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Denote by $i \in \{1, \ldots, n\}$ an economic agent. Let $\mathbf x \in \mathbb R^n$ denote a vector of actions and $x_i \in \mathbf x$ a typical element. Let further $f_i : \mathbb R^n \to \mathbb R$ denote the objective function. The vector $\mathbf x^*$ constitutes a Nash equilibrium if the following holds \begin{align} \forall i : f_i(\mathbf x^*) = \max_{x_i} f_i(x_i, \mathbf x_{-i}^*) \end{align} where $\mathbf x_{-i}^* := (\ldots, x_{i-1}^*, x_{i+1}^*, \ldots)$ are the equilibrium actions of the opponents. The $n$ first order conditions read \begin{align} \frac{\partial f_i(x_i, \mathbf x_{-i}^*)}{\partial x_i}\Bigg|_{x_i = x_i^*} = 0. \end{align} The second order conditions read \begin{align} \frac{\partial^2 f_i(x_i, \mathbf x_{-i}^*)}{\partial x_i^2}\Bigg|_{x_i = x_i^*} < 0. \end{align} Assume symmetry such that $x_j^* = x^* ~ \forall j$. Since the problem is quite complex I cannot derive the second order partial derivative. I was thus thinking to define the first derivative in a single argument (the symmetric equilibrium actions) \begin{align} \forall i : f'(x^*) = f'_i(x^*) := \frac{\partial f_i(x_i,\mathbf x_{-i}^*)}{\partial x_i} \Bigg|_{x_j^* = x^* ~ \forall j} \end{align} and then take the derivative of the the simplier first order derivative and check wether it's negative. In order to make the simplifictaion in between I must somehow make sure that no information gets lost and that's why I'm wondering if the follwoing relation holds in general \begin{align} \text{sign}\left[f''(x^*)\right] ~~ \substack{? \\ =} ~~ \text{sign}\left[\frac{\partial^2 f_i(x_i,\mathbf x_{-i}^*))}{\partial x_i^2} \Bigg|_{x_j^* = x^* ~ \forall j}\right]. \end{align}

I was thinking that there may exists a general theorem from optimization for symmetric actions.

Example Consider text book Cournot duopol with $i,j \in \{1,2\}$, $i \neq j$ and \begin{align} f_i(x_i, x_j) = (1-x_i-x_j)x_i. \end{align} Now we readily get the following partial derivatives \begin{align} &\frac{\partial f_i(x_i, x_j^*)}{\partial x_i}\Bigg|_{x_i = x_i^*} = 1 - 2x_i^* - x_j^* \\[2mm] &\frac{\partial^2 f_i(x_i, x_j^*)}{\partial x_i^2}\Bigg|_{x_i = x_i^*} = -2\\[2mm] &f'(x^*) = 1-3x^*\\[2mm] &f''(x^*) = -3\\ \end{align} So the sign is the same, but the values differ.

Solution It seems that the idea was published in the latest Theoretical Economics and the approach is valid. The author actually discusses the super simple Cournot example as well.

The following statement is adapted from Theorem 1 of Hefti (2017): "Equilibria in symmetric games: Theory and applications", Theoretical Economics 12, pp. 979–1002

Suppose $f_i(x_i, \mathbf x_{-i})$ is strongly quasiconcave in $x_i$. Define $f'(x)$ as above.

Let $x^* \in \mathbb R$ solve $f'(x^*) = 0$. If $f''(x^*) < 0$ then there exists a unique symmtric equilibrium.

Further Issues In footnote 7 the author argues

The assumption of a strongly quasiconcave payoff function in own strategies is mainly for convenience, because it is a sufficient condition for the existence of a (possibly differentiable) best-reply function. However, many results require only differentiability of best replies (and do not otherwise hinge on quasiconcavity), and some results do not require that best replies are everywhere differentiable.

This statment is particular useful, because I have a problem in mind where I cannot say anything about (quasi)concavity of the objective function. I'm, however, not sure if I can apply Theorem 1, because the author does not specify when quasiconcavity is imposed for a result.

How would I know wether the theorem is applicable?

Note that I cannot(!) derive a best response function $x_i^* = \phi_i(\mathbf x_{-i}^*)$, but only the symmetric equilibrium actions $x_j = x^* ~ \forall j$.

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  • $\begingroup$ The question might be better suited for math stack exchange, because it really is a question on technique only. I'd be happy to transfer. $\endgroup$ – clueless Jan 5 '18 at 13:17
  • $\begingroup$ It is possible that I misunderstand something, because based on your definitions I don't see the difference between $f''_i(x^*)$ and $\frac{\partial^2 f_i(x_i,\mathbf x_{-i}^*))}{\partial x_i^2} \Bigg|_{x_j^* = x^* ~ \forall j}$. Perhaps you meant to take the full derivative somewhere? $\endgroup$ – Giskard Jan 5 '18 at 18:55
  • $\begingroup$ @denesp if I’ve understood correctly, $f^{\prime \prime}_i \left (x^* \right)$ is more akin to a “total” derivative, not a partial one. $\endgroup$ – Theoretical Economist Jan 5 '18 at 20:01
  • $\begingroup$ @clueless I think you’re right that this is probably better suited for MSE. My suspicion is that this is not true in general, but it might hold true when there is sufficient symmetry in $f_i$ across its arguments. I don’t have a counter-example or a proof, however. $\endgroup$ – Theoretical Economist Jan 5 '18 at 20:03
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    $\begingroup$ @denesp The difference is when you impose symmetry. Considering $f''_i(x^*)$ you impose symmetry after the first partial derivative of $f_i(x_i, \mathbf x^*_{-i})$ w.r.t to $x_i$. While for $\frac{\partial^2 f_i(x_i, \mathbf x^*_{-i})}{\partial x_i^2}\Bigg|_{x_j = x^* \forall j}$ you first take the second partial derivative and then impose symmetry. Please, check the example for clarifaction. $\endgroup$ – clueless Jan 6 '18 at 12:06

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