4.5. Generation-Based Infectious Disease Models

When we switch to a generation-based framework, we allow a wider range of offspring distributions. Now each infected individual transmits to \(k\) others, where \(k\) is a random variable and the recipients are chosen randomly from the population. If the recipient is susceptible, he becomes infected. Otherwise, it has no effect.

As before, while \(S/N\) is approximately \(1\), we can treat this as if it is a generation-based Galton-Watson process. Thus all of our results for Galton-Watson processes apply to \(I(g)\).

4.5.1. Size distribution of small outbreaks

The derivation we did showing that the probability a tree has \(j\) nodes is

\[ \mathbb{P}[\text{tree has $j$ nodes}] = \frac{1}{j} \left[x^{j-1}\right] \left(\mu(x)^j\right) \]

applies immediately here. The only difference is that for this model, each node represents a distinct infected individual. Thus \(j\) is the size of the small outbreak.

Note to self - make the result above a theorem

Example 4.3 (Continuous-time case from a generation-based perspective)

Consider the continuous-time SIS or SIR model with rates \(\beta\) and \(\gamma\). We will convert this to a generation-based framework and use this to calculate the probability distribution of outbreak sizes.

The probability the next event from an individual is a recovery is \(p = \gamma/(\beta+\gamma)\) and the probability of transmission is \(1-p\). The probaiblity of exactly \(k\) transmissions before the recovery is

\[ p_k = p(1-p)^k \]

By looking at the PGF for a geometric distribution, we could find \(\mu(x)\), but for practice we calculate it directly

\[\begin{align*} \mu(x) &= \sum_k p_k x^k\\ &= p \sum_k (x(1-p))^k\\ &= \frac{p}{1-x(1-p)} \end{align*}\]

Thus the probability of exactly \(j\) infections in the outbreak is

\[\begin{align*} \mathbb{P}[j \text{ infections}] &= \frac{1}{j} \left[x^{j-1}\right] \left(\frac{p}{1-(1-p)x}\right)^j\\ &= \frac{1}{j} \left[x^{j-1}\right]\left(\sum_{N=0}^\infty \binom{N+j-1}{N} p^j (1-p)^N x^N\right)\\ &= \frac{1}{j} \binom{2j-2}{j-1} p^j (1-p)^{j-1} \end{align*}\]

(we used the expansion for the negative binomial distribution here, from Example 1.11). Substituting \(p=\gamma/(\beta+\gamma)\) we find

\[\begin{align*} \mathbb{P}[j \text{ infections}] &= \frac{1}{j} \binom{2j-2}{j-1} \frac{\beta^j\gamma^{j-1}}{(\beta+\gamma)^{2j-1}}\\ &= \frac{1}{j} \binom{2j-2}{j-1} \frac{\mathcal{R}_0^j}{(1+\mathcal{R}_0)^{2j-1}} \end{align*}\]

which is what we derived for the continuous-time case.