<p>I'm reading a game-theory related <a href="https://www.google.co.uk/search?hl=en&amp;q=Effort%20Levels%20in%20Contests%20with%20Two%20Asymmetric%20Players&amp;meta=" rel="nofollow noreferrer">paper</a>*, and I'm not following the derivation of some property of the best-response functions. </p>

<p>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 $argmax_{x_1}(\pi_1)$:</p>

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

<blockquote>
  <p>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]:</p>
  
  <p>$$\frac{dr_1(x_2)}{dx_2}=\frac{h'(x_1)h'(x_2)(\sigma h(x_1)-h(x_2))}{h(x_1)[h''(x_2)(\sigma
&gt; h(x_1)+h(x_2)-2h'(x_2))^2]}$$</p>
</blockquote>

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

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

<p>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.</p>

<hr>

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

CC BY-SA 3.0
2017-12-19T10:07:53.357Z
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
Maarten Punt
CC BY-SA 3.0
2017-12-19T21:47:48.417Z
Solve $\frac{\partial \frac{\partial \pi_1}{\partial x_1}}{\partial x_2} = 0$ for $\frac{\partial x_1}{\partial x_2}$.
clueless
<microeconomics><mathematical-economics><game-theory>
Deriving a Best Response Function in Baik (1994)
pafnuti
# Deriving a Best Response Function in Baik (1994)

(Asked by: username_0 on 2017-12-18)

I'm reading a game-theory related [paper](https://www.google.co.uk/search?hl=en&q=Effort%20Levels%20in%20Contests%20with%20Two%20Asymmetric%20Players&meta=)\*, 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.

**Tags:** microeconomics, mathematical-economics, game-theory

> **Comments:**
> **username_1**: 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  
> **username_2**: Solve $\frac{\partial \frac{\partial \pi_1}{\partial x_1}}{\partial x_2} = 0$ for $\frac{\partial x_1}{\partial x_2}$.  

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## 0 answers

