Patent_Race_Model - Atlas of Economic Models
 

Patent Race Model

A model in which several firms 'race' in order to obtain a patent. Originally developed in the late 1970s, early 1980s (e.g. \citet*{loury_1979}) and one of the first fruits of applying game-theoretic tools in industrial organization. It provides some nice comparative static results, and is particularly noteworthy for demonstrating the possibility of too much R&D under free competition thanks to the pooling externality effect.

The Model

  • There are n firms, all ex-ante symmetric competing to for a discovery.
  • The first firm to make the discovery is awarded a patent with private value $$V$$. The social value of the discovery is $$W \geq V$$.
  • Each firm has access to the same R&D technology $$c(x)$$. This is a memoryless, poisson, process where expenditure of $$c(x)$$ results in a probability of discovery per unit time of $$x$$. (Equivalently, but less conveniently, one could have a cost of $$x$$ and probability of discovery of $$h(x)$$ ($$h = c^{-1}$$).

    • Without loss of generality $$c(0) = 0$$. In addition $$V - c'(0) > 0$$ (otherwise will never be worth doing R&D!).

    • Diminishing returns to effort: $$c > 0$$.

      • This is not strictly necessary: could have initially increasing returns, but certainly require $$lim_{x \rightarrow \infty} c > 0$$.

    • [Optional]: can also be useful to have a fixed cost of entry into R&D race: $$f > 0$$ (i.e. pay $$f$$ if $$x > 0$$).

  • Firms discount the future at exponential rate $$r$$.

Comments:

  • Firms are competing to discover a solution to known problem not just randomly searching.

  • With a deterministic technology, due to non-continuous payoff functions (0 if don't win, V/2 if share the prize and V if win) there will be no Nash Equilibrium. There will be a Stackelberg equilibrium in which the first mover just breaks even and all other firms do no R&D.

  • Poisson R&D: this is essential for tractability. Without it things get a lot more complicated.

  • In order to endogenize entry ($$n$$) need either initially increasing returns to effort or a fixed cost of entry. Without these combination of diminishing returns and no fixed cost mean that entry will always be profitable ($$n+1c'(nx/n+1) < n c'(x)$$).

Solving

Focus on firm $$i$$. For terminological convenience let $$y = x_{i}$$ and $$X = X_{-i} = \sum_{j \neq i} x_{j}$$ (total expenditures of all other firms). Then profits of firm $$i$$ are:

$$ \begin{eqnarray} \Pi(y,X) & = & \int_{0}^{\infty} e^{-rt} \mathbb{P}(\textrm{Continue to time t}) \cdot \textrm{(Expected payoff in dt)} \\ & = & \int_{0}^{\infty} e^{-rt} e^{-(y+X)t} (yV - c(y)) dt \\ & = & \frac{yV -c(y)}{y+X+r} \end{eqnarray} $$

First order condition $$ \frac{d\Pi}{dy} = 0$$ gives:

$$ \begin{eqnarray} 0 & = & (V-c')(y+X+r) - (yV -c) \\ & = & V (X + r) - c'(y)(y+X+r) + c(y) \\ & = & f(y,X,r,V) \end{eqnarray} $$

Obviously without knowing the specific functional form of $$c$$ it is not possible to solve directly for firm i's best response function, $$y^{*}(X)$$. However, we can still make progress. First, note that:

$$ \[ \frac{df}{dX} = V - c'(y) \] $$

This must be positive for all values considered by the firm since, if it were negative, profits would also be negative. Hence this is a game of Strategic Complements. This guarantees the existence of Nash Equilibrium and various other nice properties regarding comparative statics (see references below for more details). However we can derive most of this directly. For example, note that:

$$ \[ f' = \frac{df}{dy} = -c''(y) \cdot (y+X+r) \] $$

Convex costs (diminishing returns to effort) mean $$c'' > 0$$ and hence $$f' < 0$$. Thus, given that $$y^{*}$$ is defined by $$f(y,\cdot) = 0$$, any change in another variable which increases (decreases) $$f$$ implies an increase (decrease) in $$y^{*}$$ (i.e. if $$\frac{df}{dz} > 0 $$ for some $$z$$ then $$\frac{dy^{*}}{dz} > 0$$). Thus, for example, as just discussed $$\frac{df}{dX} > 0$$ hence $$\frac{dy^{*}}{dX} > 0$$ -- in words: an increase in other firms R&D efforts increases this firms R&D efforts.

Let us try to solve for a symmetric Nash Equilibrium, i.e. where $$y = x_{i} = x_{j} = x \forall j$$. In this case we need to solve for zeros of the function $$g(x,\cdot)$$:

$$ \[ g(x,\cdot) = f(x,(n-1)x,\cdot) = V((n-1)x + r) - c' (nx + r) + c \]

Noting that $$V - c'(0) > 0$$ and that $$c(0) = 0$$ we have that $$g(0) > 0$$, while $$lim_{x \rightarrow \infty} g(x) < 0$$. Hence given continuity of $$g$$ (follows from continuity of $$f$$) by intermediate value theorem must have an $$x^{*} \in (0,\infty)$$ such that:

$$ \[ g(x^{*}) = 0 \] $$

Furthermore by the same techniques as used for $$f$$ above we can show that firms effort increase in the value of the prize $$V$$, the discount rate $$r$$ and the number of firms in the race $$n$$. (Formally: $$g' = (n-1)(V-c') - nc'' - rc''$$ and at a stable equilibrium $$g' < 0$$).

Pooling Externality and Inefficient Entry

One interesting fact about the patent race model is that entry under free competition can be inefficiently high. This is because an individual firm does not take account of its effect on other firms of its efforts: a marginal increase in R&D by one firm not only increases its probability of winning but decreases the probabilities of other firms winning (the externality). This is just like the original description of the 'pooling' externality in relation to oil-drilling where extraction from one well impacts on extraction from neighbouring well.

Formally, we can show this by considering the social planner's problem (clearly they will have symmetric expenditures by the n firms):

$$ \[ \max_{x} \frac{n x W - nc(x)}{nx +r} = \frac{xV - c}{x + \frac{r}{n}} \] $$

If $$V \approx W$$ (i.e. social and private values are close) the planner faces an equivalent problem to the private firm but with an nth of the discount rate (i.e. they are more patient). This lower discount rate reflects the planner's internalization of the externality and will obviously result in lower level of expenditure per firm (easy to check formally: in above formula for $$f$$, $$\frac{df}{dr} > 0$$ and so a decrease in $$r$$ reduces $$x$$).