Provisioning of Public Health Can Be Designed to Anticipate Public Policy Responses

Abstract

Public health policies can elicit strong responses from individuals. These responses can promote, reduce, and even reverse the expected benefits of the policies. Therefore, projections of individual responses to policy can be important ingredients in policy design. Yet our foresight of individual responses to public health investment remains limited. This paper formulates a population game describing the prevention of infectious disease transmission when community health depends on the interactions of individual and public investments. We compare three common relationships between public and individual investments and explain how each relationship alters policy responses and health outcomes. Our methods illustrate how identifying system interactions between nature and society can help us anticipate policy responses.

Appendices

Appendix 1: Existence, Uniqueness, and Evolutionary Stability

The mathematical analysis in the paper revolves around the analysis of the utility function, which is the sum over all future times of the probability of being in each state times the value of that state. The values of that states in the future are discounted so that they are worth less than the present values. Mathematically,

$$\begin{aligned} U(c_s,\overline{c}_s;c_t) := \int _{0}^{\infty } \mathrm{e}^{-ht} \mathbf {v} \cdot \mathbf {p}(t) \mathrm {d}t = \int _{0}^{\infty } \mathrm{e}^{-ht} \mathbf {v} \mathrm{e}^{\widetilde{\mathbf {Q}}t} \mathbf {p}_0 \mathrm {d}t \end{aligned}$$

Here, we have made use of our model and assumptions to replace $\mathbf {p}(t)$ with its matrix-exponential representation. The discount rate h describes how much less future returns are worth, compared to present returns. In economics, h may be an interest rate or inflation rate. In evolutionary biology, h is the exponential rate of population growth. Performing the needed integration, we determine that Eq.(5) is given as

$$\begin{aligned}&U(c_s,\overline{c}_s;c_t)\\&\;\;\;\;\;= \mathbf {v} \left( h \mathbf {I} - \widetilde{\mathbf {Q}}\right) ^{-1} \mathbf {p}_0 \\&\;\;\;\;\;= \begin{bmatrix} u(j- c_t) - c_s, u(j- c_t) - c_i \end{bmatrix} \left( h \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} - \begin{bmatrix} - \sigma (c_s,c_t) \widetilde{\lambda }&\gamma \\ \sigma (c_s,c_t) \widetilde{\lambda }&-\gamma \end{bmatrix} \right) ^{-1} \begin{bmatrix} 1 \\ 0 \end{bmatrix} \\&\;\;\;\;\;= \begin{bmatrix} u(j- c_t) - c_s, u(j- c_t) - c_i \end{bmatrix}\\&\quad \left( \frac{1}{h(h+\gamma +\widetilde{\lambda }\sigma (c_s,c_t))} \begin{bmatrix} h + \gamma&\gamma \\ \sigma (c_s,c_t) \widetilde{\lambda }&h + \widetilde{\lambda }\sigma \end{bmatrix} \right) \begin{bmatrix} 1 \\ 0 \end{bmatrix}\\&\;\;\;\;\;= \frac{u(j-c_t)}{h} - \frac{ ( h+ \gamma ) c_s+\widetilde{\lambda }(\overline{c}_s,c_t) \sigma \left( c_s,c_t \right) c_i }{ h \left[ h+\gamma + \widetilde{\lambda }(\overline{c}_s,c_t) \sigma \left( c_s,c_t \right) \right] } \end{aligned}$$

Theorem 1

If the relative exposure rate is smooth, convex, and decreasing in individual investment, then there is always a unique best response for individuals, and this best response increases with both the cost of disease and the infection pressure.

Proof

First, we repeat our argument for the existence of a unique best response when the relative exposure rate is convex and decreasing.

An individual’s best response $c_s^B$ maximizes the utility under given the typical investment $\overline{c}_s$ and public investment $c_t$ is the $c_s^B(\overline{c}_s,c_t) := {{\mathrm{\hbox {argmax}}}}_{c_s\ge 0} U(c_s,\overline{c}_s;c_t).$ The best response is chosen so that the marginal cost of preventive investment equals the marginal benefit of less frequent infection. Differentiation of U by $c_s$ leads to the geometric condition

$$\begin{aligned} (c_s^B - c_i) \frac{\partial \sigma }{\partial c_s} = \frac{h + \gamma }{\widetilde{\lambda }} + \sigma (c_s^B,c_t). \end{aligned}$$

(16)

Note that $\widetilde{\lambda }$ is fixed—it depends on the typical investment $\overline{c}_s$, not the individual investment $c_s$. Since $\overline{c}_s$ appears in Eq. (16) only implicitly through $\widetilde{\lambda }$, while $c_t$ appears in $\sigma$ and $\widetilde{\lambda }$, we will represent the best response’s dependencies by $c_s^B(c_t,\widetilde{\lambda }(\overline{c}_s,c_t))$.

The right-hand side of Eq. (16) is always positive, so equality requires $c_s^B\in [0,c_i)$. Furthermore, any line relating $c_s$ to $\sigma$ through the point $(c_i,-(h+\gamma )/\widetilde{\lambda })$ is a solution. For a fixed public investment rate $c_t$, if the best response $c_s^B>0$, then the line drawn between $( c_s^B, \sigma (c_s^B,c_t))$ and $(c_i,-(h+\gamma )/\widetilde{\lambda })$ must be tangent to the curve $\sigma (c_s,c_t)$ at $c_s=c_s^B$ on the plane of $c_s$ versus $\sigma$. Depending on the shape of $\sigma$ (see Fig. 1), there may be several points satisfying this necessary condition, possibly including the boundary point $c_s=0$. If the relative exposure rate is a convex function of the individual investment, then we can see geometrically that there is always a unique best response (see the left subplot in Fig. 1). If infection cost is small enough, then no tangent line satisfying our criteria will exist, and $c_s^B = 0$ will be the unique best response.

The observations that the best response increases with the cost of disease and the risk of infection can be shown with calculus when $\sigma$ is smooth. Rearranging (16) and differentiating with respect to $c_i$, we obtain

$$\begin{aligned} -\frac{\partial \sigma (c_s^B, c_t)}{\partial c_s}+(c_s^B-c_i)\frac{\partial ^2 \sigma (c_s^B, c_t)}{\partial c_s^2}\frac{\partial c_s^B}{\partial c_i}=0. \end{aligned}$$

(17)

By the monotonicity and convexity of the function $\sigma (c_s,c_t)$ in $c_s$, we know that

$$\begin{aligned} \frac{\partial \sigma (c_s^B, c_t)}{\partial c_s}<0 \;\;\; \text {and} \;\;\; \frac{\partial ^2 \sigma (c_s^B, c_t)}{\partial c_s^2}\ge 0. \end{aligned}$$

(18)

These imply that $\frac{\partial c_s^B}{\partial c_i}>0$. Similarly, differentiating Eq. (16) with respect to $\widetilde{\lambda }$ and rearranging,

$$\begin{aligned} (c_s^B-c_i)\frac{\partial ^2 \sigma (c_s^B,c_t)}{\partial c_s^2}\frac{\partial c_s^B}{\partial \widetilde{\lambda }}=-\frac{h+\gamma }{\widetilde{\lambda }^2}. \end{aligned}$$

(19)

By checking the signs of the both sides in the above equation, we know that $\frac{\partial c_s^B}{\partial \widetilde{\lambda }}>0$. $\square$

Theorem 2

If $\sigma (\overline{c}_s,c_t)$ is decreasing in both $\overline{c}_s$ and $c_t$, then $\widetilde{\lambda }(\overline{c}_s,c_t)$ and $\sigma (\overline{c}_s,c_t) \widetilde{\lambda }(\overline{c}_s,c_t)$ are decreasing or flat in both $\overline{c}_s$ and $c_t$.

Proof

Suppose we fix $\overline{c}_s$. Then $\sigma (\overline{c}_s,c_t)$ is decreasing in $c_t$, so $\gamma /\sigma (\overline{c}_s,c_t)$ is increasing and $\widetilde{\lambda }= \max \{0, \beta - \gamma /\sigma (\overline{c}_s,c_t)\}$ must be decreasing in $c_t$. Since both $\widetilde{\lambda }$ and $\sigma (\overline{c}_s,c_t)$ are nonnegative and decreasing or flat in $c_t$, the product $\sigma (\overline{c}_s,c_t) \widetilde{\lambda }$ must also be decreasing or flat in $c_t$ for fixed $\overline{c}_s$. The argument for $\overline{c}_s$ is the same.

Theorem 3

If $\sigma (\overline{c}_s,c_t)$ is decreasing and convex in $\overline{c}_s$, then there is a unique game equilibrium $c_s^*(c_t)$ for every public investment rate $c_t\ge 0$.

Proof

First, Theorem 1 states that increasing the infection pressure increases the best response ($\partial c_s^B / \partial \widetilde{\lambda }\ge 0$). From Theorem 2, increasing the population’s investment $\overline{c}_s$ decreases the infection pressure, so if $c_s^B(c_t,\widetilde{\lambda }(0,c_t)) = 0$, then $c_s^B(c_t,\widetilde{\lambda }(\overline{c}_s,c_t)) = 0$ for all $\overline{c}_s> 0$. So $c_s^*=0$ must be the only strategy that is a best response to itself, i.e., the game equilibrium, and therefore is unique.

On the other hand, suppose $c_s^B(c_t,\widetilde{\lambda }(0,c_t)) > 0$. We observe that $\widetilde{\lambda }< \beta -\gamma$ for all $\overline{c}_s$, implying

$$\begin{aligned} c_s^B(c_t,\widetilde{\lambda }(\overline{c}_s,c_t)) \le c_s^B(c_t,\beta -\gamma ) < \infty . \end{aligned}$$

(20)

We know that the latter inequality holds because we have shown that $c_s^B\in [0, c_i)$ and the cost of infection $c_i$ is generally finite. Then, we have,

$$\begin{aligned} \lim _{\overline{c}_s\rightarrow \infty } \left[ c_s^B(c_t, \widetilde{\lambda }(\overline{c}_s,c_t))-\overline{c}_s\right]< 0 < c_s^B(c_t, \widetilde{\lambda }(0,c_t))-0. \end{aligned}$$

(21)

Since $c_s^B(c_t, \widetilde{\lambda }(\overline{c}_s,c_t))$ is continuous in $\overline{c}_s$, by the intermediate value theorem of continuous functions, there must be at least one solution to

$$\begin{aligned} c_s^B(c_t,\widetilde{\lambda }(c_s^*,c_t)) = c_s^*. \end{aligned}$$

(22)

Since $\partial c_s^B/\partial \widetilde{\lambda }\ge 0$ and $\partial \widetilde{\lambda }/\partial \overline{c}_s\le 0$, $c_s^B(c_t,\widetilde{\lambda }(\overline{c}_s,c_t))$ must be decreasing in $\overline{c}_s$. By the monotonicity of $c_s^B(c_t,\widetilde{\lambda }(\overline{c}_s,c_t))$ with respect to $\overline{c}_s$, there can be no more than one solution to Eq. (22). Thus, there is a unique game equilibrium for $c_s^B(c_t,\widetilde{\lambda }(0,c_t)) > 0$.

We conclude that there is always a unique global game equilibrium for individual behavior under the given assumptions. $\square$

Theorem 4

If $\sigma (\overline{c}_s,c_t)$ is decreasing and convex in $\overline{c}_s$ for $c_t\ge 0$, then the equilibrium strategy always has invasion potential and hence is an evolutionarily stable strategy.

Proof

The argument for invasion potential is less straight forward than that for the Nash condition. Since our argument is independent of $c_t$, we will simplify our notation by omitting it henceforth. We begin working from our known information. First, since $\lambda (c_s)$ is a nonnegative decreasing function,

$$\begin{aligned} \frac{\overline{c}_s-c_s^*}{\lambda (\overline{c}_s)} \ge \frac{\overline{c}_s-c_s^*}{\lambda (c_s^*)}. \end{aligned}$$

(23)

Now, since $c_s^*$ is a Nash equilibrium (Theorem 3), we know $U(c_s,c_s^*) - U(c_s^*,c_s^*) \le 0$. This implies, after a fair bit of algebra, that

$$\begin{aligned} \left( \frac{c_i - c_s}{\sigma \left( c_s\right) } - \frac{c_i - c_s^*}{\sigma \left( c_s^* \right) } \right) + \frac{(h+\gamma )(c_s^*-c_s)}{ \sigma \left( c_s\right) \sigma \left( c_s^* \right) \widetilde{\lambda }(c_s^*) } \le 0. \end{aligned}$$

(24)

With just a change of sign,

$$\begin{aligned} \left( \frac{c_i - c_s^*}{\sigma \left( c_s^* \right) } - \frac{c_i - c_s}{\sigma \left( c_s\right) } \right) + \frac{(h+\gamma )(c_s-c_s^*)}{ \sigma \left( c_s^* \right) \sigma \left( c_s\right) \widetilde{\lambda }(c_s^*) } \ge 0. \end{aligned}$$

(25)

Now, an equilibrium is an evolutionarily stable strategy (ESS) if it has global invasion potential, in the sense that it improves on any alternative typical behavior. Mathematically, $U(c_s^*,\overline{c}_s) - U(\overline{c}_s,\overline{c}_s) \ge 0.$ Well,

$$\begin{aligned}&U(c_s^*,\overline{c}_s) - U(\overline{c}_s,\overline{c}_s) = - \frac{ ( h+ \gamma ) c_s^* +\widetilde{\lambda }(\overline{c}_s) \sigma \left( c_s^* \right) c_i }{ h \left[ h+\gamma + \widetilde{\lambda }(\overline{c}_s) \sigma \left( c_s^* \right) \right] } \nonumber \\&\quad + \frac{ ( h+ \gamma ) \overline{c}_s+\widetilde{\lambda }(\overline{c}_s) \sigma \left( \overline{c}_s\right) c_i }{ h \left[ h+\gamma + \widetilde{\lambda }(\overline{c}_s) \sigma \left( \overline{c}_s\right) \right] } \end{aligned}$$

(26)

From this, we can show that $c_s^*$ invades universally as long as

$$\begin{aligned} \left( \frac{c_i - c_s^*}{\sigma \left( c_s^* \right) } - \frac{c_i - \overline{c}_s}{\sigma \left( \overline{c}_s\right) } \right) + \frac{(h+\gamma )(\overline{c}_s-c_s^*)}{ \sigma \left( \overline{c}_s\right) \sigma \left( c_s^* \right) \widetilde{\lambda }(\overline{c}_s) } \ge 0. \end{aligned}$$

(27)

After we substitute $\overline{c}_s$ for $c_s$, Eq. (25) differs from Eq. (27) only in the infection-pressure term. A substitution using Eq. (23) shows us that Eq. (25) implies Eq. (27). So if $c_s^*$ is a Nash equilibrium, it also has invasion potential. Since the strategy satisfies both the Nash condition and the invasion condition, it is an evolutionarily stable strategy. $\square$

Appendix 2: Equilibrium Calculation and Bounds

It is often impossible to identify a closed-form representation of the game equilibrium $c_s^*$ from Eq. (9). For these cases, one can still identify the game equilibrium using either numerical or geometric approaches when the relative exposure rate function $\sigma$ is given. First, we can numerically locate the strategy that is a best response to itself directly using the formula for the expected utility, Eq. (5). This is sure to return the unique game equilibrium for individual behavior if function $\sigma$ satisfies the conditions in Theorem 3. Alternatively, the equilibrium can be located using a phase-plane approach as follows. Equation (9) can be read as a first-order differential equation for $\sigma$ in terms of $c_s^*$, with implicit solutions

$$\begin{aligned} C = \frac{(\sigma \beta )^{\frac{\gamma }{h}} (c_i-c_s^*)}{(\sigma \beta +h)^{1+\frac{\gamma }{h}}} \;\;\text {if } h>0 \text { or} \; \frac{(c_i-c_s^*)}{\sigma \beta \mathrm{e}^{\gamma /(\sigma \beta )}} \;\;\text {if } h=0. \end{aligned}$$

(28)

We can now draw level curves representing solutions of the necessary condition for a game equilibrium, Eq. (9). We can also plot $\sigma (c_s,c_t)$ as a function of $c_s$. By Theorem 3, there must be a point $(c_s^*, \sigma (c_s^*))$ where Eq. (9) holds. Geometrically, this point is the tangent point between $\sigma (c_s,c_t)$ and the curves representing the necessary condition (Fig. 6).

Fig. 6

(Color figure online) Points where the orbits of Eq. (9) are tangent to the relative exposure rate $\sigma$. The dashed line represents the given function $\sigma$, while the solid lines represent a family of the orbits to Eq. (9) (left) or contours of constant utility (right). The dot is the Nash equilibrium point $c_s^*$ where $\sigma$ and the solutions of Eq. (9) are tangent (left) or the socially optimal strategy $\overline{c}_s^*$ (right). This figure shows a special case when $\mathcal {R}_0=6$ and $\sigma (c_s,c_t)=\exp ({-2.5 c_s})$, so $c_s^* \approx 0.34$ and $\overline{c}_s^* \approx 0.58$

We now see that the nature of the game equilibria for individual behavior depends to some degree on how different investments reduce risk, as specified by the shape of $\sigma (c_s,c_t)$. Still, under the same assumption on the relative exposure rate described in Theorem 3, the game equilibrium will be bounded.

In order to prove the boundedness of the game equilibrium, we will first claim the following theorem, which tells us that for a given relative exposure rate function we can construct a piecewise linear relative exposure rate function sharing the same game equilibrium.

Lemma 5

For any given relative exposure rate function $\sigma (c_s, c_t)$ with the properties that $\sigma$ is decreasing and convex in $c_s$ for fixed $c_t$, there exists a piecewise linear function $\sigma _L(c_s):=\max (1-mc_s,\epsilon )$, $(m\ge 0, \epsilon \ge 0)$ with the same game equilibrium as $\sigma$.

Proof

We will explicitly construct the function of $\sigma _L$. For the given function $\sigma$, we know that it guarantees the unique existence of game equilibrium $c_s^*$ by Theorem 3. If $c_s^*=0$, then any $\sigma _L$ with $\epsilon = 1$ which has any value of $c_s$ including $c_s=0$ as its equilibria since $\sigma _L$ is flat in $c_s$ making the value of $c_s=0$ a best response to any $\overline{c}_s$.

If $c_s^* > 0$, we can construct the first piece of $\sigma _L$ by connecting the points of (0, 1) and $(c_s^*, \sigma (c_s^*))$ in the plane of $c_s$ versus $\sigma$ for any fixed $c_t$. The second piece will be the horizontal ray starting at point $(c_s^*, \sigma (c_s^*))$. Therefore, take $-m$ to be the slope of the line between points (0, 1) and $(c_s^*, \sigma (c_s^*))$, and $\epsilon =\sigma (c_s^*)$. We then have a piecewise linear function $\sigma _L=\max (1-mc_s,\epsilon )$ (Fig. 7).

Now we show that $\sigma _L$ guarantees the same game equilibrium as $\sigma$. By the construction of $\sigma _L$, we know that $\sigma _L$ is located in the convex hull of the set $\{ (c_s, \sigma (c_s)) : c_s\ge 0 \}$ which allows $\sigma _L$ to have the same tangent property as $\sigma$ at point $(c_s^*, \sigma (c_s^*))$. This implies that, as described in geometrical approach, $\sigma _L$ will also be tangent with the same solution curve to

$$\begin{aligned} \sigma (c_s^*,c_t) + (c_i-c_s^*)\frac{\partial \sigma }{\partial c_s}(c_s^*,c_t) = \frac{-(h+\gamma )}{\beta -\gamma /\sigma (c_s^*,c_t)} \end{aligned}$$

(29)

as $\sigma$ at point $(c_s^*, \sigma (c_s^*))$. In other words, $c_s^*$ is also the game equilibrium for $\sigma _L$ (Fig. 7) $\square$

Thus, by Lemma 5, if we find the set of possible game equilibria for all relative exposure rate functions of the form of $\sigma _L$, then this is also the set of possible game equilibria for all $\sigma$. We can use a geometric argument similar to that presented for smooth functions $\sigma$ (Fig. 6) to find the game equilibria for the piecewise linear functions $\sigma _L(c_s)$. For convenience, we introduce

$$\begin{aligned} \hat{m} := - \left. \frac{\partial \sigma }{\partial c_s}\right| _{c_s=c_t=0,\sigma =1} = \frac{h+\beta }{c_i(\beta -\gamma )}. \end{aligned}$$

(30)

as notation for the minimum efficiency below which no internal equilibrium exists, based on Eq. (29).

Now for any piecewise linear function $\sigma _L=\max (1-mc_s, \epsilon )$ with $(m\ge 0, \epsilon \ge 0)$, let us consider its possible game equilibria. If $m<\hat{m}$, then there are no points where $\sigma _L$ is tangent to any of the phase-plane orbits of Eq. (29). As such, $c_s^*=0$ is the only game equilibrium. If $m>\hat{m}$, we will begin with a ray starting at the point (0, 1) with slope $\hat{m}$ and then locate the tangent point where the ray is tangent with any Nash equilibrium solution curve of Eq. (29). (The existence of the tangent point follows from a property of the solution curves of Eq. (29). The slope of the solution curves decreases to the negative infinity when $\sigma \mathcal {R}_0=1$ as $c_s$ increases, as shown in Fig. 6. By the intermediate value theorem, there is a point where the solution curve has a slope with m.). This point $(\hat{c}_s, \hat{\sigma }(\hat{c}_s))$ is the solution to the system

$$\begin{aligned} \hat{\sigma }&=1-m \hat{c}_s, \quad \hat{\sigma }+(c_i-\hat{c}_s)(-m) = \frac{-(h+\gamma )}{\beta -\gamma /\hat{\sigma }}. \end{aligned}$$

(31)

We determine an alternate piecewise linear function $\hat{\sigma }_L(c_s; m):=\max (1-mc_s, \hat{\epsilon }(m))$ where $\hat{\epsilon }(m):=\hat{\sigma }(\hat{c}_s)$. If $\epsilon \le \hat{\epsilon }$, by the construction of $\hat{\sigma }_L$, we know that the game equilibrium will be the point $(\hat{c}_s, \hat{\sigma }(\hat{c}_s))$ since $\sigma _L$ and $\hat{\sigma }_L$ share this point of tangency to the orbits of Eq. (29). If $\epsilon >\hat{\epsilon }$, the game equilibrium can only possibly correspond to the corner point, $((1-\epsilon )/m, \epsilon )$, since $\sigma _L$ can only be tangent (in a geometric sense) to one of the orbits at this corner point. Therefore, $c_s^*\in [0,(1-\epsilon )/m]$. So far, we have identified the set of possible game equilibria for piecewise linear functions in the form of $\sigma _L$. This set will be bounded by the curve of $(\hat{c}_s, \hat{\sigma }(\hat{c}_s))$ determined by Eq. (31) (see the left subplot of Fig. 8). Again, since Lemma 5 shows that every game equilibrium is also an equilibrium for some $\sigma _L$, this bound also holds for any $\sigma$ and $c_t$.

Fig. 7

(Color figure online) Left the demonstration to show the determination of the general bound of the game equilibria. Extreme bounds of the set of possible game equilibria for $\mathcal {R}_0= 6$ (dashed line). Extreme equilibria occur on lines through the point $c_s=0,\sigma =1$ (dotted line) are tangent to orbits of the necessary differential condition for game equilibria (solid curves). Right the plot of the bound on game equilibrium for different values of $\mathcal {R}_0=\beta /\gamma$ when $h=0$ (Eq. (32)). Each contour is labeled with the value of $\mathcal {R}_0$ for which it is the bound. Equilibria can exist at each point with the same or smaller value of $\mathcal {R}_0$ for which they are being calculated

Fig. 8

Plot demonstrating the construct of $\sigma _L$ as used in the proof of Lemma 5. The solid line is the given $\sigma (c_s)$ with fixed $c_t$ and known game equilibrium $c_s^*$. The dashed line is $\sigma _L(c_s)$. By the construction, $c_s^*$ is always a game equilibrium under $\sigma _L$

Summarizing the above discussion, we have the following conclusion.

Theorem 6

The unique equilibrium $c_s^*$ found in Theorem 3 when $\sigma$ is convex in $c_s$ is always bounded in the sense that

$$\begin{aligned} 0 \le \frac{c_s^*}{c_i} \le \frac{\sigma \gamma -\gamma -\beta \sigma ^2+\beta \sigma }{ \sigma h+\sigma \gamma -\gamma +\beta \sigma } \le \frac{(1-\sigma ) (\sigma \mathcal {R}_0-1)}{\sigma \mathcal {R}_0-1 + \sigma }. \end{aligned}$$

(32)

Proof

The argument initiated by Lemma 5 leads us to Eq. (31) as the bounding condition on equilibria. From the first part, $m = (1-\hat{\sigma })/\hat{c}_s$. When we substitute for m in the second part,

$$\begin{aligned}&\hat{\sigma }-(c_i-\hat{c}_s) (1-\hat{\sigma })/\hat{c}_s= \frac{-(h+\gamma )}{\beta -\gamma /\hat{\sigma }}\\&\quad - \left( \frac{c_i}{\hat{c}_s} - 1 \right) (1-\hat{\sigma }) = - \frac{(h+\gamma )}{\beta -\gamma /\hat{\sigma }} - \hat{\sigma }\\&\quad \left( \frac{c_i}{\hat{c}_s} - 1 \right) = \frac{(h+\gamma )}{(\beta -\gamma /\hat{\sigma })(1-\hat{\sigma })} + \frac{\hat{\sigma }}{ (1-\hat{\sigma }) }\\&\quad \frac{c_i}{\hat{c}_s} = \frac{(h+\gamma )}{(\beta -\gamma /\hat{\sigma })(1-\hat{\sigma })} + \frac{\hat{\sigma }}{ (1-\hat{\sigma }) } +1\\&\quad \frac{c_i}{\hat{c}_s} = \frac{(h+\gamma )}{(\beta -\gamma /\hat{\sigma })(1-\hat{\sigma })} + \frac{1}{ (1-\hat{\sigma }) }\\&\quad \frac{\hat{c}_s}{c_i} = \frac{(\beta -\gamma /\hat{\sigma })(1-\hat{\sigma })}{(h+\gamma ) + (\beta -\gamma /\hat{\sigma }) }. \end{aligned}$$

Our preceding argument showed that this had to be an upper bound on the game equilibrium for a given $\hat{\sigma }$, so we now know

$$\begin{aligned} \frac{c_s^*}{c_i} \le \frac{\sigma \gamma -\gamma -\beta \sigma ^2+\beta \sigma }{ \sigma h+\sigma \gamma -\gamma +\beta \sigma }. \end{aligned}$$

The right-hand side is decreasing in h, so the special case of $h=0$ provides a weaker upper bound that applies for all discount rates. When we take $h=0$ and substitute $\mathcal {R}_0 = \beta /\gamma$, we find the parsimonious upper bound

$$\begin{aligned} \frac{c_s^*}{c_i} \le \frac{(1-\sigma ) (\sigma \mathcal {R}_0-1)}{\sigma \mathcal {R}_0-1 + \sigma }. \end{aligned}$$

(33)

$\square$

Appendix 3: Theorems on Free-Riding and Policy Effects

First, we provide a free-riding theorem.

Theorem 7

Let typical utility $W(\overline{c}_s;c_t):= U(\overline{c}_s,\overline{c}_s;c_t)$. When $\sigma$ is convex in $c_s$, the best public investment rate $\overline{c}_s^*={{\mathrm{\hbox {argmax}}}}_{\overline{c}_s} W(\overline{c}_s;c_t)$ is always greater than the game equilibrium investment ($c_s^* \le \overline{c}_s^*$) and for every $\overline{c}_s\in (0,\overline{c}_s^*)$, $\dfrac{\partial W}{\partial \overline{c}_s} > 0.$

Proof

The utility of the typical investment rate is given by

$$\begin{aligned} W(\overline{c}_s;c_t) = {\left\{ \begin{array}{ll} {\frac{u-c_t}{h}} -{\frac{ c_i [\beta \sigma \left( \overline{c}_s,c_t \right) - \gamma ]+\overline{c}_s(h+\gamma ) }{h \left[ h+\beta \sigma \left( \overline{c}_s,c_t \right) \right] }} &{} \text {if}\quad \sigma (\overline{c}_s,c_t) \mathcal {R}_0> 1,\\ {\frac{u-c_t-\overline{c}_s}{h}} &{} \text {if}\quad \sigma (\overline{c}_s,c_t) \mathcal {R}_0\le 1. \end{array}\right. } \end{aligned}$$

(34)

By inspection, W is decreasing in $\overline{c}_s$ if $\sigma \mathcal {R}_0\le 1$. So the maximum occurs for some $\overline{c}_s$ such that $\sigma \mathcal {R}_0\ge 1$. Differentiating W with respect to $\overline{c}_s$ when $\sigma \mathcal {R}_0> 1$, we find

$$\begin{aligned} \frac{\partial W}{\partial \overline{c}_s} = \frac{-(\gamma +h)}{h[h+\sigma (\overline{c}_s,c_t) \beta ]} \left( 1+ \frac{(c_i - \overline{c}_s) \beta }{[h+\sigma (\overline{c}_s,c_t) \beta ]} \frac{\partial \sigma }{\partial c_s} \right) . \end{aligned}$$

(35)

Since $\sigma$ is monotone decreasing, $\dfrac{\partial W}{\partial \overline{c}_s}$ can change sign no more than once for $\overline{c}_s\in [0,c_i)$.

Using the same geometric approach applied for best responses, the local maximum occurs at points on $\sigma$ where the tangent lines pass through the point $(c_i, -h/\beta )$. The geometry shows that if $\sigma (c_s,c_t)$ is convex in $c_s$, then $\overline{c}_s^*$ is always uniquely defined (possibly, $\overline{c}_s^*=0$). So $\overline{c}_s^*$ is equal to

$$\begin{aligned} \min \left\{ c_s: 1 = \sigma (c_s,c_t) \beta /\gamma ,\;\overline{c}_s: 1 = \frac{-\beta (c_i - \overline{c}_s) }{h+\beta \sigma (\overline{c}_s,c_t)} \frac{\partial \sigma (\overline{c}_s,c_t)}{\partial c_s} \right\} . \end{aligned}$$

(36)

If $0 \le \overline{c}_s< \overline{c}_s^*$, ${\partial W}/{\partial \overline{c}_s} > 0$. Since $\gamma > 0$ implies $-h/\beta > -(h+\gamma )/\widetilde{\lambda }$, the geometry also shows $\overline{c}_s^* \ge c_s^B$ for all best responses $c_s^B$, and in particular, $\overline{c}_s^* \ge c_s^*$. $\square$

Theorem 8

Assume the relative exposure rate function $\sigma (c_s,c_t)$ satisfies the following conditions:

(H1):

$\sigma$ is decreasing in $c_s$ and $c_t$, smooth and convex with respect to $c_s$, and $\frac{\partial ^2 \sigma }{\partial c_t \partial c_s}>0$;

(H2):

$\sigma (c_s^*(c_t),c_t) \in \left( \frac{\gamma }{\beta },\frac{\gamma +\sqrt{\gamma ^2+h\gamma }}{\beta }\right)$.

Then, increased taxation and public reinvestment decreases equilibrium individual investment in self-protection ( $\mathrm {d}c_s^* / \mathrm {d}c_t \le 0$ ).

Note that $\sigma (c_s^*(c_t),c_t) \in \left( \frac{\gamma }{\beta },\frac{\gamma +\sqrt{\gamma ^2+h\gamma }}{\beta }\right)$ implies $1 \le \sigma \mathcal {R}_0 \le 2$, so this theorem is slightly stronger than the version given in the main text.

Proof

Based on Eq. (29), a game equilibrium $c_s^*$ satisfies

$$\begin{aligned} (c_s^*-c_i)\frac{\partial \sigma }{\partial c_s}=\frac{\sigma (c_s^*, c_t)(h+\beta \sigma (c_s^*, c_t))}{\beta \sigma (c_s^*, c_t)-\gamma }. \end{aligned}$$

(37)

For convenience, we define the right-hand side as a function

$$\begin{aligned} \phi (\sigma ) :=\frac{\sigma (h+\beta \sigma )}{\beta \sigma -\gamma }. \end{aligned}$$

(38)

Note that this is hyperbola in $\sigma$, with two linear asymptotes: one local minimum and one local maximum. The conditions of (H2) specify that $\sigma$ is in the range where $\phi$ is positive and decreasing.

We proceed by differentiating Eq. (37) with respect to $c_t$. We find

$$\begin{aligned} \frac{\mathrm {d}c_s^*}{\mathrm {d}c_t}\frac{\partial \sigma }{\partial c_s}+(c_s^*-c_i)\Big (\frac{\partial ^2 \sigma }{\partial c_s^2}\frac{\mathrm {d}c_s^*}{\mathrm {d}c_t}+\frac{\partial ^2 \sigma }{\partial c_s\partial c_t}\Big )=\frac{\mathrm {d}\phi }{\mathrm {d}\sigma }\Big (\frac{\partial \sigma }{\partial c_s}\frac{\mathrm {d}c_s^*}{\mathrm {d}c_t}+\frac{\partial \sigma }{\partial c_t}\Big ), \end{aligned}$$

(39)

which can be rearranged to the form

$$\begin{aligned} \frac{\mathrm {d}c_s^*}{\mathrm {d}c_t}=\frac{(c_i-c_s^*)\frac{\partial ^2 \sigma }{\partial c_s\partial c_t}+\frac{\mathrm {d}\phi }{\mathrm {d}\sigma }\frac{\partial \sigma }{\partial c_t}}{(c_s^*-c_i)\frac{\partial ^2 \sigma }{\partial c_s^2}+(1-\frac{\mathrm {d}\phi }{\mathrm {d}\sigma })\frac{\partial \sigma }{\partial c_s}}. \end{aligned}$$

(40)

Next, we can calculate

$$\begin{aligned} \frac{\mathrm {d}\phi }{\mathrm {d}\sigma }=\frac{(\sigma \beta )(\beta \sigma - 2\gamma )-h\gamma }{(\beta \sigma -\gamma )^2}. \end{aligned}$$

(41)

Assumption (H2) implies that

$$\begin{aligned} \frac{\mathrm {d}\phi }{\mathrm {d}\sigma } < 0. \end{aligned}$$

(42)

This, together with Assumption (H1), implies that the denominator of the right-hand side of (40) is negative, the numerator is positive, and finally that $\mathrm {d}c_s^* / \mathrm {d}c_t \le 0$. $\square$

Corollary 1

If Theorem 7 holds, then a small increase in public investment that increases public good ( ${\partial W}/{\partial c_t} > 0$ ) will also suffer from policy resistance:

$$\begin{aligned} \frac{\partial W}{\partial c_t} \Delta c_t > 0 \quad \text {and} \quad \frac{\partial W}{\partial c_s^*} \frac{\partial c_s^*}{\partial c_t} \Delta c_t < 0. \end{aligned}$$

(43)

Proof

For a small increase in public investment, $\Delta c_t > 0$. By assumption, then,

$$\begin{aligned} \frac{\partial W}{\partial c_t} \Delta c_t > 0. \end{aligned}$$

Now, we also know that for any Nash equilibrium, $c_s^* \in (0,\overline{c}_s^*)$, so by Theorem 5,

$$\begin{aligned} \frac{\partial W}{\partial c_s^*} > 0. \end{aligned}$$

and from Thereorem 6,

$$\begin{aligned} \frac{\partial c_s^*}{\partial c_t} < 0. \end{aligned}$$

The conclusion follows by inspection. $\square$

Theorem 9

If the effects of government and individual interventions are independent, such that

$$\begin{aligned} \sigma (c_s,c_t) = \sigma _s(c_s) \sigma _t(c_t), \end{aligned}$$

and $\sigma _s(c_s)$ is smoothly decreasing and convex, then increased public investment decreases equilibrium individual investment in self-protection ($\mathrm {d}c_s^* / \mathrm {d}c_t \le 0$).

Proof

Since the game equilibrium $c_s^*(c_t)$ solves Eq. (22),

$$\begin{aligned} \frac{\partial c_s^B}{\partial c_t} + \frac{\partial c_s^B}{\partial \widetilde{\lambda }} \frac{\partial \widetilde{\lambda }}{\partial c_t}+ \frac{\partial c_s^B}{\partial \widetilde{\lambda }} \frac{\partial \widetilde{\lambda }}{\partial \overline{c}_s} \frac{\partial \overline{c}_s}{\partial c_s^*} \frac{\partial c_s^*}{\partial c_t} = \frac{\partial c_s^*}{\partial c_t}, \end{aligned}$$

(44)

with $\overline{c}_s= c_s^*$. We can rearrange the equation and show

$$\begin{aligned} \frac{\partial c_s^*}{\partial c_t} = \left( \frac{\partial c_s^B}{\partial c_t} + \frac{\partial c_s^B}{\partial \widetilde{\lambda }} \frac{\partial \widetilde{\lambda }}{\partial c_t} \right) \left( 1- \frac{\partial c_s^B}{\partial \widetilde{\lambda }} \frac{\partial \widetilde{\lambda }}{\partial \overline{c}_s} \right) ^{-1}. \end{aligned}$$

(45)

The proof proceeds by showing the right-hand side of Eq. (45) is never positive.

If $\sigma (c_s,c_t) = \sigma _s(c_s) \sigma _t(c_t)$, then Eq. (16) reduces to

$$\begin{aligned} (c_s-c_i)\frac{\partial \sigma _s}{\partial c_s}-\sigma _s(c_s)=\frac{h+\gamma }{\widetilde{\lambda }(\overline{c}_s,c_t)\sigma _t(c_t)}, \end{aligned}$$

(46)

and then, the best response satisfies

$$\begin{aligned} (c_s^B-c_i)\frac{\partial \sigma _s}{\partial c_s}-\sigma _s(c_s^B)=\frac{h+\gamma }{\widetilde{\lambda }(\overline{c}_s,c_t)\sigma _t(c_t)}, \end{aligned}$$

(47)

Differentiating the above equation with respect to $c_t$ and rearranging, we have

$$\begin{aligned} \Big (\frac{\partial c_s^B}{\partial c_t}+\frac{\partial c_s^B}{\partial \widetilde{\lambda }}\frac{\partial \widetilde{\lambda }}{\partial c_t}\Big )\Big ((c_s^B-c_i)\frac{\partial ^2\sigma _s}{\partial c_s^2}\Big ) =-\frac{(h+\gamma )(\frac{\partial \widetilde{\lambda }}{\partial c_t}\sigma _t(c_t)+\widetilde{\lambda }\frac{\partial \sigma _t}{\partial c_t})}{\big (\widetilde{\lambda }(\overline{c}_s, c_t)\sigma _t(c_t)\big )^2} \end{aligned}$$

(48)

By inspection, we know that

$$\begin{aligned} -\frac{(h+\gamma )(\frac{\partial \widetilde{\lambda }}{\partial c_t}\sigma _t(c_t)+\widetilde{\lambda }\frac{\partial \sigma _t}{\partial c_t})}{\big (\widetilde{\lambda }(\overline{c}_s, c_t)\sigma _t(c_t)\big )^2}\ge 0 \end{aligned}$$

(49)

and

$$\begin{aligned} (c_s^B-c_i)\frac{\partial ^2\sigma _s}{\partial c_s^2}\le 0, \end{aligned}$$

(50)

Hence,

$$\begin{aligned} \frac{\partial c_s^B}{\partial c_t}+\frac{\partial c_s^B}{\partial \widetilde{\lambda }}\frac{\partial \widetilde{\lambda }}{\partial c_t}\le 0. \end{aligned}$$

(51)

From this, ${\partial c_s^B}/{\partial \widetilde{\lambda }}>0$ in Remark 1, and ${\partial \widetilde{\lambda }}/{\partial \overline{c}_s}\le 0$ in Theorem 2, we can see by inspection of Eq. (45) that ${\partial c_s^*}/{\partial c_t}\le 0$. $\square$