<p>True to the saying that a picture is worth a thousand words, let's plot the profit function under the Stackelberg game and see what we're dealing with.</p>

<p>You correctly observed (or copied from the Baye, Shin paper) that the reaction function of player 2 is:</p>

<p>$$
R(x_1)=\frac{2\mathit{x_1}-\sqrt{2}\,\sqrt{2\mathit{x_1}-{{\mathit{x_1}}^{2}}}}{2\mathit{x_1}-2}\
$$</p>

<p>Inserting this reaction function in the profit function of player 1 gives us a complicated expression:</p>

<p>$$
\pi(x_1)=-\frac{{{2}^{\frac{3}{2}}}\,\sqrt{2-\mathit{x_1}}\,{{\mathit{x_1}}^{2}}-2{{\mathit{x_1}}^{\frac{3}{2}}}-3\sqrt{2}\,\sqrt{2-\mathit{x_1}}\,\mathit{x_1}+4\sqrt{\mathit{x_1}}}{{{2}^{\frac{3}{2}}}\,\sqrt{2-\mathit{x_1}}\,\left( \mathit{x_1}-1\right) }
$$</p>

<p>(Perhaps more advanced symbolic math programs can further simplify this, mine won't). In any case this is  a function of one variable $\mathit{x_1}$, and can be plotted. For completeness I have inserted the original equilibrium as well.</p>

<p><a href="https://i.sstatic.net/cTy7H.jpg" rel="noreferrer"><img src="https://i.sstatic.net/cTy7H.jpg" alt="SB profit function and symmetric NE"></a></p>

<p>As you can see the original NE is no longer a local maximum, and player 1 can increase its profits by increasing their effort.</p>

CC BY-SA 3.0
2017-11-08T12:26:28.693Z
2017-11-08T12:26:28.693Z
Maarten Punt
<p>The following is taken from <a href="https://www.jstor.org/stable/117041" rel="nofollow noreferrer">Baye, Shin (1999)</a></p>
<p>Consider a contest over a prize valued at 1 with symmetric players <span class="math-container">$1$</span> and <span class="math-container">$2$</span> who exert a level of effort <span class="math-container">$x_1$</span> and <span class="math-container">$x_2$</span> respectively. Effort cannot exceed <span class="math-container">$2/3$</span>. Profit (<span class="math-container">$\pi$</span>) of player 1 is</p>
<p><span class="math-container">$$\pi_1=\frac{x_1-\frac{x_1x_2}{2}}{x_1+x_2-x_1x_2}-x_1$$</span></p>
<p>The simultaneous-move equilibrium is (denoted with superscript <span class="math-container">$*$</span>)</p>
<p><span class="math-container">$$x_1^*=x_2^*=1-\frac{\sqrt{2}}{2}$$</span></p>
<p>The profit for each is</p>
<p><span class="math-container">$$\pi_1^*(x_1^*,x_2^*)=\pi_2^*(x_1^*,x_2^*)=\frac{\sqrt{2}-1}{2}$$</span></p>
<p>And we have that the best response of player <span class="math-container">$2$</span></p>
<p><span class="math-container">$$R_2(x_1)=\frac{2x_1-\sqrt{(-2x_1^2+4x_1)}}{2(x_1-1)}$$</span></p>
<p>However, suppose that player <span class="math-container">$1$</span> 'moves' before player <span class="math-container">$2$</span>. Then, player <span class="math-container">$1$</span> would deviate (upwards) from <span class="math-container">$x_1^*$</span>, presumably, because player <span class="math-container">$2$</span> would reduce it's level of effort, and it would receive more of the prize. How does one show this?</p>
<hr />
<p>In the article, Baye and Shin consider the new Stackelberg equilibrium <span class="math-container">$\pi_1^s$</span> where they show that <span class="math-container">$\pi_1^s(x_1^*+\epsilon)&gt;0$</span> for some <span class="math-container">$\epsilon&gt;0$</span>. Specifically</p>
<p><span class="math-container">$$\pi_1^s(x_1^*+\epsilon)-\pi_1^*(x_1^*+x_2^*)=\frac{\sqrt{1+2\epsilon\sqrt{2}-2\epsilon^2}-1-\epsilon\sqrt{2}+2\epsilon^2}{\sqrt{2}-2\epsilon}&gt;0$$</span></p>
<hr />
<p>I've tried replicating their result by seeing how <span class="math-container">$x_2$</span> changes with <span class="math-container">$x_1+\epsilon$</span> and substituting <span class="math-container">$x_1+\epsilon$</span> and the new value of <span class="math-container">$x_2^s$</span> into <span class="math-container">$1$</span>'s profit function, but I fail to reach their result.</p>

<microeconomics><mathematical-economics><game-theory>
Subgame Perfect Equilibrium in Baye, Shin (1999)
pafnuti
# Subgame Perfect Equilibrium in Baye, Shin (1999)

(Asked by: username_0 on 2017-11-08)

The following is taken from [Baye, Shin (1999)](https://www.jstor.org/stable/117041)

Consider a contest over a prize valued at 1 with symmetric players $1$ and $2$ who exert a level of effort $x\_1$ and $x\_2$ respectively. Effort cannot exceed $2/3$. Profit ($\pi$) of player 1 is

$$\pi\_1=\frac{x\_1-\frac{x\_1x\_2}{2}}{x\_1+x\_2-x\_1x\_2}-x\_1$$

The simultaneous-move equilibrium is (denoted with superscript $\*$)

$$x\_1^\*=x\_2^\*=1-\frac{\sqrt{2}}{2}$$

The profit for each is

$$\pi\_1^\*(x\_1^\*,x\_2^\*)=\pi\_2^\*(x\_1^\*,x\_2^\*)=\frac{\sqrt{2}-1}{2}$$

And we have that the best response of player $2$

$$R\_2(x\_1)=\frac{2x\_1-\sqrt{(-2x\_1^2+4x\_1)}}{2(x\_1-1)}$$

However, suppose that player $1$ 'moves' before player $2$. Then, player $1$ would deviate (upwards) from $x\_1^\*$, presumably, because player $2$ would reduce it's level of effort, and it would receive more of the prize. How does one show this?

---

In the article, Baye and Shin consider the new Stackelberg equilibrium $\pi\_1^s$ where they show that $\pi\_1^s(x\_1^\*+\epsilon)>0$ for some $\epsilon>0$. Specifically

$$\pi\_1^s(x\_1^\*+\epsilon)-\pi\_1^\*(x\_1^\*+x\_2^\*)=\frac{\sqrt{1+2\epsilon\sqrt{2}-2\epsilon^2}-1-\epsilon\sqrt{2}+2\epsilon^2}{\sqrt{2}-2\epsilon}>0$$

---

I've tried replicating their result by seeing how $x\_2$ changes with $x\_1+\epsilon$ and substituting $x\_1+\epsilon$ and the new value of $x\_2^s$ into $1$'s profit function, but I fail to reach their result.

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


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## 1 answers

### Answer by username_1 on 2017-11-08. Score: 6 | Accepted answer

True to the saying that a picture is worth a thousand words, let's plot the profit function under the Stackelberg game and see what we're dealing with.

You correctly observed (or copied from the Baye, Shin paper) that the reaction function of player 2 is:

$$
R(x\_1)=\frac{2\mathit{x\_1}-\sqrt{2}\,\sqrt{2\mathit{x\_1}-{{\mathit{x\_1}}^{2}}}}{2\mathit{x\_1}-2}\
$$

Inserting this reaction function in the profit function of player 1 gives us a complicated expression:

$$
\pi(x\_1)=-\frac{{{2}^{\frac{3}{2}}}\,\sqrt{2-\mathit{x\_1}}\,{{\mathit{x\_1}}^{2}}-2{{\mathit{x\_1}}^{\frac{3}{2}}}-3\sqrt{2}\,\sqrt{2-\mathit{x\_1}}\,\mathit{x\_1}+4\sqrt{\mathit{x\_1}}}{{{2}^{\frac{3}{2}}}\,\sqrt{2-\mathit{x\_1}}\,\left( \mathit{x\_1}-1\right) }
$$

(Perhaps more advanced symbolic math programs can further simplify this, mine won't). In any case this is a function of one variable $\mathit{x\_1}$, and can be plotted. For completeness I have inserted the original equilibrium as well.

[![SB profit function and symmetric NE](https://i.sstatic.net/cTy7H.jpg)](https://i.sstatic.net/cTy7H.jpg)

As you can see the original NE is no longer a local maximum, and player 1 can increase its profits by increasing their effort.


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