C}}, r_{ {R}})\right|_{\nabla_2\Pi_{\rm M}=0} =& - \alpha \beta \gamma \left( \int_0^{\frac{p + r_{{R}} - w}{p+r_{ {R}}-v}} z d \xi + \frac{w-v}{p+r_{{R}}-v} y^* \right) \\ & + \left[(w-c) y^* - (r_{{R}} + \beta r_{{C}}) \left( \int_0^{\frac{p + r_{{R}} - w}{p+r_{{R}}-v}} z d \xi + \frac{w-v}{p+r_{{R}}-v} y^* \right)\right]\\ & (\alpha^2 \theta - \alpha^2 \gamma^2) \ \end{aligned}$$

(f) For $p + r_{{R}} > v$:

$$\begin{aligned} \left.\nabla^2_{33}\Pi_{\rm M}(p, r_{{C}}, r_{ {R}})\right|_{\nabla_3\Pi_{\rm M}=0} =& -\frac{1}{p+r_{{R}}-v} \left[\int_0^{\frac{p + r_{{R}} - w}{p+r_{ {R}}-v}} z d \xi + \frac{w-v}{p+r_{{R}}-v} y^*\right] \\ &- \left[(w-c) - (r_{{R}}+\beta r_{{C}}) \frac{w-v}{p+r_{ {R}}-v}\right] \frac{f h^\prime}{(p+r_{{R}}-v)^2 h^3} \\ &+ \left[-2 \frac{w-v}{p+r_{{R}}-v} + (r_{{R}}+\beta r_{{C}}) \frac{w-v}{(p+r_{{R}}-v)^2}\right] \frac{f }{(p+r_{{R}}-v) h} < 0 \ \end{aligned}$$

(g) For $p + r_{{R}} > v$:

$$\begin{aligned} \left.\nabla^2_{23}\Pi_{\rm M}(p, r_{{C}}, r_{ {R}})\right|_{\nabla_2\Pi_{\rm M}=0} &=& - \frac{w-v}{p+r_{{R}}-v} \frac{\beta f} {(p+r_{{R}}-v) h} < 0 \ \end{aligned}$$

(h) For $p + r_{{R}} > v$:

$$\begin{aligned} \left.\nabla^2_{13}\Pi_{\rm R}(p, r_{{C}}, r_{ {R}})\right|_{\nabla_1\Pi_{\rm R}=0} &=& \frac{w-v}{p+r_{{R}}-v} \nabla_1 y^* - \gamma \int_0^{\frac{p + r_{{R}} - w}{p+r_{{R}}-v}} z d \xi<0 \ \end{aligned}$$

(i) For $p + r_{{R}} > v$:

$$\begin{aligned} \left.\nabla^2_{13}\Pi_{\rm M}(p, r_{{C}}, r_{ {R}})\right|_{\nabla_3\Pi_{\rm M}=0} =& - \left[(w-c) - (r_{{R}} + \beta r_{{C}}) \frac{w}{p+r_{ {R}}-v}\right] \frac{f h^\prime}{(p+r_{{R}}-v)^2 h^3} \\ &- \left[(2w-c) - 2 (r_{{R}} + \beta r_{{C}}) \frac{w-v}{p+r_{ {R}}-v}\right] \frac{f} {(p+r_{{R}}-v)^2 h} < 0 \end{aligned}$$

(j) For $p + r_{{R}} > v$:

$$\begin{aligned} \left.\nabla^2_{12}\Pi_{\rm R}(p, r_{{C}}, r_{ {R}})\right|_{\nabla_1\Pi_{\rm R}=0} &=& \alpha (p+r_{{R}}-v) \gamma' \int_0^{\frac{p + r_{{R}} - w}{p+r_{{R}}-v}} z d \xi > 0 \ \end{aligned}$$

Proof of Lemma 6

Proof of (a) The result follows from partial differentiation of $\Pi_{\rm R}(p, r_{{C}}, r_{{R}})$ (defined by (5)) with respect to p and substituting for $\nabla_1z$ using Lemma 1(a).

Proof of (b) The result follows by partial differentiation of $\Pi_{\rm M}(p, r_{{C}}, r_{{R}})$ (defined by (6)) with respect to $r_{{C}}$ and substituting for $\nabla_2z$ and $\nabla_2y^*$ from Lemma 1(b) and from Lemma 2(b).

Proof of (c) The result follows by partial differentiation of $\Pi_{\rm M}(p, r_{{C}}, r_{{R}})$ (defined by (6)) with respect to $r_{{R}}$ and substituting for $\nabla_3y^*$ from Lemma 2(c).

Proof of (d) The second partial of $\Pi_{\rm R}(p, r_{{C}}, r_{{R}})$ with respect to p is given, after substituting for $\nabla_1z$ and $\nabla^2_{11}z$ using Lemma 1(a) and (c), by

$$\begin{aligned} \nabla^2_{11}\Pi_{\rm R}(p, r_{{C}}, r_{{R}}) =& - 2 \gamma \int_0^{\frac{p + r_{{R}} - w}{p+r_{{R}}-v}}z d \xi + \frac{w-v}{p+r_{{R}}-v} \nabla_1y^* - \frac{w-v}{p+r_{{R}}-v} \gamma y^* \\ &+ (p+r_{{R}}-v) \theta \int_0^{\frac{p + r_{{R}} - w}{p+r_{{R}}-v}} z d \xi\ \end{aligned}$$

Thus, when $\nabla_1\Pi_{\rm R} =0,$ using part (a) of the lemma, we have

$$\begin{aligned} \nabla^2_{11}\Pi_{\rm R}(p, r_{{C}}, r_{{R}}) = - \gamma \int_0^{\frac{p + r_{{R}} - w}{p+r_{{R}}-v}} z d \xi + \frac{w-v}{p+r_{{R}}-v} \nabla_1y^* \hskip 8pt&\\+ (p+r_{{R}}-v) \left(\theta- \gamma^2 \right) \int_0^{\frac{p + r_{{R}} - w}{p+r_{{R}}-v}} z d \xi, \ \end{aligned}$$

which is strictly negative, by Lemma 3 and since $\gamma > 0$ (by (A3)) and $\gamma' = \gamma^2-\theta \geq 0$ (by (A4)).

Proof of (e) The second partial of $\Pi_{\rm M}(p, r_{{C}}, r_{{R}})$ with respect to $r_{{C}}$ is given, after substituting for $\nabla_2z$, $\nabla^2_{22}z$, $\nabla_2y^*$ and $\nabla^2_{22}y^*$ from Lemma 1(b) and (d), and from Lemma 2(b) and (e), by

$$\begin{aligned} \nabla^2_{22}\Pi_{\rm M}(p, r_{{C}}, r_{{R}}) = (w-c)\alpha^2 \theta y^* - \left[2 \alpha \beta \gamma +\right.\\ \left.(r_{{R}} + \beta r_{{C}}) \alpha^2 \theta \right] \left( \int_0^{\frac{p + r_{{R}} - w}{p+r_{{R}}-v}} z d \xi + \frac{w-v}{p+r_{{R}}-v} y^* \right) \ \end{aligned}$$

Thus, when $\nabla_2\Pi_{\rm M} =0,$ using part (b) of the lemma, we have

$$\begin{aligned} \nabla^2_{22}\Pi_{\rm M}(p, r_{{C}}, r_{{R}}) = &- \alpha \beta \gamma \left( \int_0^{\frac{p + r_{{R}} - w}{p+r_{{R}}-v}} z d \xi +\right. \left.\frac{w-v}{p+r_{{R}}-v} y^* \right)\\ &+ \left[(w-c) y^* - (r_{{R}} + \beta r_{{C}}) \left( \int_0^{\frac{p + r_{{R}} - w}{p+r_{{R}}-v}} z d \xi + \frac{w-v}{p+r_{{R}}-v} y^* \right)\right] (\alpha^2 \theta - \alpha^2 \gamma^2), \ \end{aligned}$$

which is strictly negative since $\gamma >0$ (by (A3)), $\gamma' = \gamma^2-\theta \geq 0$ (by (A4)) and the term in brackets is $\Pi_{\rm M}$ which should be positive when $\nabla_2\Pi_{\rm M} =0$.

Proof of (f) The second partial of $\Pi_{\rm M}(p, r_{{C}}, r_{{R}})$ with respect to $r_{{R}}$ is given, after substituting for $\nabla_3y^*$ and $\nabla^2_{33}y^*$ from Lemma 2(c) and (f), by

$$\begin{aligned} \nabla^2_{33}\Pi_{\rm M}(p, r_{{C}}, r_{{R}}) = &\left[(w-c) - (r_{{R}}+\beta r_{{C}}) \frac{w-v}{p+r_{{R}}-v}\right] \left[-\frac{f}{(p+r_{{R}}-v)^2 h} - \frac{f h^\prime}{(p+r_{{R}}-v)^2 h^3}\right] \\ &+ \left[-2 \frac{w-v}{p+r_{{R}}-v} + (r_{{R}}+\beta r_{{C}}) \frac{w-v}{(p+r_{{R}}-v)^2}\right] \frac{f}{(p+r_{{R}}-v) h} . \ \end{aligned}$$

Thus, when $\nabla_3\Pi_{\rm M} =0,$ using part (c) of the lemma, we have

$$\begin{aligned} \nabla^2_{33}\Pi_{\rm M}(p, r_{{C}}, r_{{R}}) =& - \frac{1}{p+r_{{R}}-v} \left[\int_0^{\frac{p + r_{{R}} - w}{p+r_{{R}}-v}} z d \xi + \frac{w-v}{p+r_{{R}}-v} y^*\right] \\ &-\left[(w-c) - (r_{{R}}+\beta r_{{C}}) \frac{w-v}{p+r_{{R}}-v}\right] \frac{f h^\prime}{(p+r_{{R}}-v)^2 h^3} \\ &+\left[-2 \frac{w-v}{p+r_{{R}}-v} + (r_{{R}}+\beta r_{{C}}) \frac{w-v}{(p+r_{ {R}}-v)^2}\right]\frac{f}{(p+r_{ {R}}-v) h} . \ \end{aligned}$$

In order to show $\left. \nabla^2_{33}\Pi_{\rm M}(p, r_{{C}}, r_{{R}}) \right|_{\nabla_3\Pi_{\rm M}=0}<0$, first note that, by Lemma 6 (c), if $\nabla_3\Pi_{\rm M}(p, r_{{C}}, r_{{R}})=0$, then we must have $(w-c) - (r_{{R}}+\beta r_{{C}}) \frac{w-v}{p+r_{{R}}-v}>0$, in which case we will also have $-2 \frac{w-v}{p+r_{{R}}-v} + (r_{{R}}+\beta r_{{C}}) \frac{w-v}{(p+r_{{R}}-v)^2}<0$. (This can be verified through some algebra.) After making these observations, the desired result now follows since $h'>0$ by assumption (A2).

Proof of (g) It can be verified that the cross-partial $\nabla_{23}\Pi_{\rm M}(p, r_{{C}}, r_{{R}})$ is given, after substituting for $\nabla_2 z$ from Lemma 1(b) and for $\nabla_2y^*, \nabla_3y^*$ and $\nabla_{23}y^*$ from Lemma 2(b), (c) and (g), by

$$\begin{aligned} \nabla^2_{23}\Pi_{\rm M}(p, r_{{C}}, r_{{R}}) = &- \left[(w-c) - (r_{{R}} + \beta r_{{C}}) \frac{w-v}{p+r_{{R}}-v}\right] \alpha \gamma \frac{f} {(p+r_{{R}}-v) h} \\ &- \alpha \gamma \left(\int_0^{\frac{p + r_{{R}} - w}{p+r_{{R}}-v}} z d \xi + \frac{w-v}{p+r_{{R}}-v} y^* \right) - \beta \frac{w-v}{p+r_{{R}}-v} \frac{f}{(p+r_{{R}}-v) h} \ \end{aligned}$$

Thus, when $\nabla_3\Pi_{\rm M} =0,$ using part (c) of the lemma, we have

$$\begin{aligned} \nabla^2_{23}\Pi_{\rm M}(p, r_{{C}}, r_{{R}}) &=& - \beta \frac{w-v}{p+r_{{R}}-v} \frac{f}{(p+r_{{R}}-v) h} \ \end{aligned}$$

Proof of (h) It can be verified that $\nabla^2_{13}\Pi(p_c, r_{{R}})$ is given, after substituting for $\nabla_1z$ from Lemma 1(a) and $\nabla_3y^*$ from Lemma 2(c), by

$$\begin{aligned} \nabla^2_{13}\Pi_{\rm R}(p, r_{{C}}, r_{{R}}) &=& \frac{w-v}{p+r_{ {R}}-v} \left(-\gamma y^* + \frac{f}{(p+r_{{R}}-v) h} \right) - \gamma \int_0^{\frac{p + r_{{R}} - w}{p+r_{ {R}}-v}} z d \xi\ \end{aligned}$$

Now, from part (a) of Lemma 2, we note that $- \gamma y^* + \frac{f}{h(p+r_{{R}}-v)} = \nabla_1y^*$. The desired conclusion on the sign follows from $\gamma >0$ (by (A3)) and Lemma 3.

Proof of (i) It can be verified that $\nabla_{13}\Pi_{\rm M}(p, r_{{C}}, r_{{R}})$ is given, after substituting for $\nabla_1z$ from Lemma 1(a) and for $\nabla_1y^*$, $\nabla_3y^*$, and $\nabla_{13}y^*$ from Lemma 2(a), (c) and (h), by

$$\begin{aligned} \nabla^2_{13}\Pi_{\rm M}(p, r_{{C}}, r_{{R}}) =& \textstyle \left[(w-c) - (r_{{R}} + \beta r_{{C}}) \frac{w-v}{p+r_{{R}}-v}\right] \left[ -\gamma \frac{f} {(p+r_{{R}}-v) h} - \frac{f}{(p+r_{{R}}-v)^2 h} - \frac{f h^\prime}{(p+r_{{R}}-v)^2 h^3} \right] \\ &+ \gamma \int_0^{\frac{p + r_{{R}} - w}{p+r_{{R}}-v}} z d \xi - \frac{w-v}{p+r_{{R}}-v} \left[-\gamma y^* + \frac{f}{(p+r_{{R}}-v) h}\right] \\ &+ (r_{{R}} + \beta r_{{C}}) \frac{w-v}{(p+r_{{R}}-v)^2}\frac{f}{(p+r_{{R}}-v) h} \ \end{aligned}$$

Now, using part (c) of the lemma and the above expression, one can verify through some algebra that the following is true when $\nabla_3\Pi_{\rm M} =0$:

$$\begin{aligned} \nabla^2_{13}\Pi_{\rm M}(p, r_{{C}}, r_{{R}}) =& - \left[(w-c) - (r_{ {R}} + \beta r_{{C}}) \frac{w-v}{p+r_{{R}}-v}\right] \frac{f h^\prime}{(p+r_{{R}}-v)^2 h^3} \\ &- \left[(2w-c) - 2 (r_{{R}} + \beta r_{{C}}) \frac{w-v}{p+r_{{R}}-v}\right] \frac{f} {(p+r_{{R}}-v)^2 h} \ \end{aligned}$$

In order to show that $\left. \nabla^2_{13}\Pi_{\rm M}(p, r_{{C}}, r_{{R}}) \right|_{\nabla_3\Pi_{\rm M}=0}<0$, note that, by part (c) of the lemma, if $\nabla_3\Pi_{\rm M}(p, r_{{C}}, r_{{R}})=0$, then we must have $(w-c) - (r_{{R}}+\beta r_{{C}}) \frac{w}{p+r_{{R}}-v}>0$, in which case we will also have $(2w-c) - (r_{{R}}+\beta r_{{C}}) \frac{w}{p+r_{{R}}-v}>0$. The desired result now follows since $h'>0$ by assumption (A2).

Proof of (j) It can be verified that $\nabla^2_{12}\Pi_{\rm R}(p, r_{{C}}, r_{{R}})$ is given, after substituting for $\nabla_2z$ and $\nabla^2_{12}z$ from Lemma 1(b) and (e) and for $\nabla_2y^*$ from Lemma 2(b) by

$$\begin{aligned} \nabla^2_{12}\Pi_{\rm R}(p, r_{{C}}, r_{{R}}) = \alpha \gamma \int_0^{\frac{p + r_{{R}} - w}{p+r_{{R}}-v}} z d \xi + \frac{w-v}{p+r_{{R}}-v} \alpha \gamma y^* - \alpha \theta (p+r_{{R}}-v) \int_0^{\frac{p + r_{{R}} - w}{p+r_{{R}}-v}} z d \xi \ \end{aligned}$$

Now, when $\nabla_1\Pi_{\rm R}=0,$ the following relationship can be verified through algebra, using part (a) of the lemma and the above expression:

$$\begin{aligned} \nabla^2_{12}\Pi_{\rm R}(p, r_{{C}}, r_{{R}}) = \alpha (\gamma^2 - \theta) (p+r_{{R}}-v) \int_0^{\frac{p + r_{{R}} - w}{p+r_{{R}}-v}} z d \xi, \ \end{aligned}$$

which is strictly positive since $\gamma' = \gamma^2 - \theta > 0$ by virtue of (A4).

Proof of Proposition 1 Using Lemma 6(a), one can verify that $\nabla_1 \Pi_{\rm R}(w - r_{{R}}, r_{{C}}, r_{{R}}) > 0$. Again using Lemma 6(a), one can also verify that $\nabla_1 \Pi_{\rm R}(p, r_{{C}}, r_{{R}}) < 0$ as $p \rightarrow \infty$. Given these observations, the result now follows from Lemma 4(a) and Lemma 6(d).

Proof of Proposition 2

Proof of (a) Given p and $r_{{R}}$, using Lemma 6(b), one can verify that $\nabla_2 \Pi_{\rm M}(p, r_{{C}}, r_{{R}}) < 0$ as $r_{{C}} \rightarrow \infty$. From Lemma 6(e), we know that $\nabla^2_{22} \Pi_{\rm M}(p, r_{{C}},r_{{R}}) |_{\nabla_2\Pi_{\rm M}=0}<0$. The result now follows by applying parts (a) and (b) of Lemma 4.

Proof of (b) We can focus on $r_{{R}}$ such that $p+r_{{R}} \geq w$ (and, hence, $p+r_{{R}} \geq v$), since the retailer would stock zero units otherwise, and the manufacturer would make zero profits. Given p and $r_{{C}}$, using Lemma 6(c), one can verify that $\nabla_3 \Pi_{\rm M}(p, r_{{C}}, r_{{R}}) < 0$ as $r_{{R}} \rightarrow \infty$. From Lemma 6(f), we know that $\nabla^2_{33} \Pi_{\rm M}(p, r_{{C}},r_{{R}}) |_{\nabla_3\Pi_{\rm M}=0}<0$. The result now follows by applying parts (a) and (b) of Lemma 4.

Proof of (c) When $\nabla_2\Pi_{\rm M}(p, r_{{C}}, r_{{R}}) = 0$, we can use Lemma 6(b) to write

$$\begin{aligned}\beta \int_0^{\frac{p + r_{{R}} - w}{p+r_{{R}}-v}} z d \xi + \beta \frac{w-v}{p+r_{{R}}-v} y^* =& \left[(w-c) - (r_{{R}} + \beta r_{{C}}) \frac{w-v}{p+r_{{R}}-v}\right] \alpha \gamma y^* \\ &- (r_{{R}} + \beta r_{{C}}) \alpha \gamma \int_0^{\frac{p + r_{{R}} - w}{p+r_{{R}}-v}} z d \xi. \ \end{aligned}$$

Note that for the above equality to hold, we need to have $(w-c) - (r_{{R}} + r_{{C}}) \frac{w-v}{p+r_{{R}}-v} > 0$ (since $\gamma > 0$ by assumption (A3)). Similarly, when $\nabla_3\Pi_{\rm M}(p, r_{{C}}, r_{{R}}) = 0$, we can use Lemma 6 (c) to write

$$\begin{aligned} \int_0^{\frac{p + r_{{R}} - w}{p+r_{{R}}-v}} z d \xi + \frac{w-v}{p+r_{{R}}-v} y^* &= \left[(w-c) - (r_{{R}} + \beta r_{{C}}) \frac{w-v}{p+r_{{R}}-v}\right] \frac{f}{(p+r_{{R}}-v) h}. \ \end{aligned}$$

Again, note that for the above equality to hold, we need to have $(w-c) - (r_{{R}} + r_{{C}}) \frac{w-v}{p+r_{{R}}-v} > 0$. By using the last two equalities, we obtain:

$$\begin{aligned} - (r_{{R}} + \beta r_{{C}}) \alpha \gamma \int_0^{\frac{p + r_{{R}} - w}{p+r_{{R}}-v}} z d \xi = \left[(w-c) - (r_{{R}} + \beta r_{{C}}) \frac{w-v}{p+r_{{R}}-v}\right] \left[\frac{\beta f}{(p+r_{{R}}-v) h} - \alpha \gamma y^* \right] \ \end{aligned}$$

Now, note that the second term in brackets on the right-hand side of the equality above is $- \nabla_2y^*+ \beta \nabla_3y^*$ (from parts (b) and (c) of Lemma 2). Also, as noted above, we must have $(w-c) - (r_{{R}} + r_{{C}}) \frac{w-v}{p+r_{{R}}-v} > 0$. The term on the left-hand side is negative (since $\gamma>0$ by assumption (A3)). The desired result now follows.

Proof of Proposition 3 Under an equilibrium solution $(\tilde{p}, \tilde r_{{C}}, \tilde r_{{R}})$ with $\tilde r_{{C}} > 0$, we need to have $\nabla_2\Pi_{\rm M}(\tilde{p}, \tilde r_{{C}}, \tilde r_{{R}}) = \nabla_1\Pi_{\rm R}(\tilde{p}, \tilde r_{{C}}, \tilde r_{{R}}) = 0$ (by Proposition 1 and part (a) of Proposition 2). Since $\nabla_2\Pi_{\rm M}(\tilde{p}, \tilde r_{{C}}, \tilde r_{{R}})=0$, we know from Lemma 6(b) that

$$\begin{aligned} \beta \frac{w-v}{{\tilde{p}}+\tilde r_{{R}}-v} y^* + \beta \int_0^{\frac{\tilde{p} + \tilde r_{{R}} - w}{\tilde{p}+\tilde r_{{R}} -v}} z d \xi =&(w-c) \alpha \gamma y^* \\ &- (\tilde r_{{R}} + \beta \tilde r_{{C}}) \alpha \gamma \left( \int_0^{\frac{\tilde{p} + \tilde r_{ {R}} - w}{\tilde{p}+\tilde r_{{R}} -v}} z d \xi + \frac{w-v}{\tilde{p}+\tilde r_{{R}}-v} y^*\right), \\ &= \alpha \gamma \Pi_{\rm M}(\tilde{p}, \tilde r_{{C}}, \tilde r_{{R}}) \mbox\,{\rm by \,\,(6)}\end{aligned}$$

((10))

Also, since $\nabla_1\Pi_{\rm R}(\tilde{p}, \tilde r_{{C}}, \tilde r_{{R}})=0$, we know from Lemma 6 (a) that

$$\begin{aligned} \frac{w-v}{\tilde{p}+\tilde r_{{R}} -v} y^* + \int_0^{\frac{\tilde{p} + \tilde r_{{R}} - w}{\tilde{p}+\tilde r_{{R}}-v}} z d \xi &= (\tilde{p}+ \tilde r_{{R}} -v) \gamma \int_0^{\frac{\tilde{p} + \tilde r_{{R}} - w}{\tilde{p}+\tilde r_{{R}} -v}} z d \xi, \\ &= \gamma \Pi_{\rm R}(\tilde{p}, \tilde r_{ {C}}, \tilde r_{{R}}) \mbox{\rm \,by\, (5)}\end{aligned}$$

((11))

Now, (10) and (11) together allow us conclude $\frac{\Pi_{\rm M}(\tilde{p}, \tilde r_{{C}}, \tilde r_{{R}})} {\Pi_{\rm R}(\tilde{p}, \tilde r_{{C}}, \tilde r_{{R}})} = \frac{\beta}{\alpha}$.

Proof of Proposition 4

Proof of (a) Throughout the proof, recall that $p^*(r_{{R}})$ will satisfy $\nabla_1\Pi_{\rm R}(p^*(r_{{R}}), 0, r_{{R}}) = 0$ at any given $r_{{R}}$ (by Proposition 1). By implicit differentiation of this identity with respect to $r_{{R}}$, we obtain $\frac{dp^*(r_{{R}})}{dr_{{R}}} = - \frac{\nabla^2_{13}\Pi_{\rm R}(p^*(r_{{R}}), 0, r_{ {R}})}{\nabla^2_{11}\Pi_{\rm R}(p^*(r_{{R}}), 0, r_{{R}})}.$ Hence, we will conclude the proof of part (a) if we can show that $\nabla^2_{11}\Pi_{\rm R}(p^*(r_{{R}}), 0, r_{{R}}) \leq \nabla^2_{13}\Pi_{\rm R}(p^*(r_{{R}}), 0, r_{{R}}) < 0$. From Lemma 6(d) and (h), we know that $\nabla^2_{11}\Pi_{\rm R}(p^*(r_{{R}}), 0, r_{{R}})<0$ and $\nabla^2_{13}\Pi_{\rm R}(p^*(r_{{R}}), 0, r_{{R}})<0$. Again, from Lemma 6(d) and (h), note that:

$$\begin{aligned} \nabla^2_{11}\Pi_{\rm R}(p^*(r_{{R}}), 0, r_{{R}}) - \nabla^2_{13}\Pi_{\rm R}(p^*(r_{{R}}), 0, r_{{R}})= - (p+r_{{R}}-v) \gamma' \int_0^{\frac{p + r_{{R}} - w}{p+r_{{R}}-v}}z \leq 0,\end{aligned}$$

((12))

where the inequality follows from $\gamma' \geq 0$ (by (A4)). Thus, we are able to conclude that

$$\nabla^2_{11}\Pi_{\rm R}(p^*(r_{{R}}), 0, r_{{R}}) \leq \nabla^2_{13}\Pi_{\rm R}(p^*(r_{{R}}), 0, r_{{R}}) < 0,$$

which concludes the proof of part (a).

Proof of (b) Given p, w and $r_{{C}}=0$, it follows from Proposition 2(b) that either $r_{{R}}^*(p)=0$ or $r_{{R}}^*(p)>0$ in which case $r_{{R}}^*(p)$ satisfies $\nabla_3\Pi_{\rm M}(p, 0, r_{{R}}^*(p)) = 0$. If $r_{{R}}^*(p)=0$ for all $p>0$, then part (b) holds trivially. Suppose now there exists a p at which $r_{{R}}^*(p) > 0$ and satisfies $\nabla_3\Pi_{\rm M}(p, 0, r_{{R}}^*(p)) = 0$. By implicit differentiation of this identity with respect to p, we obtain $\frac{dr_{{R}}^*(p)}{dp} = - \frac{\nabla^2_{13}\Pi_{\rm M}(p, 0, r_{ {R}}^*(p))}{\nabla^2_{33}\Pi_{\rm M}(p, 0, r_{{R}}^*(p))}.$ We already know from Lemma 6(f) and (i) that $\nabla^2_{33}\Pi_{\rm M}(p, 0, r_{{R}}^*(p))<0$ and $\nabla^2_{13}\Pi_{\rm M}(p, 0, r_{{R}}^*(p))<0$. Furthermore, again from Lemma 6(f) and (i), one can verify that

$$\begin{aligned} \left.\nabla^2_{13}\Pi_{\rm M}(p, r_{{C}}, r_{ {R}})\right|_{\nabla_3\Pi_{\rm M}=0} =\,& \left.\nabla^2_{33}\Pi_{\rm M}(p, r_{{C}}, r_{ {R}})\right|_{\nabla_3\Pi_{\rm M}=0} + \frac{w-v}{p+r_{{R}}-v} \frac{f}{(p+r_{{R}}-v) h} \\ &- \frac{1}{p+r_{{R}}-v} \left\{-\int_0^{\frac{p + r_{{R}} - w}{p+r_{{R}}-v}} z d \xi - \frac{w-v}{p+r_{{R}}-v} y^* \right. \\ &+ \left. \left[(w-c) - (r_{{R}} + \beta r_{{C}}) \frac{w-v}{p+r_{{R}}-v}\right] \frac{f}{(p+r_{{R}}-v) h}\right\} \ \end{aligned}$$

From Lemma 6(c), we observe that the term in curly brackets above is in fact $\nabla_3\Pi_{\rm M}(p, r_{{C}}, r_{{R}})$. Therefore, from the above expression, we obtain:

$$\begin{aligned} \left.\nabla^2_{13}\Pi_{\rm M}(p, r_{{C}}, r_{ {R}})\right|_{\nabla_3\Pi_{\rm M}=0} =& \left.\nabla^2_{33}\Pi_{\rm M}(p, r_{{C}}, r_{{R}})\right|_{\nabla_3\Pi_{\rm M}=0} + \frac{w-v}{p+r_{ {R}}-v} \frac{f}{(p+r_{{R}}-v) h} \ \end{aligned}$$

Hence, from the last equality, we conclude that $\nabla^2_{33}\Pi_{\rm M}(p, 0, r_{{R}}^*(p)) < \nabla^2_{13}\Pi_{\rm M}(p, 0, r_{ {R}}^*(p))$, which, along with $\nabla^2_{33}\Pi_{\rm M}(p, 0, r_{{R}}^*(p))<0$ and $\nabla^2_{13}\Pi_{\rm M} (p, 0, r_{{R}}^*(p))<0$, allows us to conclude that $-1< \frac{dr_{{R}}^*(p)}{dp} < 0$. Recall that we assumed p is such that $r_{{R}}^*(p)>0$. For some $p^\prime$, we will have $r_{{R}}^*(p^\prime)=0$, and $r_{{R}}^*(p)$ will remain zero for all $p > p^\prime$, and hence $\frac{dr_{{R}}^*(p)}{dp}$ will be zero for all $p > p^\prime$. (If $r_{{R}}^*(p)$ were to become positive for some $p^{\prime\prime} > p^\prime$, this would be a contradiction to the result that $\frac{dr_{{R}}^*(p)}{dp} < 0$ when $r_{{R}}^*(p) >0$.)

Proof of (c) The existence of the Nash equilibrium follows from Lemma 5(a), Proposition 1 and Proposition 2(b). The uniqueness of the Nash equilibrium follows from Lemma 5(b) and parts (a) and (b) of this proposition. (Note that, in order to apply Lemma 5, we need upper bounds on the decision variables of the retailer and the manufacturer, p and $r_{{R}}$, respectively. We could satisfy this requirement by picking arbitrarily large numbers to bound the feasible choices for p and $r_{{R}}$.)

Proof of Proposition 5

Throughout the proof, let $p^*(r_{{R}})$ denote the optimal retail price chosen by the retailer at a given $r_{{R}}$ when $r_{{C}}=0$.

Proof of (a) Note that $p_o = p^*(0)$ whereas $\tilde{p}=p^*(\tilde r_{{R}})$. Therefore, $\tilde{p}-p_o = \int_0^{\tilde r_{{R}}} \frac{dp^*(r_{{R}})}{dr_{{R}}} dr_{{R}}$. By Proposition 4(a), $-1< \frac{dp^*(r_{{R}})}{dr_{{R}}}<0$. The desired result follows.

Proof of (b) Note that $y_o = y^*(p^*(0), 0, 0)$ whereas $\tilde{y}=y^*(p^*(\tilde r_{{R}}), 0, \tilde r_{{R}})$. Now, $\tilde{y} - y_o = \int_0^{\tilde r_{{R}}}\frac{dy^*(p^*(r_{{R}}),0,r_{ {R}})}{dr_{ {R}}}dr_{{R}}$. Therefore, we will conclude the proof if we can show that $\frac{dy^*(p^*(r_{{R}}),0,r_{{R}})}{dr_{{R}}}>0$. Note that $\frac{dy^*(p^*(r_{{R}}),0,r_{{R}})}{dr_{{R}}} = \nabla_3y^*(p^*(r_{{R}}), 0,r_{{R}}) + \frac{dp^*(r_{{R}})}{dr_{{R}}} \nabla_1y^*(p^*(r_{{R}}), 0,r_{{R}})$. Now, $\nabla_3y^*(p^*(r_{{R}}), 0,r_{{R}}) > 0$ from Lemma 2(c), $\frac{dp^*(r_{{R}})}{dr_{{R}}} < 0$ from of Proposition 4(a) and $\nabla_1y^*(p^*(r_{{R}}), 0,r_{{R}}) < 0$ from Lemma 3. (To see why Lemma 3 can be applied here, recall that $\nabla_1\Pi_{\rm R}(p^*(r_{{R}}), 0, r_{{R}}) = 0$ by Proposition 1 since $p^*(r_{{R}})$ optimizes $\Pi_{\rm R}$.) These observations imply that $\frac{dy^*(p^*(r_{{R}}),0,r_{{R}})}{dr_{{R}}}>0$, which yields the desired result.

Proof of (c) Note that $\Pi_{\rm SC}(\tilde{p}, 0, \tilde r_{{R}}) = \Pi_{\rm SC}(p^*(\tilde r_{{R}}), 0, \tilde r_{{R}})$ and $\Pi_{\rm SC}(p_o, 0, 0) = \Pi_{\rm SC}(p^*(0), 0, 0)$. Therefore, $\Pi_{\rm SC}(\tilde{p}, 0, \tilde r_{{R}}) - \Pi_{\rm SC}(p_o, 0, 0) = \int_0^{\tilde r_{{R}}}\frac{d\Pi_{\rm SC}(p^*(r_{{R}}), 0, r_{{R}}))}{dr_{{R}}} dr_{{R}}$. Hence, if we can show that $\Pi_{\rm SC}(p^*(r_{{R}}), 0, r_{{R}})$ is increasing in $r_{{R}}$ for $r_{{R}} \leq w-c$, then the desired result will follow. Hence, we want to show that

$$\frac{d\Pi_{\rm SC}(p^*(r_{{R}}), 0, r_{{R}})}{dr_{{R}}} = \frac{dp^*(r_{ {R}})}{dr_{ {R}}} \nabla_1\Pi_{\rm SC}(p^*(r_{{R}}), 0, r_{{R}}) + \nabla_3\Pi_{\rm SC}(p^*(r_{{R}}), 0, r_{{R}})$$

is positive. The following equalities can be verified using (5) and (6):

$$\begin{aligned} \nabla_1\Pi_{\rm SC}(p, r_{{C}}, r_{{R}}) =& \nabla_1\Pi_{\rm R}(p, r_{{C}}, r_{{R}}) +\nabla_1\Pi_{\rm M}(p, r_{{C}}, r_{{R}})\\ =& \nabla_1\Pi_{\rm R}(p, r_{{C}}, r_{ {R}}) + \left[w-c-(r_{{R}} + \beta r_{{C}})\frac{w-v}{p + r_{{R}}-v}\right] \nabla_1y^* \\ &- (r_{{R}}+ \beta r_{ {C}}) \int_0^{\frac{p + r_{{R}} - w}{p+r_{{R}}-v}} \nabla_1z d \xi\\ \nabla_3\Pi_{\rm SC}(p, r_{{C}}, r_{{R}}) &= \nabla_3\Pi_{\rm R}(p, r_{{C}}, r_{{R}}) +\nabla_3\Pi_{\rm M}(p, r_{{C}}, r_{{R}})\\ &= \left[w-c-(r_{ {R}} + \beta r_{{C}})\frac{w-v}{p + r_{{R}}-v}\right] \nabla_3y^* \ \end{aligned}$$

Note that $\nabla_1\Pi_{\rm R}(p^*(r_{{R}}), 0, r_{{R}})=0$ by definition of $p^*(r_{{R}})$ and Proposition 1. Thus, after substitution and rearranging terms, we get

$$\begin{aligned} \frac{d\Pi_{\rm SC}(p^*(r_{{R}}), 0, r_{ {R}})}{dr_{{R}}} =& \left(1 + \frac{dp^*(r_{{R}})}{dr_{{R}}}\right) \left[w-c-r_{{R}}\frac{w-v}{p^*(r_{{R}}) + r_{{R}} -v}\right] \nabla_3y^* \\ &+ \frac{dp^*(r_{{R}})}{dr_{{R}}} \left\{\left[w-c-r_{{R}}\frac{w-v}{p^*(r_{{R}}) + r_{{R}} -v}\right](\nabla_1y^* - \nabla_3y^*) \right. \\ &- \left. r_{{R}} \int_0^{\frac{p^*(r_{{R}}) + r_{{R}} - w}{p^*(r_{ {R}})+r_{ {R}}-v}} \nabla_1z d \xi\right\} \ \end{aligned}$$

By Lemma 2(c) and Proposition 4(a), the first term above is positive. We show that the second term is also positive, to conclude the proof. Since $\frac{dp^*(r_{{R}})}{dr_{{R}}}<0$, all we need to show is

$$\begin{aligned} &\left[w-c-r_{{R}}\frac{w-v}{p^*(r_{{R}}) + r_{ {R}}-v}\right](\nabla_1y^*- \nabla_3y^*) - r_{{R}} \int_0^{\frac{p^*(r_{{R}}) + r_{{R}} - w}{p^*(r_{{R}})+r_{{R}}-v}} \nabla_1z d \xi< 0.\end{aligned}$$

((13))

Now, for $\xi \leq \frac{p^*(r_{{R}})+r_{{R}}-w}{p^*(r_{{R}})+r_{{R}}-v}$,

$$\begin{aligned} \nabla_1y^*- \nabla_3y^* &= - \gamma y^* \\ &= \nabla_1z \left(p^*(r_{ {R}}),0, \frac{p^*(r_{{R}}) + r_{{R}} - w}{p+r_{{R}}-v}\right) \\ &< \nabla_1z(p^*(r_{{R}}),0, \xi) \ \end{aligned}$$

The first equality follows from Lemma 2(a) and (c), the second from Lemma 1(a), and the inequality holds because $\xi \leq \frac{p^*(r_{{R}})+r_{{R}}-w}{p^*(r_{{R}})+r_{{R}}-v}$. Thus, using $r_{{R}} \leq w-c$, (13) holds.

Proof of Proposition 6

Proof of (a) Note that $p^*(r_{{C}})$ will satisfy $\nabla_1\Pi_{\rm R}(p^*(r_{{C}}), r_{{C}}, 0) = 0$ at any given $r_{{C}}$ (by Proposition 1). By implicit differentiation of this identity with respect to $r_{{C}}$, we obtain $\frac{dp^*(r_{{C}})}{dr_{{C}}} = - \frac{\nabla^2_{12}\Pi_{\rm R}(p^*(r_{{C}}), r_{ {C}}, 0)} {\nabla^2_{11}\Pi_{\rm R}(p^*(r_{{C}}), r_{{C}}, 0)}$. We know from Lemma 6(d) and (j) that

$$\nabla^2_{11}\Pi_{\rm R}(p^*(r_{{C}}), r_{{C}}, 0)<0 \mbox{ \,\,and \,} \nabla^2_{12}\Pi_{\rm R}(p^*(r_{{C}}), r_{{C}}, 0)>0.$$

Therefore, it follows that $\frac{dp^*(r_{{C}})}{dr_{{C}}}>0$. Furthermore, from Lemma 6(d) and (j), we can write:

$$\begin{aligned} \left.\nabla^2_{12}\Pi_{\rm R}(p, r_{{C}}, r_{ {R}})\right|_{\nabla_1\Pi_{\rm R}=0} =&- \alpha \left.\nabla^2_{11}\Pi_{\rm R}(p, r_{{C}}, r_{{R}})\right|_{\nabla_1\Pi_{\rm R}=0} \\&+ \alpha \frac{w-v}{p+r_{{R}}-v} \nabla_1y^* - \alpha \gamma \int_0^{\frac{p + r_{{R}} - w}{p+r_{{R}}-v}} z d \xi \ \end{aligned}$$

From the equality above, since $\nabla_1y^*<0$ when $\nabla_1\Pi_{\rm R}=0$ (from Lemma 3) and $\gamma > 0$ (by (A3)), we have $\left.\nabla^2_{12}\Pi_{\rm R}(p, r_{{C}}, r_{{R}})\right|_{\nabla_1\Pi_{\rm R}=0} < - \alpha \left.\nabla^2_{11}\Pi_{\rm R}(p, r_{{C}}, r_{{R}})\right|_{\nabla_1\Pi_{\rm R}=0}$. Therefore, we have

$$\nabla_{12}\Pi_{\rm R}(p^*(r_{{C}}), r_{{C}}, 0) < - \alpha \nabla_{11}\Pi_{\rm R}(p^*(r_{{C}}), r_{{C}},0).$$

This observation yields $\frac{dp^*(r_{{C}})}{dr_{{C}}} < \alpha$.

Proof of (b) The existence of the Nash equilibrium follows from Lemma 5(a), Proposition 1 and Proposition 2(a). (Note that, in order to apply Lemma 5, we need upper bounds on the decision variables of the retailer and the manufacturer, p and $r_{{C}}$, respectively. We could satisfy this requirement by picking arbitrarily large numbers to bound the feasible choices for p and $r_{{C}}$.)

Proof of Proposition 7

Throughout the proof, let $p^*(r_{{C}})$ denote the optimal retail price chosen by the retailer at a given $r_{{C}}$ when $r_{{R}}=0$.

Proof of (a) Note that $p_o = p^*(0)$ whereas $\tilde{p}=p^*(\tilde r_{{C}})$. Therefore, $\tilde{p}-p_o = \int_0^{\tilde r_{{C}}} \frac{dp^*(r_{{C}})}{dr_{{C}}} dr_{{C}}$. By Proposition 6, $0<\frac{dp^*(r_{{C}})}{dr_{{C}}}<\alpha$. The desired result follows.

Proof of (b) Note that $y_o = y^*(p^*(0), 0, 0)$ whereas $\tilde{y}=y^*(p^*(\tilde r_{{C}}), \tilde r_{{C}},0)$. Now, $\tilde{y} - y_o = \int_0^{\tilde r_{{C}}}\frac{dy^*(p^*(r_{{C}}),r_{{C}}, 0)}{dr_{ {C}}}dr_{{C}}$. We will conclude the proof if we can show that $\frac{dy^*(p^*(r_{{C}}),r_{{C}}, 0)}{dr_{{C}}}>0$. Note that $\frac{dy^*(p^*(r_{{C}}),r_{{C}},0)}{dr_{{C}}} = \nabla_2y^*(p^*(r_{{C}}), r_{ {C}},0) + \frac{dp^*(r_{{C}})}{dr_{{C}}} \nabla_1y^* (p^*(r_{{C}}), r_{{C}},0)$. Since $0<\frac{dp(r_{{C}})}{dr_{{C}}} < \alpha$ by Proposition 6 and $\nabla_1y^*(p^*(r_{{C}}), r_{{C}},0)<0$ by Lemma 3, we obtain $\frac{dy^*(p^*(r_{{C}}),r_{{C}},0)}{dr_{{C}}} > \nabla_2y^*(p^*(r_{{C}}), r_{{C}},0) + \alpha \nabla_1y^*(p^*(r_{{C}}), r_{{C}},0)$. Using this last inequality and substituting for $\nabla_1y^*(p^*(r_{{C}}), r_{{C}},0)$ from Lemma 2(a) and for $\nabla_2y^*(p^*(r_{{C}}), r_{{C}},0)$ from Lemma 2(b), we can deduce that $\frac{dy^*(p^*(r_{{C}}),r_{{C}},0)}{dr_{{C}}} >0$, which concludes the proof of this part.

Proof of (c) As in the proof of part (c) of Proposition 5, we will show that $\Pi_{\rm SC}(p^*(r_{{C}}), r_{{C}},0)$ is increasing in $r_{{C}}$ for $r_{{C}} \leq w-c$ when $\alpha \geq \beta$. The desired result would then follow. Now, the following equalities can be verified by partial differentiation of (5) and (6):

$$\begin{aligned} \frac{d\Pi_{\rm SC}(p^*(r_{{C}}), r_{{C}}, 0)}{dr_{{C}}} =& \frac{dp^*(r_{{C}})}{dr_{{C}}} \nabla_1\Pi_{\rm SC}(p^*(r_{{C}}), r_{{C}}, 0) + \nabla_2\Pi_{\rm SC}(p^*(r_{{C}}), r_{{C}}, 0) \\ =& \frac{dp^*(r_{{C}})}{dr_{{C}}} \nabla_1\Pi_{\rm R}(p^*(r_{{C}}), r_{{C}}, 0)\\ &+ \left(w-c-\beta r_{{C}} \frac{w-v}{p^*(r_{{C}})-v}\right) \left(\nabla_2y^* + \frac{dp^*(r_{{C}})}{dr_{{C}}} \nabla_1y^*\right)\\ &+ \beta r_{{C}} \left(\int_0^{\frac{p^*(r_{{C}}) - w}{p^*(r_{{C}})-v}} \nabla_2z d \xi - \frac{dp^*(r_{ {C}})}{dr_{{C}}} \int_0^{\frac{p^*(r_{{C}}) - w}{p^*(r_{{C}})-v}} \nabla_1 z d \xi\right) \\ &+ p^*(r_{{C}}) \int_0^{\frac{p^*(r_{{C}}) - w}{p^*(r_{{C}})-v}} \nabla_2z d \xi \\ &- \beta \left( \int_0^{\frac{p^*(r_{{C}}) - w}{p^*(r_{ {C}})-v}} z d \xi + \frac{w-v}{p^*(r_{{C}})-v} y^* \right) \ \end{aligned}$$

Now, the first term is zero, by definition of $p^*(r_{{C}})$. The second term is positive, because, as in the proof of part (b), $\nabla_2y^*(p^*(r_{{C}}), r_{{C}},0) +$ $\frac{dp^*(r_{{C}})}{dr_{{C}}} \nabla_1y^* (p^*(r_{{C}}), r_{{C}},0)>0$, and $r_{{C}} \leq w-c$. The third term is positive by virtue of Lemma 1(a)-(b) and Proposition 6(a). Using Lemma 6 (a), Lemma 1(a)-(b) and the fact that $\nabla_1\Pi_{\rm R}(p^*(r_{{C}}),r_{{C}},0)=0$ we get that

$$p^*(r_{{C}}) \int_0^{\frac{p^*(r_{{C}}) - w-v}{p^*(r_{{C}})-v}} \nabla_2z d \xi = \alpha \left( \int_0^{\frac{p^*(r_{{C}}) - w}{p^*(r_{{C}})-v}} z d \xi + \frac{w-v}{p^*(r_{{C}})-v} y^* \right).$$

Thus, the sum of the last two terms can be written as

$$\begin{aligned} (\alpha - \beta) \left( \int_0^{\frac{p^*(r_{{C}}) - w}{p^*(r_{ {C}})-v}} z d \xi + \frac{w-v}{p^*(r_{{C}})-v} y^* \right), \ \end{aligned}$$

which is positive because $\alpha \ge \beta$.

Proof of (d) The proof of this part is almost identical to the analogous result in Proposition 3. Set $\tilde r_{{R}}=0$ and the proof follows the same line of argument.