To Confine or Not to Confine: A Mean Field Game Analysis of the End of an Epidemic

Abstract

We analyze a mean field game where the players dynamics follow the SIR model. The players are the members of the population and the strategy consists in choosing the probability of being exposed to the infection, i.e., its confinement level. The goal of each player is to minimize the sum of the confinement cost, which is linear and decreasing on its strategy, and a cost of infection per unit time. We formulate this problem as a mean field game and we investigate the structure of a mean field equilibrium. We study the behavior of agents during the end of the epidemic, where the proportion of infected population is decreasing. Our main results show that: (a) when the cost of infection is low, a mean field equilibrium consists of never getting confined, i.e., the probability of being exposed to the infection is always one and (b) when the cost of infection is large, a mean field equilibrium consists of being confined at the beginning and, after a given time, being exposed to the infection with probability one.

Appendices

Proof of Lemma 1

We first note that $V_I^*(T-1)=c_I+(1-\rho )V_I^*(T)=c_I,$ which clearly verifies the desired condition.

We now assume that $V_I^*(t+1)=c_I\sum _{i=0}^{T-2-t}(1-\rho )^i,$ for $t<T-1$ and we verify the following:

$$\begin{aligned} V_I^*(t)&=c_I+(1-\rho )c_I\sum _{i=0}^{T-2-t}(1-\rho )^i=c_I+c_I\sum _{i=1}^{T-1-t}(1-\rho )^i\\&= c_I\sum _{i=0}^{T-1-t}(1-\rho )^i =c_I\frac{1-(1-\rho )^{T-t}}{\rho }. \end{aligned}$$

And the desired result follows.

Proof of Lemma 2

Let us first observe that the desired result holds for $t=T$ since $V_S^*(T)=0$ and $V_I^*(T)=0$. We now note that, from Lemma 1, we have that $V_I^*(T-1)=c_I$. Besides, using that when $t=T-1$ the best response to $\pi$ is always one, it follows that $V_S^*(T-1)=c_L-1$. As a consequence, $V_S^*(T-1)\ge V_I^*(T-1)$ when $c_I\le c_L-1$, i.e., the desired result is also satisfied when $t=T-1$. We now assume that $V_S^*(t+1)\ge V_I^*(t+1)$ for $t<T-1$ and we verify that $V_S^*(t)\ge V_I^*(t)$ when $c_I\le c_L-1$.

$$\begin{aligned} V_S^*(t)&=\min _{\pi ^0(t)\in [0,1]} \left[ c_L-\pi ^0(t)+(1-\gamma m_I(t)\pi ^0(t))V_S^*(t+1)+\gamma m_I(t)\pi ^0(t)V_I^*(t+1)\right] \\&=\min _{\pi ^0(t)\in [0,1]}\left[ c_L-\pi ^0(t)+V_S^*(t+1)+\gamma m_I(t)\pi ^0(t)\left( V_I^*(t+1)-V_S^*(t+1)\right) \right] \\&=c_L-1+V_S^*(t+1)+\gamma m_I(t)\left( V_I^*(t+1)-V_S^*(t+1)\right) , \end{aligned}$$

where the last equality holds since $V_I^*(t+1)-V_S^*(t+1)\le 0$ and, therefore, the value that minimizes $c_L-\pi ^0(t)+(1-\gamma m_I(t)\pi ^0(t))V_S^*(t+1)+\gamma m_I(t)\pi ^0(t)V_I^*(t+1)$ is $\pi ^0(t)=1$. Since $V_I^*(t)=c_I+(1-\rho )V_I^*(t+1)<c_I+V_I^*(t+1)$, it follows that

$$\begin{aligned} V_I^*(t)-V_S^*(t)&<c_I+V_I^*(t+1)-\left( c_L-1+V_S^*(t+1)+\gamma m_I(t)\left( V_I^*(t+1)-V_S^*(t+1)\right) \right) \\&=c_I-c_L+1 +(1-\gamma m_I(t))\left( V_I^*(t+1)-V_S^*(t+1)\right) , \end{aligned}$$

which is clearly non-positive because $c_I\le c_L-1$ and $\gamma m_I(t)<1$ and since we assumed that $V_S^*(t+1)\ge V_I^*(t+1)$. And the desired result follows.

Proof of Lemma 3

Since the best response at time t is one,

$$ V_S^*(t)= c_L-1+(1-\gamma m_I(t))V_S^*(t+1)+\gamma m_I(t)V_I^*(t+1). $$

Besides,

$$ V_I^*(t)=c_I+(1-\rho )V_I^*(t+1)<c_I+V_I^*(t+1). $$

As a result,

$$\begin{aligned} V_I^*(t)-V_S^*(t)&< c_I+V_I^*(t+1)-c_L+1-(1-\gamma m_I(t))V_S^*(t+1)-\gamma m_I(t)V_I^*(t+1)\\&< c_I+V_I^*(t+1)-c_L+1-\gamma m_I(t)V_I^*(t+1)\\&<c_I+V_I^*(t+1)-c_L+1\\&=c_I\left( 1+\frac{1-(1-\rho )^{T-t-1}}{\rho }\right) -c_L+1\\&<c_I\left( 1+\frac{1-(1-\rho )^{T-1}}{\rho }\right) -c_L+1. \end{aligned}$$

And the desired result follows.

Proof of Proposition 2

We know that the best response to $\pi$ is one when $t=T-1$ because the costs at time T are zero. Therefore, we only need to show that the best response to any $\pi$ is one for all $t<T-1$.

We assume that the best response to any $\pi$ is one at time $t+1$. Therefore, from Lemma 3, it follows that

$$ V_S^*(t+1)-V_I^*(t+1)<c_I\left( 1+\frac{1-(1-\rho )^{T-1}}{\rho }\right) -c_L+1. $$

As a result,

$$ \gamma m_I(t) \left( V_I^*(t+1)-V_S^*(t+1)\right) <\gamma m_I(t)\left( c_I\left( 1+\frac{1-(1-\rho )^{T-1}}{\rho }\right) -c_L+1 \right) . $$

From Assumption 1, it follows that $m_I(t)<m_I(0)$ and therefore, the rhs of the above expression is upper bounded by

$$ \gamma m_I(0)\left( c_I\left( 1+\frac{1-(1-\rho )^{T-1}}{\rho }\right) -c_L+1 \right) $$

because $c_I\left( 1+\frac{1-(1-\rho )^{T-1}}{\rho }\right) -c_L+1 >0$ since $c_I>c_L-1$ and $\rho >0$. Since we have that

$$ \gamma m_I(0)\left( c_I\left( 1+\frac{1-(1-\rho )^{T-1}}{\rho }\right) -c_L+1 \right) \le 1, $$

from the above reasoning, it follows that

$$ \gamma m_I(t) \left( V_I^*(t+1)-V_S^*(t+1)\right) <1, $$

which, according to (COND-BR=1), it implies that the best response to any $\pi$ at time t is one. And the desired result follows.

Proof of Proposition 4

Let us recall that the best response to any $\pi$ at time $T-1$ is one. We aim to show that, when (COND1-JUMP) holds, the best response to $\pi$ does not have more than one jump. For a fixed strategy $\pi$, let $t_0$ be the first time (starting from T) such that the best response to $\pi$ is zero. The desired result follows if we show that for all $t\le t_0$ the best response to $\pi$ is zero. Using an induction argument, we assume that, there exists a $\bar{t}\le t_0$ such that the best response to $\pi$ at time $\bar{t}$ is zero, which according to (COND-BR=0) is achieved when

$$\begin{aligned} \gamma m_I(\bar{t}) (V_I^*(\bar{t}+1)-V_S^*(\bar{t}+1))>1 \end{aligned}$$

(5)

and we aim to show that

$$\begin{aligned} \gamma m_I(\bar{t}-1) (V_I^*(\bar{t})-V_S^*(\bar{t}))>1 \end{aligned}$$

(6)

i.e., that the best response to $\pi$ at time $\bar{t}-1$ is zero as well. Since the best response to $\pi$ at time $\bar{t}$ is zero, it follows that

$$ V_S^*(\bar{t})=c_L+V_S^*(\bar{t}+1), $$

and we also have that $V_I^*(\bar{t})=c_I(1-\rho )^{T-\bar{t} -1 }+V_I^*(\bar{t}+1)$. As a result,

$$ V_I^*(\bar{t})-V_S^*(\bar{t})=c_I(1-\rho )^{T-\bar{t} -1 }-c_L+V_I^*(\bar{t}+1)-V_S^*(\bar{t}+1). $$

From (5), we obtain that $V_I^*(\bar{t}+1)-V_S^*(\bar{t}+1)>\frac{1}{\gamma m_I(\bar{t}) }$, therefore

$$ V_I^*(\bar{t})-V_S^*(\bar{t})>c_I(1-\rho )^{T-\bar{t} -1 }-c_L+\frac{1}{\gamma m_I( \bar{t})} $$

From (COND1-JUMP), we derive that $c_I(1-\rho )^{T-\bar{t} -1 }-c_L>0$, which means that the rhs of the above expression is lower bounded by

$$ V_I^*(\bar{t})-V_S^*(\bar{t})>\frac{1}{m_I( \bar{t})}. $$

We multiply both sides by $\gamma m_I(\bar{t}-1)$:

$$ \gamma m_I(\bar{t}-1) (V_I^*(\bar{t})-V_S^*(\bar{t}))>\frac{m_I(\bar{t}-1)}{m_I( \bar{t})}. $$

We now notice that $\frac{m_I(\bar{t}-1)}{m_I( \bar{t})}>1$ due to Assumption 1 and, therefore, (6) holds which implies that the desired result follows.

Proof of Lemma 4

1.1 $\widetilde{m}_I(t)\ge m_I(t)$

We first show that $\widetilde{m}_I(t)\ge m_I(t)$ for $t\ge 0.$ We note that, at time zero, both values coincide, i.e., $\widetilde{m}_I(0)= m_I(0)$. We now assume that for $t \ge 0$, $\widetilde{m}_I(t)\ge m_I(t)$, and we aim to show that $\widetilde{m}_I(t+1)\ge m_I(t+1)$. Using (1) and also that $\bar{\pi }(t)$ is formed by all zeros,

$$ m_I(t+1)=m_I(t)(1+\gamma m_S(t)\bar{\pi }(t)-\rho )=m_I(t)(1-\rho ), $$

whereas for $\widetilde{\pi }$ we have

$$ \widetilde{m}_I(t+1)=\widetilde{m}_I(t)(1+\gamma \widetilde{m}_S(t) \widetilde{\pi }(t)-\rho ) \ge \widetilde{m}_I(t)(1-\rho )\ge m_I(t)(1-\rho )=m_I(t+1). $$

And the desired result follows.

1.2 $V_I^*(t)\ge V_S^*(t)$

We now focus on the proof of $V_I^*(t)\ge V_S^*(t)$ for all $t>t_0$. We first note that $V_I^*(T)=V_S^*(T)=0$ and, therefore, the desired result follows at time T. We now assume that $V_I^*(t+1)\ge V_S^*(t+1)$ for $t+1>t_0$ and we aim to show that $V_I^*(t)\ge V_S^*(t)$. Since the best response to $\bar{\pi }$ at time t is one, we have that

$$ V_S^*(t)=c_L-1+(1-\gamma m_I(t))V_S^*(t+1)+\gamma m_I(t)V_I^*(t+1) $$

and taking into account that $V_I^*(t)=c_I(1-\rho )^{T-1-t}+V_I^*(t+1)$, we have that

$$ V_I^*(t)-V_S^*(t)=c_I(1-\rho )^{T-1-t}-c_L+1+(1-\gamma m_I(t))(V_I^*(t+1)-V_ S^*(t+1))\ge 0, $$

which holds since $V_I^*(t+1)\ge V_S^*(t+1)$ and $c_I \ge \frac{c_L}{(1-\rho )^{T-1}}>\frac{c_L}{(1-\rho )^{T-t-1}}$. And the desired result follows.

1.3 $\widetilde{V}_S^*(t)\ge V_S^*(t)$

Finally, we show that $\widetilde{V}_S^*(t)\ge V_S^*(t)$ for all $t>t_0.$ We know that $\widetilde{V}_S^*(T)= V_S^*(T)=0$ since the cost at the end is zero. Therefore, we assume that $\widetilde{V}_S^*(t+1)\ge V_S^*(t+1)$ for $t>t_0$ and we aim to show that $\widetilde{V}_S^*(t)\ge V_S^*(t)$.

We know that the best response to $\bar{\pi }$ for $t>t_0$ is one. Therefore, $V_S^*(t)=c_L-1+V_S^*(t+1)+\gamma m_I(t)(V_I^*(t+1)-V_S^*(t+1))$. For $\widetilde{V}_S^*(t)$, we denote by a the best response to $\widetilde{\pi }$ at time t. Thus,

$$ \widetilde{V}_S^*(t)=c_L-a+(1-\gamma \widetilde{m}_I(t)a)\widetilde{V}_S^*(t+1)+\gamma \widetilde{m}_I(t)aV_I^*(t+1). $$

Using that $\widetilde{V}_S^*(t+1)\ge V_S^*(t+1)$ and because $1-\gamma \widetilde{m}_I(t)a$ is positive, we get

$$\begin{aligned} \widetilde{V}_S^*(t)\ge c_L-a+(1-\gamma \widetilde{m}_I(t)a)V_S^*(t+1)+\gamma \widetilde{m}_I(t)aV_I^*(t+1). \end{aligned}$$

(7)

Therefore, $\widetilde{V}_S^*(t)\ge V_S^*(t)$ is true when

$$\begin{aligned} c_L-a+(1-\gamma \widetilde{m}_I(t)a) V_S^*(t+1)+\gamma \widetilde{m}_I(t)aV_I^*(t+1)\\ \ge c_L-1+V_S^*(t+1)+\gamma m_I(t)(V_I^*(t+1)-V_S^*(t+1)). \end{aligned}$$

Simplifying

$$ 1-a\ge \gamma (m_I(t)-\widetilde{m}_I(t)a)(V_I^*(t+1)-V_ S^*(t+1)). $$

Using that $\widetilde{m}_I(t)\ge m_I(t)$ and since $V_I^*(t+1)-V_ S^*(t+1)$ is non-negative, the rhs of the above expression is smaller or equal than $\gamma (1-a)m_I(t)(V_I^*(t+1)-V_ S^*(t+1))$. Therefore, a sufficient condition for the desired result to hold is

$$ 1-a\ge \gamma (1-a)m_I(t)(V_I^*(t+1)-V_ S^*(t+1)). $$

We now differentiate two cases: (a) when $a=1$, we have zero in both sides of the expression and therefore, the condition is satisfied; (b) when $a\ne 1$, we divide by $1-a$ both sides of the expression and we get

$$ 1\ge \gamma m_I(t)(V_I^*(t+1)-V_ S^*(t+1)), $$

which is also satisfied from (COND-BR=1) because the best response to $\bar{\pi }$ at time t is one.

Proof of Lemma 6

We aim to show that

$$ V_I^*(t)-\widetilde{V}_S^*(t)\ge V_I^*(t+1)-\widetilde{V}_S^*(t+1), $$

for $t>t_0$. From Lemma 5 we know that the best response to $\widetilde{\pi }$ is one for $t>t_0$, because (COND1-JUMP) is satisfied, and therefore,

$$ \widetilde{V}_S^*(t)=c_L-1+\widetilde{V}_S^*(t+1)+\gamma m_I(t)(V_I^*(t+1)-\widetilde{V}_S^*(t+1)). $$

From Lemma 1, we have that

$$ V_I^*(t)=c_I(1-\rho )^{T-t-1}+V_I^*(t+1). $$

Therefore,

$$ V_I^*(t)-\widetilde{V}_S^*(t)=c_I(1-\rho )^{T-t-1}-c_L+1+(1-\gamma m_I(t))(V_I^*(t+1)-\widetilde{V}_S^*(t+1)). $$

Hence, using the above expression, we get that

$$\begin{aligned} V_I^*(t)-\widetilde{V}_S^*(t)\ge V_I^*(t+1)-\widetilde{V}_S^*(t+1) \iff \\ c_I(1-\rho )^{T-t-1}-c_L+1-\gamma m_I(t)(V_I^*(t+1)-\widetilde{V}_S^*(t+1))\ge 0. \end{aligned}$$

We now note that when $c_I\ge \frac{c_L}{(1-\rho )^{T-1}},$ we get $c_I(1-\rho )^{T-t-1}>c_L$, and as the best response to $\widetilde{\pi }$ is one we have (COND-BR=1). Therefore,

$$\begin{aligned} c_I(1-\rho )^{T-t-1}-c_L+1-\gamma m_I(t)(V_I^*(t+1)-\widetilde{V}_S^*(t+1))\\> 1-\gamma m_I(t)(V_I^*(t+1)-\widetilde{V}_S^*(t+1))\ge 0 \end{aligned}$$

and the desired result follows.