Finite-Population Depletion and the Stochastic Final Epidemic Size

Why the Final Size Depends on Susceptible Depletion

In a finite epidemic, each infection removes one person from the susceptible group. Even if the biological transmission and recovery parameters stay fixed, the infection hazard therefore changes as the epidemic proceeds. The final number of cases is random: it depends on the order in which infection and recovery events occur, not just on their average rates. For a small population, we can calculate its distribution exactly by multiplying the probabilities along possible event paths.

Consider a closed population of size $N$. On a probability space $(\Omega,\mathcal{F},\mathbb{P})$, let $(S_t,I_t,R_t)_{t\ge0}$ be a continuous-time Markov chain, adapted to its natural filtration $(\mathcal{F}_t)_{t\ge0}$, with state space $\{(s,i,r):s+i+r=N\}$. The coordinates count susceptible, infectious, and removed people. Assume no births, deaths, or reinfections; transmission parameter $\beta\ge0$; and recovery parameter $\gamma>0$. From state $(s,i,r)$, infection changes the state to $(s-1,i+1,r)$ at rate $\beta si/N$, while recovery changes it to $(s,i-1,r+1)$ at rate $\gamma i$. These are the only possible transitions. The generator’s diagonal entry is the negative sum of its outgoing rates, so each generator row sums to zero.

Write the initial state as $(S_0,I_0,R_0)$. Let $K=S_0-S_\infty$ be the number of new infections after time zero, where $S_\infty$ is the susceptible count once no infectious people remain. We call $F=I_0+K$ the final epidemic size, counting the initially infectious people as cases but not counting people already removed at time zero.

Definition: For a state with $i>0$, the next event is an infection with probability $a_s=\frac{\beta s}{\beta s+\gamma N}$, and a recovery with probability $1-a_s=\frac{\gamma N}{\beta s+\gamma N}$. These probabilities are defined because $\gamma>0$. When $\beta=0$, $a_s=0$. The factor $i$ cancels from the ratio of the infection rate $\beta si/N$ to the total rate $\beta si/N+\gamma i$.

To obtain the next-event probability, the two competing event times, conditional on the current state, are exponential with rates $\lambda=\beta si/N$ and $\mu=\gamma i$. If $\lambda+\mu>0$, the probability infection occurs first is $\int_0^\infty \lambda e^{-(\lambda+\mu)t}\,dt=\lambda/(\lambda+\mu)$. Here $\lambda+\mu>0$ whenever $i>0$, since $\gamma>0$. This gives the formula above, including the case $\lambda=0$.

Depletion Changes the Event Probabilities

At fixed $N,\beta,\gamma$, the probability of infection at the next event depends on the current susceptible count $s$, not just on the number infectious. If $\beta>0$, then for $s\ge0$,

a_{s+1}-a_s =\frac{\beta\gamma N}{(\beta(s+1)+\gamma N)(\beta s+\gamma N)} >0.

Thus, among states with infectious people present, reducing $s$ reduces the probability that the next event is an infection. The infection rate itself, $\beta si/N$, also falls with $s$ when $i$ is held fixed. This does not mean every later event is less likely to be an infection than every earlier one: recoveries change $i$, and infection events change both $s$ and $i$. The comparison concerns the next-event probability at a fixed parameter set and current state.

Worked Example: The Next Event Before and After Depletion

Take $N=4$ and $\beta=\gamma$. Compare states $(3,1,0)$ and $(1,1,2)$. At $(3,1,0)$, the infection and recovery rates are $3\beta/4$ and $\beta$, respectively, so the total outgoing rate is $7\beta/4$; the diagonal generator entry is $-7\beta/4$, and that row sums to zero. The probability the next event is an infection is therefore $(3\beta/4)/(7\beta/4)=3/7$.

At $(1,1,2)$, the rates are $\beta/4$ and $\beta$. Their sum is $5\beta/4$, so the diagonal entry is $-5\beta/4$, again giving row sum zero. The next-event infection probability is $(\beta/4)/(5\beta/4)=1/5$. With the same one infectious person and the same parameters, the smaller susceptible count lowers this probability from $3/7$ to $1/5$.

Check: The general formula gives $a_3=3/(3+4)=3/7$ and $a_1=1/(1+4)=1/5$, as computed from the rates.

A Recursion for the Final-Size Distribution

For $s,i\ge0$, define $H_{s,i}(k)$ as the probability of exactly $k$ additional infections before extinction, starting with $s$ susceptible and $i$ infectious people. The population size $N$ and parameters $\beta,\gamma$ are held fixed, and $k$ is a nonnegative integer. When $i=0$, no further events occur and $H_{s,0}(0)=1$, while $H_{s,0}(k)=0$ for $k>0$. If $s=0$, only recoveries can occur, so the same values hold for every $i$.

Theorem (Finite-population final-size recursion): Suppose $N$ is finite, $\beta\ge0$, and $\gamma>0$. Starting from any state, extinction occurs with probability one. For $s,i>0$, the final-size probabilities satisfy $$ H_{s,i}(k)=a_sH_{s-1,i+1}(k-1)+(1-a_s)H_{s,i-1}(k), $$ where $H_{s-1,i+1}(-1)=0$. Consequently, from initial state $(S_0,I_0,R_0)$, $\mathbb{P}(F=I_0+k)=H_{S_0,I_0}(k)$.

Proof. First consider extinction. Put $c=\gamma/(\beta+\gamma)>0$. In any state, $s\le N$, so the probability that the next event is a recovery is $\gamma N/(\beta s+\gamma N)\ge c$. Also, the number infectious is at most $N$. From any nonabsorbed state, the probability of successive recoveries until extinction is therefore at least $c^N$: at most $N$ recoveries are needed, and a product of at most $N$ factors, each at least $c$, is at least $c^N$. Extinction thus has probability at least $c^N$ within the next $N$ events, conditional on any nonabsorbed state. By the Markov property, the probability of avoiding extinction through $m$ such blocks is at most $(1-c^N)^m$, which tends to zero as $m$ tends to infinity. Hence extinction occurs almost surely.

For the recursion, condition on the first event from $(s,i)$, with $s,i>0$. With probability $a_s$ it is an infection, leaving state $(s-1,i+1)$; exactly $k$ further infections then requires exactly $k-1$ infections after that event. With probability $1-a_s$ it is a recovery, leaving $(s,i-1)$, after which exactly $k$ further infections are required. The Markov property gives the displayed equation; if $k=0$, the infection branch contributes zero, as encoded by $H_{s-1,i+1}(-1)=0$. Finally, the total cases are the $I_0$ initially infectious people plus the $k$ new infections, so the stated distribution of $F$ follows. $\square$

Equivalently, a particular finite event path has probability equal to the product of its successive next-event probabilities. Paths with the same number of infections can be added to find the probability of that final size. The recursion organizes the same calculation without requiring a separate list of every path. It is finite: there can be at most $s$ infections, and at most $i+s$ recoveries before extinction.

Worked Example: All Final Sizes for a Three-Person Epidemic

Let $N=3$, $\beta=\gamma$, and start from $(S_0,I_0,R_0)=(2,1,0)$. Initially the infection rate is $2\beta/3$ and the recovery rate is $\beta$; the total is $5\beta/3$, so the diagonal rate is $-5\beta/3$ and the row sum is zero. Thus the first event is an infection with probability $2/5$, and a recovery with probability $3/5$.

If the first event is a recovery, there are no more infections. This gives $\mathbb{P}(K=0)=3/5$, or final size $F=1$. If the first event is an infection, the state becomes $(1,2,0)$. There, the infection rate is $2\beta/3$, the recovery rate is $2\beta$, and the total is $8\beta/3$, giving diagonal rate $-8\beta/3$. The next event is an infection with probability $1/4$ and a recovery with probability $3/4$.

From $(1,2,0)$, no further infection occurs only if both next events are recoveries: after the first recovery the state is $(1,1,1)$, where the recovery probability is still $3/4$. Therefore $\mathbb{P}(K=1)=(2/5)(3/4)(3/4)=9/40$, corresponding to $F=2$. At least one more infection occurs with probability $1-(3/4)^2=7/16$ from $(1,2,0)$; it uses the last susceptible person. Hence $\mathbb{P}(K=2)=(2/5)(7/16)=7/40$, corresponding to $F=3$.

Check: The three final-size probabilities sum to $3/5+9/40+7/40=24/40+9/40+7/40=1$. The full distribution is therefore $F=1,2,3$ with probabilities $3/5,9/40,7/40$, respectively.

Worked Example: A Specific Transition Path

Take $N=5$, $\beta=\gamma$, and initial state $(3,1,1)$. Consider the event path infection, recovery, infection, recovery, recovery. Its successive states are $(3,1,1)\to(2,2,1)\to(2,1,2)\to(1,2,2)\to(1,1,3)\to(1,0,4)$. It has two new infections, so on this path the final epidemic size is $F=1+2=3$.

At $(3,1,1)$, infection and recovery rates are $3\beta/5$ and $\beta$, total $8\beta/5$, with diagonal entry $-8\beta/5$. The required infection probability is $3/8$. At $(2,2,1)$, the rates are $4\beta/5$ and $2\beta$, total $14\beta/5$, diagonal entry $-14\beta/5$, and recovery probability $5/7$. At $(2,1,2)$, the rates are $2\beta/5$ and $\beta$, total $7\beta/5$, diagonal entry $-7\beta/5$, and infection probability $2/7$. At $(1,2,2)$, the rates are $2\beta/5$ and $2\beta$, total $12\beta/5$, diagonal entry $-12\beta/5$, and recovery probability $5/6$. Finally, at $(1,1,3)$, the rates are $\beta/5$ and $\beta$, total $6\beta/5$, diagonal entry $-6\beta/5$, and recovery probability $5/6$. Each listed diagonal entry is the negative outgoing-rate sum, so the corresponding generator row sums to zero.

Multiplying the probabilities for this ordered path gives $(3/8)(5/7)(2/7)(5/6)(5/6)=750/14112=125/2352$. Check: The path has five events, two of which are infections and three recoveries; the final state has zero infectious people and $S_\infty=1$, consistent with two new infections from the initial three susceptible people. This is the probability of this particular path, not the probability of every path with final size three.

Common Pitfall: Holding the Susceptible Count Fixed

A common error is to use the initial infection probability for every event. In the three-person example, after the first infection the susceptible count is $1$, not $2$. An incorrect calculation of the probability of exactly one new infection might use the initial recovery probability $3/5$ for both recoveries, giving $(2/5)(3/5)^2=18/125$. That is wrong because the relevant recovery probability after the first infection is $3/4$, not $3/5$. The correct probability is $(2/5)(3/4)^2=9/40$. The next-event probabilities must be recalculated from the current state after every transition.

For larger finite populations, the same method applies: set boundary values for $i=0$ or $s=0$, then use the recursion to obtain the distribution of $K$, and shift by $I_0$ to obtain the distribution of total cases $F$. Exact path calculations show why final size is not determined by the early epidemic growth rate alone: susceptible depletion changes the transition probabilities along each path.

Check Your Understanding

Use the current state, rather than the initial state, whenever you calculate a next-event probability.

For $N=6$, $\beta=2$, and $\gamma=1$, calculate the probability that the next event is an infection when $s=2$ and $i=3$.
In a closed population with $S_0=4$ and $I_0=2$, express the final epidemic size in terms of $S_\infty$.
For $N=3$, $\beta=\gamma$, and state $(1,2,0)$, write the two possible next-event probabilities and identify the states they lead to.
Explain why the final-size recursion assigns zero probability to $k=0$ on its infection branch.
Why does the path probability in the five-person example not by itself give the probability that the final epidemic size is three?

Finite-Population Depletion and the Stochastic Final Epidemic Size

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