Licensing a common value innovation when signaling strength may backfire

Abstract

This paper reconsiders the licensing of a common value innovation to a downstream duopoly, assuming firms observe imperfect signals of the cost reduction induced by the innovation. The innovator adopts a direct revelation mechanism and awards an unrestricted license to the firm that reports the highest signal and a royalty contract to the other. Firms may signal strength to their rivals through exaggerated messages, which may however backfire, and give rise to higher royalty payments. We provide sufficient conditions for truthful implementation, and for the profitability of adding royalty contracts to what is otherwise a first-price license auction.

Appendix

1.1 Supplement to the proofs of Lemma 1 and Proposition 2

1.1.1 Proof of Step a) in Lemma 1

Proof

By definition, $q_{1_b}(x,z,y)$ satisfies the first-order condition of the best-reply problem (11) for all signals, which yields the identity in $x,z, y$:

$$\begin{aligned} P'(q_{1_b}(x,z,y)+q_{2_b}(y))q_{1_b}(x,z,y)+P(\cdot )-c+\theta (y,x)-\theta (y,z) \equiv 0. \end{aligned}$$

Differentiating w.r.t. $z$ and $x$, respectively, gives

$$\begin{aligned} \left( 2P'(\cdot )+P^{''}(\cdot ) q_{1_b}(\cdot )\right) \partial _zq_{1_b}(\cdot ) -\partial _z \theta (y,z)&= 0\\ \left( 2P'(\cdot )+P^{''}(\cdot ) q_{1_b}(\cdot )\right) \partial _xq_{1_b}(\cdot ) +\partial _x \theta (y,x)&= 0. \end{aligned}$$

Adding the two equations and setting $z=x$ gives

$$\begin{aligned} \left( 2P'(\cdot )+P^{''}(\cdot ) q_{1_b}(\cdot )\right) \left( \partial _xq_{1_b}(\cdot )+ \partial _zq_{1_b}(\cdot )\right) = 0. \end{aligned}$$

Since marginal revenue is decreasing, $(2P'(\cdot )+P^{''}(\cdot ) q_{1_b}(\cdot ))<0$; hence, $\partial _xq_{1_b}(\cdot )+ \partial _zq_{1_b}(\cdot )=0$. Using this result together with (28), we conclude:

$$\begin{aligned} \begin{aligned} \partial _{zx} \left. \Pi _b(x,z,y)\right| _{z=x}+\partial _{zz} \left. \Pi _b(x,z,y)\right| _{z=x}&= - (1-t) \left( \partial _xq_{1_b}(\cdot )+ \partial _zq_{1_b}(\cdot )\right) \\&= 0, \end{aligned} \end{aligned}$$

as asserted.$\square$

1.1.2 Monotonicity of $\mathbf \beta _n$ (Proposition 2)

Here we show that assumptions (4), (5) concerning the $P$ and the $F$ functions assure the monotonicity of $\beta , \beta _n$.

We show, in several steps, that $\frac{d}{dx}\left( \frac{F(x)}{f(x)}\left. \partial _z \Pi _a(x,z)\right| _{z=x}\right) >0$.

Proof

1) For the assumed probability distributions one can check

$$\begin{aligned}&\frac{f'(x)}{f(x)}-\frac{f(x)}{F(x)}-\left( \frac{f(x)}{F(x)}-\frac{F(x)}{\int _{d}^{x}F(y)dy}\right) \end{aligned}$$

(31)

$$\begin{aligned}&=\frac{\lambda \varepsilon ^\lambda }{x-d+\varepsilon (c-d)}\frac{((\lambda +1)u+2\varepsilon )\epsilon ^\lambda +((\lambda -1)u-2\varepsilon )(u+\epsilon )^\lambda }{((u+\varepsilon )^\lambda -\varepsilon ^\lambda )((u+\varepsilon )^{\lambda +1}-((\lambda +1)u\!+\!\varepsilon )\epsilon ^\lambda )}\!\ge \! 0,\qquad \quad \end{aligned}$$

(32)

which is equivalent to $\mu ''(x)\ge 0$ as we will show below.¹⁷

Since $\mu '(x)=\frac{f(x)}{F(x)}(x-\mu (x))$ is positive, $\mu ''(x)\ge 0$ if and only if $\frac{d}{dx}\ln \mu '(x) \ge 0$. Because

$$\begin{aligned} \mu '(x)=\frac{f(x)}{F(x)}(x-\mu (x))=\frac{f(x)}{F(x)}\left( \frac{1}{F(x)}\int \limits _{d}^{x}F(y)dy\right) , \end{aligned}$$

$\ln \mu '(x)=\ln f(x)-\ln F(x)-(\ln F(x)-\ln \int _{d}^{x}F(y)dy)$. Therefore, nonnegativity of $\mu ''(x)$ is equivalent to nonnegativity of (31).

2) By the assumed log-concavity of the distribution function, $F(x)/f(x)$ is strictly increasing. Therefore, we only need to show that $\left. \partial _z \Pi _a(x,z)\right| _{z=x}$ is increasing.

By (9), $\Pi _{a}(x,z) = \pi _1\left( q_{1_a}(x,z), q_{2_a}(z);c-\theta _1(x,z)\right)$, and by the envelope theorem,

$$\begin{aligned} \left. \partial _{z} \Pi _a(x,z)\right| _{z=x}=\left( \left. \left( P'(\cdot ) q_{2_a}'(z)+ \partial _z \theta _1(x,z)\right) q_{1_a}(x,z)\right) \right| _{z=x}. \end{aligned}$$

Since $\left. q_{1_a}(x,z)\right| _{z=x}$ and $\left. \partial _z \theta _1(x,z)\right| _{z=x} = (1-t)\mu '(x)$ are increasing, it is sufficient to show that $(P'(\cdot ) q_{2_a}'(z) q_{1_a}(x,z))|_{z=x}$ is increasing.

3) The first-order conditions of the oligopoly subgame $\Gamma _a=\Gamma \left( c-\theta (z,\mu (z)),c\right)$ are nonlinear. We employ the following algorithm to compute the solution of this subgame $(q_1(z), q_{2_a}(z))$:

Adding up the two first-order conditions gives $2(1-c)-(2+\alpha ) Q^\alpha +\theta (z,\mu (z)) = 0,$ where $\theta (z,\mu (z))= t z +(1-t) \mu (z)$. Therefore, the aggregate equilibrium output is

$$\begin{aligned} Q(z)=\left( \frac{2(1-c)+\theta (z,\mu (z))}{2+\alpha }\right) ^{\frac{1}{\alpha }}. \end{aligned}$$

Plug $Q(z)$ into the first-order conditions for $q_{2_a}(z)$, and then set $z=x$. This gives

$$\begin{aligned} q_{2_a}(x)&= \frac{\alpha (1-c) -\theta (x,\mu (x))}{\alpha \left( 2(1-c)+\theta (x,\mu (x))\right) }Q(x)\\ q_{1_a}(x,x)&= \frac{\alpha (1-c) +(1+\alpha )\theta (x,\mu (x))}{\alpha \left( 2(1-c)+\theta (x,\mu (x))\right) }Q(x). \end{aligned}$$

4) Since $P'(Q(x))=-\alpha Q(x)^{\alpha -1}$, we have:

$$\begin{aligned}&\frac{d}{dx} \left( P'(Q(x))q_{2_a}'(x)q_{1_a}(x,x)\right) =Q(x)^{1+\alpha }\alpha ^{-3}\left( 2(1-c)+\theta (x,\mu (x))\right) ^{-4}\\&\qquad \times \left( \left( a_0+a_1 \theta +a_2 \theta ^2 \right) \left( t+(1-t)\mu '(x)\right) ^2 \!+\! \left( b_0 \!+\! b_1 \theta \!+\!b_2 \theta ^2 \right) (1-t)\mu ''(x)\right) \!,\qquad \end{aligned}$$

where $\theta$ is a short-hand expression for $\theta (x,\mu (x))$, and the coefficients $a_i$ and $b_i$ are constants with $b_i>0$, $a_0=\alpha ^2(5+3\alpha )(1-c)^2$, $a_1=-\alpha (\alpha ^3+\alpha ^2-4\alpha -6)(1-c)$, and $a_2=1+\alpha$. One can show that $a_0+a_1 \theta +a_2 \theta ^2>0$ for $\theta \in [d,c]$ and for $c<.5$ if $\alpha <4.4$.$\square$

1.2 Proof of Lemma 3, Part 1)

1) We show that $\partial _{zx}\bar{\Pi }_{\text {up} }^n(x,z)\ge 0$, as follows:

$$\begin{aligned} \begin{aligned} \partial _{zx}\bar{\Pi }_{\text {up} }^n(x,z)&=\partial _{zx} \Pi _{\text {up} }^n(x,z)\\&= \partial _x \left( \partial _z \Pi _a(x,z)F(z)+ \left( \Pi _a(x,z)-\left. \Pi _b^n(y)\right| _{y=z} \right) f(z) \right) \\&= F(z) \partial _{zx}\Pi _a(x,z) +f(z) \partial _x \Pi _a(x,z). \end{aligned} \end{aligned}$$

Using the envelope theorem and the fact that $q_{2_a}$ is independent of $x$, one has

$$\begin{aligned} \partial _x \Pi _a(x,z)&= q_{1_a}(x,z)\partial _x \theta _1(x,z). \end{aligned}$$

Note that

$$\begin{aligned} \partial _x \theta _1(x,z) = (2t-1) \frac{F(x)}{F(z)} + 1-t>0 \quad \text {since } t\ge 1/2. \end{aligned}$$

(33)

Therefore,

$$\begin{aligned} \begin{aligned} \partial _{zx}\Pi _a(x,z)&= \partial _z q_{1_a}(x,z)\partial _x \theta _1(x,z) + q_{1_a}(x,z) \partial _{zx}\theta _1(x,z) \\&> q_{1_a}(x,z) \partial _{zx}\theta _1(x,z) \quad \text {(as explained below)}\\&= -q_{1_a}(x,z) (2t-1)\frac{F(x)f(z)}{F(z)^2}, \end{aligned} \end{aligned}$$

where the inequality follows from (33) together with the fact that $\partial _z q_{1_a}(x,z)>0$ (the loser reduces his output when the winner has signalled strength, and the winner takes advantage of this by increasing his own output).

Combining the above, gives

$$\begin{aligned} \begin{aligned} \partial _{zx}\bar{\Pi }_{\text {up} }^n(x,z)&> q_{1_a}(x,z)f(z) \left( -(2t-1)\frac{F(x)}{F(z)} + (2t-1)\frac{F(x)}{F(z)} +1-t\right) \\&= q_{1_a}(x,z)f(z) (1-t) >0. \end{aligned} \end{aligned}$$

2) Similarly, we show that $\partial _{zx}\bar{\Pi }_{\text {dn} }^n(x,z)\ge 0$, as follows:

$$\begin{aligned} \begin{aligned} \partial _{zx}\bar{\Pi }_{\text {dn} }^n(x,z)&= F(z) \partial _{zx}\Pi _c(x,z) +f(z) \partial _x \Pi _c(x,z). \end{aligned} \end{aligned}$$

Using the envelope theorem and the fact that $q_{2_c}$ is independent of $x$, one has

$$\begin{aligned} \begin{aligned} \partial _x \Pi _c(x,z)&= q_{1_c}(x,z) \partial _x \theta (x,\mu (z)) \\&= q_{1_c}(x,z) t>0. \end{aligned} \end{aligned}$$

Therefore,

$$\begin{aligned} \partial _{zx} \Pi _{c}(x,z) = t \partial _z q_{1_c}(x,z) >0. \end{aligned}$$

Combining the above confirms that $\partial _{zx}\bar{\Pi }_{\text {dn} }^n(x,z)>0$ everywhere.

1.3 Proof of Lemma 3, Part 2)

By definition of $\Pi _{\text {up} }, \Pi _{\text {up} }^n$ one has

$$\begin{aligned} \Pi _{\text {up} }(x,z)&= \Pi _{\text {up} }^n(x,z)+\int \limits _{z}^{c} \left( \Pi _b(x,z,y)-\Pi _b^n(y)\right) dF(y). \end{aligned}$$

Therefore, using $\Pi _{\text {up} }^n$ and the fact that $\partial _x\Pi _b(x,z,y)=(1-t)q_{1_b}(x,z,y)$,

$$\begin{aligned} \partial _{zx}\bar{\Pi }_{\text {up} }(x,z)&= \partial _{zx} \Pi _{\text {up} }(x,z)\\&= \partial _z\left( \partial _x\Pi _{\text {up} }(x,z)\right) \\&= \partial _z\left( \partial _x\Pi _{\text {up} }^n(x,z) + \int \limits _{z}^{c} \left( \partial _x \Pi _b(x,z,y)-\partial _x \Pi _b^n(y) \right) dF(y) \right) \\&= \partial _z\left( \partial _x\Pi _{\text {up} }^n(x,z) + (1-t)\int \limits _{z}^{c} q_{1_b}(x,z,y) dF(y) \right) \\&= \partial _{zx}\Pi _{\text {up} }^n(x,z) \!+\!(1-t)\left( \int \limits _{z}^{c} \partial _z q_{1_b}(x,z,y)dF(y) \!-\! q_{1_b}(x,z,z)f(z)\right) \\&= (1-t)q_{1_a}(x,z)f(z) + \left( (2t-1)\frac{F(x)}{F(z)}+1-t\right) F(z)\partial _z q_{1_a}(x,z) \\&- (1-t)q_{1_b}(x,z,z)f(z)+(1-t)\int \limits _{z}^{c} \partial _z q_{1_b}(x,z,y)dF(y)\\&= (1-t)f(z)\left( q_{1_a}(x,z)-q_{1_b}(x,z,z)\right) \\&+ \left( (2t-1)\frac{F(x)}{F(z)}+1-t\right) F(z)\partial _z q_{1_a}(x,z) \\&+(1-t)^2\int \limits _{z}^{c} \frac{1}{2P'(\cdot )+P^{''}(\cdot )q_{1_b}}dF(y) \end{aligned}$$

To explain the last step, which involves substituting $\partial _z q_{1_b}(\cdot )$, recall the first-order condition concerning $q_{1_b}$:

$$\begin{aligned} P\left( q_{1_b}(x,z,y)+q_{2_b}(y)\right) + P'(\cdot )q_{1_b}(\cdot )-c+\theta (y,x)-\theta (y,z) = 0. \end{aligned}$$

Differentiating w.r.t. $z$ gives

$$\begin{aligned} \partial _z q_{1_b}(\cdot ) = \frac{1-t}{2P'(\cdot )+P^{''}(\cdot )q_{1_b}(\cdot )}. \end{aligned}$$

Obviously, if $t=1$ the above cross derivative is positive. Whereas if $t<1$, it cannot be positive everywhere for $d=0$. Because if $d=0$, one finds that the first two terms in the last equation vanish as $x$ and $z$ approach zero, whereas the last term is negative, so that $\partial _{zx}\bar{\Pi }_{\text {up} }(x,z)<0$.

If $d$ is positive the above cross derivative is positive for all sufficiently large $t$, because in that case the effect of an upward deviation on the royalty rate (i.e., the last term) becomes insignificant, whereas the output difference between winner and loser does not vanish relative to the last term as $t$ is increased towards 1. This shows that the second order conditions are satisfied if $d$ is positive and $t$ is sufficiently close to 1.

Next consider downward deviations, $z\le x$ and the associated payoff function $\Pi _{\text {dn} }(x,z)$. By definition of $\Pi _{\text {dn} }, \Pi _{\text {dn} }^n$ one has

$$\begin{aligned} \Pi _{\text {dn} }(x,z) \!=\! \Pi _{\text {dn} }^n(x,z)\!-\!\!\int \limits _{z}^{c}\!\Pi _d^n(y)dF(y)\!+\!\!\int \limits _{z}^{x}\! \Pi _d(x,z,y)dF(y)\!+\!\!\int \limits _{x}^{c} \!\Pi _e^n(x,z,y)dF(y). \end{aligned}$$

Therefore, using $\Pi _{\text {dn} }^n$ and the facts that $\partial _x\Pi _d(x,z,y)\!=\!tq_{1_d}(x,z,y)$, and $\partial _x\Pi _e(x,z,y)=(1-t)q_{1_e}(x,z,y)$, one has

$$\begin{aligned}&\partial _{zx}\bar{\Pi }_{\text {dn} }(x,z) =\partial _{zx} \Pi _{\text {dn} }(x,z)\\&\qquad = \partial _z\left( \partial _x\Pi _{\text {dn} }(x,z)\right) \\&\qquad = \partial _z\left( \! \partial _x\Pi _{\text {dn} }^n(x,z) \!+\!\! \int \limits _{z}^{x}\! \partial _x \Pi _d(x,z,y)dF(y)\!+\!\! \int \limits _{x}^{c} \partial _x \Pi _e(x,z,y) dF(y)\!\! \right) \\&\qquad = \partial _{zx}\Pi _{\text {dn} }^n(x,z) - tq_{1_d}(x,z,z)f(z) + t\int \limits _{z}^{x} \partial _z q_{1_d}(x,z,y)dF(y) \\&\qquad \quad +(1-t)\int \limits _{x}^{c} \partial _z q_{1_e}(x,z,y)dF(y) \\&\qquad = tF(z)\partial _z q_{1_c}(x,z)+tq_{1_c}(x,z)f(z) - tq_{1_d}(x,z,z)f(z) \\&\qquad \quad + t\int \limits _{z}^{x} \partial _z q_{1_d}(x,z,y)dF(y) +(1-t)\int \limits _{x}^{c} \partial _z q_{1_e}(x,z,y)dF(y) \\&\qquad = tF(z)\partial _z q_{1_c}(x,z)\!+\!tf(z)\left( q_{1_c}(x,z) \!-\! q_{1_d}(x,z,z)\right) \\&\qquad \quad + (1\!-\!t)\left( \int \limits _{z}^{x}\! \frac{t}{2P'(\cdot )\!+\!P^{''}(\cdot )q_{1_d}} dF(y) \!+\!\int \limits _{x}^{c}\! \frac{1-t}{2P'(\cdot )\!+\!P^{''}(\cdot )q_{1_e}}dF(y)\!\!\right) \!\!>\!0 \end{aligned}$$

for $d>0$ and $t$ sufficiently close to 1, because the first two terms are positive, the second term is bounded away from 0, and the third term vanishes as $t$ approaches 1.

1.4 Proof of Proposition 3

• 1) Proof that (29) $\Rightarrow$ $\Delta >0$:

  $$\begin{aligned} \Delta&= -2(1-t)\int \limits _{d}^{c}\int \limits _{d}^{x}\int \limits _{y}^{c}q_{L}^{*}(s)dF(s)dydF(x)+ 2\int \limits _{d}^{c}\theta (x,\mu (x)) q_{L}^{*}(x)F(x)dF(x) \\&> -2(1-t)\int \limits _{d}^{c}\int \limits _{d}^{x}\int \limits _{d}^{c}q_{L}^{*}(s)dF(s)dydF(x)+ 2\int \limits _{d}^{c}\theta (x,\mu (x))q_{L}^{*}(x)F(x)dF(x) \\&= -2(1-t)\int \limits _{d}^{c}\int \limits _{d}^{x}\bar{q}_{L}^{*}dydF(x)+\int \limits _{d}^{c} \theta (x,\mu (x))q_{L}^{*}(x)dF(x)^2\\&\ge -2(1-t)\bar{q}_{L}^{*}\int \limits _{d}^{c}\int \limits _{d}^{x}1 dydF(x)+d \int \limits _{d}^{c}q_{L}^{*}(x)dF(x)^2 \\&= -2(1-t)\bar{q}_{L}^{*}\int \limits _{d}^{c}(x-d)dF(x)+d\cdot \hat{q}_{L}^{*},\text { where } \hat{q}_L^*:=E[q_L^*(X_{(1:2)})] \\&= -2(1-t)\bar{q}_{L}^{*}E(X-d)+d\cdot \hat{q}_{L}^{*} \\&= \left( 2(1-t)\bar{q}_{L}^{*}+\hat{q}_{L}^{*}\right) d-2(1-t)\bar{q}_{L}^{*}E(X) \\&\ge 0, \quad \text {by} (29). \end{aligned}$$

• 2) Proof that (30) $\Rightarrow \,\Delta >0$ (using the first two lines from 1) above)

  $$\begin{aligned} \Delta&> -2(1-t)\int \limits _{d}^{c}\int \limits _{d}^{x}\bar{q}_{L}^{*}dy dF(x)+\int \limits _{d}^{c} \theta (x,\mu (x))q_{L}^{*}(x)dF(x)^2 \\&= -2(1-t)\bar{q}_{L}^{*}\int \limits _{d}^{c}\int \limits _{d}^{x}1dydF(x)+\int \limits _{d}^{c} \theta (x,\mu (x))q_{L}^{*}(x)dF(x)^2 \\&= -2(1-t)\bar{q}_{L}^{*}\int \limits _{d}^{c}(x-d)dF(x)+\int \limits _{d}^{c}\theta (x,\mu (x))q_{L}^{*}(x)dF(x)^2 \\&= -2(1-t)\bar{q}_{L}^{*}\int \limits _{d}^{c}(x-d)dF(x)\\&+\left( \hat{q} _{L}^{*}\int \limits _{d}^{c}\theta (x,\mu (x))dF(x)^2+\text {Cov }\left( \theta (X_{(1)},\mu (X_{(1)})),q_{L}^{*}(X_{(1)})\right) \right) \\&= -2(1-t)\bar{q}_{L}^{*}\left( E(X)-d\right) +\hat{q}_{L}^{*}E\left( \theta (X_{(1)},\mu (X_{(1)}))\right) \\&+\,\text {Cov }\left( \theta (X_{(1)},\mu (X_{(1)})),q_{L}^{*}(X_{(1)})\right) \\&= -2(1-t)\bar{q}_{L}^{*}\left( E(X)-d\right) +\hat{q}_{L}^{*}E\left( tX_{(1)}+(1-t)\mu (X_{(1)})\right) \\&+\,\text {Cov }\left( \theta (X_{(1)},\mu (X_{(1)})),q_{L}^{*}(X_{(1)})\right) \\&\ge 0, \quad \text {by} (30). \end{aligned}$$

1.5 Model with linear demand and uniformly distributed signals

Here we summarize the technical details of the linear example sketched in Sect. 5.

As one can easily confirm, in this case one has

$$\begin{aligned} \Pi _{\text {up} }^n&= \frac{1}{432(c-d)}\left( \frac{4}{1+t}\left( \left( (3+t)c+(1-t)d-2\right) ^{3}-\left( (1+t)z+(1-t)d\right. \right. \right. \\&\!\!\left. \left. \left. -\,2(1\!-\!c)\right) ^{3}\right) \!+\!\frac{3}{z-d}\left( 4(1-c)z\!+\!(1+4t)z^2 \!+\!6(1-t)xz\!+\!3(2t-1)x^{2}\right. \right. \\&\!\!\left. \left. -\,4(1-t)d^{2}\right) ^{2}\right) \\ \Pi _{\text {dn} }^n \!&= \! \frac{1}{108(c\!-\!d)}\Big (\!\frac{1}{1\!+\!t}\!\left( \left( (3\!+\!t)c\!+\!(1\!-\!t)d\!-\!2\right) ^{3}\!-\!\left( (1\!+\!t)z\!+\!(1\!-\!t)d\!-\!2(1\!-\!c)\right) ^{3}\!\right) \\&+\,3(z-d)\left( 2(1-c)+2(1-t)d+3tx+(2-t)z\right) ^{2}\Big )\\ \Pi _{\text {up} }&= \frac{1}{432(c-d)}\left( \frac{4}{1+t}\left( \left( (3+t)c+(1-t)\left( 3(z-x)+d\right) -2\right) ^{3}\right. \right. \\&-\,\left. \left( 2(2-t)z+(1-t)\left( d-3x\right) -2(1-c)\right) ^{3}\right) \\&+\,\frac{3}{z-d}\left( 4(1-c)(z-d)+(1+4t)z^{2}-2t(3x+z)d \right. \\&+\,\left. \left. 2(1-t)(3xz-2d^{2})+3(2t-1)x^{2}\right) ^{2}\right) \\ \Pi _{\text {dn} }&= \frac{1}{108(c-d)}\left( \frac{1}{1+t}\left( \left( (3+t)c+(1-t)\left( 3(z-x)+d\right) -2\right) ^{3}\right. \right. \\&-\,\left. \left( 2(2t-1)x+(1-t)(3z+d)-2(1-c)\right) ^{3}\right) \\&+\,\frac{1}{7t-2}\left( \left( 2(2t-1)x+(1-t)(3z+d)-2(1-c)\right) ^{3}\right. \\&-\,\left. \left( (4t+1)z-3tx+(1-t)d-2(1-c)\right) ^{3}\right) \\&+\,\left. 3(z-d)\left( 2(1-c)+2(1-t)d+3tx+(2-t)z\right) ^{2}\right) \\ \beta _n(x)&= \frac{1}{108}\Big ( d^2(5-t)(1+t)+6d(1-c)(7-2t) \\&+ \,\, 2\big ( d(5-t)(2-t)+3(1-c)(5+2t) \big )x + (11-t)(1+t)x^2\Big )\\ \beta (x)&= \beta _n(x) + \Delta _\beta (x) \\ \Delta _\beta (x)&= \frac{1-t}{36} \Big (2d^2(2\!-\!t)\!-\!6d\!-\!3c^2(5\!+\!t)\!+\!6c(2\!+\!dt) \!+\!\big ( 2d(2\!-\!t)\!-\!6(1\!-\!c) \big )x\\&+ \,(1+t)x^2 \Big ). \end{aligned}$$

1.5.1 Proof of Lemma 4

Proof

1) The fact that $\Pi _{\text {up} }^{n}$ is pseudoconcave can be proved by showing that the cross derivative, $\partial _{zx}\Pi _{\text {up} }^{n}$, is decreasing in $t$ and positive at $t=1$. In fact one can check that

$$\begin{aligned} \partial _{t}\left( \partial _{zx}\Pi _{\text {up} }^{n}\right) \!=\!\frac{-2(1\!-\!c\!-\!z)\!-\!(8t\!-\!1)(z\!-\!x)\!-\!6\left( td\!+\!(1\!-\!t)x\right) }{6(c\!-\!d)}\!-\! \frac{(2t\!-\!1)(x\!-\!d)^{3}}{(c\!-\!d)(z\!-\!d)^{2}}. \end{aligned}$$

Since $1-c>z$ and $z>x$ (upward deviation), one has $\partial _{t}\left( \partial _{zx}\Pi _{\text {up} }^{n}\right) <0$. The value of $\partial _{zx}\Pi _{\text {up} }^{n}$ at $t=1$ is

$$\begin{aligned} \partial _{zx}\left. \Pi _{\text {up} }^{n}\right| _{t=1}=\frac{(x-d)(2d^{2}+6dx-3x^{2}-10dz+5z^{2})}{ 12(c-d)(z-d)^{2}}. \end{aligned}$$

The second term in the numerator is an increasing function of $z$ and takes the value $2(x-d)^{2}$ when $z=x$. Hence $\partial _{zx}\Pi _{\text {up} }^{n}$ at $t=1$ is positive for $z>x.$

The fact that $\Pi _{\text {dn} }^{n}$ is pseudoconcave can be proved by checking that the cross derivative, $\partial _{zx}\Pi _{\text {dn} }^{n}$, is positive:

$$\begin{aligned} \partial _{zx}\Pi _{\text {dn} }^{n}=\frac{t\left( 2(1-c)+4z+t(3x-2z-d)\right) }{ 6(c-d)}>0. \end{aligned}$$

2) The fact that $\Pi _{\text {up} }$ is pseudoconcave for $d\ge c/5$ can be proved by checking that (i) the cross derivative, $\partial _{zx}\Pi _{\text {up} }$, is a concave quadratic function of $t$, (ii) $\partial _{zx}\Pi _{\text {up} }$ takes the same (positive) value as $\partial _{zx}\Pi _{\text {up} }^{n}$ at $t=1$  and is decreasing in $t$ at $t=1$, and (iii) $\Pi _{\text {up} }$ takes the value $(5z-c)/8(c-d)$ at $t=1/2$. The cross derivative is

$$\begin{aligned} \partial _{zx}\Pi _{\text {up} }&= \frac{1}{12(c-d)(z-d)^{2}}\left( 2\left( 3-(2-t)t\right) d^{3}-3(2t-1)^{2}x^{3}\right. \\&-6c(1-t)^{2}(z-d)^{2}+(2t-1)(4t+1)xz^{2}+2(8-t)(1-t)z^{3} \\&+2d^{2}\left( 5\left( 1-(1-t)t\right) z-\left( 5-(17-14t)t\right) x\right) +9(2t-1)^{2}dx^{2} \\&\left. -2(2t-1)(4t+1)dxz-\left( 29-8(4-t)t\right) dz^{2}\right) . \end{aligned}$$

The derivative of $\partial _{zx}\Pi _{\text {up} }$ w.r.t. $t$ at $t=1$ is

$$\begin{aligned} \left. \partial _{t}\left( \partial _{zx}\Pi _{\text {up} }\right) \right| _{t=1} =\frac{-6d-7(z-x)}{ 6(c-d)}-\frac{(x-d)^{3}}{(c-d)(z-d)^{2}}<0. \end{aligned}$$

3) This follows from the proof of 2) and the continuity of $\partial _{zx}\Pi _{\text {up} }$. In fact, if $\partial _{zx}\Pi _{\text {up} }>0$ for some $t^{*}$, it is positive for all $t>t^{*}$.$\square$

1.5.2 Proof of Proposition 4

Proof

We show that the difference of the innovator’s expected profit, $\Delta (t):=G-G_{n}$, is a concave quadratic function of $t$, which takes positive values at $t=\frac{1}{2}$ and 1. One can easily find that

$$\begin{aligned} \Delta (t)&\!=-\frac{(c\!-\!d)^{2}}{9}t^{2}\!+\!\frac{c\!-\!d}{36}(12\!-\!15c\!-\!13d)t\nonumber \\&+\frac{1}{36 }\left( 5c^{2}\!-\!2c(2\!+\!9d)\!+\!(16\!-\!11d)d\right) \!. \end{aligned}$$

(34)

Hence,

$$\begin{aligned} \Delta (1)&= \frac{1}{18}\left( c(4-7c)+2(1-2c)d-d^{2}\right) \\ \Delta (1/2)&= \frac{1}{72}\left( c(4-7c)+10(2-3c)d-11d^{2}\right) . \end{aligned}$$

Both are concave functions of $d$, taking positive values at $d=0$ and $d=c$;

$$\begin{aligned} \left. \Delta (1)\right| _{d=0}=\frac{c(4-7c)}{18}>0, \quad \left. \Delta (1/2)\right| _{d=0}=\frac{c(4-7c)}{72}>0 \\ \left. \Delta (1)\right| _{d=c}=\left. \Delta (1/2)\right| _{d=c}=\frac{c(1-2c)}{3}>0. \end{aligned}$$

$\square$

1.5.3 Proof of Proposition 5

Proof

1) We first prove the asserted comparative statics in $t$ (using the fact that $d<c<1/2$):

$$\begin{aligned} \partial _t \beta _n(x) = \frac{x-d}{54}\left( 6-6c-2d +dt+(5-t)x \right) > 0. \end{aligned}$$

(35)

2) We show that $\partial _t \Delta _\beta (x)<0, \forall x$, in several steps:

$$\begin{aligned} \partial _{xx}(\partial _{t}\Delta _\beta (x))&= -t/18 <0 \quad \text {(concavity)} \end{aligned}$$

(36)

$$\begin{aligned} \partial _t \Delta _\beta (d)&= \frac{c-d}{6}\left( -2+2c+2d+(c-d)t \right) \nonumber \\&< \frac{c-d}{6}\left( -2+2c+2d+(c-d) \right) < 0 \end{aligned}$$

(37)

$$\begin{aligned} \partial _t \Delta _\beta (c)&= \frac{c-d}{18}\left( 3(c+d-1)+2(c-d)t\right) \nonumber \\&< \frac{c-d}{18}\left( 5 c + d - 3)\right) < 0\end{aligned}$$

(38)

$$\begin{aligned} \partial _x\left( \partial _{t} \Delta _\beta (c)\right)&= \frac{1}{18}\left( 3(1-c-d)-2(c-d)t\right) >0. \end{aligned}$$

(39)

Combining (36)–(39) the assertion 2) follows immediately.

3) Because $\beta _n$ is strictly increasing in $t$ and $\beta _n - \beta$ is strictly decreasing in $t$, it follows that both $\beta$ and $\beta _n$ are strictly increasing in $t$.

Similarly one can show that $\partial _{d} \beta _n(x)>0$ and $\partial _{d} \Delta _\beta (x)<0$ and the asserted comparative statics concerning $d$ follows immediately.$\square$

1.5.4 Proof of Proposition 6

Proof

The innovator’s expected profits are equal to

$$\begin{aligned} G_n&= \frac{1}{216} \Big ( c(40\!-\!70d)\!-\!29c^2\!+\!d(104\!+\!27d)\!+\!2(c\!-\!d)(8\!-\!3c\!-\!d)t \!-\!(c\!-\!d)^2 t^2 \Big )\\ G&= G_n+ \Delta , \end{aligned}$$

where $\Delta$ is defined in the proof of Proposition 4.

Taking partial derivatives with respect to $t$ and $d$ one finds:

$$\begin{aligned} \partial _t G_n&= \frac{c-d}{108} \Big ( 8-3c-d-(c-d)t \Big ) > \frac{c-d}{108} \Big ( 8-3c-d-(c-d)\Big ) > 0 \\ \partial _t G&= \frac{c\!-\!d}{108} \Big ( 44\!-\!48c\!-\!40d\!-\!25(c\!-\!d)t \Big )\\&> \frac{c\!-\!d}{108} \Big ( 44\!-\!48c\!-\!40d\!-\!25(c\!-\!d)\Big ) \!>\!0 \\ \partial _d G_n&= \frac{1}{108} \Big ( 52-35c+27d -(8-2c-2d)t + (c-d) t^2 \Big )\\ \partial _d G&= \frac{1}{108} \Big ( 100-89c -39d - (44-8c-80d)t + 25(c-d)t^2\Big ) \\ \partial _t \Delta&= \frac{c-d}{36}\Big (12-15c-13d-8(c-d)t\Big ) \\ \partial _d \Delta&= \frac{1}{18}\Big (8-9c-11d+(c+13d-6)t +4(c-d)t^2\Big ). \end{aligned}$$

Both functions $\partial _d G_n$ and $\partial _d G$ are strictly convex in $t$ and have a minimum at $t>1$. Moreover, both $\partial _d G_n$ and $\partial _d G$ are positive at $t=1$; therefore, they are positive valued for all $t \in [1/2,1]$. The assertions concerning the comparative statics of $\Delta$ follow easily.$\square$