This maybe a trivial question to most. I am fairly new to game theory.
The usual cournot duopoly (same constant marginal cost for both players) is solved using pure strategies. Are there mixed strategy solutions also? If not, is it obvious or is that solving would give p=1 for the pure strategy solution? What's the economic logic for (not) having mixed strategy equilibria?
EDIT: Let there be two players $A$ and $B$ who produce, respectively, $q_a$ and $q_b$ quantities of the same good. The market price of the good is determined by the inverse demand function: $P=\alpha - \beta q$; where $q\equiv q_a + q_b$. The marginal cost of production is $c$ for both $A$ and $B$.
So the profit function is: $\pi_i=Pq_i-cq_i$; $\forall i=a,b$
The best response function, $R_i(q_j)$ is obtained by $\partial \pi_i/ \partial q_i=0$
The pure strategy equilibrium is intersection point of the two response functions.
It is enough to show that a best-response of player $i\in\{A,B\}$ to any mixed strategy by player $j, j\neq i$ is a pure-best response. This implies that there cannot be a mixed strategy NE.
To see this, suppose that player $j$ uses a mixed strategy described by a distribution $F$, that is $q_j\sim F$, such that $\mathbb{E}(q_j)$ exists. Let $\bar{q}_j \equiv\mathbb{E}(q_j)$. Then, the problem of player $i$, $i\neq j$, is given by:
$$ max_{q_i} [(\alpha-\beta \mathbb{E}(q)] q_i - cq_i) $$
But, $\mathbb{E}(q)=q_i+\bar{q}_j$ and so the F.O.C. to the above problem is $$ \alpha - \beta(2q_i+\bar{q}_j)-c=0$$
Rearranging, we obtain $$ q_i = \frac{\alpha-\beta \bar{q}_j-c}{2\beta}$$
From this equation, we see that there is a unique quantity $q_i$ that is the best-response to player $j's$ mixed-strategy. It follows that players will always use pure-strategies.