A Dynamic Collective Choice Model with an Advertiser

Abstract

This paper studies a dynamic collective choice model in the presence of an advertiser, where a large number of consumers are choosing between two alternatives. Their choices are influenced by the group’s aggregate choice and an advertising effect. The latter is produced by an advertiser making investments to convince as many consumers as possible to choose a specific alternative. In schools, for example, teenagers’ decisions to smoke are considerably affected by their peers’ decisions, as well as the ministry of health campaigns against smoking. We model the problem as a Stackelberg dynamic game, where the advertiser makes its investment decision first, and then the consumers choose one of the alternatives. On the methodological side, we use the theory of mean field games to solve the game for a continuum of consumers. This allows us to describe the consumers’ individual and aggregate behaviors, and the advertiser’s optimal investment strategies. When the consumers have sufficiently diverse a priori opinions toward the alternatives, we show that a unique Nash equilibrium exists between them, which predicts the distribution of choices over the alternatives, and the advertiser can always make optimal investments. For a certain uniform distribution of a priori opinions, we give an explicit form of the advertiser’s optimal investment strategy and of the consumers’ optimal choices.

Appendix A

1.1 Quantities Related to the mA Nash Equilibrium

For all $x\in L_2([0,T],\mathbb {R}^n)$,

$$\begin{aligned} \varDelta (x) =\frac{Mq}{r}(p_{1}-p_{2})' \int _T^0 \! \int _T^\eta \; \phi (\eta , T)'B^{(2)}\phi (\eta , \sigma )x(\sigma ) \, \mathrm {d}\sigma \mathrm {d}\eta , \end{aligned}$$

(24)

where $\phi$ is the unique solution of $\frac{d}{dt}\phi (t,s)=\left( \frac{1}{r}\varGamma (t) B^{(2)}-A'\right) \phi (t,s)$, $\phi (s,s)=I_n$, and

$$\begin{aligned} \dot{\varGamma }&=\frac{1}{r} \varGamma B^{(2)}\varGamma -\varGamma A-A'\varGamma -q I_{n},\qquad \varGamma (T) = MI_n \nonumber \\ \beta&= M\phi (0,T)(p_{2}-p_{1})\nonumber \\ \delta&= \frac{1}{2}M(\Vert p_{2}\Vert ^2-\Vert p_{1}\Vert ^2) +\frac{M^{2}}{2r}p_{2}' \int _T^0 \Big (\phi (\eta , T)'B\Big )^{(2)} \mathrm {d}\eta \,p_{2} \nonumber \\&\quad -\frac{M^{2}}{2r}p_{1}' \int _T^0 \Big (\phi (\eta , T)'B\Big )^{(2)} \mathrm {d}\eta \,p_{1}. \end{aligned}$$

Proof of Theorem 2

The cost functional $\bar{J}_0$ is positive and coercive with respect to $v\in L_2([0,T])$, i.e., $\lim \limits _{\Vert v\Vert _2 \rightarrow \infty }\bar{J}_0(v)/\Vert v\Vert _2=\infty$. If we show that $\bar{J}_0$ is continuous in the reflexive Banach space $L_2([0,T])$ with respect to v, then by Tonelli’s existence theorem [7, Theorem 5.51], $\bar{J}_0$ has a finite minimum. Thus, we need only show that $\bar{J}_0$ is continuous. The state y is continuous with respect to v. The fixed points $\lambda (y)$ of F are continuous with respect to v. In fact, consider v and $v'$ in $L_2([0,T])$ and denote by $y, y'$ the corresponding MA trajectories and by $\lambda$ and $\lambda '$ the corresponding fixed points. We have

$$\begin{aligned} |\lambda - \lambda '|&=|F(\lambda ,y)- F(\lambda ',y')| \le |F(\lambda ,y)- F(\lambda ',y)| +|F(\lambda ',y) -F(\lambda ',y')|\\&\le \sup \limits _{s \in [0,1]}\Big |\frac{dF}{d\lambda } (s,y)\Big | |\lambda - \lambda '|+ |F(\lambda ',y) -F(\lambda ',y')|. \end{aligned}$$

Therefore, $\Big (1-\sup \limits _{s \in [0,1]}\Big |\frac{dF}{d\lambda } (s,y)\Big |\Big ) |\lambda - \lambda '|\le |F(\lambda ',y) -F(\lambda ',y')|.$ Under Assumption 3, $\sup \limits _{s \in [0,1]}\Big |\frac{dF}{d\lambda } (s,y)\Big |<1$. Under Assumption 2, $\bar{F}$ is continuous. Moreover, $\varDelta$ is continuous with respect to the $L_2$ norm $\Vert \Vert _2$. Hence, F is continuous with respect to y, and $|F(\lambda ',y)-F(\lambda ',y')|$ converges to zero as $\Vert y-y'\Vert _2$ converges to zero. Therefore, the fixed points $\lambda$ of F are continuous. In view of (12) and the continuity of the fixed points $\lambda$, $\bar{x}(T)$ is continuous. Therefore, $\bar{J_0}$ is continuous.$\square$

Proof of Theorem 3

We derive the condition on $v^*$ (17) by studying the first variation of the cost functional in (13) with respect to a perturbation $v=v^*+\eta \delta v$, where $\eta \in \mathbb {R}$, and $\delta v \in L_2([0,T],\mathbb {R}^{m_1})$. To this end, we need to derive at first an explicit form of the constraint on $\bar{x}_v$. We have that $\bar{x}_v=\bar{x}^\lambda$ defined in (12), where $\lambda$ is the unique fixed point of $\lambda \mapsto F(\lambda ,y)$. By taking the derivative of $\bar{x}^\lambda$ with respect to time, we obtain that,

$$\begin{aligned} \dot{\bar{x}}_v=\mathcal {L}(\bar{x}_v,y)(t), \qquad \bar{x}(0)=\mu _0, \end{aligned}$$

(25)

where

$$\begin{aligned} \mathcal {L}(\bar{x}_v,y)(t)&= \left( A- \frac{1}{r} B^{(2)}\varPi \right) \bar{x}_v -\frac{q}{r}B^{(2)}\int _T^t \varPhi (\sigma ,t)'K(p_2)y(\sigma )\mathrm {d}{\upsigma }\nonumber \\&\quad + \frac{M}{r}B^{(2)}\varPhi (T,t)'\bar{F}\circ \varDelta \big (\alpha \bar{x}_v+K(p_2)y)\big )(p_1-p_2)+ \frac{M}{r}B^{(2)}\varPhi (T,t)' p_2. \end{aligned}$$

We compute now the Gâteaux derivatives [7] of y and $\bar{x}$ at $v^*$ in the direction $\delta v$:

$$\begin{aligned} \begin{aligned} \frac{d}{d\eta }y_{v^*+\eta \delta v} \Big |_{\eta =0}&:=\delta y\\ \frac{d}{d\eta }\bar{x}_{v^*+\eta \delta v} \Big |_{\eta =0}&:=\delta \bar{x}, \end{aligned} \end{aligned}$$

(26)

where,

$$\begin{aligned} \frac{d}{dt}\delta y&= A_0\delta y + B_0\delta v, \;\; \delta y(0)=0\\ \frac{d}{dt} \delta \bar{x}&= \mathcal {L}_1 (\delta y)(t) + \mathcal {L}_2 (\delta \bar{x})(t), \;\; \delta \bar{x}(0)=0, \end{aligned}$$

and $\mathcal {L}_1$ (resp. $\mathcal {L}_2$) is a continuous linear operator from the Hilbert space $L_2([0,T],\mathbb {R}^{n_1})$ (resp. $L_2([0,T],\mathbb {R}^{n})$) to $L_2([0,T],\mathbb {R}^{n})$ such that for all $z_1 \in L_2([0,T],\mathbb {R}^{n_1})$ and $z_2 \in L_2([0,T],\mathbb {R}^{n})$,

$$\begin{aligned} \mathcal {L}_1 (z_1)(t)&=-\frac{q}{r}B^{(2)}\int _T^t \varPhi (\sigma ,t)'K(p_2)z_1(\sigma )\mathrm {d}{\upsigma }\\&\quad +\frac{M}{r}\xi ^*\varDelta \left( K(p_2) z_1 \right) B^{(2)} \varPhi (T,t)' (p_1-p_2),\\ \mathcal {L}_2 (z_2)(t)&=\left( A-\frac{1}{r} B^{(2)}\varPi \right) z_2(t) + \frac{M\alpha }{r}\xi ^*\varDelta \left( z_2 \right) B^{(2)}\varPhi (T,t)' (p_1-p_2). \end{aligned}$$

Using Fubini–Tonelli’s theorem [23], one can show that the adjoint operators of $\mathcal {L}_1$ and $\mathcal {L}_2$ are, respectively, $\mathcal {L}_1^*$ and $\mathcal {L}_2^*$ defined in (16). We recall from [24] that the adjoint operator of a linear continuous operator $\mathcal {G}$ defined from the Hilbert space $(H_1,\langle ,\rangle _1)$ into the Hilbert space $(H_2,\langle ,\rangle _2)$ is the linear continuous operator $\mathcal {G}^*$ defined from the Hilbert space $(H_2,\langle ,\rangle _2)$ into the Hilbert space $(H_1,\langle ,\rangle _1)$ and satisfying for all $x\in H_1$ and $y \in H_2$ $\langle \mathcal {G}(x), y \rangle _2 = \langle x, \mathcal {G}^*(y) \rangle _1.$ Here, we use the explicit form of the operator $\varDelta$ (24). The Gâteaux derivative of $\bar{J}_0$ is

$$\begin{aligned} \delta \bar{J}_0&=\frac{d}{d\eta } \bar{J}_0 \Big (v^*+\eta \delta v,\bar{x}_{v^*+\eta \delta v}\Big ) \bigg |_{\eta =0} \\&= r_0 \Big \langle v^*, \delta v \Big \rangle +M_0(\bar{x}_{v^*}(T)-p_2)'\delta \bar{x}(T). \end{aligned}$$

We have

$$\begin{aligned} \frac{d}{dt}(\delta y'P)&=\delta v'B_0'P-\delta y' \mathcal {L}_1^*(Q)(t)\nonumber \\ \frac{d}{dt}(\delta \bar{x}'Q)&= \mathcal {L}_1(\delta y)(t)'Q + \mathcal {L}_2(\delta \bar{x})(t)'Q-\delta \bar{x}' \mathcal {L}_2^*(Q)(t). \end{aligned}$$

(27)

By integrating (27) from 0 to T we get $0=\Big \langle B_0'P,\delta v \Big \rangle -\Big \langle \mathcal {L}_1^* (Q)(t),\delta y \Big \rangle$. Similarly, we have

$$\begin{aligned} M_0\delta \bar{x}(T)' (\bar{x}_{v^*}(T)-p_2)&=\Big \langle \mathcal {L}_1(\delta y)(t),Q \Big \rangle +\Big \langle \mathcal {L}_2(\delta \bar{x})(t),Q \Big \rangle -\Big \langle \delta \bar{x}, \mathcal {L}_2^*(Q)(t) \Big \rangle \\&=\Big \langle \mathcal {L}_1^*(Q)(t),\delta y \Big \rangle . \end{aligned}$$

Therefore, $\delta \bar{J}_0= \Big \langle B_0'P,\delta v \Big \rangle + r_0 \Big \langle v^*, \delta v \Big \rangle .$ By optimality, $\delta \bar{J}_0=0$ for all $\delta v \in L_2([0,T])$. Hence, $v^*=-\frac{1}{r_0}B_0'P$.$\square$

Proof of Lemma 2

The idea of the proof is to replace the term

$$\begin{aligned} \int _0^T \varPhi (T,\sigma ) B^{(2)} Q(\sigma ) \mathrm {d} \sigma \end{aligned}$$

(28)

in the expression of $\mathcal {L}_2^*(Q)$ by an assumed known constant $K_1$. Equation (15) is then a linear differential equation parameterized by $K_1$ whose solution is a linear operator of $K_1$. By replacing this solution in the term (28), and by requiring that $K_1$ is equal to (28), one can show that the unique solution of (15) is $Q(t)= \varPhi (T,t)'\Big (\alpha \xi ^*\int _t^T \varPhi (\sigma ,T)'H(\sigma )\mathrm {d \sigma }Y +M_0(\bar{x}_{v^*}(T)-p_2)\Big )$, where Y is the unique solution of the following linear algebraic equation:

$$\begin{aligned} (I_{n}-\xi ^*&\varSigma )Y=M_0\int _0^T \Big (\varPhi (T,\sigma )B\Big )^{(2)}\mathrm {d \sigma }(\bar{x}^*(T)-p_2). \end{aligned}$$

$\square$

Proof of Theorem 4

Let v and $v'$ in $L_2([0,T])$, and denote by $y, y'$ the corresponding MA trajectories, and by $\lambda$ and $\lambda '$ the corresponding fixed points. We have,

$$\begin{aligned} |\lambda - \lambda '|&=|F(\lambda ,y)- F(\lambda ',y')| \le |F(\lambda ,y)- F(\lambda ',y)|+|F(\lambda ',y) -F(\lambda ',y')|\\&\le \frac{\alpha }{c}\Big |\varDelta \big (\bar{R}(t)(p_1-p_2)\big )\Big | |\lambda - \lambda '| +|F(\lambda ',y) -F(\lambda ',y')|. \end{aligned}$$

The rest of the proof is similar to the proof of Theorem 2.$\square$

Proof of Lemma 3

The uniqueness follows from Assumption 5. Let $v\in L_2([0,T])$. The path $\bar{x}_v$ defined in (12) is uniformly bounded with c (with respect to the $L_2$ norm). Therefore, the optimal cost $\bar{J}_0(v^*,\bar{x}_{v^*})\le \bar{J}_0(v,\bar{x}_{v})$ of the MA optimal control problem defined in (13) is uniformly bounded with c. Hence, the optimal control law $v^*$ and the optimal state $y_{v^*}$ are uniformly bounded with c. Consequently, the term $\varDelta \left( K(p_2)y_{v^*}+ \alpha \bar{x}^\lambda \right)$, where $\bar{x}^\lambda$ is defined in (12), is uniformly bounded with c by a positive constant $L_1$. This means that $-L_1\le \varDelta \left( K(p_2)y_{v^*}+ \alpha \bar{x}^\lambda \right) \le L_1$. Hence, $\bar{F}(-L_1) \le F(\lambda ,y) \le \bar{F}(L_1)$. If we choose $-L_1>a-c/2$ and $L_1<a+c/2$, that is, $c>\max (2(a+L_1),2(-a+L_1)):=c_0$, then the map F takes its values in (0, 1). Therefore, F has a unique fixed point $\lambda \in (0,1)$.$\square$

Proof of Theorem 5

By Theorem 4 we know that there exists $v^*$ an optimal investment strategy. Moreover, we know that $v^*$ and $\bar{x}_{v^*}$ should be equal to $v^*=-\frac{1}{r_0}B_0'P^\lambda$ and $\bar{x}_{v^*}=\bar{x}^\lambda$, where $P^\lambda$ and $\bar{x}^\lambda$ are defined in (21)–(22) for a fixed point $\lambda$ of $F_u$. It remains to show that $F_u$ has a unique fixed point $\lambda _*$. Let $\lambda$ and $\lambda '$ be two distinct fixed points of $F_u$. Then, $\lambda$ and $\lambda '$ are, respectively, the fixed points of $s\mapsto F(s,y^\lambda )$ and $s\mapsto F(s,y^{\lambda '})$, where F is defined above (11) and $y^\lambda$ in (21). Following Lemma 3, $\lambda$ and $\lambda '$ belong to (0, 1). But, $\beta 'x^0 - \delta$ has a uniform distribution, which implies that $F_u$ has a shape similar to that of the cumulative distribution function of a uniform distribution. Thus, all the real numbers in the interval [0, 1] are fixed points of $F_u$. In particular, $\lambda =0$ is a fixed point of $s \mapsto F(s,y^{\lambda =0})$. This leads to a contradiction and shows that $F_u$ has a unique fixed point. $\square$