<p>Let $F(p)$ denote the distribution of prices in a market, $\pi(p, F)$ are profits choosing $p$ given distribution $F$. $E\pi(p,F)$ is defined to be</p>

<p>$$ E \pi(p, F) = R(p) \psi(p, F)$$</p>

<p>where $R(p) = p D(p)$ is the revenue per customer, and $\psi(p, F)$ denotes the expected share of customers that a seller with price $p$, given distribution $F$ will face.</p>

<p><strong>Definition (2.1)</strong> in <a href="http://www.sciencedirect.com/science/article/pii/0167718794004749" rel="nofollow">Stahl (1996)</a> (just below eq. 2.3) defines the symmetric Nash equilibrium as:</p>

<p>$$ E \pi(p,F) \leq \max_p p E \pi(p, F) \, \forall p$$</p>

<p>My question is why $(2.1)$ is defined as such? I would have thought that a Nash equilibrium requires choice being optimal, hence no incentive to change profits:</p>

<p>$$ E \pi(p,F) \leq \max_p E \pi(p,F) \, \forall p$$</p>

CC BY-SA 3.0
2015-03-31T08:33:41.903Z
To me it looks like a typo. as R(p) = p D(p) we already have the p in expectation. Have you tried checking the proofs? How it is used in there?
The Almighty Bob
CC BY-SA 3.0
2015-03-31T15:03:24.633Z
@TheAlmightyBob On page 350 he says "*[...] These are necessary conditions that stem from the requirement that $E\pi(\bar p, F) \geq E\pi(p, F)$*". This almost makes me think it is a typo, but (i) there's no maximization operator on the RHS here, so I may just overlook something (ii) it would be a typo in the fundamental definition of the paper, and that's somewhat hard to believe.
FooBar
CC BY-SA 3.0
2015-04-01T07:12:31.793Z
I have to say that I thought this was a typo too. The inequality sign is a bit confusing, because the condition holds with equality almost everywhere. The condition $E\pi(p,F)=\max_p E\pi(p,F)\forall p$ looks much more like a standard mixed strategy equilibrium condition. The $E\pi(p,F)= E\pi(\overline{p},F)\forall p$ condition makes sense because the way one often solves for these distributions is to set profits from an arbitrary $p$ to those from the maximum of the support. Doing this ensures that $F(\overline{p})=1$ on the RHS so there is only one $F$ term (on the LHS) to solve for.
Ubiquitous
<game-theory><industrial-organisation><nash-equilibrium>
Symmetric Nash Equilibrium in Stahl (1996)
FooBar
# Symmetric Nash Equilibrium in Stahl (1996)

(Asked by: username_0 on 2015-03-30)

Let $F(p)$ denote the distribution of prices in a market, $\pi(p, F)$ are profits choosing $p$ given distribution $F$. $E\pi(p,F)$ is defined to be

$$ E \pi(p, F) = R(p) \psi(p, F)$$

where $R(p) = p D(p)$ is the revenue per customer, and $\psi(p, F)$ denotes the expected share of customers that a seller with price $p$, given distribution $F$ will face.

**Definition (2.1)** in [Stahl (1996)](http://www.sciencedirect.com/science/article/pii/0167718794004749) (just below eq. 2.3) defines the symmetric Nash equilibrium as:

$$ E \pi(p,F) \leq \max\_p p E \pi(p, F) \, \forall p$$

My question is why $(2.1)$ is defined as such? I would have thought that a Nash equilibrium requires choice being optimal, hence no incentive to change profits:

$$ E \pi(p,F) \leq \max\_p E \pi(p,F) \, \forall p$$

**Tags:** game-theory, industrial-organisation, nash-equilibrium
> **Comments:**
> **username_1**: To me it looks like a typo. as R(p) = p D(p) we already have the p in expectation. Have you tried checking the proofs? How it is used in there?  
> **username_0**: @TheAlmightyBob On page 350 he says "*[...] These are necessary conditions that stem from the requirement that $E\pi(\bar p, F) \geq E\pi(p, F)$*". This almost makes me think it is a typo, but (i) there's no maximization operator on the RHS here, so I may just overlook something (ii) it would be a typo in the fundamental definition of the paper, and that's somewhat hard to believe.  
> **username_2**: I have to say that I thought this was a typo too. The inequality sign is a bit confusing, because the condition holds with equality almost everywhere. The condition $E\pi(p,F)=\max_p E\pi(p,F)\forall p$ looks much more like a standard mixed strategy equilibrium condition. The $E\pi(p,F)= E\pi(\overline{p},F)\forall p$ condition makes sense because the way one often solves for these distributions is to set profits from an arbitrary $p$ to those from the maximum of the support. Doing this ensures that $F(\overline{p})=1$ on the RHS so there is only one $F$ term (on the LHS) to solve for.  

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## 0 answers

