Climb on the Bandwagon: Consensus and Periodicity in a Lifetime Utility Model with Strategic Interactions

Abstract

What is the emergent long-run equilibrium of a society, where many interacting agents bet on the optimal energy to put in place in order to climb on the Bandwagon? In this paper, we study the collective behavior of a large population of agents being either Left or Right: The core idea is that agents benefit from being with the winner party, but, on the other hand, they suffer a cost in changing their status quo. At the microscopic level, the model is formulated as a stochastic, symmetric dynamic game with N players. In the macroscopic limit as $N \rightarrow +\,\infty$, the model can be rephrased as a mean field game, whose equilibria describe the “rational” collective behavior of the society. It is of particular interest to detect the emerging long time attractors, e.g., consensus or oscillating behavior. Significantly, we discover that bandwagoning can be persistent at the macrolevel: We provide evidence, also on the basis of numerical simulations, of endogenously generated periodicity.

Appendices

A Derivation of the Mean Field HJB Equation

Define

$$\begin{aligned} V(\sigma ,t) := \sup _{u}{{\mathbb {E}}}_{\sigma ,t}\left[ \int _t^{+\,\infty } e^{-\lambda (r-t)} R_u(r) \mathrm{d}r \right] , \end{aligned}$$

where

$$\begin{aligned} R_u(t) :=\sigma (t) m(t) - \frac{1}{2\mu (\sigma (t), m(t))} u^2(t), \end{aligned}$$

u(t) is the rate at which $\sigma (t)$ flips to $-\sigma (t)$, and $u = (u(t))_{t \ge 0}$ ranges over right-continuous nonnegative closed-loop controls, i.e., $u(t) = \varphi _u(t,\sigma (t))$ with $\varphi _u : [0,+\,\infty ) \times \{-1,1\} \rightarrow [0,+\,\infty ]$ right continuous in t. Let

$$\begin{aligned} J_{\sigma ,t}(u) := {{\mathbb {E}}}_{\sigma ,t}\left[ \int _t^{+\,\infty } e^{-\lambda (r-t)} R_u(r) \mathrm{d}r \right] . \end{aligned}$$

If $u^*$ is an optimal control for (2), then by the Bellman principle $V(\sigma ,t) = J_{\sigma ,t}(u^*)$ for all $\sigma ,t$. For t fixed and $h >0$, denote by $u^{h,\alpha }$ the control defined on $[t,+\,\infty )$,

$$\begin{aligned} u^{h,\alpha }(s) = \left\{ \begin{array}{ll} \alpha &{} \text{ for } t \le s < t+h \\ u^*(s) &{} \text{ for } s \ge t+h. \end{array} \right. \end{aligned}$$

Observe that

$$\begin{aligned} J_{\sigma ,t}(u^{h,\alpha })= & {} {{\mathbb {E}}}_{\sigma ,t}\left[ \int _t^{t+h} R_{u^{h,\alpha }}(r) dr + e^{-\lambda h} V(\sigma (t+h),t+h) \right] \nonumber \\= & {} h\left[ \sigma (t) m(t) - \frac{1}{2\mu (\sigma (t), m(t))} \alpha ^2 \right] \nonumber \\&\qquad +\, {{\mathbb {E}}}_{\sigma ,t}\left[ e^{-\lambda h} V(\sigma (t+h),t+h) \right] + o(h) . \end{aligned}$$

(7)

Moreover

$$\begin{aligned} J_{\sigma ,t}(u^{h,\alpha }) \ge V(\sigma ,t) \end{aligned}$$

(8)

for every $\alpha$, while

$$\begin{aligned} J_{\sigma ,t}(u^{h,u^*(t, \sigma )} )= V(\sigma ,t) + o(h), \end{aligned}$$

(9)

where right continuity is used in this last estimate. It follows that

$$\begin{aligned}&- \lim _{h \downarrow 0} \frac{{{\mathbb {E}}}_{\sigma ,t}\left[ e^{-\lambda h} V(\sigma (t+h),t+h) - V(\sigma ,t) \right] }{h} \nonumber \\&\qquad \ge \sigma (t) m(t) - \frac{1}{2\mu (\sigma (t), m(t))} \alpha ^2\,, \end{aligned}$$

(10)

for every $\alpha \ge 0$, with the equality being attained at $\alpha = u^*(t, \sigma )$. By standard results on continuous time Markov chains

$$\begin{aligned}&\lim _{h \downarrow 0} \frac{{{\mathbb {E}}}_{\sigma ,t}\left[ e^{-\lambda h} V(\sigma (t+h),t+h) - V(\sigma ,t) \right] }{h} \\&\qquad = \frac{\partial V}{\partial t}(\sigma ,t) + \alpha \left[ V(-\sigma ,t) - V(\sigma ,t)\right] - \lambda V(\sigma ,t), \end{aligned}$$

and (3) follows.

We, now, show that, if $(V(\sigma ,t), m(t))$ solve (3) coupled to the second equation in (4) and $V(\sigma ,t)$ is bounded, then

$$\begin{aligned} u^*(t) = \mu (\sigma ,m(t)) \left[ \nabla V(\sigma ,t)\right] ^+ \end{aligned}$$

maximizes (2). Note that the equation for m(t) guarantees that ${{\mathbb {E}}}(\sigma ^*(t)) = m(t)$, so $u^*$ is an equilibrium control. To show that $u^*$ maximizes (2) observe that

$$\begin{aligned} \begin{aligned} 0&= -\,\lambda V(\sigma ,t)+u^*(t) \nabla V(\sigma ,t) -\frac{1}{2 \mu (\sigma ,m(t))}(u^*(t))^2 \\&\quad ~+\,\frac{\partial V}{\partial t}(\sigma ,t)+\sigma m(t) \\&= -\,\lambda V(\sigma ,t) + \sup _a \left[ a \nabla V(\sigma ,t) - \frac{1}{2 \mu (\sigma ,m(t))}a^2 \right] \\&\quad ~+\, \frac{\partial V}{\partial t}(\sigma ,t)+\sigma m(t) . \end{aligned} \end{aligned}$$

(11)

Consider now an arbitrary feedback control u, and denote by $\sigma (t)$ the process with control u. A standard application of Ito’s rule for Markov chains yields, for every $t>0$,

$$\begin{aligned}&{{\mathbb {E}}}\,\Bigg \{e^{-\lambda t} V(\sigma (t),t) - V(\sigma (0),0) - \int _0^t \Big [ - \lambda e^{- \lambda s} V(\sigma (s),s) \nonumber \\&\qquad +\, e^{- \lambda s} \frac{\partial V}{\partial s}(\sigma (s),s) - e^{-\lambda s} u(s) \nabla V(\sigma (s),s) \Big ]ds \Bigg \} = 0 \end{aligned}$$

(12)

Using (11):

$$\begin{aligned} u(s) \nabla V(\sigma (s),s) \le \lambda V(\sigma (s),s) +\, \frac{1}{2 \mu (\sigma ,m(s))} u^2(s) - \frac{\partial V}{\partial s}(\sigma (s),s) - \sigma (s) m(s), \end{aligned}$$

which, inserted in (12) gives

$$\begin{aligned}&{{\mathbb {E}}}\,\Bigg \{e^{-\lambda t} V(\sigma (t),t) - V(\sigma (0),0) \nonumber \\&\qquad +\, \int _0^t e^{- \lambda s} \left[ \sigma (s) m(s) - \frac{1}{2 \mu (\sigma ,m(s))} u^2(s) \right] ds \Bigg \} \le 0, \end{aligned}$$

(13)

where equality is attained for $u = u^*$. Letting $t \rightarrow +\,\infty$ and using the boundedness of V, we obtain

$$\begin{aligned} J(u) \le {{\mathbb {E}}}[V(\sigma (0),0)] = J(u^*), \end{aligned}$$

and the proof is complete.

B Derivation of Other Facts

Proof of Facts related to the constant-mobility model We first observe that (5), besides the origin O, admits two other equilibria P and Q, symmetric with respect to the origin: $\pm \left( -\left( {\sqrt{\lambda ^2 + 4 \mu } - \lambda }\right) /{\mu }, 1 \right)$. Linear analysis shows that P and Q are saddle points for all values of the parameters; the origin O is linearly unstable:

• for $\mu \le \frac{\lambda ^2}{8}$ it is repellent, i.e., the eigenvalues of the linearized system are both negative reals;

• for $\mu > \frac{\lambda ^2}{8}$ is an unstable spiral, i.e., the eigenvalues of the linearized system have both negative real part, but nonzero imaginary part.

In order to perform a global analysis, we first consider the nullcline ${\mathcal {N}}$ given by the equation $\frac{\mu }{2}z|z| + \lambda z + 2m=0$. Off the nullcline, solutions to (5) have trajectories that are locally graphs of a function $m = m(z)$. By implicit differentiation, assuming $(z,m) \in [0,+\,\infty ) \times [-1,1]$, it turns out that $m''(z) > 0$ if and only if $\phi ^-(z)<m<\phi ^+(z)$, with

$$\begin{aligned} \phi ^{\pm }(z) = - \frac{z}{4} \left[ \lambda \mp \sqrt{\lambda ^2 - 8 \mu + 6 \lambda \mu z + 4 \mu ^2 z^2} \right] . \end{aligned}$$

For $(z,m) \in (-\infty ,0) \times [-1,1]$, similar convexity conditions are obtained by reflection w.r.t. the origin. Consider the fixed point Q and its stable manifold ${\mathcal {M}}_s$, i.e., the trajectory of a solution of (5) converging to Q.

Low-mobility regime: $\mu \le \frac{\lambda ^2}{8}$. In this case the graphs of $\phi ^+$ and $\phi ^-$ meet at the origin (see Fig. 4, top panel). Moreover, the graph of $\phi ^-$ meets the nullcline ${\mathcal {N}}$ at the equilibrium point Q. A linear analysis at Q and the study of the direction of the vector field of (5) at the points of the graph of $\phi ^-$ show that ${\mathcal {M}}_s$ is at the left of the graph of $\phi ^-$. In particular ${\mathcal {M}}_s$ is concave, so it cannot intersect the nullcline ${\mathcal {N}}$, that can be intersected only vertically by a solution of (5). It follows that ${\mathcal {M}}_s$ is within the area between ${\mathcal {N}}$ and the graph of $\phi ^-$. Since the origin is stable for the time-reversal of (5), necessarily ${\mathcal {M}}_s$ joins the origin with Q. Moreover, in the area between ${\mathcal {N}}$ and the graph of $\phi ^-$, it easily checked that $\frac{dm}{dz} = \frac{{\dot{m}}}{{\dot{z}}} < 0$, so it is the graph of a strictly decreasing function. Thus, for every $m_0 \in (-1,0)$, there is a unique point of ${\mathcal {M}}_s$ with $m=m_0$, which is the starting point of a solution of (5) converging to Q; in particular $m(t) \rightarrow -1$ as $t \rightarrow +\,\infty$. It is actually the only bounded solution starting from a point of the form $(m_0,z)$. This can be seen as follows. The point $(m_0,z)$, with $m_0<0$, cannot belong to the stable manifold of P, which is the image of ${\mathcal {M}}_s$ under reflection w.r.t the origin. Thus the solution starting from $(m_0,z)$ cannot converge to any fixed point. Moreover, since the divergence of the vector field driving (5) is constantly equal to $\lambda >0$, then periodic orbits are not allowed. Thus, by the Poincaré-Bendixon Theorem, the solution starting from $(m_0,z)$ must be unbounded.

Fig. 4

Low-mobility regime (top panel) and high-mobility regime (bottom panel)

High-mobility regime: $\mu > \frac{\lambda ^2}{8}$. In this case the graphs of $\phi ^+$ and $\phi ^-$ do not reach the origin (see Fig. 4, bottom panel). As in the low-mobility regime, the stable manifold ${\mathcal {M}}_s$, as departing from Q, forms a concave curve between ${\mathcal {N}}$ and the graph of $\phi ^-$. If we show that ${\mathcal {M}}_s$ gets arbitrarily close to the origin, then the previous linear analysis implies that it must spiral around the origin, in particular it is not the graph of an injective function.

Thus we are left to show that ${\mathcal {M}}_s$ gets arbitrarily close to the origin. This amounts to show that the solution $({\hat{z}}(t),{\hat{m}}(t))$ of the time-reversed system starting from a point in ${\mathcal {M}}_s$ close to Q, converges to the origin as $t \rightarrow +\,\infty$. Due to the spiraling around the origin, $({\hat{z}}(t),{\hat{m}}(t))$ cannot converge to the origin following the graph of a monotone function. Thus it must intersect first the positive z-axis and then the positive m axis at some $m^*>0$. Suppose $m^*<1$. Note that ${\mathcal {M}}_s$ intersects the m-axis horizontally, so, again by convexity, after having touched $(0,m^*)$ it continues downward. Since ${\mathcal {M}}_s$, in the half-plane $z<0$ cannot touch the stable manifold of P, it follows it is trapped in a bounded region. Due to the absence of periodic orbits, necessarily $({\hat{z}}(t),{\hat{m}}(t)) \rightarrow (0,0)$ as $t \rightarrow +\,\infty$.

Finally, we need to show that $m^*<1$. By continuity from the low-mobility regime, this is certainly true for $\mu - \frac{\lambda ^2}{8}$ sufficiently small. If our claim is false, then there must be a value of $\mu$ for which $m^* = m^*(\mu ) = 1$. In this situation, ${\mathcal {M}}_s$ continuous horizontally up to P. It follows that the union of ${\mathcal {M}}_s$ with the stable manifold of P forms a closed curve, tangent to the vector field driving (5); this is impossible by the Divergence Theorem.

Sketch of the proof of Facts related to the crowding effects model

We first observe that (6) has three equilibria: the origin O, whose linear properties are identical to those of the constant-mobility model treated in the previous section, and the points P and Q with coordinates $\pm \left( -\frac{2}{\lambda },1 \right)$. Both P and Q are easily seen to be saddle points, for all values of the parameters. Similarly to the constant-mobility case, the manifolds of P and Q can be proved to be monotone functions in the low-mobility regime, while they spiral around the origin in the moderate-mobility regime. What fails here is that the divergence of the driving vector field is not of constant sign, so that limit cycles cannot be ruled out. Although we do not have a full proof about the existence of a limit cycle, we provide clear evidence based on arguments derived by numerical inspection. Our analysis suggests that the m coordinate of the first intersection of the stable manifold of Q with the m-axis is increasing in $\mu$, and it equals 1 at some $\mu = {\hat{\mu }}$. Then, the manifold continues horizontally to reach P (as depicted in Fig. 2). Thus, by symmetry, the two stable manifolds join to form a separatrix. By increasing $\mu$ further, a periodic orbit bifurcates from the separatrix through a homoclinic bifurcation.