<p>The monotonicity result that they prove in their lemma 1 (p. 556) is stronger than you think: it states that whenever $q&gt;q'$, the support of $R_1(q)$ is "uniformly below" the support of $R_1(q')$. In other words, it is impossible to find <em>any</em> realizations $(r,r')$ of $R_1(q)$ and $R_1(q')$, respectively such that $r&gt;r'$.</p>

<p>The proof of this result relies on the idea that firms decisions are strategic substitutes in the Cournot duopoly model: the more firm 2 produces, the less firm 1 wants to produce. Thus, if firm 1 finds it optimal to produce an amount $r'$ with positive probability when firm 2 produces $q'$, it cannot find it optimal to produce a larger amount ($r&gt;r'$) when firm 2 itself produces more ($q&gt;q')$.</p>

<p>Consider therefore a production level $q$ by firm 2 such that $0$ is a realization of $R_1(q)$. Take $\epsilon&gt;0$ and suppose that there exists $r&gt;0$ that is a realization of $R_1(q+\epsilon)$. The condition
\begin{equation*}
r &gt; 0 \text{ and } q&lt;q+\epsilon
\end{equation*}
violates their lemma 1, since $r$ is a realization of $R_1(q+\epsilon)$ and $q$ is a realization of $R_1(q)$. This proves that $R_1(q+\epsilon)$ has all its weight on 0.</p>

CC BY-SA 3.0
2016-01-20T20:15:15.580Z
2016-01-20T20:15:15.580Z
Oliv
<p>The full paper can be found here: <a href="http://www.dklevine.com/archive/refs4397.pdf" rel="noreferrer">http://www.dklevine.com/archive/refs4397.pdf</a>. I refer here to the Lemma 2 in page 557.</p>

<p>My doubt is regarding the first statement of the proof: "Because reaction functions are nonincreasing and $0$ is a realization of $R^1(q)$, firm 1 must react to any quantity above $q$ by setting $0$ with probability $1$". Why is that so? If $0$ is a <em>possible</em> realization of $R^1(q)$ (if I understood correctly what the authors mean by <em>realization</em> -- since the reaction functions can be understood as random variables --, and maybe that is where I am troubled), what guarantees the fact that <em>any</em> small quantity above $q$ will make firm 1 set its reaction function equal $0$?   </p>

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2016-01-11T22:39:16.760Z
I'm very familiar with this paper, but never really gave importance to Lemma 2. All it does is extend the results to consider the case when the reaction functions could be random variables instead of deterministic (which isn't very realistic anyway). It's more there for completeness and theoretical rigor. You could omit it by simply assuming the Rs aren't random variables. You won't need it to understand the rest of the paper. So, if time constrained and not going into pure theory I would skip over it, especially the first few times you read it.
BB King
CC BY-SA 3.0
2016-01-12T02:45:45.040Z
It seemed that it wouldn't too important ahead, yes. But I'm not time constrained so I really would like to know what the authors meant by that statement.
John Doe
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2016-01-19T19:31:38.997Z
This question could be made more searchable, useful to others and easier to answer by typing up lemma 2 and the associated proof within your question.
BKay
CC BY-SA 3.0
2016-01-20T18:51:25.947Z
@Bkay, I will do that as soon as I have more time.
John Doe
<game-theory><industrial-organisation>
Lemma 2 in Tirole and Maskin's Dynamic Oligpoly I (1988)
John Doe
# Lemma 2 in Tirole and Maskin's Dynamic Oligpoly I (1988)

(Asked by: username_0 on 2016-01-11)

The full paper can be found here: <http://www.dklevine.com/archive/refs4397.pdf>. I refer here to the Lemma 2 in page 557.

My doubt is regarding the first statement of the proof: "Because reaction functions are nonincreasing and $0$ is a realization of $R^1(q)$, firm 1 must react to any quantity above $q$ by setting $0$ with probability $1$". Why is that so? If $0$ is a *possible* realization of $R^1(q)$ (if I understood correctly what the authors mean by *realization* -- since the reaction functions can be understood as random variables --, and maybe that is where I am troubled), what guarantees the fact that *any* small quantity above $q$ will make firm 1 set its reaction function equal $0$?

**Tags:** game-theory, industrial-organisation
> **Comments:**
> **username_1**: I'm very familiar with this paper, but never really gave importance to Lemma 2. All it does is extend the results to consider the case when the reaction functions could be random variables instead of deterministic (which isn't very realistic anyway). It's more there for completeness and theoretical rigor. You could omit it by simply assuming the Rs aren't random variables. You won't need it to understand the rest of the paper. So, if time constrained and not going into pure theory I would skip over it, especially the first few times you read it.  
> **username_0**: It seemed that it wouldn't too important ahead, yes. But I'm not time constrained so I really would like to know what the authors meant by that statement.  
> **username_2**: This question could be made more searchable, useful to others and easier to answer by typing up lemma 2 and the associated proof within your question.  
> **username_0**: @Bkay, I will do that as soon as I have more time.  

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## 1 answers

### Answer by username_3 on 2016-01-20. Score: 5 | Accepted answer

The monotonicity result that they prove in their lemma 1 (p. 556) is stronger than you think: it states that whenever $q>q'$, the support of $R\_1(q)$ is "uniformly below" the support of $R\_1(q')$. In other words, it is impossible to find *any* realizations $(r,r')$ of $R\_1(q)$ and $R\_1(q')$, respectively such that $r>r'$.

The proof of this result relies on the idea that firms decisions are strategic substitutes in the Cournot duopoly model: the more firm 2 produces, the less firm 1 wants to produce. Thus, if firm 1 finds it optimal to produce an amount $r'$ with positive probability when firm 2 produces $q'$, it cannot find it optimal to produce a larger amount ($r>r'$) when firm 2 itself produces more ($q>q')$.

Consider therefore a production level $q$ by firm 2 such that $0$ is a realization of $R\_1(q)$. Take $\epsilon>0$ and suppose that there exists $r>0$ that is a realization of $R\_1(q+\epsilon)$. The condition
\begin{equation\*}
r > 0 \text{ and } q


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