Academic commercialization and changing nature of academic cooperation

Abstract

Recent economic policies emphasize the role of academic science in technological innovation and economic growth and encourage universities and individual academics to engage in commercial activities. In this trend of academic commercialization, a growing concern has been expressed that its potential incompatibility with the traditional norms of open science could undermine the cooperative climate in academia. Drawing on the framework of evolution of the cooperation, this study examines the changing nature of academic cooperation under the current policy trend. In an ideal state of open science, academics are supposed to cooperate gratis and unconditionally. However, results predict that the commercialized regime could compromise underlying mechanisms of cooperation and allow defectors to prevail. As the trend further grows, academics would become more demanding of direct reward in exchange for cooperation, and they would refrain from engaging in cooperation but would prefer to work independently. Some interventions (e.g., centralized rewarding) could mitigate the problem but require delicate system design.

Mathematical appendix

1.1 Prediction 1

$g=\frac{1}{2-q{z}_3\left(1-{x}_0\right)}$ by solving g ₂ = g ₀ = 1 − g, g ₃ = 1 − (1 − q)g, and g = x ₂ g ₂ + x ₃ g ₃ + x ₀ g ₀. With (4a) and (4b), ${P}_3-{P}_2=\frac{q\left\{qb{z}_3\left(1-{x}_0\right)-c\right\}}{2-q{z}_3\left(1-{x}_0\right)}$. The solution of P ₃ − P ₂ = 0 for z ₃ gives ${z}_3^{*}=\frac{c}{qb\left(1-{x}_0\right)}$. $\frac{d{z}_3^{*}}{d{x}_0}=\frac{c}{qb{\left(1-{x}_0\right)}^2}>0$.

1.2 Prediction 2

I assume that recipients of self-regarding types (PAY and COM) accept paying bilateral rewards, but that DISC does not because it violates the norms of open science.¹⁷ Thus, PAY donors cooperate with PAY and COM recipients with the probability of p but never cooperate with DISC. With this setting, the equilibrium reputation of PAY is given by g ₄ = (1 − g ₃)x ₃ + {pg ₄ + (1 − p)(1 − g ₄)}x ₄ + {pg ₀ + (1 − p)(1 − g ₀)}x ₀, where g ₀ = 1 − g, g ₃ = 1 − (1 − q)g and g = x ₃ g ₃ + x ₄ g ₄ + x ₀ g ₀.

Formally, Prediction 2 states $\frac{d{z}_3^{*}}{d{x}_0}>0$, where z ^*₃ is the solution of P ₃ − P ₄ = 0 for z ₃. From (5a) and (5b), P ₃ − P ₄ = b(g ₃ − g ₄)qx ₃ − p(β − c)x ₀ − p(b − γ + β − c)x ₄ − cgq. This is rearranged as $\frac{f\left({z}_3,{x}_0\right)}{h\left({z}_3,{x}_0\right)}$, where f and h > 0 are polynomials of x ₀ and z ₃.¹⁸ Since $\frac{d{z}_3^{*}}{d{x}_0}>0$ is not analytically provable, I indirectly show this by simulation. From the whole parameter regions, $c\in \left(0,b\right),q\in \left(c,1\right),p\in \left(0,\frac{q}{2-q}\right)$,¹⁹ $\beta \in \left[c,b\right],\ \gamma \in \left[0,\ \beta \right],$ and x ₀ ∈ [0, 1), I randomly choose a set of parameters, with which f(z ₃, x ₀) = 0 is numerically solved for z ₃. If a solution is found in (0,1), the same equation is solved again with the same set of parameters except that x ₀ is replaced by x ₀ + ε, where $\varepsilon =\frac{1}{10000}$. The first solution is denoted by z ^*₃ and the second by z ^* *₃ . This computation is repeated 100,000 times. Approximately 70 % of the time, no solution was found in (0,1). For the rest, a single solution was found in (0,1),²⁰ where z ^* *₃  > z ^*₃ always holds. This implies $\frac{d{z}_3^{*}}{d{x}_0}>0$. Because f is a cubic polynomial of z₃ whose leading coefficient >0 and f(0, x ₀) < 0, z ^*₃ is the unstable equilibrium; i.e., P ₃ − P ₄ < 0 if z ₃ < z ^*₃ and P ₃ − P ₄ > 0 if z ₃ > z ^*₃ .■

1.3 Prediction 3 (DISC vs. ABST)

The game involving ABST is played as follows. Two players are randomly chosen from a population of DISC, ABST, and COM. When ABST is chosen as a donor, he always defects, so payoffs for both sides are zero. When ABST is chosen as a recipient, he does not ask for cooperation, where the payoff for ABST is σ while that for a donor is zero. As the donor neither defects nor cooperates, his reputation does not change. Because the reputations of DISC and COM are unaffected by games with ABST recipients, reputation is computed only within non-ABST players; i.e., g ₃ = 1 − (1 − q)g _− 5 and g ₀ = 1 − g _− 5, where ${g}_{-5}=\frac{x_3{g}_3+{x}_0{g}_0}{x_3+{x}_0}$.

Prediction 3 is formally $\frac{d{z}_3^{*}}{d{x}_0}>0$, where z ^*₃ is the solution of P ₃ − P ₅ = 0 for z ₃. From the reputation equations, (6a), and (6b), ${P}_3-{P}_5=\frac{q\left\{\left(b-c\right){\left(1-{x}_0\right)}^2{z}_3^2+\left(b+q-2c\right)\left(1-{x}_0\right){x}_0{z}_3-c{x}_0^2\right\}}{\left(2-q\right)\left(1-{x}_0\right){z}_3+2{x}_0}-\sigma$. Let k(z ₃, x ₀) = P ₃ − P ₅. Since k(z ^*₃ , x ₀) = 0, $\frac{d{z}_3^{*}}{d{x}_0}=-\frac{\partial k}{\partial {x}_0}\ /\frac{\partial k}{\partial {z}_3}$. As $\frac{\partial k}{\partial {z}_3}>0$ is easily shown, proving $\frac{\partial k}{\partial {x}_0}<0$ suffices. $\frac{\partial k}{\partial {x}_0}$ is rearranged as $\frac{k_2\left({z}_3,{x}_0\right)}{k_1\left({z}_3,{x}_0\right)}$, where k ₁ > 0 and k ₂ are polynomials of x ₀ and z ₃.²¹ As $\frac{\partial {k}_2}{\partial {x}_0}<0$ is easily shown, it follows that $\frac{\partial k}{\partial {x}_0}<0\ \forall {x}_0\Leftarrow {k}_2\left({z}_3,0\right)<0\iff {z}_3>\frac{\left(qb-2c\right)\left(1-q\right)}{\left(2-q\right)\left(b-c\right)}$. Since k(z ^*₃ , x ₀) = 0, the sufficient condition for $\frac{\partial k}{\partial {x}_0}<0\ \forall {x}_0$ is ${\left.{z}_3^{*}\right|}_{y=0}=\frac{\left(2-q\right)\sigma }{q\left(b-c\right)}>\frac{\left(qb-2c\right)\left(1-q\right)}{\left(2-\mathrm{q}\right)\left(b-c\right)}\iff c>\frac{qb}{2}$ or $\sigma >\frac{q\left(1-q\right)\left(qb-2c\right)}{{\left(2-q\right)}^2}$. Otherwise, ∃ x ^*₀  ∈ (0, 1) s.t. $\frac{\partial k}{\partial {x}_0}<0\left({x}_0>{x}_0^{*}\right)$ and $\frac{\partial k}{\partial {x}_0}>0\ \left({x}_0<{x}_0^{*}\right)$.²² In sum, if c or σ is sufficiently large, COM offers a favorable condition for ABST regardless of COM’s frequency. Otherwise, with a minimal frequency of COM, ABST gains advantage over DISC.

1.4 Prediction 3 (PAY vs. ABST)

Since DISC is not present, reputation does not play a role. From (6b) and (6c), P ₄ − P ₅ = p(β − c)(x ₄ + x ₀) + p(b − γ)x ₄ − σ. Solving P ₄ − P ₅ = 0 for z ₄, ${z}_4^{*}=\frac{\sigma -p\left(\beta -c\right){x}_0}{p\left(b-\gamma +\beta -c\right)\left(1-{x}_0\right)}$. $\frac{d{z}_4^{*}}{d{x}_0}=\frac{\sigma -p\left(\beta -c\right)}{p\left(b-\gamma +\beta -c\right){\left(1-{x}_0\right)}^2}$ . $\frac{d{z}_4^{*}}{d{x}_0}>0$ if σ > p(β − c). Thus, the invasion of COM is favorable for ABST when the matching rate (p) or the value of return payment (β) is sufficiently small.