Consumer store choice in Asian markets

Abstract

Consumers in Asian markets often need to make choices between highuncertainty stores and low-uncertainty stores. For example, an iPhone may be purchased from an Apple authorized retailer (a low-uncertainty store) or from an unauthorized seller (a high-uncertainty one). We build a game-theoretical model to capture the crucial roles of store uncertainty and consumer risk aversion in store choice. A model-based explanation is provided for the coexistence of charging a higher retail price and having a higher product demand. Interestingly, our finding shows that store uncertainty and risk aversion have the potential to enhance the overall market profitability under the full market coverage, whereas such profitability will be reduced under the partial coverage.Moreover, we find that consumer learning might not be beneficial to the market as a whole. While consumer learning increases the expected overall market profitability under the partial coverage, such learning actually decreases the expected overall profitability if the market is fully covered. Not only is the precision of signals during learning relevant, but also the number of signals per se plays an indispensable role in the market. In addition, we find that the second-mover advantage possibly enjoyed by a high-uncertainty store is an important force that changes market outcomes.

Appendix

1.1 Proof of Lemma 1

Adding the two first-order conditions (the second-order conditions are checked as well), we have p ₁ + p ₂ = 2t; and subtracting leads to $p_{1}-p_{2}=\frac {\varphi r}{3} .$ Therefore, the equilibrium prices are $p_{1}^{\ast }=t+\frac {\varphi r}{6}$ at the LS, and $p_{2}^{\ast }=t-\frac { \varphi r}{6}$ at the other one. Hence, $p_{1}^{\ast }$ increases with t and φ r, whereas $p_{2}^{\ast }$ increases with t but decreases with φ r. Here, we have the price premium $p_{1}^{\ast }>p_{2}^{\ast }$ and $p_{1}^{\ast }-p_{2}^{\ast }=\frac {\varphi r}{3}$.

Based on the results of $p_{1}^{\ast }$ and $p_{2}^{\ast }$ , the equilibrium demands for store 1 and store 2 are $q_{1}^{\ast }=\frac {1}{2t} (t+\frac {\varphi r}{6})$ and $q_{2}^{\ast }=\frac {1}{2t}(t-\frac {\varphi r}{6 })$, respectively. Hence, $q_{1}^{\ast }$ increases with φ r but decreases with t, whereas $q_{2}^{\ast }$ increase with t but decreases with φ r. Here, the LS has a demand advantage $q_{1}^{\ast }>q_{2}^{\ast }$, and $q_{1}^{\ast }-q_{2}^{\ast }=\frac {\varphi r}{6t}$. Then, equilibrium profits can be obtained: ${\Pi }_{1}^{\ast }=\frac {1 }{2t}(t+\frac {\varphi r}{6})^{2}$ and ${\Pi }_{2}^{\ast }=\frac {1}{2t}(t-\frac { \varphi r}{6})^{2}$, where ${\Pi }_{1}^{\ast }>{\Pi }_{2}^{\ast }$ and ${\Pi }_{1}^{\ast }-{\Pi }_{2}^{\ast }=\frac {\varphi r}{3}$. ${\Pi }_{1}^{\ast }$ increases with t and φ r, whereas ${\Pi }_{2}^{\ast }$ increases with t but decreases with φ r. Notice that we focus on the more interesting case where φ r < 6t, such that the price and demand at store 2 are positive. However, if this condition is not satisfied (corresponding to the situation where the store uncertainty and the risk aversion are too high), there will be only one seller (i.e., store 1) in the market.

1.2 Proof of Proposition 1

• (a) Under the full market coverage. Assume that φ r < 6t, and denote the total profit with store uncertainty and risk aversion (i.e., φ r > 0) as ${\Pi }_{full}^{\ast }|_{\varphi r>0}={\Pi }_{1}^{\ast }+{\Pi } _{2}^{\ast }=\frac {1}{2t}(t+\frac {\varphi r}{6})^{2} +\frac {1}{2t}(t-\frac { \varphi r}{6})^{2}$. The total profit in the benchmark cases is ${\Pi }_{full}^{\ast }|_{\varphi r=0}=\frac {1}{2t}t^{2} +\frac {1}{2t}t^{2}=t$. Hence, ${\Delta }_{full}={\Pi }_{full}^{\ast }|_{\varphi r>0}-{\Pi }_{full}^{\ast }|_{\varphi r=0}=\frac {1}{t}(\frac {\varphi r}{6})^{2}>0$, where Δ _{f u l l} increases with φ r and decreases with t.

• (b) Under the partial market coverage. Based on the locations of the marginal consumers, the demands are $q_{1}=\widehat {x}_{1}=\frac {1}{t}(v- \frac {r}{2}\sigma ^{2}-p_{1})$ and $q_{2}=\widehat {x}_{2}=\frac {1}{t}(v- \frac {r}{2}(\sigma ^{2}+\varphi )-p_{2})$. Solving the maximization problem of each store $\max \limits _{p_{i}}q_{i}p_{i}$, we get the optimal retail prices as $p_{1}=\frac {1}{2}(v-\frac {r}{2}\sigma ^{2})$ and $p_{2}=\frac {1}{2 }(v-\frac {r}{2}(\sigma ^{2}+\varphi ))$. The demands and profits are $q_{1}= \frac {1}{4t}\left (2v-\sigma ^{2}r\right )$, $q_{2}=\frac {1}{4t}(2v-\sigma ^{2}r-\varphi r)$, ${\Pi }_{1}=\frac {1}{16t}\left (2v-\sigma ^{2}r\right )^{2}$ and ${\Pi }_{2}=\frac {1}{16t}(2v-\sigma ^{2}r-\varphi r)^{2}$. Under the partial coverage, the condition q ₁ + q ₂ < 1 must be satisfied, and this means that $v<\frac {r}{2}(\sigma ^{2}+\frac {1}{2}\varphi )+t$. Thus, we get the price premium p ₁ > p ₂ and demand advantage q ₁ > q ₂ under the partial coverage. The total profit for two stores under the partial coverage is ${\Pi }_{partial}^{\ast }=\frac {1}{16t}\left (2v-\sigma ^{2}r\right )^{2}+ \frac {1}{16t}(2v-\sigma ^{2}r-\varphi r)^{2}.$ Hence, the total profit under the partial coverage is lower that in the benchmark cases of φ r = 0: ${\Delta }_{partial}={\Pi }_{partial}^{\ast }|_{\varphi r>0}-{\Pi }_{partial}^{\ast }|_{\varphi r=0}<0$, where |Δ _{p a r t i a l} | increases with φ r and v, but decreases with t and σ ². In the partially-covered retail market, notice that we will have two stores if $v>\frac {r}{2}(\sigma ^{2}+\varphi )$, only store 1 if $\frac {r}{2}\sigma ^{2}<v\leq \frac {r}{2} (\sigma ^{2}+\varphi )$, and no store if $v\leq \frac {r}{2}\sigma ^{2}$.

1.3 Proof of Proposition 2

• (a) Under the full market coverage. Suppose that the marginal consumer who is indifferent between shopping at the two stores is located at ${x_{1}^{L}}$, where the superscript denotes the scenario with consumer learning. This means that

  $$v-\frac{r}{2}\sigma^{2}-t{x_{1}^{L}}-p_{1}=E[\widetilde{\widetilde{v}} |s_{1},{\ldots} ,s_{n}]-\frac{r}{2}V[\widetilde{\widetilde{v}}|s_{1},\ldots ,s_{n}]-t\left(1-{x_{1}^{L}}\right)-p_{2}. $$

  Hence, we have ${x_{1}^{L}}$=$\frac {1}{2t}(t+\frac {r}{2(1/\varphi +n/\sigma _{\varepsilon }^{2})}-\frac {n/\sigma _{\varepsilon }^{2}}{1/\varphi +n/\sigma _{\varepsilon }^{2}}\frac {{\Sigma }_{i=1}^{n}s_{i}}{n}-p_{1}+p_{2}).$ Since $E[\frac {{\Sigma }_{i=1}^{n}s_{i}}{n}]=E[\widetilde {v}^{\prime }]=0$, the average value of signals $\frac {{\Sigma }_{i=1}^{n}s_{i}}{n}$ is an unbiased estimator of the true value $\widetilde {v}^{\prime }$, and then the expected location of ${x_{1}^{L}}$ is $E[{x_{1}^{L}}]=\frac {1}{2t}(t+\frac { \varphi ^{L}r}{2}-p_{1}+p_{2})$, where we define $\varphi ^{L}\equiv \frac {1 }{1/\varphi +n/\sigma _{\varepsilon }^{2}}<\varphi$.

  Following similar analyses as the basic model, we find that the expected price premium with learning $E[p_{1}^{\ast L}-p_{2}^{\ast L}]=\frac {\varphi ^{L}r}{3}<p_{1}^{\ast }-p_{2}^{\ast }=\frac {\varphi r}{3}$ (the price premium without learning). The expected demand advantage with learning $E[q_{1}^{\ast L}-q_{2}^{\ast L}]=\frac {\varphi ^{L}r}{6t}<$ $q_{1}^{\ast }-q_{2}^{\ast }=\frac {\varphi r}{6t}$ (the demand advantage without learning). As $\varphi ^{L}=\frac {1}{1/\varphi +n/\sigma _{\varepsilon }^{2}}$, the decreases in the price premium and demand advantage depend critically on the number of signals n and the corresponding signal precision $1/\sigma _{\varepsilon }^{2}$.

  The expected total profit for the two stores with learning $E[{\Pi }_{full}^{\ast L}]={\Pi }_{1}^{\ast L}+{\Pi }_{2}^{\ast L}=\frac {1}{2t}(t+\frac { \varphi ^{L}r}{6})^{2}$ $+\frac {1}{2t}(t-\frac {\varphi ^{L}r}{6})^{2}<{\Pi }_{full}^{\ast }=\frac {1}{2t}(t+\frac {\varphi r}{6})^{2}$ $+\frac {1}{2t}(t- \frac {\varphi r}{6})^{2}$ (the total profit without learning). Through partial differentiations, we get that the absolute value of the difference $|E[{\Pi }_{full}^{\ast L}]-{\Pi }_{full}^{\ast }|$, which increases with φ, r, n, and $1/\sigma _{\varepsilon }^{2},$ but decreases with t. Define ${\Delta }_{full}^{L}\equiv E[{\Pi }_{full}^{\ast L}|_{\varphi r>0}-{\Pi }_{full}^{\ast L}|_{\varphi r=0}]=\frac {1}{t}(\frac {\varphi ^{L}r}{6} )^{2}>0$, and we have ${\Delta }_{full}^{L}<$ Δ _{f u l l} . Thus the qualitative result in Proposition 1 remains unchanged, while the improvement of the overall profitability gets smaller.

• (b) Under the partial market coverage. The expected price premium with learning is $E[{p_{1}^{L}}-{p_{2}^{L}}]=\frac {1}{2}(v-\frac {r}{2}\sigma ^{2})-\frac {1}{2}(v-\frac {r}{2}(\sigma ^{2}+\varphi ^{L}))=\frac {\varphi ^{L}r}{4}$ which is less than the price premium without learning (i.e., $\frac {\varphi r}{4}$); and the expected demand advantage is $E[{q_{1}^{L}}-{q_{2}^{L}}]=\frac {1}{2t}\left (v-\frac {r}{2}r\sigma ^{2}\right ) - \frac {1}{2t}(v-\frac {r}{2}(\sigma ^{2}+\varphi ^{L}))=\frac {\varphi ^{L}r}{4t }$ which is less than the demand advantage without learning (i.e., $\frac { \varphi r}{4t}$). Furthermore, the expected joint profit under the partial coverage with learning is $E[{\Pi }_{partial}^{\ast L}]=\frac {\left (2v-r\sigma ^{2}\right )^{2}+(2v-r\sigma ^{2}-r\varphi ^{L})^{2}}{16t}$, which is greater than the joint profit without learning ${\Pi }_{partial}^{\ast }=\frac {\left (2v-r\sigma ^{2}\right )^{2}+(2v-r\sigma ^{2}-r\varphi )^{2}}{16t}$, as φ ^L < φ. Hence, we have $E[{\Pi }_{partial}^{\ast L}]-{\Pi }_{partial}^{\ast }>0$. Through partial differentiations, we get that the difference $E[{\Pi }_{partial}^{\ast L}]-{\Pi }_{partial}^{\ast }$ increases with n and $1/\sigma _{\varepsilon }^{2},$ but decreases with σ ² and t. In addition, $\frac {\partial E[{\Pi } _{partial}^{\ast L}]-{\Pi }_{partial}^{\ast }}{\partial r}=\frac {\left (\varphi -\varphi ^{L}\right ) \left (2v-\varphi r-\varphi ^{L}r-2\sigma ^{2}r\right ) }{8t}\geq 0$ if r is sufficiently low, i.e., $r\leq \frac {2v}{ 2\sigma ^{2}+\varphi +\varphi ^{L}}$. This implies that the learning tends to be most impactful when the consumers’ risk aversion is in an intermediate range. Define ${\Delta }_{partial}^{L}\equiv E[{\Pi }_{partial}^{\ast L}|_{\varphi r>0}-{\Pi }_{partial}^{\ast L}|_{\varphi r=0}]<0$ and we have $|{\Delta }_{partial}^{L}|<|{\Delta }_{partial}|$. Thus the qualitative result in Proposition 1 remains unchanged, while the decrease of the overall profitability under the partial market coverage becomes less pronounced.

1.4 Proof of Proposition 3

We solve the Stackelberg game through backward induction. In the second stage of the game, store 2’s profit maximization problem is $\max \limits _{p_{2}}q_{2}p_{2}=\frac {1}{2t}(t-\frac {r}{2}\varphi +p_{1}-p_{2})p_{2}$, where the first-order condition is $t-\frac {r}{2} \varphi +p_{1}-2p_{2}=0$. We focus on the full market coverage situation, as the sequence of stores’ moves does not impact the market outcomes under the partial coverage. Hence, we have the best response function for store 2: $p_{2}(p_{1})=\frac {1}{2}(t-\frac {r}{2}\varphi +p_{1})$. In the first stage of the game, store 1 sets its price p ₁ to maximize its profit, taking into account store 2’s best response function:

$$\max\limits_{p_{1}}q_{1}p_{1}=\frac{1}{2t}\left(t+\frac{r}{2}\varphi -p_{1}+p_{2}(p_{1})\right)p_{1}, $$

where the first-order condition is $3t+\frac {r}{2}\varphi -2p_{1}=0$. This leads to the equilibrium price $p_{1}^{\ast }=\frac {3t}{2}+\frac {\varphi r}{4 }$. Hence, we get $p_{2}^{\ast }=\frac {5t}{4}-\frac {\varphi r}{8}$, and then the price premium is $p_{1}^{\ast }-p_{2}^{\ast }=\frac {t}{4}+\frac {3\varphi r}{8}$, which is greater than $\frac {\varphi r}{3}$ (i.e., the price premium under Nash). Consequently, the equilibrium demands are $q_{1}^{\ast }=\frac { 6t+\varphi r}{16t}$ and $q_{2}^{\ast }=\frac {10t-\varphi r}{16t}$.

Notice that the price and demand at store 2 are positive when φ r < 10t under the Stackelberg structure (cf. φ r < 6t under Nash). The demand and price at store 2 are zero under Nash, if φ r≥6t. This implies that the HS is more likely to be financially viable as a follower under the Stackelberg structure than as a competitor under Nash, when φ r is quite high (i.e., 6t ≤ φ r < 10t). Then, we have: (a) the LS has demand advantage $q_{1}^{\ast }-q_{2}^{\ast }=\frac {\varphi r-2t}{ 8t}>0$ if 2t < φ r < 6t, and (b) the HS has demand advantage $q_{1}^{\ast }-q_{2}^{\ast }<0$ if φ r < 2t. For the special case when φ r = 2t, neither store has the advantage. In addition, we have ${\Pi }_{Stackelberg}^{\ast }=$ $\frac {(6t+\varphi r)^{2}}{64t}+\frac {(10t-\varphi r)^{2}}{128t}$, thus we define

$${\Delta}_{S}\equiv {\Pi}_{Stackelberg}^{\ast }|_{\varphi r>0}-{\Pi}_{Stackelberg}^{\ast }|_{\varphi r=0}=\frac{\varphi r(4t+3\varphi r)}{128t}. $$

As we have proved previously that ${\Delta }_{N}\equiv {\Pi }_{Nash}^{\ast }|_{\varphi r>0}-{\Pi }_{Nash}^{\ast }|_{\varphi r=0}=\frac {1}{t}(\frac { \varphi r}{6})^{2}$, ${\Delta }_{S}-{\Delta }_{N}=\frac {\varphi r(36t-5\varphi r) }{1152t}$. Therefore, Δ _S > Δ _N if $\varphi r<\frac {36t}{5}$ . Recall that φ r < 6t for $p_{2}^{\ast }>$ 0 and $q_{2}^{\ast }>0$ under the Nash game. Hence Δ _S > Δ _N for φ r < 6t.