Quantifying and mitigating inefficiency in information acquisition under competition

Abstract

This paper analyzes a horizontally differentiated product market in which firms acquire costly information about the stochastic market. Our results provide guidelines to government agencies on regulating information acquisition. We show that firms overinvest in acquiring information only when information acquisition is particularly cost-effective. Otherwise, underinvestment could occur even under very intense horizontal competition. Moreover, it is underinvestment in information acquisition that is more damaging to firms. Using a linear cost function, we demonstrate that the loss in return on investment caused by horizontal competition can be at least one-third of the first-best return on investment. If the degree of competitive intensity and demand variation decrease, or the marginal cost increases, information acquisition will become increasingly inefficient. We further find that firms benefit from agreeing in advance to exert the same investment level and strategically invest less than the competitive equilibrium level, which can benefit consumers as well. Industry associations are therefore recommended to facilitate effective communication between firms.

Appendix

Proof of Lemma 1

As shown in Raju and Roy (2000), the conditional expectation of $\varepsilon$ given the firm’s forecast can be expressed as follows.

$$\begin{aligned} E\left[ {\varepsilon \left| {{s_i}} \right. } \right] = \left( {1 - \frac{{\sigma _0^2}}{{\sigma _0^2 + \sigma _i^2}}} \right) E\left[ \varepsilon \right] + \frac{{\sigma _0^2}}{{\sigma _0^2 + \sigma _i^2}}{s_i}. \end{aligned}$$

Since $E\left[ \varepsilon \right] =0$ and ${v_i} = {{\sigma _0^2}/{\sigma _i^2}}$, we have $E\left[ {\varepsilon \left| {{s_i}} \right. } \right] = {{{v_i}{s_i}} /{\left( {1 + {v_i}} \right) }}$. In addition, we have

$$\begin{aligned} E\left[ {{s_j}\left| {{s_i}} \right. } \right] = \left( {1 - \frac{{\sigma _0^2 + {s_{ij}}}}{{\sigma _0^2 + \sigma _i^2}}} \right) E\left[ \varepsilon \right] + \frac{{\sigma _0^2 + {s_{ij}}}}{{\sigma _0^2 + \sigma _i^2}}{s_i}, \end{aligned}$$

where $s_{ij}$ is the covariance between forecasts $s_{i}$ and $s_{j}$. Since we assume that the forecast errors of two firms are uncorrelated, i.e., $s_{ij}=0$, we obtain (2). By Ericson (1969), we have (3). The proposed model structure also implies that

$$\begin{aligned} Var\left[ {{s_i}} \right] = E\left[ {s_i^2} \right] = \sigma _0^2 + \sigma _i^2 = {{\sigma _0^2\left( {1 + {v_i}} \right) } /{{v_i}}}. \end{aligned}$$

$\square$

Proof of Lemma 2

Given observed signals $\mathbf{{s}} = \left( {{s_1},{s_2}, \ldots ,{s_n}} \right)$ and investment levels $\mathbf{{v}} = \left( {{v_1},{v_2}, \ldots ,{v_n}} \right)$, the centralized system’s expected profit can be transformed as

$$\begin{aligned} E\left[ {{\varPi ^{I}}\left| \mathbf{{s},\mathbf {v}} \right. } \right]= & {} \left( {{K_0} + E\left[ {\varepsilon \left| \mathbf{{s},\mathbf {v}} \right. } \right] } \right) Q - b{Q^2} \\&- \left( {1 - b} \right) \left( {q_1^2 + q_2^2 + \cdots + q_n^2} \right) - \sum \limits _{i = 1}^n {C\left( {{v_i}} \right) }, \end{aligned}$$

where $Q=q_{1}+q_{2}+\cdots +q_{n}$. Based on average value inequality, we have $q_1^2 + q_2^2 + \cdots + q_n^2 \ge \frac{{{Q^2}}}{n}$. The equality holds if and only if $q_{1}=q_{2}=\cdots =q_{n}$. Given Q, $\mathrm{{E}}\left[ {{\varPi ^{I}}\left| \mathbf{{s},\mathbf {v}} \right. } \right]$ reaches the maximum value if and only if $q_{1}=q_{2}=\cdots =q_{n}$. $\square$

Proof of Proposition 1

We have

$$\begin{aligned} E\left[ {{\varPi ^I}} \right]= & {} E\left[ {E\left[ {{\varPi ^I}\left| \mathbf{{s}} \right. } \right] } \right] = \frac{{K_0^2}}{{4\left( {b + \frac{{1 - b}}{n}} \right) }} + \frac{{\sigma _0^2}}{{4\left( {b + \frac{{1 - b}}{n}} \right) }} \cdot \frac{{\sum \nolimits _{i = 1}^n {{v_i}} }}{{1 + \sum \nolimits _{i = 1}^n {{v_i}} }} \nonumber \\&- \sum \nolimits _{i = 1}^n {C\left( {{v_i}} \right) }. \end{aligned}$$

(A1)

Following the same steps in Shin and Tunca (2010), we find that in the optimum, $v_{i}=v_{j}=v^{I}$ for all i, j. Consequently, we obtain

$$\begin{aligned} E\left[ {{\varPi ^I}} \right] = \frac{{K_0^2}}{{4\left( {b + \frac{{1 - b}}{n}} \right) }} + \frac{{\sigma _0^2}}{{4\left( {b + \frac{{1 - b}}{n}} \right) }} \cdot \frac{{n{v^I}}}{{1 + n{v^I}}} - nC\left( {{v^I}} \right) . \end{aligned}$$

(A2)

Thus, taking the first derivative of (A2) with respect to v, we obtain (7). $\square$

Proof of Proposition 2

This proof follows the same steps in Shin and Tunca (2010). Let ${q_j}\left( {{s_j}} \right) = {\alpha _{0j}} + {\alpha _{sj}}{s_j}$, ${\alpha _{0j}},{\alpha _{sj}} \in R$, $v_{i} \in R_{+}$, for all $i\ne j$. Taking expectation given $s_{i}$, we obtain expected profit for firm i after observing $s_{i}$

$$\begin{aligned} E\left[ {\pi _i^D\left| {{s_i}} \right. } \right] = {q_i}\left( {{K_0} + \frac{{{v_i}{s_i}}}{{1 + {v_i}}} - {q_i} - b\sum \limits _{j \ne i} {\left( {{\alpha _{0j}} + {\alpha _{sj}}E\left[ {{s_j}\left| {{s_i}} \right. } \right] } \right) } } \right) - C\left( {{v_i}} \right) .\nonumber \\ \end{aligned}$$

(A3)

Note that the second order condition for $q_{i}$ indicates that there is a unique equilibrium on production quantity. Using (2), the first order condition for $q_{i}$ is written as

$$\begin{aligned} {q_i} = \frac{1}{2}\left( {{K_0} - b\sum \limits _{j \ne i} {{\alpha _{0j}}} + \frac{{{v_i}{s_i}}}{{1 + {v_i}}}\left( {1 - b\sum \limits _{j \ne i} {{\alpha _{sj}}} } \right) } \right) . \end{aligned}$$

(A4)

Submitting (A4) into (A3) and taking expectation, we obtain

$$\begin{aligned} E\left[ {\pi _i^D} \right] = \frac{1}{4}{\left( {{K_0} - b\sum \limits _{j \ne i} {{\alpha _{0j}}} } \right) ^2} + \frac{{\sigma _0^2{v_i}}}{{4\left( {1 + {v_i}} \right) }}{\left( {1 - b\sum \limits _{j \ne i} {{\alpha _{sj}}} } \right) ^2} - C\left( {{v_i}} \right) . \end{aligned}$$

(A5)

The first order condition for $v_{i}$ from (A5) is $\frac{{\sigma _0^2}}{{4{{\left( {1 + {v_i}} \right) }^2}}}{\left( {1 - b\sum \limits _{j \ne i} {{\alpha _{sj}}} } \right) ^2} - C'\left( {{v_i}} \right)$, and the second order condition $\frac{{ - \sigma _0^2}}{{2{{\left( {1 + {v_i}} \right) }^3}}}{\left( {1 - b\sum \limits _{j \ne i} {{\alpha _{sj}}} } \right) ^2} - C''\left( {{v_i}} \right) < 0$. From (A4) and the first order condition of $v_{i}$, we have

$$\begin{aligned} {\alpha _{0i}} = \frac{1}{2}\left( {{K_0} - b\sum \limits _{j \ne i} {{\alpha _{0j}}} } \right) , {\alpha _{si}} = \frac{{{v_i}}}{{2\left( {1 + {v_i}} \right) }}\left( {1 - b\sum \limits _{j \ne i} {{\alpha _{sj}}} } \right) , \end{aligned}$$

(A6)

and

$$\begin{aligned} C'\left( {{v_i}} \right) = \frac{{\sigma _0^2}}{{4{{\left( {1 + {v_i}} \right) }^2}}}{\left( {1 - b\sum \limits _{j \ne i} {{\alpha _{sj}}} } \right) ^2}. \end{aligned}$$

(A7)

By summing over $\alpha _{0i}$ for all i, $\sum \nolimits _i {{\alpha _{\mathrm{{0}}i}}} = {{n{K_\mathrm{{0}}}} /{\left( {\mathrm{{2 + }}nb - b} \right) }}$. Substituting this into (A6), and simplifying, we obtain ${\alpha _{0i}} = {\alpha _0} = \frac{{{K_0}}}{{2 + b\left( {n - 1} \right) }}$. From (A6), it follows

$$\begin{aligned} {\alpha _{si}} = \frac{{{v_i}\left( {1 - b\sum \nolimits _{j = 1}^n {{\alpha _{sj}}} } \right) }}{{2\left( {1 + {v_i}} \right) - b{v_i}}}. \end{aligned}$$

(A8)

Substituting (A8) into (A7), we have

$$\begin{aligned} C'\left( {{v_i}} \right) = \frac{{\sigma _0^2{{\left( {1 - b\sum \nolimits _{j = 1}^n {{\alpha _{sj}}} } \right) }^2}}}{{{{\left( {2\left( {1 + {v_i}} \right) - b{v_i}} \right) }^2}}}, \end{aligned}$$

(A9)

for all i. Observe that since (A9) holds for all i, the first order condition for all $v_{i}$ is identical. Further, since the difference between the two sides of (A9) is strictly monotonic for $v_{i}>0$, it can have at most one solution in $v_{i}$. Thus, $v_{i}=v$, for some $v\ge 0$, for all i. Adding up (A8) for all i, and plugging in $v_{i}=v^{D}$, we then have

$$\begin{aligned} \sum \limits _{i = 1}^n {{\alpha _{si}}} = \frac{{n{v^D}}}{{2 + \left( {2 + \left( {n - 1} \right) b} \right) {v^D}}}. \end{aligned}$$

(A10)

Substituting (A10) into (A6) and simplifying, we obtain ${\alpha _{si}} = {\alpha _{s}} = \frac{{{v^D}}}{{2 + \left( {2 + \left( {n - 1} \right) b} \right) {v^D}}}$. Finally, substituting (A10) into (A9), we obtain (8). Since $C(\cdot )$ is convex, non-decreasing and non-identically zero, the left hand side of (8) is decreasing in v and becomes strictly negative as $v \rightarrow + \infty$. Consequently, there exists an unique v that satisfies (8) if and only if $C'(0)<\sigma _{0}^{2}/4$, with $v^{D}=0$ otherwise. This confirms $v^{D}$ and completes the proof. $\square$

Proof of Proposition 3

It can be verified that $v^{I}<v^{D}$ in the case that the products are perfectly substitutes, i.e., $b=1$. From Propositions 1 and 2, firm’s investment level v can be viewed as a function of b. It is easy to verify that

$$\begin{aligned} \frac{\partial }{{\partial b}}\left( {4\left( {b + \frac{{1 - b}}{n}} \right) {{\left( {1 + nv} \right) }^2}} \right) > \frac{\partial }{{\partial b}}{\left( {2 + \left( {2 + nb - b} \right) v} \right) ^2}. \end{aligned}$$

Thereby it can be proved that $\left| {\frac{{\partial {v ^I}}}{{\partial b}}} \right| > \left| {\frac{{\partial {v ^D}}}{{\partial b}}} \right|$. Actually, this is the proof of Lemma 3. This means that the level of investment in information acquisition in horizontally centralized setting would fall more dramatically than in the decentralized setting with the increase of substitutability. When $v _0^{I} \le v _0^{D}$, $v^{I}(b) \le v^{D}(b)$ always holds for all $b\in (0,1)$. On the other hand, if $v _0^{I} > v _0^{D}$, there must exist an unique ${b^*} \in \left( {0,1} \right)$ that makes ${v ^{I}}\left( {{b^*}} \right) = {v ^{D}}\left( {{b^*}} \right)$. This completes the proof. $\square$

Proof of Proposition 4

When $C(v)=c_{f}\cdot v$, by Propositions 1 and 2, we have

$$\begin{aligned} {v^I}= & {} \sqrt{\frac{{\sigma _0^2}}{{4n{c_f}\left( {1 + nb - b} \right) }}} - \frac{1}{n}, \end{aligned}$$

(A11)

$$\begin{aligned} {v^D}= & {} \frac{2}{{2 + nb - b}}\left( {\sqrt{\frac{{\sigma _0^2}}{{4{c_f}}}} - 1} \right) . \end{aligned}$$

(A12)

We find that

$$\begin{aligned} \frac{{{\varDelta ^D}}}{{{\varDelta ^I}}}\mathrm{{ = }}{\left( {\frac{{{v^D}}}{{{v^I}}}} \right) ^2} \cdot \frac{1}{n}. \end{aligned}$$

(A13)

From (A13), the result in part (i) can be obtained.

For part (ii), it is easy to verify that ${{{v^D}} /{{v^I}}}$ increases with b by Lemma 3. So we have ${{{v^D}} /{{v^I}}} \ge {{v_0^D} /{v_0^I}}$. Let $G = \sqrt{{{C'\left( {v_0^D} \right) } /{C'\left( {v_0^I} \right) }}}$. Then we have

$$\begin{aligned} \frac{{v_0^D}}{{v_0^I}} = \frac{{\sqrt{n} }}{G} - \frac{1}{{v_0^I}} + \frac{1}{{\sqrt{n} Gv_0^I}}. \end{aligned}$$

(A14)

When $v_0^D \le v_0^I$, $0 < G \le 1$. ${{v_0^D} /{v_0^I}}$ is minimized at $G=1$, which is attainable for $C\left( v \right) = {c_f}v$, $c_{f}>0$. Plugging in (A14), we obtain the lower bound given in (13). Following similar steps, the upper bound can be derived.

Finally,

$$\begin{aligned}&\frac{\partial }{{\partial \sigma _\mathrm{{0}}^\mathrm{{2}}}}\left( {\frac{{{v^D}}}{{{v^I}}}} \right) \\&\quad = \frac{{2n\sqrt{\frac{{\sigma _0^2}}{{{c_f}}}} \left( { - b(n - 1)\sqrt{\frac{{\sigma _0^2}}{{b{c_f}{n^2} - b{c_f}n + {c_f}n}}} - \sqrt{\frac{{\sigma _0^2}}{{b{c_f}{n^2} - b{c_f}n + {c_f}n}}} + \sqrt{\frac{{\sigma _0^2}}{{{c_f}}}} } \right) }}{{\sigma _0^2(b(n - 1) + 1)(b(n - 1) + 2)\sqrt{\frac{{\sigma _0^2}}{{b{c_f}{n^2} - b{c_f}n + {c_f}n}}} {{\left( {n\sqrt{\frac{{\sigma _0^2}}{{b{c_f}{n^2} - b{c_f}n + {c_f}n}}} - 2} \right) }^2}}}. \end{aligned}$$

Note that

$$\begin{aligned} \frac{\partial }{{\partial b}}\left( { - b(n - 1)\sqrt{\frac{{\sigma _0^2}}{{b{c_f}{n^2} - b{c_f}n + {c_f}n}}} - \sqrt{\frac{{\sigma _0^2}}{{b{c_f}{n^2} - b{c_f}n + {c_f}n}}} + \sqrt{\frac{{\sigma _0^2}}{{{c_f}}}} } \right) < 0 \end{aligned}$$

and

$$\begin{aligned} - b(n - 1)\sqrt{\frac{{\sigma _0^2}}{{b{c_f}{n^2} - b{c_f}n + {c_f}n}}} - \sqrt{\frac{{\sigma _0^2}}{{b{c_f}{n^2} - b{c_f}n + {c_f}n}}} + \sqrt{\frac{{\sigma _0^2}}{{{c_f}}}} = 0 \end{aligned}$$

when $b=1$. Thereby it can be proved that $\frac{\partial }{{\partial \sigma _0^2}}\left( {\frac{{{v^D}}}{{{v^I}}}} \right) > 0$, which means that $\varPhi ^{D}/\varPhi ^{I}$ increases with $\sigma _{0}^{2}$. Other results in part (iii) can be obtained by following the similar steps. $\square$

Proof of Proposition 5

From (16), we have

$$\begin{aligned} {\left. {\frac{{d{\pi ^S}}}{{dv}}} \right| _{v = {v^D}}} = \frac{{\sigma _0^2\left( {2 + \left( {2 + b - nb} \right) {v^D}} \right) }}{{{{\left( {2 + \left( {2 + nb - b} \right) {v^D}} \right) }^3}}} - C'\left( {{v^D}} \right) . \end{aligned}$$

(A15)

Substituting (8) into (A15) and simplifying, we obtain

$$\begin{aligned} {\left. {\frac{{d{\pi ^S}}}{{dv}}} \right| _{v = {v^D}}} = \frac{{2b\sigma _0^2\left( {1 - n} \right) {v^D}}}{{{{\left( {2 + \left( {2 + nb - b} \right) {v^D}} \right) }^3}}} < 0. \end{aligned}$$

$\square$

Proof of Proposition 6

Substituting (1) into (17), consumer surplus can be rewritten as $CS = \frac{1}{2}\left( {q_1^2 + \cdots + q_n^2} \right) + b\sum \limits _{j \ne i} {{q_i}{q_j}}$. Based on the results above, if firms exert a fixed investment level v in advance, the expected consumer surplus is

$$\begin{aligned} ECS(v) = \frac{{n\left( {2 + nb - b} \right) }}{4}\left( {\frac{{K_0^2}}{{{{\left( {2 + nb - b} \right) }^2}}} + \frac{{v\left( {1 + v} \right) \sigma _0^\mathrm{{2}}}}{{{{\left( {\mathrm{{2}} + \left( {\mathrm{{2}} + nb - b} \right) v} \right) }^2}}}} \right) . \end{aligned}$$

(A16)

From Proposition 2, the expected consumer surplus in horizontally decentralized setting equals $ECS(v^{D})$. Furthermore, we have

$$\begin{aligned} {\left. {\frac{d}{{dv}}ECS\left( v \right) } \right| _{v = {v^D}}} = \frac{{n\left( {2 + nb - b} \right) }}{4} \cdot \frac{{\sigma _0^\mathrm{{2}}\left( {2 + \left( {2 + b - nb} \right) {v^D}} \right) }}{{{{\left( {\mathrm{{2}} + \left( {\mathrm{{2}} + nb - b} \right) {v^D}} \right) }^3}}}. \end{aligned}$$

It is easy to verified that $\frac{{dECS\left( v \right) }}{{dv}}\left| {_{v = {v^D}}} \right.< 0 \Leftrightarrow {2 + \left( {2 + b - nb} \right) {v^D}<0}$. Therefore, expected consumer surplus can also be improved through employing the approach provided by Proposition 5. $\square$