An Anonymous Buyer of Intangible Property

Abstract

In a market for ideas, an intermediary often masks buyers’ information that is not to be exposed to the public. This article shows that such obscuration plays an important role in removing excess inertia because it prevents the emergence of a bandwagon effect among buyers. This model applies to outsourcing-type R&D competition. Although innovation is spurred, welfare implications are ambiguous because the competition among buyers becomes harsher than it is in the absence of the intermediary.

Appendix: Proofs

Lemma 1

Let us define $\phi (a_{1},a_{2}):=U(a_{1},a_{1})+U(a_{2},a_{2})-U(a_{1},a_{2})-U(a_{2},a_{1})$, where $a_{i} \in \{n,w,d\}$ and U is the payoff function of a firm. Then, $\phi (n,w) = \lambda (1 + D - M)$, $\phi (w,d) = (1 - \lambda ) (1-D+\theta _{i}+\rho )$ and $\phi (n,d) = 1+D-M$ are non-negative, which prove supermodularity. In addition, the payoff function has decreasing differences in $(a_{i},\theta _{i})$. Hence, on the basis of Topkis (1978), a lower-type firm employs a higher-order strategy.

Proposition 1

When $\theta _{i} = 1$, n dominates w and d for firm i because $M-1 < D < 1$. Comparing n and w, we can see that $\theta _{i} < D$ chooses w or d. Suppose, first, that some type chooses w or d. For a firm such that n and w are indifferent, $\bar \theta = D$. In addition, $(1-\bar \theta )(M-\underline \theta )+ (\bar \theta -\underline \theta )(D-\underline \theta ) = (1-\underline \theta ) \cdot 1$ if w and d are indifferent. However, the solutions are $\underline \theta < 0$ or $\underline \theta > D = \bar \theta$, which is a contradiction. Next, suppose that w is not chosen. Then, the threshold between n and d occurs at $\theta _{i} = \frac {M-1}{M-D} \in (0,D)$, but in the range between this threshold and D, a firm becomes better off by changing its strategy from n to w. Thus, d is excluded.

Lemma 2

Recall that an interior solution is assumed in Eqs. 1 and 2. These equations being transformed, we see that the thresholds must satisfy

$$\begin{array}{@{}rcl@{}} (1-\bar\theta)(M-\bar\theta-1) + \left[\bar\theta + \underline\theta\left(\frac{1}{\lambda}-1\right)\right](D-\bar\theta) = \left[(1-\underline\theta) + \frac{\underline\theta}{\lambda}\right]\rho \end{array} $$

(3)

$$\begin{array}{rll} (1-\bar\theta) (M-\underline\theta-1) + (\bar\theta-\underline\theta)(D-\underline\theta-1) = (1-\underline\theta)\rho. \end{array} $$

(4)

By subtracting Eq. 4 from Eq. 3, we obtain

$$\begin{array}{*{20}l} \lambda\underline\theta + (1-\lambda)\bar\theta = D-\rho. \end{array} $$

(5)

The value function can be rewritten as

$$\begin{array}{*{20}l} V(\lambda,\rho) = \rho[\underline\theta-(1-\bar\theta)(1+D-M)] \end{array} $$

using Eqs. 4 and 5. For the agent to maximize its profit, $\bar \theta$ and $\underline \theta$ should be large. In addition to this, for any $\rho$, $\bar \theta$ and $\underline \theta$ increase as $\lambda$ increases in Eq. 5. Thus, $\lambda = 1$ in equilibrium. For this proof, an intersection of Eqs. 4 (a quadratic curve that is downwardly convex) and 5 (a line) is checked in the $\underline \theta -\bar \theta$ plane. The line always passes through the point $(D-\rho ,D-\rho )$ located under the rising quadratic curve. A larger. $\lambda$ makes the slope of the line steeper (Fig. 1). The other intersection, $\theta _{i} < 0$, should not be a solution.

Fig. 1

Graphical Illustration for the Proof in Lemma 2

Now, w and d are indifferent for the firms, so we will consider that n and d meet at $\bar \theta = \underline \theta$ determined as follows: $(1-\bar \theta ) M + \bar \theta D - \bar \theta - \rho = (1-\bar \theta ) \cdot 1 + \bar \theta \cdot 0$. By substitution of $\bar \theta$, the optimization problem $\max _{\rho } V|_{\lambda = 1} = \rho \bar \theta$ is transformed into $\max _{\rho } \rho (M-1-\rho )$. The solution is $\rho = \frac {M-1}{2}$.

Proposition 2

By Lemma 2, $\lambda = 1$ and $\rho = \frac {M-1}{2}$. At $\underline \theta$,

$$\begin{array}{*{20}l} \underline\theta \cdot 1 + (1 - \underline\theta) \cdot 0 = \underline\theta M + (1 - \underline\theta) D, \end{array} $$

so the threshold $\underline \theta = \frac {M-1}{2(M-D)}$.