Theoretical background

From SIR to final size with homogeneous mixing

The dynamics of the proportion of susceptible and infectious individuals over time in the standard susceptible-infectious-recovered (SIR) epidemic model are given by the following ordinary differential equations: \[ \frac{d S(t)}{d t} = -\beta S(t) I(t) \\ \frac{d I(t)}{d t} = \beta S(t) I(t) -\gamma I(t) \] where \(\beta\) is the transmission rate and \(\gamma\) is the rate of recovery.

To calculate the proportion infected, we need to derive an expression for the proportion of susceptibles who got infected during the epidemic, i.e. \(\phi = 1- S(\infty)/S(0)\). Rather than simulating the model over a very long time period (i.e. \(t \rightarrow \infty\)), we can instead calculate \(dI/dS\) and integrate:

\[ \frac{d I}{d S} = - 1 + \frac{\gamma}{\beta S} \\ I(t) = - S(t) + \frac{\gamma}{\beta} \log S(t) + c \] where \(c\) depends on initial conditions. Under the assumption the population is fully susceptible initially (i.e. \(S(0)=1\)), and \(I(t) \rightarrow 0\) as \(t \rightarrow \infty\), we obtain the following: \[ \log S(\infty) = - R_0 [1-S(\infty) ] \\ \mathrm{log}~\phi -R_0 (1-\phi) = 0 = F(\phi) \] This equation doesn’t have an analytical solution, but there are various methods available to allow it to be solved numerically to find the unique solution in (0,1]. For example, we can use Newton’s method (solver = "newton" in the final_size() function) to find the final epidemic size:

1. Select an arbitrary initial estimate \(x_0\) and tolerance \(T\) ;
2. Solve \(\frac{dF}{d \phi} (x_0) \Delta x = −F(x_0) ~ \text{for } ~ \Delta x\) ;
3. Set \(x_{i+1} = x_i + \Delta x\);
4. Repeat until \(x_i\) converges sufficiently (i.e. difference in each step is less than \(T\))

Varying susceptibility

We can extend the above formulation to account for variable susceptibility among individuals, following the methods outlined in Miller (2012). First, we define \(p(x)\) to be the probability density for a randomly chosen individual to be of susceptibility type \(x\). If we define the probability an individual of type \(x\) to be ultimately infected as \(\phi(x)\), then the overall proportion of the population to be infected is defined as follows, depending on whether \(p(x)\) is a continuous or discrete distribution:

\[ \hat\phi = \int_0^\infty \phi(x) p(x) dx \\ \text{or} \\ \hat\phi = \sum_x \phi(x) p(x) dx \\ \]

If we define \(x\hat\phi\) to be the expected number of transmissions to a given individual of type \(x\), we can therefore calculate the probability that this individual remains susceptible. This gives the following relationship between \(\phi\) and \(\hat\phi\):

\[ 1-\phi(x) = S(x,0) e^{-x \hat\phi } \\ \] And hence we can combine the above to give: \[ \hat\phi = \int_0^\infty (1- S(x,0) e^{-x \hat\phi }) p(x) dx \\ \hat\phi = 1- \int_0^\infty S(x,0) e^{-x \hat\phi } p(x) dx \]

Under the assumption that \(S(x,0)\) is near 1 initially (i.e. f), we therefore obtain: \[ \hat\phi = 1- \int_0^\infty p(x) e^{-x \hat\phi } dx \]

Final size equation with heterogeneous mixing

We can extend the same derivation for the above homogenous model for populations with multiple age groups, following Andreasen (2011). The age-dependent ODEs are defined as follows:

\[ \frac{d S(a,t)}{d t} = -S(a,t) \sum_b \beta_{ab} I(b,t) \\ \frac{d I(a,t)}{d t} = S(a,t) \sum_b \beta_{ab} I(b,t) -\gamma I(a,t) \] Where \(\beta_{a,b}\) is the transmission rate from group \(b\) to group \(a\). If we define \(R_{a,b} =\beta_{a,b}/\gamma\) to be group specific reproduction number, we can rearrange the above and integrate to derive an expression for the final size in each age group:

\[ - \gamma \sum_b R_{a,b} I(b,t) = \frac{d S(a,t)}{d t} \frac{1}{S(a,t)} \\ - \gamma \sum_b R_{a,b} \int_0^\infty I(b,t) ~dt = \int_0^\infty \frac{d S(a,t)}{d t} \frac{1}{S(a,t)} ~dt = \mathrm{log}~S(a,\infty) -\mathrm{log}~S(a,0) = \mathrm{log}~\frac{S(a,\infty)}{S(a,0)} \\ -\gamma \int_0^\infty I(a,t) ~dt = \int_0^\infty \frac{d S(a,t)}{d t} +\frac{dI(a,t)}{dt} ~dt = S(a,\infty) -S(a,0) ~. \] and hence:

\[ \mathrm{log}~\frac{S(a,\infty)}{S(a,0)} = - \sum_b R_{a,b} [ S(b,\infty) -S(b,0)] \]

If we define \(\underline\phi = (\phi_1, ... \phi_n)\) to be a vector of final sizes in each of \(n\) age group, we can rewrite the above as

\[ \mathrm{log}~\underline\phi -\mathbf{R} (1-\underline\phi) = 0 = F(\underline\phi) \] where \(\mathbf{R}\) is a matrix with entries \(R_{i,j}\). Once again, we can use Newton’s method, this time with \(\underline\phi\) as a vector, and hence we need to solve a set of linear equations in step 2.