<p>Well it says in the paper on p. 423:</p>

<p>"Thus the sufficiency assumption implies a unique pure strategy equilibrium. And if the sufficiency assumption is violated, at least one player will receive a negative payoff; <em>that player can improve by reducing his or her effort to zero</em>, which will destroy the equilibrium." [emphasis mine]</p>

<p>So that player will not make any effort because he/she cannot win. I leave it to you to figure out what the best reply from the other player is in that case. </p>

<p>I guess to be completely mathematically correct one would have to formulate these kind of games with non-negativity constraints, but because it complicates the analysis they're often left out and an interior equilibrium is either assumed or, as here, checked afterwards.</p>

CC BY-SA 3.0
2017-11-27T10:27:45.357Z
IMO Your middle paragraph is not necessary for this answer and creates confusion.
Giskard
CC BY-SA 3.0
2017-11-27T10:14:13.143Z
2017-11-27T10:14:13.143Z
Maarten Punt
<p>The following is taken from <a href="https://www.jstor.org/stable/30024495" rel="nofollow noreferrer">Nti (1999)</a> </p>

<p>Consider a 2 player game in which each exerts effort in attempt to win a prize. Let $V_1$ be player $1$'s valuation of the prize and let $V_2$ be player $2$'s valuation, where $0&lt;V_1&lt; V_2$. Player $1$ and $2$ expend efforts $x_1$ and $x_2$ respectively to win prize $p=1$. The cost of effort is $1$ per unit. Given a profile of effort levels ($x_1$, $x_2$),the probability that player i wins the prize takes the symmetric form</p>

<p>$$p_i(x_1,x_2)=\frac{x_i^r}{x_1^r+x_2^r}$$</p>

<p>$r$ is the returns to scale parameter associated with effort $r&gt;0$. Profit is</p>

<p>$$\pi_i(x_1,x_2)=v_i\frac{x_i^r}{x_1^r+x_2^r}-x_1$$</p>

<p>In equilibrium (Nash-Cournot) player $i$ exerts effort $x_i^*$: $$x_i^*=\frac{rv_i^{r+1}v_{-i}^r}{(v_1^r+v_2^r)^2}$$</p>

<p>If both follow the equilibrium strategy, $i$'s probability of winning is
$P_i=\frac{v_i^r}{v_1+v_2^r}$. Inserting these two into the second equation for player 1 (with simplification) yields</p>

<p>$$\pi_1(x_1^*,x_2^*)=\frac{v_1^{r+1}}{(v_1^r+v_2^r)^2}\big[v_1^r+v_2^r-rv_2^r\big]$$</p>

<p>The second order (sufficiency) condition $\frac{\partial^2\pi_1}{\partial{x_1^2}}=\frac{rv_1^{1-r}x_2^rx_1^{2r-2}}{(x_1^r+x_2^r)^3}\big[rv_2^r-v_2^r-v_1^r-rv_1^r\big]$ is satisfied iff $$rv_2^r-v_2^r-v_1^r-rv_1^r&lt;0$$</p>

<p>However, it is clear that for $x_1^*$ to be a Nash Equilibrium, $\pi_1(x_1^*,x_2^*)\geq0$ and hence that $$v_1^r+v_2^r&gt;rv_2^r$$ </p>

<p>which is a stronger requirement than the second order condition. Hence, it is possible that the second order condition is met but not that $v_1^r+v_2^r&gt;rv_2^r$. My question is what happens when this occurs? Is there no Nash equilibrium here? Does the player who makes negative profit drop out, if so what does the remaining player do?</p>

<microeconomics><mathematical-economics><game-theory><public-economics>
Contest game: second order condition satisfied, but negative profits?
pafnuti
# Contest game: second order condition satisfied, but negative profits?

(Asked by: username_0 on 2017-11-26)

The following is taken from [Nti (1999)](https://www.jstor.org/stable/30024495)

Consider a 2 player game in which each exerts effort in attempt to win a prize. Let $V\_1$ be player $1$'s valuation of the prize and let $V\_2$ be player $2$'s valuation, where $0

$$p\_i(x\_1,x\_2)=\frac{x\_i^r}{x\_1^r+x\_2^r}$$

$r$ is the returns to scale parameter associated with effort $r>0$. Profit is

$$\pi\_i(x\_1,x\_2)=v\_i\frac{x\_i^r}{x\_1^r+x\_2^r}-x\_1$$

In equilibrium (Nash-Cournot) player $i$ exerts effort $x\_i^\*$: $$x\_i^\*=\frac{rv\_i^{r+1}v\_{-i}^r}{(v\_1^r+v\_2^r)^2}$$

If both follow the equilibrium strategy, $i$'s probability of winning is
$P\_i=\frac{v\_i^r}{v\_1+v\_2^r}$. Inserting these two into the second equation for player 1 (with simplification) yields

$$\pi\_1(x\_1^\*,x\_2^\*)=\frac{v\_1^{r+1}}{(v\_1^r+v\_2^r)^2}\big[v\_1^r+v\_2^r-rv\_2^r\big]$$

The second order (sufficiency) condition $\frac{\partial^2\pi\_1}{\partial{x\_1^2}}=\frac{rv\_1^{1-r}x\_2^rx\_1^{2r-2}}{(x\_1^r+x\_2^r)^3}\big[rv\_2^r-v\_2^r-v\_1^r-rv\_1^r\big]$ is satisfied iff $$rv\_2^r-v\_2^r-v\_1^r-rv\_1^r<0$$

However, it is clear that for $x\_1^\*$ to be a Nash Equilibrium, $\pi\_1(x\_1^\*,x\_2^\*)\geq0$ and hence that $$v\_1^r+v\_2^r>rv\_2^r$$

which is a stronger requirement than the second order condition. Hence, it is possible that the second order condition is met but not that $v\_1^r+v\_2^r>rv\_2^r$. My question is what happens when this occurs? Is there no Nash equilibrium here? Does the player who makes negative profit drop out, if so what does the remaining player do?

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

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## 1 answers

### Answer by username_1 on 2017-11-27. Score: 3 | Accepted answer

Well it says in the paper on p. 423:

"Thus the sufficiency assumption implies a unique pure strategy equilibrium. And if the sufficiency assumption is violated, at least one player will receive a negative payoff; *that player can improve by reducing his or her effort to zero*, which will destroy the equilibrium." [emphasis mine]

So that player will not make any effort because he/she cannot win. I leave it to you to figure out what the best reply from the other player is in that case.

I guess to be completely mathematically correct one would have to formulate these kind of games with non-negativity constraints, but because it complicates the analysis they're often left out and an interior equilibrium is either assumed or, as here, checked afterwards.

> **Comments:**
> **username_2**: IMO Your middle paragraph is not necessary for this answer and creates confusion.  

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