Extinction and outbreak probabilities
At the beginning of an epidemic, only a few infectious individuals may be present. Even when transmission is strong enough for infection to grow on average, chance can still determine whether the first transmission chains disappear or become established.
Why can infection die out when \(R_0>1\)?
Suppose \(R_0=2\). This does not mean that every infectious person infects exactly two others. It means that the average number of secondary infections is 2 under the assumptions defining \(R_0\).
One person may infect nobody, another one person, and another four people. Therefore an early transmission chain can still disappear by chance.
Two possible outcomes from the same model
What do we mean by extinction?
For a simple closed epidemic without imported infection, extinction means that no infectious individuals remain:
\[\boxed{I(t)=0}.\]If state 0 is absorbing, infection cannot restart from that state.
For an SEIR model, \(I=0\) alone may not mean extinction if exposed individuals remain. A more appropriate disease-free condition is
\[E=0\quad\text{and}\quad I=0.\]What do we mean by an outbreak?
“Outbreak” must be defined before its probability is calculated. A useful definition is
\[\boxed{\text{outbreak}=\{I(t)\text{ reaches }K\text{ before }0\}},\]where \(K\) is a chosen threshold.
Early transmission as a branching process
When only a few people are infected, almost everyone is still susceptible. Under suitable assumptions, each infectious individual can be viewed as starting a new transmission branch.
The offspring distribution
Let \(X\) be the number of secondary infections caused by one infectious individual during the early phase:
\[P(X=k)=p_k,\qquad k=0,1,2,\ldots\]The mean is
\[E[X]=\sum_{k=0}^{\infty}kp_k.\]In a simple branching approximation, this mean plays the role of the reproduction number.
Why the full distribution matters
Two pathogens can have the same mean number of secondary infections but different variability. If many individuals cause no infections while a few cause many, extinction behaviour can differ markedly from a model in which transmission is much more even.
The probability-generating function
Define
\[\boxed{G(s)=p_0+p_1s+p_2s^2+p_3s^3+\cdots}.\]Equivalently, \(G(s)=E[s^X]\).
Why does extinction satisfy \(q=G(q)\)?
Let
\[q=P(\text{eventual extinction from one initial infectious individual}).\]If the first person produces exactly \(k\) new infectious individuals, all \(k\) descendant branches must eventually die out. Under the branching approximation, that has probability
\[q^k.\]Weighting over all possible values of \(k\) gives
\[q=p_0+p_1q+p_2q^2+p_3q^3+\cdots,\]so
\[\boxed{q=G(q)}.\]Seeing the extinction fixed point
The graph is generated directly from the function \(G(s)=e^{2(s-1)}\), not drawn by eye.
Why do we choose the smaller solution?
The equation \(q=G(q)\) always has \(q=1\) as a solution because \(G(1)=1\). If the mean offspring number is at most 1, extinction occurs with probability 1 under the standard branching assumptions.
If the mean is greater than 1, a second fixed point may appear below 1. The extinction probability is the smallest solution in \([0,1]\).
A simple birth–death epidemic approximation
Suppose each infectious individual generates new infections at rate \(\lambda\) and recovers at rate \(\mu\). The next event for one infectious person is either infection or recovery.
The probabilities are
\[P(\text{recovery first})=\frac{\mu}{\lambda+\mu},\qquad P(\text{infection first})=\frac{\lambda}{\lambda+\mu}.\]If recovery occurs first, the chain ends. If infection occurs first, there are two infectious branches, and both must eventually die out for extinction. Therefore
\[q=\frac{\mu}{\lambda+\mu}+\frac{\lambda}{\lambda+\mu}q^2.\]This gives
\[(q-1)(\lambda q-\mu)=0.\]Hence
\[\boxed{q=\begin{cases}1,&\lambda\le\mu,\\\mu/\lambda,&\lambda>\mu.\end{cases}}\]Connection with \(R_0\)
For this particular simple birth–death approximation,
\[R_0=\frac{\lambda}{\mu}.\]Therefore, when \(R_0>1\),
\[\boxed{q=\frac1{R_0}},\qquad\boxed{P(\text{major outbreak})\approx1-\frac1{R_0}}.\]How outbreak probability changes with \(R_0\)
Near \(R_0=1\), extinction is still very likely. As \(R_0\) increases, establishment becomes more likely, but extinction from a small introduction remains possible.
Starting with several infectious individuals
If the extinction probability from one initial infectious individual is \(q\), and the initial transmission branches can be approximated as independent, then
\[\boxed{P(\text{extinction}\mid I(0)=i_0)\approx q^{i_0}}.\]Therefore
\[\boxed{P(\text{major outbreak}\mid I(0)=i_0)\approx1-q^{i_0}}.\]Finite-population hitting probabilities
For a finite birth–death CTMC, define
\[h_i=P(\text{reach }K\text{ before }0\mid I(0)=i).\]The boundary conditions are
\[h_0=0,\qquad h_K=1.\]For \(1\le i\le K-1\), conditioning on the next jump gives
\[h_i=\frac{b(i)}{b(i)+d(i)}h_{i+1}+\frac{d(i)}{b(i)+d(i)}h_{i-1}.\]Equivalently,
\[\boxed{b(i)[h_{i+1}-h_i]+d(i)[h_{i-1}-h_i]=0}.\]Monte Carlo estimation
When an analytical calculation is difficult, simulate the stochastic model many times. For each run define
\[Y_r=\begin{cases}1,&\text{outbreak threshold reached first},\\0,&\text{extinction reached first}.\end{cases}\]Then
\[\boxed{\widehat P(\text{outbreak})=\frac1n\sum_{r=1}^{n}Y_r}.\]With enough simulations, this estimate approaches the corresponding model probability, subject to Monte Carlo sampling error.
What each quantity tells us
| Quantity | Question answered |
|---|---|
| \(R_0\) | Is average early growth possible? |
| extinction probability | What is the chance the chain eventually disappears? |
| outbreak probability | What is the chance a specified outbreak criterion is reached? |
| Monte Carlo estimate | How can we estimate these probabilities for a complicated model? |