Optimal licensing under incomplete information: the case of the inside patent holder

Abstract

We reconsider the inside patent holders’ optimal licensing problem of non-drastic and (super-) drastic innovations under incomplete information, taking into account restrictions concerning royalty rates and the use of exclusive licenses implied by antitrust rules. We employ methods developed in the analysis of license auctions with downstream interaction and optimal control theory. Our analysis differs from the literature which assumed particular patterns of cost reductions across firms induced by the innovation and either complete information or particular probability distributions.

A Appendix

1.1 A.1 Proof of Proposition 1

As a preliminary, note that it is optimal to adopt the highest possible fixed fee. This allows us to replace the participation constraint (13) by

$$\begin{aligned} f(\hat{x})=\pi _2(q_1^*(\hat{x}), q_2^*(\hat{x});\hat{x}+r(\hat{x}))- \Pi _2^N. \end{aligned}$$

(A.1)

Step 1:   Consider the patentee’s maximization problem:

$$\begin{aligned} \max _{r,f, \hat{x}} \Pi _1= & {} \int _{\underline{x} }^{\hat{x} } \left( \pi _1^*(x) + r(x) q_2^*(x)+f(x)\right) dG(x) \nonumber \\&+ \pi _{1}\left( q_{1}^N, q_{2}^N;d\right) (1-G(\hat{x} )) \end{aligned}$$

(A.2)

s.t. (10), (14), (15), (A.1).

We solve (A.2) as an optimal control problem for given $\hat{x}$, then characterize the optimal $\hat{x}$, and confirm that the ignored constraint is never binding.

To translate (A.2) into an optimal control problem define the control variable, $u(x){:}=r'(x)$, the state variables r(x), f(x), the constraint functions $g_1, g_2, k_1, k_2$, the co-state variables, $\lambda _1(x)$, $\lambda _2(x)$, $\eta _1(x)$, $\eta _2(x)$ and the Hamiltonian, $H(x, r,f,u,\lambda _1, \lambda _2, \eta _1, \eta _2)$:

$$\begin{aligned}&H(x,r,f,u,\lambda _1,\lambda _2,\eta _1, \eta _2): = h(x, r, f) + \lambda _1(x) g_1 + \lambda _2(x) g_2\nonumber \\&\quad +\, \eta _1(x)k_1 + \eta _2(x)k_2 \end{aligned}$$

(A.3)

$$\begin{aligned}&h {:}= \left( \left( \frac{1-2d+x+r(x)}{3}\right) ^2 +r(x) \frac{1-2 x - 2 r(x)+d }{3} +f(x)\right) G'(x)\qquad \qquad \end{aligned}$$

(A.4)

$$\begin{aligned}&g_1{:}= u(x), \quad g_2 {:}= - \frac{1}{9} \left( 1 + d - 2 x - 2 r(x) \right) (1 + 4 u(x)) \end{aligned}$$

(A.5)

$$\begin{aligned}&k_1{:}=r(x), \quad k_2{:}= c- \frac{3 \hat{x}+x}{4}-r(x). \end{aligned}$$

(A.6)

There, $g_2$ corresponds to the incentive compatibility conditions (10), and $k_1, k_2$ to the constraints, (14), (15). The endpoints $f(\hat{x})$ and $r(\hat{x})$ are not free (see A.1) and the endpoints $f(\underline{x})$, $r(\underline{x})$ are free.

The maximum principle requires that the following conditions are satisfied:²²

$$\begin{aligned} 0= & {} \partial _u H = \lambda _1(x) - \frac{4 \lambda _2(x)}{9} \left( 1 + d - 2 x - 2 r(x) \right) \end{aligned}$$

(A.7)

$$\begin{aligned} \lambda _1'(x)= & {} - \partial _r H = -\eta _1+\eta _2 - \frac{2 \lambda _2(x) (1+ 4 u(x)) + (5-d-4 x -10 r(x)) G'(x)}{9} \nonumber \\ \end{aligned}$$

(A.8)

$$\begin{aligned} \lambda _2'(x)= & {} - \partial _{f} H = - G'(x) \quad \end{aligned}$$

(A.9)

$$\begin{aligned} r'(x)= & {} \partial _{\lambda _1} H = u(x) \end{aligned}$$

(A.10)

$$\begin{aligned} f'(x)= & {} \partial _{\lambda _2} H = - \frac{1}{9} \left( 1 + d - 2 x - 2 r(x) \right) (1 + 4 u(x)) \end{aligned}$$

(A.11)

$$\begin{aligned} f(\hat{x})= & {} \left( \frac{1-2(\hat{x}+r(\hat{x}))+d}{3} \right) ^2 - \left( \frac{1-2 c+d}{3} \right) ^2 \end{aligned}$$

(A.12)

$$\begin{aligned} \lambda _1(\underline{x})= & {} \lambda _2(\underline{x}) = 0 \end{aligned}$$

(A.13)

$$\begin{aligned} \eta _1(x)\ge & {} 0, \quad k_1=r(x) \ge 0, \quad \eta _1(x)k_1 = 0. \end{aligned}$$

(A.14)

$$\begin{aligned} \eta _2(x)\ge & {} 0, \quad k_2=c- \frac{3 \hat{x}+x}{4}-r(x) \ge 0, \quad \eta _2(x)k_2 = 0, \end{aligned}$$

(A.15)

Step 2: Deriving r(x). (A.9) and (A.13) yield $\lambda _2(x)= -G(x)$. Inserting this into (A.7) yields

$$\begin{aligned} \lambda _1(x) = -\frac{4 G(x)}{9} \left( 1 + d - 2 x - 2 r(x) \right) . \end{aligned}$$

(A.16)

Differentiating (A.16) and equating with (A.8) yields:

$$\begin{aligned} r(x) = \frac{1-5 d+4 x}{2} + \frac{6 G(x)+9 (\eta _1- \eta _2)}{2 G'(x)}. \end{aligned}$$

(A.17)

Define $\eta {:}=(\eta _1,\eta _2)$ and:

(A.18)

Consider the function . Because , by (18), the equation has a unique solution denoted by $x_1$ and:²³

(A.19)

Similarly, consider the function . Because , the equation has a unique solution denoted by $x_2$ (which may be above $\hat{x}$ or even $\bar{x}$) and:

(A.20)

By the definition of $x_1$ we have $r_0(x_1)=0$; hence

(A.21)

which implies $x_1 < x_2$.

Also we have

(A.22)

(A.23)

We now derive the asserted r(x) function.

r(x) is zero on $[\underline{x},x_1]$, increasing on $[x_1,x_2]$, which confirms that the ignored constraint is not binding there, and decreasing on $[x_2,\hat{x}]$, as asserted.

Step 3: Deriving f(x). Denote the unique solution to the equation $c-x=r_0(x)$ by $x_0$, then we have

(A.24)

We distinguish three cases: Case 1: $\varvec{\hat{x}} \ge \varvec{x}_0$. In this case we have $x_1\le x_2\le \hat{x}$ by (A.24) and $r(\hat{x})=c-\hat{x}$. Thus $f(\hat{x})=0$.

$$\begin{aligned} x \in [x_2, \hat{x}]\Rightarrow & {} 1+4 r'(x) =0 \Rightarrow f'(x)=0 \Rightarrow f(x) = f(\hat{x}) \\&- \int _{x}^{\hat{x}}f'(y)dy= f(\hat{x})= 0\\ x \in [x_1, x_2]\Rightarrow & {} f'(x)=-\frac{1}{9} (1+d-2 x-2 r_0(x)) \left( 1+4 r_0'(x)\right) \\ {}= & {} -q_2^*(x) \left( 3+4 \Big (\frac{G(x)}{G'(x)}\Big )' \right) \\\Rightarrow & {} f(x)=f(x_2)+\int _{x}^{x_2} q_2^*(y) \left( 3+4 \Big (\frac{G(y)}{G'(y)}\Big )' \right) dy \\ {}= & {} \int _{x}^{x_2} q_2^*(y) \left( 3+4 \Big (\frac{G(y)}{G'(y)}\Big )' \right) dy\\ x \in [\underline{x},x_1]\Rightarrow & {} r(x)=r'(x) =0 \Rightarrow f'(x)=- \frac{1}{9} \left( 1 + d - 2 x\right) =-\frac{q_2^*(x)}{3}<0 \\\Rightarrow & {} f(x) = f(x_1) - \int _{x}^{x_1}f'(y)dy\\= & {} f(x_1)+\frac{1}{9}(x_1-x)(1+d-x_1-x). \end{aligned}$$

In this case f(x) is positive and decreasing on $[\underline{x}, x_2)$ and equal to zero on $[x_2,\hat{x}]$.

Case 2: $\varvec{\hat{x}} <\varvec{x}_0~\text {and}~\varvec{\hat{x}}>\varvec{x}_1$. In this case we have $x_1< \hat{x} <x_2$ by (A.24). Thus, $k_2(\hat{x})>0$, which implies $r(\hat{x})<c-\hat{x}$, and $f(\hat{x})=\pi _2(q_1^*(\hat{x}), q_2^*(\hat{x});\hat{x}+r(\hat{x}))- \Pi _2^N>0$.

$$\begin{aligned} x \in [x_1, \hat{x})\Rightarrow & {} f'(x)=-\frac{1}{9} (1+d-2 x-2 r_0(x)) \left( 1+4 r_0'(x)\right) \\= & {} -q_2^*(x) \left( 3+4 \Big (\frac{G(x)}{G'(x)}\Big )' \right) \\\Rightarrow & {} f(x)=f(\hat{x})+\int _{x}^{\hat{x}} q_2^*(y) \left( 3+4 \Big (\frac{G(y)}{G'(y)}\Big )' \right) dy \\ x \in [\underline{x},x_1]\Rightarrow & {} r(x)=r'(x) =0 \Rightarrow f'(x)=- \frac{1}{9} \left( 1 + d - 2 x\right) \\\Rightarrow & {} f(x) = f(x_1) - \int _{x}^{x_1}f'(y)dy\\= & {} f(x_1)+\frac{1}{9}(x_1-x)(1+d-x_1-x). \end{aligned}$$

In this case f(x) is positive and decreasing on $[\underline{x}, \hat{x}]$.

Case 3: $\varvec{\hat{x}} <\varvec{x}_0~\text {and}~\varvec{\hat{x}}\le \varvec{x}_1~(\le \varvec{x}_2)$. In this case we have $f(\hat{x})=\pi _2(q_1^*(\hat{x}), q_2^*(\hat{x});\hat{x}+r(\hat{x}))- \Pi _2^N$.

$$\begin{aligned} x \in [\underline{x},\hat{x}]\Rightarrow & {} r(x)=r'(x) =0 \Rightarrow f'(x)=- \frac{1}{9} \left( 1 + d - 2 x\right) \\\Rightarrow & {} f(x) = f(\hat{x}) - \int _{x}^{\hat{x}}f'(y)dy =f(\hat{x}) +\frac{1}{9}(\hat{x}-x)(1+d-\hat{x}-x). \end{aligned}$$

In this case f(x) is positive and decreasing.

Step 4: Global incentive compatibility and second-order conditions.

Next, we confirm incentive compatibility.²⁴

• (a) For $x,z \ge x_2$, we have $\Pi _2(x,x) = \Pi _2(x,z)=\frac{1}{36} (2-4 c+2 d+3 \hat{x}-3 x)^2$.

• (b) For $x,z \in [x_1, x_2]$, we have $\partial _{zx}\Pi _2(x,z)=\frac{1}{6} \left( 1+4 r'(z)\right) >0$.

• (c) For $x_1\le z<x_2\le x$, we have $\Pi _2(x_2,x_2)-\Pi _2(x_2,z)\ge 0$ from (b) and

  $$\begin{aligned} \left( \Pi _2(x,x)-\Pi _2(x,z)\right) -\left( \Pi _2(x_2,x_2)-\Pi _2(x_2,z)\right) = \frac{2}{3} (x-x_2)k_2(z) \ge 0, \end{aligned}$$

  which implies $\Pi _2(x,x)-\Pi _2(x,z)\ge 0$.

• (d) For $x_1 \le x<x_2\le z$, we have $\Pi _2(x,x)\ge \Pi _2(x,x_2)=\Pi _2(x,z)$.

• (e) For $x,z \le x_1$, we have $\Pi _2(x,z) =\frac{1}{36} (2+2 d-3 x-z)^2-f(x_1)-\frac{1}{9}(x_1-z)(1+d-x_1-z)$. Thus, $\partial _{zx} \Pi _2(x,z) ={1}/{6}>0$.

• (f) For $x\le x_1 <z \le x_2$, we have $\Pi _2(x_1,x_1)-\Pi _2(x_1,z)\ge 0$ from (b) and

  $$\begin{aligned}&\left( \Pi _2(x,x)-\Pi _2(x,z)\right) -\left( \Pi _2(x_1,x_1)-\Pi _2(x_1,z)\right) \\&= \frac{x_1-x}{12}\left( 8r_0(z)+ 2z-x_1-x\right) \ge 0, \end{aligned}$$

  which implies $\Pi _2(x,x)-\Pi _2(x,z)\ge 0$.

• (g) For $z< x_1 \le x \le x_2$, we have $\Pi _2(x,x)\ge \Pi _2(x,x_1)=\Pi _2(x,z) +\frac{x_1-z}{12}\left( 2x-x_1-z\right) > \Pi _2(x,z)$.

• (h) For $x\le x_1, z \ge x_2$, we have $\Pi _2(x_1,x_1)-\Pi _2(x_1,z)\ge 0$ from (d) and

  $$\begin{aligned}&\left( \Pi _2(x,x)-\Pi _2(x,z)\right) -\left( \Pi _2(x_1,x_1)-\Pi _2(x_1,z)\right) \\&= \frac{x_1-x}{12}\left( 8r_2(z)+ 2z-x_1-x\right) \ge 0, \end{aligned}$$

  which implies $\Pi _2(x,x)-\Pi _2(x,z)\ge 0$.

• (i) For $x\ge x_2, z\le x_1$, we have $\Pi _2(x_2,x_2)-\Pi _2(x_2,z)\ge 0$ from (b) and (g) and

  $$\begin{aligned} \left( \Pi _2(x,x)-\Pi _2(x,z)\right) -\left( \Pi _2(x_2,x_2)-\Pi _2(x_2,z)\right) = \frac{x-x_2}{6} (4 c-3 \hat{x}-z) \ge 0, \end{aligned}$$

  which implies $\Pi _2(x,x)-\Pi _2(x,z)\ge 0$.

Therefore, incentive compatibility is satisfied.

The Arrow Sufficiency Theorem (Seierstad and Sydstæter 1977, Theorem 3) applies, because, at the optimum, $\lambda _1 g_1 + \lambda _2 g_2 =0$ and $H^*{:}=\max _u H$ is concave in r and f.

Step 5: $\hat{x}> \underline{x}$. Denote the maximum expected payoff of the patentee by $\Pi _1^*(\hat{x})$. We now prove that a contract is awarded with positive probability by showing

(A.25)

We consider three cases: (i) $\underline{x} \ge x_2$, (ii) $\underline{x} \in [x_1,x_2)$, (iii) $\underline{x}<x_1$.

Case (i): If $\underline{x} \ge x_2$, then $r(\underline{x})=r_2(\underline{x})$ and we have

(A.26)

Case (ii): If $\underline{x} \in [x_1,x_2)$, then $r(\underline{x})=r_0(\underline{x})$ and we have

(A.27)

because the second term in the big bracket is a concave function of d which takes the minimum value 0 at $d=2c-1$ on $[2c-1,c]$.

Case (iii): If $\underline{x} <x_1$, then $r(\underline{x})=0$ and we have

(A.28)

Depending upon the parameters, the optimal $\hat{x}$ is either a corner solution, $\hat{x}= \bar{x}$ (no-exclusion), or an interior solution (exclusion). $\square$

1.2 A.2 Supplement to the proof of Proposition 2

Here, we supplement the proof of Proposition 2 and show that $\Pi _1^*(\hat{x})$ increases on $[x_1,x_2)$.

On the interval $[x_1,x_2)$ one has $r(x)=r_0(x)$. First, suppose $x_1 < \underline{x}$. Then, the patentee’s expected revenue can be written as

$$\begin{aligned} \Pi _1^*(\hat{x})&=\int _{\underline{x}}^{\hat{x}}\Big (r_0(y)q_2^*(y) + f(y) + \pi _1^*(y) \Big ) dG(y) + (1 - G(\hat{x}))\Pi _1^N. \end{aligned}$$

(A.29)

Its derivative with respect to $\hat{x}$ is

$$\begin{aligned} {\Pi _1^*}'(\hat{x})= & {} \frac{1}{36G'(\hat{x})}\Big (36 G(\hat{x})^2-72 (d-\hat{x}) G(\hat{x}) G'(\hat{x}) \Big . \nonumber \\&\Big .+\big (1-20 c^2+25 d^2+8 c (1+4 d)+36 \hat{x}^2-2 d (5+36 \hat{x})\big ) G'(\hat{x})^2\Big )\nonumber \\= & {} \frac{1}{36G'(\hat{x})}\Big (36 \big (G(\hat{x})- (d-\hat{x}) G'(\hat{x}) \big )^2 \nonumber \\&+\,(1 + 10 c - 11 d) (d- 2 c +1) G'(\hat{x})^2 \Big ), \end{aligned}$$

(A.30)

which is positive because $d>2c-1$.

Next, suppose $x_1 \ge \underline{x}$. Then, $r(x)=r_1(x)$ on $[\underline{x},x_1]$, and $\Pi _1$ can be written as

$$\begin{aligned} \Pi _1^*(\hat{x})= & {} \int _{\underline{x}}^{x_1}\Big (r_1(y)q_2^*(y) + f(x_1) + \pi _1^*(y) \Big ) dG(y) \nonumber \\&+ \int _{x_1}^{\hat{x}}\Big (r_0(y)q_2^*(y) + f(y) + \pi _1^*(y) \Big ) dG(y) + (1 - G(\hat{x}))\Pi _1^N.\qquad \quad \end{aligned}$$

(A.31)

The derivative of $\Pi _1^*(\hat{x})$ with respect to $\hat{x}$ is exactly the same as above, which completes the proof. $\square$

1.3 A.3 Supplement to the proof of Proposition 3

By Proposition 1 it is sufficient to show that $x_2 \le \underline{x}$. This holds true if and only if , which confirms as follows:

(A.32)

$\square$