Game-Theoretical Model of Retroactive Hepatitis B Vaccination in China

Abstract

Hepatitis B (HepB) is one of the most common infectious diseases affecting over two billion people worldwide. About one third of all HepB cases are in China. In recent years, China made significant efforts to implement a nationwide HepB vaccination program and reduced the number of unvaccinated infants from 30 to 10%. However, many individuals still remain unprotected, particularly those born before 2003. Consequently, a catch-up retroactive vaccination is an important and potentially cost-effective way to reduce HepB prevalence. In this paper, we analyze a game theoretical model of HepB dynamics that incorporates government-provided vaccination at birth coupled with voluntary retroactive vaccinations. Given the uncertainty about the long-term efficacy of the HepB vaccinations, we study several scenarios. When the waning rate is relatively high, we show that this retroactive vaccination should be a necessary component of any HepB eradication effort. When the vaccine offers long-lasting protection, the voluntary retroactive vaccination brings the disease incidence to sufficiently low levels. Also, we find that the optimal vaccination rates are almost independent of the vaccination coverage at birth. Moreover, it is in an individual’s self-interest to vaccinate (and potentially re-vaccinate) at a rate just slightly above the vaccine waning rate.

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Acknowledgements

AC, SM, MN, and VP worked on this manuscript as part of the course MATH/BIOL 380—Introduction to mathematical biology. They wish to acknowledge the help and support of their classmates. The authors would also like to thank the anonymous reviewers for their comments and suggestions that helped to improve the manuscript.

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Appendix A: Equilibria of the Dynamics

Appendix A: Equilibria of the Dynamics

We have to solve the following system of (algebraic) equations.

$$\begin{aligned} \frac{\mathrm{d}S}{\mathrm{d}t}&= 0=\mu \omega (1-vC)+\psi V-S(\mu _0+\gamma _3+\beta I+ \epsilon \beta C) \end{aligned}$$
(23)
$$\begin{aligned} \frac{\mathrm{d}V}{\mathrm{d}t}&=0=\mu (1-\omega )+\gamma _3S-V(\mu _0+\psi ) \end{aligned}$$
(24)
$$\begin{aligned} \frac{\mathrm{d}L}{\mathrm{d}t}&=0= (\beta I+\epsilon \beta C)S-L(\mu _0+\sigma ) \end{aligned}$$
(25)
$$\begin{aligned} \frac{\mathrm{d}I}{\mathrm{d}t}&=0=\sigma L-I(\mu _0+\gamma _1) \end{aligned}$$
(26)
$$\begin{aligned} \frac{\mathrm{d}R}{\mathrm{d}t}&=0=(1-q)\gamma _1 I+\gamma _2 C-R \mu _0 \end{aligned}$$
(27)
$$\begin{aligned} \frac{\mathrm{d}C}{\mathrm{d}t}&=0 = q \gamma _1 I + \mu \omega r C - C (\mu _0 + \mu _1 + \gamma _2) \end{aligned}$$
(28)

Disease Free Equilibrium

In the disease free equilibrium when \(0=L_0=I_0=C_0=R_0\), the system (23)–(28) reduces to

$$\begin{aligned} 0&= \mu \omega +\psi V_0-S_0(\mu _0+\gamma _3) \end{aligned}$$
(29)
$$\begin{aligned} 0&=\mu (1-\omega )+\gamma _3S_0-V_0(\mu _0+\psi ). \end{aligned}$$
(30)

Adding (29) and (30) together yields

$$\begin{aligned} S_0+V_0 = \frac{\mu }{\mu _0}. \end{aligned}$$
(31)

By (30),

$$\begin{aligned} V_0 = \frac{\mu (1-\omega )+\gamma _3 S_0}{\mu _0 + \psi } \end{aligned}$$
(32)

and thus

$$\begin{aligned} S_0&= \frac{\mu }{\mu _0}\cdot \frac{\mu _0\omega + \psi }{\mu _0+\gamma _3+\psi } \end{aligned}$$
(33)
$$\begin{aligned} V_0&= \frac{\mu }{\mu _0}\cdot \frac{\mu _0+\gamma _3 -\mu _0\omega }{\mu _0+\gamma _3+\psi }. \end{aligned}$$
(34)

Now, we will determine the number of secondary infections caused by a single infected individual. The individual stays infected for a period \(\frac{1}{\mu _0+\gamma _1}\). During that period, at rate \(\beta S_0\), they infect susceptible individuals who become latently infected. Of those, only a fraction of \(\frac{\sigma }{\sigma +\mu _0}\) becomes acutely infected. Moreover, an infected individual can become a carrier with probability \(\frac{q\gamma _1}{\mu _0+\gamma _1}\). The individual stays in the carrier compartment for a period \(\frac{1}{\mu _0+\mu _1+\gamma _2}\). During that period, at rate \(\beta \epsilon S_0\), they infect susceptible individuals who then become latently infected. The carrier also makes other carriers at rate \(\mu \omega v\). Consequently, if l is a number of latently infected individuals caused by a single carrier, we get

$$\begin{aligned} l = \frac{1}{\mu _0+\mu _1+\gamma _2} (\beta \epsilon S_0 + \mu \omega v l) \end{aligned}$$
(35)

and thus

$$\begin{aligned} l = \frac{\beta \epsilon S_0}{\mu _0+\mu _1+\gamma _2 - \mu \omega v}. \end{aligned}$$
(36)

Putting it all together, we get, as in Zou et al. (2010b)

$$\begin{aligned} R_0 = \beta \frac{\mu }{\mu _0}\cdot \frac{\mu _0\omega + \psi }{\mu _0+\gamma _3+\psi } \cdot \frac{1}{\mu _0+\gamma _1}\cdot \frac{\sigma }{\mu _0+ \sigma } \left[ 1 + \frac{q\gamma _1\epsilon }{\mu _0+\mu _1+\gamma _2-\mu \omega v} \right] . \end{aligned}$$
(37)

The disease free equilibrium is stable when \(R_0<1\) (Zou et al. 2010b, Theorem 1).

Endemic Equilibrium

From (28), (26), and (27) we get

$$\begin{aligned} C^*&= \frac{q \gamma _1 I^*}{\mu _0+\mu _1+\gamma _2 - \mu \omega v} \end{aligned}$$
(38)
$$\begin{aligned} L^*&=\frac{\mu _0+\gamma _1}{\sigma }I^* \end{aligned}$$
(39)
$$\begin{aligned} R^*&=\frac{(1-q)\gamma _1 I^*+\gamma _2 C^*}{\mu _0}. \end{aligned}$$
(40)

From (25),

$$\begin{aligned} 0&=\beta (I^*+\epsilon C^*)S^*-L^*(\mu _0+\sigma ) \end{aligned}$$
(41)
$$\begin{aligned}&=\beta \left( I^*+\epsilon \frac{q \gamma _1}{\mu _0 + \mu _1 +\gamma _2 - \mu \omega v} I^*\right) S^*-\frac{\mu _0+\gamma _1}{\sigma }I^*(\mu _0+\sigma ) \end{aligned}$$
(42)

which, after dividing by \(I^*\ne 0\), yields

$$\begin{aligned} S^*=\frac{\mu _0+\sigma }{\sigma }(\mu _0+\gamma _1)\frac{1}{\beta \left( 1+\frac{\epsilon q \gamma _1}{\mu _0 + \mu _1 +\gamma _2 - \mu \omega v}\right) } = \frac{S_0}{R_0}. \end{aligned}$$
(43)

From (24),

$$\begin{aligned} V^*=\frac{\mu (1-\omega )+\gamma _3 S^*}{\mu _0+\psi } \end{aligned}$$
(44)

and finally, by (23)

$$\begin{aligned} \begin{aligned} 0&=\mu \omega \left( 1-v \frac{q \gamma _1}{\mu _0 + \mu _1 +\gamma _2 - \mu \omega v} I^*\right) + \psi V^* \ldots \\&\quad -S^*\left( \mu _0+\gamma _3+\beta I^*+\epsilon \beta \frac{q \gamma _1}{\mu _0 + \mu _1 +\gamma _2 - \mu \omega v}I^*\right) \end{aligned} \end{aligned}$$
(45)

which yields

$$\begin{aligned} I^*=\frac{\mu \omega +\psi V^*-S^*(\mu _0+\gamma _3)}{\mu \omega v \frac{q \gamma _1}{\mu _0 + \mu _1 +\gamma _2 - \mu \omega v}+\beta S^*+ \epsilon \beta \frac{q \gamma _1}{\mu _0 + \mu _1 +\gamma _2 - \mu \omega v} S^*}. \end{aligned}$$

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Chouhan, A., Maiwand, S., Ngo, M. et al. Game-Theoretical Model of Retroactive Hepatitis B Vaccination in China. Bull Math Biol 82, 80 (2020). https://doi.org/10.1007/s11538-020-00748-5

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Keywords

  • Nash equilibrium
  • Cost-benefits analysis
  • Vaccine waning
  • Herd immunity