Pure-Death Processes And Extinction Times

Pure-Death Dynamics and Extinction

A pure-death process models a population that can decrease by one at a time but cannot increase. Its extinction time depends on the successive waiting times spent at positive population sizes. When the death rate is proportional to the current population, the population at a fixed time has a simple binomial distribution, and extinction can also be understood as the time by which every individual has died.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$. For a fixed initial population $m$, let $(X_t)_{t\ge0}$ be a continuous-time Markov chain on $S=\{0,1,\ldots,m\}$, with natural filtration $\mathcal{F}_t=\sigma(X_s:0\le s\le t)$. State $0$ is absorbing. For $n\ge1$, the process leaves state $n$ at rate $\mu_n$, moving to $n-1$; assume each $\mu_n$ is finite and strictly positive. Thus its generator $Q=(q_{ij})$ has $q_{n,n-1}=\mu_n$, $q_{nn}=-\mu_n$, and all other entries zero; the row for state $0$ is zero. Every generator row sums to zero. The extinction time is $T_0=\inf\{t\ge0:X_t=0\}$.

Definition: A pure-death process has only transitions from $n$ to $n-1$ for positive $n$, with state $0$ absorbing. Its total departure rate from state $n$ is $\mu_n$; this need not equal a per-individual death rate unless the model specifies that relation.

Since a chain started at $m$ can make at most $m$ jumps before reaching $0$, no infinite sequence of jumps can occur in finite time. Its holding time in state $n$ is exponential with rate $\mu_n$. In the usual holding-time construction of a continuous-time Markov chain, the successive holding times are independent. This gives an immediate description of extinction.

Theorem: If $X_0=m\ge1$ and $\mu_1,\ldots,\mu_m>0$, then $$ T_0\ \overset{d}{=}\ E_m+E_{m-1}+\cdots+E_1, $$ where the $E_k$ are independent exponential random variables with rates $\mu_k$, respectively. Consequently, for $s\ge0$, $$ \mathbb{E}_m[e^{-sT_0}]=\prod_{k=1}^{m}\frac{\mu_k}{\mu_k+s},\qquad \mathbb{E}_m[T_0]=\sum_{k=1}^{m}\frac1{\mu_k},\qquad \operatorname{Var}_m(T_0)=\sum_{k=1}^{m}\frac1{\mu_k^2}. $$ For $m=0$, $T_0=0$; the sum is empty and the product is $1$.

Proof. Starting from $m$, the chain must first wait in state $m$ before moving to $m-1$; that holding time has the exponential distribution with rate $\mu_m$. It then waits in state $m-1$ for an exponential time with rate $\mu_{m-1}$, and continues in this way until its final holding time in state $1$, after which it enters $0$. The holding-time construction gives independence of these successive waiting times, so their sum is $T_0$. For an exponential random variable $E_k$ of rate $\mu_k$, direct integration of its density $\mu_k e^{-\mu_k x}$, $x\ge0$, gives $\mathbb{E}[e^{-sE_k}]=\mu_k/(\mu_k+s)$, $\mathbb{E}[E_k]=1/\mu_k$, and $\operatorname{Var}(E_k)=1/\mu_k^2$. Independence makes the transform of the sum the product of the transforms and makes its mean and variance the sums of the individual means and variances. $\square$

Worked Example: Extinction with Two Different Rates

Start with $X_0=2$, and take $\mu_2=3$ and $\mu_1=1$. In state $2$, the process waits at rate $3$; in state $1$, it waits at rate $1$. In the state order $0,1,2$, the generator is $$ Q=\begin{bmatrix} 0&0&0\\ 1&-1&0\\ 0&3&-3 \end{bmatrix}. $$ The row sums are $0$, $1-1=0$, and $3-3=0$, as required.

The extinction time is the sum of independent exponential waiting times with rates $3$ and $1$. Its mean and variance are $$ \mathbb{E}_2[T_0]=\frac13+1=\frac43,\qquad \operatorname{Var}_2(T_0)=\frac19+1=\frac{10}{9}. $$ To obtain its distribution, the probability of still being in state $2$ at time $t$ is $e^{-3t}$. The probability of being in state $1$ is the probability of a transition from $2$ at time $s$, followed by no transition out of $1$ before $t$, integrated over $0\le s\le t$: $$ \mathbb{P}_2(X_t=1)=\int_0^t 3e^{-3s}e^{-(t-s)}\,ds =\frac32e^{-t}(1-e^{-2t}). $$ Therefore $$ \mathbb{P}_2(X_t=0)=1-\mathbb{P}_2(X_t=1)-\mathbb{P}_2(X_t=2) =1-\frac32e^{-t}+\frac12e^{-3t}. $$ Because state $0$ is absorbing, this is also $\mathbb{P}_2(T_0\le t)$.

Check at $t=\log 2$. Substitution gives $e^{-t}=\tfrac12$ and $e^{-3t}=\tfrac18$. Thus the probabilities of states $2$, $1$, and $0$ are, respectively, $\tfrac18$, $\tfrac32\cdot\tfrac12(1-\tfrac14)=\tfrac9{16}$, and $1-\tfrac9{16}-\tfrac18=\tfrac5{16}$. They sum to $\tfrac2{16}+\tfrac9{16}+\tfrac5{16}=1$, and the extinction CDF formula also gives $1-\tfrac34+\tfrac1{16}=\tfrac5{16}$.

Linear Death Rates Give a Binomial Population

Now suppose the total rate in state $n$ is $\mu_n=n\mu$, where the per-individual death rate $\mu$ is strictly positive. If the initial population is $m$, each of the $m$ individuals can be assigned an independent lifetime with exponential distribution of rate $\mu$. Let $X_t$ count the individuals whose lifetimes exceed $t$. The probability that one individual is still alive at time $t$ is $e^{-\mu t}$.

Theorem: For a pure-death process with $\mu_n=n\mu$, $\mu>0$, and $X_0=m$, $$ \mathbb{P}_m(X_t=k)=\binom{m}{k}(e^{-\mu t})^k(1-e^{-\mu t})^{m-k}, \qquad k=0,1,\ldots,m,\quad t\ge0. $$ Thus $X_t$ has a binomial distribution with $m$ trials and success probability $e^{-\mu t}$.

Proof. Construct $m$ independent lifetimes, each exponential with rate $\mu$, and define $X_t$ to be the number exceeding $t$. Conditional on an individual still being alive, the probability they survive an additional time $h$ is $$ \mathbb{P}(E>t+h\mid E>t)=\frac{e^{-\mu(t+h)}}{e^{-\mu t}}=e^{-\mu h}. $$ So each living individual has an exponential residual lifetime of rate $\mu$, independent of the other living individuals. With $n$ alive, the chance that no one dies in a time interval of length $h$ is $e^{-n\mu h}$. The chance of two or more deaths in that interval is of smaller order than $h$ as $h$ decreases to zero. Hence the count process has total departure rate $n\mu$, with each departure reducing the count by one, as specified. At fixed $t$, each individual survives independently with probability $e^{-\mu t}$. The event $X_t=k$ occurs when exactly $k$ of the $m$ individuals survive; there are $\binom{m}{k}$ choices of those individuals, and each such choice has probability $(e^{-\mu t})^k(1-e^{-\mu t})^{m-k}$. Summing over the choices proves the formula. $\square$

This result includes the endpoints: at $t=0$, $e^{-\mu t}=1$, so $X_0=m$ with probability one; for $t>0$, the displayed formula applies for every $k$ from $0$ through $m$. In particular, the probability of extinction by time $t$ is $$ \mathbb{P}_m(T_0\le t)=\mathbb{P}_m(X_t=0)=(1-e^{-\mu t})^m. $$ Extinction is the time by which all $m$ lifetimes have ended, so $T_0$ is the maximum of the $m$ individual lifetimes, not their sum. Equivalently, the holding-time theorem gives a sum of independent exponentials with rates $m\mu,(m-1)\mu,\ldots,\mu$.

These descriptions also give exact moments. To see them from the holding-time representation, consider the time between successive deaths while $r$ individuals remain. The minimum of $r$ independent exponential lifetimes of rate $\mu$ is exponential with rate $r\mu$: its survival probability for duration $u$ is $(e^{-\mu u})^r=e^{-r\mu u}$. The memoryless property leaves the survivors with fresh independent exponential residual lifetimes. Thus the successive waiting times have rates $m\mu,(m-1)\mu,\ldots,\mu$, and the variance and mean of their sum are $$ \mathbb{E}_m[T_0]=\frac1\mu\sum_{r=1}^m\frac1r,\qquad \operatorname{Var}_m(T_0)=\frac1{\mu^2}\sum_{r=1}^m\frac1{r^2}. $$ For $m=0$, extinction has already occurred and both quantities are zero.

Worked Example: Three Individuals with Rate One

Let $X_0=3$ and $\mu=1$. At time $t=\log 2$, an individual survives with probability $e^{-t}=\tfrac12$, and dies by that time with probability $1-\tfrac12=\tfrac12$. The binomial formula gives $$ \mathbb{P}_3(X_t=3)=\frac18,\quad \mathbb{P}_3(X_t=2)=3\cdot\frac14\cdot\frac12=\frac38,\quad \mathbb{P}_3(X_t=1)=3\cdot\frac12\cdot\frac14=\frac38,\quad \mathbb{P}_3(X_t=0)=\frac18. $$ These probabilities sum to $\tfrac18+\tfrac38+\tfrac38+\tfrac18=1$. In particular, the extinction probability by $t=\log 2$ is $\tfrac18$, in agreement with $(1-e^{-t})^3=(1-\tfrac12)^3=\tfrac18$.

The mean extinction time is $$ \mathbb{E}_3[T_0]=1+\frac12+\frac13=\frac{11}{6}, $$ and its variance is $$ \operatorname{Var}_3(T_0)=1+\frac14+\frac19=\frac{49}{36}. $$ The sum in the variance has denominator $36$, with numerator $36+9+4=49$, verifying the stated value.

Worked Example: Two Individuals at a Faster Death Rate

Take $X_0=2$ and per-individual rate $\mu=2$. At $t=\frac{\log 3}{2}$, $e^{-\mu t}=e^{-\log 3}=\tfrac13$, so each individual is alive with probability $\tfrac13$ and dead with probability $\tfrac23$. Therefore $$ \mathbb{P}_2(X_t=2)=\left(\frac13\right)^2=\frac19,\quad \mathbb{P}_2(X_t=1)=2\cdot\frac13\cdot\frac23=\frac49,\quad \mathbb{P}_2(X_t=0)=\left(\frac23\right)^2=\frac49. $$ The probabilities sum to $\tfrac19+\tfrac49+\tfrac49=1$, and the probability of extinction by this time is $\tfrac49=(1-e^{-\mu t})^2$.

The holding-time representation gives $$ \mathbb{E}_2[T_0]=\frac1{4}+\frac1{2}=\frac34,\qquad \operatorname{Var}_2(T_0)=\frac1{16}+\frac1{4}=\frac5{16}. $$ The two waiting rates are $2\mu=4$ and $\mu=2$; their reciprocal rates and squared reciprocal rates give the displayed mean and variance.

Checks and a Common Rate Error

The population law answers a fixed-time question, while the holding-time sum answers a time-to-extinction question. These are related but not interchangeable: $X_t=0$ is the event $T_0\le t$, whereas the extinction time itself is a random duration. The formulas also rely on strictly positive death rates in every positive state. If some $\mu_k=0$, the process can remain forever in state $k$, and extinction from an initial state above $k$ is not guaranteed. For the linear-rate formula, $\mu>0$ is essential; if $\mu=0$, the population never decreases.

Common mistake: using the per-individual rate as the total rate. With $m=3$ and per-individual rate $\mu=1$, the probability that nobody has died by time $t$ is not $e^{-t}$. Each of three independent individuals survives with probability $e^{-t}$, so all three survive with probability $$ \mathbb{P}_3(X_t=3)=(e^{-t})^3=e^{-3t}. $$ At $t=\log 2$, this is $(\tfrac12)^3=\tfrac18$, not $e^{-t}=\tfrac12$. The total rate in state $3$ is $3\mu$, because any of the three individuals may die first.

For a general pure-death process, check the rate diagram or generator before using a linear-death formula: the binomial law applies specifically when $\mu_n=n\mu$ with the same $\mu$ for all positive $n$. In other cases the general sum-of-independent-exponentials representation still applies from a fixed finite initial population, but its distribution need not be binomial.

Check Your Understanding

Use the assumptions in each question explicitly, and check probabilities or moments by substitution.

A pure-death process starts from $3$, with rates $\mu_3=2$, $\mu_2=4$, and $\mu_1=1$. Write the extinction time as a sum of independent exponential variables and calculate its mean and variance.
For linear death rates $\mu_n=n\mu$, what is $\mathbb{P}_m(X_t=0)$? Explain why it equals $\mathbb{P}_m(T_0\le t)$.
Start with two individuals, each dying at rate $1$. Find the probabilities of $X_{\log 2}=0,1,2$, and verify that they sum to one.
For $X_0=4$ and per-individual death rate $\mu$, write the four rates for the successive holding times before extinction.
Explain what changes if $\mu_k=0$ for some positive state $k$. Why can the strictly positive-rate extinction-time formula not be applied unchanged?

Pure-Death Processes And Extinction Times

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