Roles of a preselling strategy under asymmetric information

Abstract

In many product markets, one firm may provide an option to buy ahead of time. In this circumstance, customers inevitably make a purchase decision under uncertainty on the future price as well as the valuation of product, which critically relies on the marginal cost. The marginal cost is usually private information of the firm in reality. There is a common misconception that the firm would profit a lot from the preselling owing to private information on the marginal cost. However, this paper finds the role of preselling under asymmetric information is disputable. We employ the framework of dynamic game to examine the impact of cost information asymmetry and signaling effect in influencing the firm?’s preselling strategy and profitability. We find the firm has no incentive to presell, and the information superiority may not benefit the firm. Likewise, the disadvantage of information may not hurt customers due to their rational expectation of cost.

Appendix

Proof of Lemma 1

By solving $\pi ^{\textsuperscript {Pre*}}(q,\tilde {c})=\pi ^{\textsuperscript {Re*}}(q,\tilde {c})$, we can obtain $\tilde {c}=\frac {q}{3}$. Note that $\pi ^{\textsuperscript {Pre*}}(q,c)-\pi ^{\textsuperscript {Re*}}(q,c)$ is strictly decreasing with c, and $\pi ^{\textsuperscript {Pre*}}(q,0)>\pi ^{\textsuperscript {Re*}}$ (q, 0). Then, we have $\pi ^{\textsuperscript {Pre*}}(q,c)>\pi ^{\textsuperscript {Re*}}(q,c)$ for $c<\tilde {c}$; $\pi ^{\textsuperscript {Pre*}}(q,c)\leq \pi ^{\textsuperscript {Re*}}(q,c)$ for $c\geq \tilde {c}$. Then, Lemma 1 holds. □

Proof of Lemma 2

By solving $\pi ^{\textsuperscript {Pre*}}(q,\hat {c})=\pi ^{\textsuperscript {Re*}}(q,\hat {c})$, we have $\tilde {c}=\frac {2\sqrt {2}-2}{3}q$. Note that $\pi ^{\textsuperscript {Pre*}}(q,c)-\pi ^{\textsuperscript {Re*}}(q,c)$ is strictly decreasing with c, and $\pi ^{\textsuperscript {Pre*}}(q,0)>\pi ^{\textsuperscript {Re*}}(q,0)$. Then, we have $\pi ^{\textsuperscript {Pre*}}(q,c)>\pi ^{\textsuperscript {Re*}}(q,c)$ for $c<\hat {c}$; $\pi ^{\textsuperscript {Pre*}}(q,c)\leq \pi ^{\textsuperscript {Re*}}(q,c)$ for $c\geq \hat {c}$. Then, Lemma 2 holds. □

Proof of Proposition 1

Noting that $\hat {c}=\frac {2\sqrt {2}-2}{3}q$ and $\tilde {c}=\frac {q}{3}$, the result immediately follows. □

Proof of Proposition 2

Since both π^S(q, c) and π^A(q, c) are piecewise functions of c, we need to consider the three scenarios: $c\in (\tilde {c},q)$, $c\in (\hat {c},\tilde {c})$ and $c\in (0,\hat {c})$: If $c\in (\tilde {c},q)$, regular selling is more profitable in both the symmetric information case and the asymmetric information case. Hence, there is no difference, namely, π^S(q, c) = π^A(q, c). If $c\in (\hat {c},\tilde {c})$, it is straightforward to have $\pi ^{S}(q,c)>\pi ^{A}(q,c)$ since $\frac {1}{2}q-\frac {(q-c)^{2}}{8q}-c>\frac {(q-c)^{2}}{4q}$ by Lemma 1. If $c\in (0,\hat {c})$, we have $\pi ^{S}(q,c)-\pi ^{A}(q,c)=\frac {(q-\hat {c}/2)^{2}}{8q}-\frac {(q-c)^{2}}{8q}$. Since $-\frac {(q-c)^{2}}{8q}$ is increasing in c for any q, there is a unique point $c^{\ast }=\frac {\hat {c}}{2}=\frac {\sqrt {2}-1}{3}q$ such that π^S(q, c^∗) = π^A(q, c^∗). Subsequently, π^S(q, c) < π^A(q, c) for 0 ≤ c < c^∗, and π^S(q, c) ≥ π^A(q, c) for $c^{\ast }\leq c\leq \hat {c}$.

Then, if c ∈ (0, c^∗), π^S < π^A; if $c\in (c^{\ast },\tilde {c})$, π^S > π^A; if $c\in (\tilde {c},q)$, π^S = π^A. □

Proof of Proposition 3

Recall that

$$\begin{array}{@{}rcl@{}} p^{S}(q,c)=\left\{ \begin{array}{ll} \frac{1}{2}q-\frac{(q-c)^{2}}{8q}, & if\ c<\tilde{c};\\ \frac{q+c}{2}, & if\ c\geq \tilde{c}. \end{array} \right. \ \ \ \text{and} \ \ \ p^{A}(q,c)=\left\{ \begin{array}{ll} \frac{9 + 4\sqrt{2}}{36}q, & if\ c<\hat{c};\\ \frac{q+c}{2}, & if\ c\geq \hat{c}. \end{array} \right. \end{array} $$

Note that $\hat {c}=\frac {2\sqrt {2}-2}{3}q$ and $\tilde {c}=\frac {q}{3}$. Let $\overline {\overline {c}}=\frac {\sqrt {2}-1}{3}q$. Then, we have

$$\begin{array}{@{}rcl@{}} &&p^{A}(q,c)-p^{S}(q,c)=\left\{ \begin{array}{cc} \frac{9 + 4\sqrt{2}}{36}q-\left[ \frac{1}{2}q-\frac{(q-c)^{2}}{8q}\right], & if\ c<\overline{\overline{c}}; \\ \frac{9 + 4\sqrt{2}}{36}q-\left[ \frac{1}{2}q-\frac{(q-c)^{2}}{8q}\right], & if\ \overline{\overline{c}}\leq c<\hat{c}; \\ \frac{q+c}{2}-\left[ \frac{1}{2}q-\frac{(q-c)^{2}}{8q}\right], & if\ \hat{c}\leq c<\tilde{c}; \\ \frac{q+c}{2}-\frac{q+c}{2}, & if\ c>\tilde{c}, \end{array} \right. \\ &=&\left\{ \begin{array}{cc} \left( \frac{q-c}{2\sqrt{2}}-\frac{2\sqrt{2}-1}{6}q\right) \left( \frac{q-c}{2\sqrt{2}}+\frac{2\sqrt{2}-1}{6}q\right), & if\ c<\overline{\overline{c}}; \\ \left( \frac{q-c}{2\sqrt{2}}-\frac{2\sqrt{2}-1}{6}q\right) \left( \frac{q-c}{2\sqrt{2}}+\frac{2\sqrt{2}-1}{6}q\right), & if\ \overline{\overline{c}}\leq c<\hat{c}; \\ \frac{(q+c)^{2}}{8q}, & if\ \hat{c}\leq c<\tilde{c}; \\ 0, & if\ c>\tilde{c}, \end{array} \right. \ \ \ \left\{ \begin{array}{cc} >0, & if\ c<\overline{\overline{c}}; \\ \leq 0, & if\ \overline{\overline{c}}\leq c<\hat{c}; \\ \geq 0, & if\ \hat{c}\leq c<\tilde{c}; \\ = 0, & if\ c>\tilde{c}. \end{array} \right. \end{array} $$

Then, the proof is completed. □

Proof of Proposition 4

In the case without asymmetric information, customers’ surplus is

$$\begin{array}{@{}rcl@{}} U^{S}(q,c) &=&\left\{ \begin{array}{cc} \frac{1}{2}q-p^{S}(q,c), & if\ c<\tilde{c};\\ \mathsf{E}\left[ (V(q)- \frac{q+c}{2})^{+} \right], & if\ c\geq\tilde{c}, \end{array} \right.=\left\{ \begin{array}{cc} \frac{(q-c)^{2}}{8q}, & if\ c<\tilde{c};\\ \frac{(q-c)^{2}}{8q}, & if\ c\geq\tilde{c}, \end{array} \right. = \frac{(q-c)^{2}}{8q}. \end{array} $$

In the case with asymmetric information, customers surplus becomes

$$\begin{array}{@{}rcl@{}} U^{A}(q,c)&=&\left\{ \begin{array}{cc} \frac{1}{2}q-p^{A}(q,c), & if\ c<\hat{c};\\ \mathsf{E}\left[ (V(q)-\frac{q+c}{2})^{+}\right] , & if\ c\geq \hat{c}, \end{array} \right.=\left\{ \begin{array}{cc} \frac{1}{2}q-\frac{9 + 4\sqrt{2}}{36}q, & if\ c<\hat{c};\\ \frac{(q-c)^{2}}{8q}, & if\ c\geq \hat{c}, \end{array} \right.\\&=&\left\{ \begin{array}{cc} \frac{9-4\sqrt{2}}{36}q, & if\ c<\hat{c};\\ \frac{(q-c)^{2}}{8q}, & if\ c \geq \hat{c}. \end{array} \right. \end{array} $$

Note that $\overline {\overline {c}}=\frac {\sqrt {2}-1}{3}q$. Then, we have

$$\begin{array}{@{}rcl@{}} &&U^{A}(q,c)-U^{S}(q,c)=\left\{ \begin{array}{cc} \frac{9-4\sqrt{2}}{36}q-\frac{(q-c)^{2}}{8q}, & if\ c<\overline{\overline{c}}; \\ \frac{9-4\sqrt{2}}{36}q-\frac{(q-c)^{2}}{8q}, & if\ \overline{\overline{c}}\leq c<\hat{c}; \\ \frac{(q-c)^{2}}{8q}-\frac{(q-c)^{2}}{8q}, & if\ c\geq \hat{c}, \end{array} \right. \\ & = &\left\{ \begin{array}{cc} -\left( \frac{q-c}{2\sqrt{2}}-\frac{2\sqrt{2}-1}{6}q\right) \left( \frac{q-c}{2\sqrt{2}}+\frac{2\sqrt{2}-1}{6}q\right) , & if\ c<\overline{\overline{c}}; \\ -\left( \frac{q-c}{2\sqrt{2}}-\frac{2\sqrt{2}-1}{6}q\right) \left( \frac{q-c}{2\sqrt{2}}+\frac{2\sqrt{2}-1}{6}q\right) , & if\ \overline{\overline{c}}\leq c<\hat{c};\!\!\!\! \\ \frac{(q-c)^{2}}{8q}-\frac{(q-c)^{2}}{8q}, & if\ c\geq \hat{c}, \end{array} \right. \ \ \ \left\{ \begin{array}{cc} <0, & if\ c<\overline{\overline{c}}; \\ >0, & if\ \overline{\overline{c}}\leq c<\hat{c}; \\ = 0, & if\ c\geq \hat{c}. \end{array} \right. \end{array} $$

Then, Proposition 4 follows. □

Proof of Lemma 3

Letting ${\Delta }^{S}(c)=\pi ^{\textsuperscript {Pre*}}(q,\alpha ,c)- \pi ^{\textsuperscript {Re*}}(q,\alpha ,c)=\frac {\alpha +(1-\alpha )q}{2}-\alpha \frac {(1 + 2\delta )(1-c)^{2}}{8}-(1-\alpha )\frac {(1 + 2\delta )(q-c)^{2}}{8q}-c$, we have

$$\begin{array}{@{}rcl@{}} \frac{d{\Delta}^{S}(c)}{dc}&=&-\alpha\frac{(1 + 2\delta)c}{4}-(1-\alpha)\frac{(1 + 2\delta)c}{4q}-\frac{3-2\delta}{4}<0. \end{array} $$

Then, Δ^S(c) is strictly decreasing with c. Note that ${\Delta }^{S}(0)=\frac {(\alpha +(1-\alpha )q)(3-2\delta )}{8}>0$. Then, there exists a unique $\tilde {c}^{UQ}$ such that $\pi ^{\textsuperscript {Pre*}}(q,\alpha ,c)>\pi ^{\textsuperscript {Re*}}(q,\alpha ,c)$ if $c<\tilde {c}^{UQ}$ and $\pi ^{\textsuperscript {Pre*}}(q,\alpha ,c)\leq \pi ^{\textsuperscript {Re*}}(q,\alpha ,c)$ if $c\geq \tilde {c}^{UQ}$. Hence, Lemma 3 holds. □

Proof of Lemma 4

Letting ${\Delta }^{A}(c)=\pi ^{\textsuperscript {Pre*}}(q,\alpha ,c)- \pi ^{\textsuperscript {Re*}}(q,\alpha ,c)=\frac {\alpha +(1-\alpha )q}{2}-\alpha \frac {(1-c/2)^{2}}{8}-(1-\alpha )\frac {(q-c/2)^{2}}{8q}-\delta \alpha \frac {(1-c)^{2}}{4}-\delta (1-\alpha )\frac {(q-c)^{2}}{4q}-c$, we have

$$\begin{array}{@{}rcl@{}} \frac{d{\Delta}^{A}(c)}{dc}&=&-\alpha\frac{(1 + 8\delta)c}{16}-(1-\alpha)\frac{(1 + 8\delta)c}{16q}-\frac{7-4\delta}{8}<0. \end{array} $$

Then, Δ^A(c) is strictly decreasing with c. Note that ${\Delta }^{A}(0)=\frac {(\alpha +(1-\alpha )q)(3-2\delta )}{8}>0$. Then, there exists a unique $\hat {c}^{UQ}$ such that $\pi ^{\textsuperscript {Pre*}}(q,\alpha ,c)>\pi ^{\textsuperscript {Re*}}(q,\alpha ,c)$ if $c<\hat {c}^{UQ}$ and $\pi ^{\textsuperscript {Pre*}}(q,\alpha ,c)\leq \pi ^{\textsuperscript {Re*}}(q,\alpha ,c)$ if $c\geq \hat {c}^{UQ}$. Hence, Lemma 4 holds. □

Proof of Proposition 5

Note that Δ^S(c) and Δ^A(c) strictly decrease with c and ${\Delta }^{S}(c)>{\Delta }^{A}(c)$ for any c. Also note that ${\Delta }^{S}(\tilde {c}^{UQ})= 0$ and ${\Delta }^{A}(\hat {c}^{UQ})= 0$. Then, there must be $\hat {c}^{UQ}<\tilde {c}^{UQ}$. Then, Proposition 5 follows. □