There are two agents $i=1,2$. Consider the following programm \begin{align} &V_1(x_0) := \max_u \int^\infty_0 e^{-\rho t}F_1(x(t),u(t),v(t))dt\\ &V_2(x_0) := \max_v \int^\infty_0 e^{-\rho t}F_2(x(t),u(t),v(t))dt\\ s.t.~&\dot x(t)=f(x(t),u(t),v(t))\\ &x(0) = x_0 \end{align} where $\rho > 0$ denotes time preference, $V_i(\cdot)$ is the value and $F_i(\cdot)$ an objective function. $x\in X = [0,2]$ is the state variable and $u\in U=[0,1]$ the control of agent 1 and $v\in V=[0,1]$ the control of agent 2 respectively. The state is governed by $f(\cdot)$. The Hamilton-Jacobi-Bellman equation for each agent is given by \begin{align} \rho V_1(x)=\max_u [F(x,u,v^*) + V_1'(x)f(x,u,v^*)],\quad \forall t\in[0,\infty)\\ \rho V_2(x)=\max_v [F(x,u^*,v) + V_2'(x)f(x,u^*,v)],\quad \forall t\in[0,\infty)\\ \end{align}

given the respective maximizers \begin{align} u^* &= \max_u [F(x,u,v^*) + V_1'(x)f(x,u,v^*)]\\ v^* &=\max_v [F(x,u^*,v) + V_2'(x)f(x,u^*,v)] \end{align}

such that the HJBs become \begin{align} \rho V_1(x)=F(x,u^*,v^*) + V_1'(x)f(x,u^*,v^*)\\ \rho V_2(x)=F(x,u^*,v^*) + V_2'(x)f(x,u^*,v^*) \end{align}

Symmetric equilibirum

A symmetric equilibirum is given at $\dot x = 0 \Leftrightarrow f(\tilde x,\tilde u,\tilde v) = 0$ with $\tilde x = 1$ and $\tilde u=\tilde v$ and $V_1(\tilde x) = V_2(\tilde x) =: V(\tilde x)$.


The equilibirum controls $\tilde u$ and $\tilde v$ can't be determinend with the information at hand. The equation \begin{align} \rho V(\tilde x)=F(\tilde x,\tilde u, \tilde v) + V'(\tilde x)\underbrace{f(\tilde x,\tilde u, \tilde v)}_{=0} \end{align} is true for every $\{(u,v)\in[0,1]\times[0,1]:u=v\}$. That is, we have multiple equilibria.

Select Equilibrium

My idea is (I made this up, haven't read anything about it) that I select the equilibrium associated with the highest value. We can determine $V(\tilde x)$ for all $\{(u,v)\in[0,1]\times[0,1]:u=v\}$. Say $V(\tilde x)$ is monoton increasing in $u$ and $v$, i.e. \begin{align} \lim_{u=v\to 0} V(\tilde x) < \lim_{u=v\to 1} V(\tilde x) \end{align}

Perosnal I would choose the fixed point $(k = 1, u = 1, v = 1)$. I'd like to know if I can motivate it formally as the unique solution.

  • Do I select the equilibrium associated with the highest value, by definition of the value function?
  • Can you point me to some literature concerning this point?

Motivating example

Let $F_1(x,u,v) = xu^2$ and $F_2(x,u,v) = (2-x)v^2$ with $f(x,u,v) = v-u$. The HJBs read (with $\rho=1$) \begin{align} V_1(x)&=\max_u [xu^2 + V_1'(x)(v^*-u)]\\ V_2(x)&=\max_v [(2-x)v^2 + V_2'(x)(v-u^*)] \end{align}

Maximizers are \begin{align} u^*&= \frac{V'_1(x)}{2x}\in[0,1]\\[2mm] v^*&=\frac{V'_2(x)}{2(2-x)}\in[0,1] \end{align}

In symmetric equilibrium we have $\tilde x = 1$ and $\tilde v = \tilde u \Leftrightarrow \dot x = 0$ which gives \begin{align} V'_1(1) = -V'_2(1) \end{align}

The HJb simplifies to \begin{align} V_1(1)=\left(\frac{V'_1(1)}{2}\right)^2 = \left(\frac{-V'_2(1)}{2}\right)^2 = V_2(1) \end{align}

Since both vlaues are equal in equilibirum we proceed with 1. We know from the control space that \begin{align} 0\leq V'_1(x) \leq 2x \end{align}

Which is in equilibirum \begin{align} 0\leq V'_1(1) \leq 2 \end{align}

We can evaluate $V(1)$ for all $V'_1(1)\in[0,2]$ or since $\tilde u=V'_1(1)/2$ for all $\tilde u \in[0,1]$. In the picture I highlighted two possible equilibria $E^A$ and $E^B$. Since the payoff is inreasing with the control we have a higher value associated with equilibirum $E^A$, i.e. $V^A(1) > V^B(1)$.

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2 Answers 2


I'm not sure I follow the logic on that equation having infinitely many solutions and steady states. In any case, in what follows are some guidelines for equilibrium selection.

It depends a lot on the context. Here are some criteria, in descending order of relevance

Clean Ways

  • Is any of the steady states unstable? If so, it's less likely to be the one that we*re observing in reality. See for example the 0-steady state in the Solow model.
  • The last point, somewhat generalized; Is there any meaningful starting value? Due to path dependence, you will only converge to one steady state from your meaningful value.
  • The last point, somewhat generalized; Do we converge more often to one than the other? Basically take a range of meaningful starting values and see whether most of them lead to the same steady state

Believe in Coordination

If all these fail... make the government powerful. Kaplan and Menzio (2015) model unemployment, and can get two steady states given their calibration. They take the lower unemployment level as the "reasonable" steady state, that they use to calibrate parameters against. I don't think they spell it out, but the argument most likely is that if there are two steady states, a powerful government could steer the economy to the steady state implying the highest welfare.

This is basically your "highest value" point - but it should be clear that there is some reason why that is achieved. That's why I said "it depends a lot on the context" - many coordination failure models hinge critically on the assumption that no higher power can steer the economy. See also the Chicken Economy model, a term I believe originally to be coined in Minnesota.


Finally, if all else fails, sun spots are what determine which equilibrium you are in. In economics, we mean by that basically a random exogenous variable, something not modeled. They're very disputed as they have no economic meaning and are hard to digest. Basically, if there was anything that had economic meaning and could determine the equilibrium, you would rather put it into your model than having a coin flip determine it.

See also Karl Shell's summary on the subject for The New Palgrave: A Dictionary of Economics.


In addition to the solutions that FooBar presented in his answer, one of the ways that people are able to get rid of multiple equilibria is incomplete information. I can't find the papers that are most relevant to your example at the moment, but the "canonical" example that I think of is Morris Shin 1998. Essentially what happens is the following:

Their model is a model of a "currency attack." There is a state of the world that determines whether a currency attack is worthwhile or not and to successfully attack a currency a certain number of investors (which depends on the state) must participate in the attack. Under complete information multiple equilibria can be supported, but with even a very small amount of noise in their signal about the underlying state of the world the equilibrium becomes unique.

One relevant paper could be "Coordinating Business Cycles." I don't remember most of the details in this paper, but it essentially has the same type of result in a more fully developed model.


  • Stephen Morris and Hyun Song Shin. *Unique Equilibrium in a Model of Self-Fulfilling Currency Attacks. The American Economic Review. 1998.
  • Edouard Schaal and Mathieu Taschereau-Dumouchel. Coordinating Business Cycles. Working Paper. 2014.
  • 1
    $\begingroup$ Sleeping over it for a night - and now that OP has given more information on his setup - I believe that we should open a new, general question on the subject, and post these answers there. $\endgroup$
    – FooBar
    Jul 17, 2015 at 9:34

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