I'm reading a game-theory related paper*, and I'm not following the derivation of some property of the best-response functions.

Suppose I have two players $1$ and $2$, whose strategies are continuous levels of effort $x_1$ and $x_2$ respectively. The first order condition is given by for a Nash equilibrium strategy is given just standardly by $argmax_{x_1}(\pi_1)$:

$$\frac{\partial \pi_i}{\partial x_1}=\frac{\alpha\sigma h'(x_1)h(x_2)}{(\sigma h(x_1)+h(x_2))^2}-1=0$$

Let $x_1=r_1(x_2)$ denote player $1$'s reaction function. Since it is derived from player $1$'s first-order condition (above), we obtain it's derivative by differentiating along [the above FOC]:

$$\frac{dr_1(x_2)}{dx_2}=\frac{h'(x_1)h'(x_2)(\sigma h(x_1)-h(x_2))}{h(x_1)[h''(x_2)(\sigma > h(x_1)+h(x_2)-2h'(x_2))^2]}$$

Note that this is not the derivative of the first equation w.r.t. $x_2$, which simply is

$$\frac{\alpha\sigma h'(x_1)h'(x_2)(\sigma h(x_1)-h(x_2))}{(\sigma h(x_1)+h(x_2))^3}=0$$

However, besides this approach, I fail to see how the equation could be obtained. Rearranging for $x_1$ to express player $1$'s BR function the conventional way seems no option here either, since the function $h$ is undefined, and the BR function would also become much messier that the one quoted.

*Some of the working out is in a technical appendix elsewhere, which I can email.

  • 1
    $\begingroup$ Given that the first order condition that determines the reaction function is an implicit function, you can use the implicit function theorem to determine that derivative. I haven't checked because it's tedious but I guess you'd arrive at the derivative that you have there $\endgroup$ Commented Dec 19, 2017 at 10:07
  • $\begingroup$ Solve $\frac{\partial \frac{\partial \pi_1}{\partial x_1}}{\partial x_2} = 0$ for $\frac{\partial x_1}{\partial x_2}$. $\endgroup$
    – clueless
    Commented Dec 19, 2017 at 21:47


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