Death process

 

Defining the death process

Consider a population where individuals only die, and nobody is born. We now study the death process under the following assumptions.

1. The probability that each individual dies in a short period of time $\delta t$ is $\mu \delta t+o(\delta t)$ where $\mu >0$ is the death rate.
2. For a population of $n$ individuals, the probability of each death is $n\mu \delta t+o(\delta t)$. We assume that the probability of multiple deaths in $\delta t$ is negligible. 

Thus, the population size $N(t)$ is a random variable that only decreases. Let $$p_n(t)=Pr(N(t)=n),$$ the probability that the population size is $n$ at time $t$. Suppose the initial population size is $N(0)=n_0$, and hence the initial condition is given as $$\begin{array}{}\text{(Eq:Init)}& p_n(t)={\delta }_{n,n_0}.\end{array}$$

A sample path of a death process is shown below:

Differential-difference equations

Now, let's find the differential-difference equations for the death process.

First, suppose that $N(t)=n_0$, the initial population size. The population can retain this size at time $t+\delta t$ only if there are no deaths during $\delta t$. Thus,

$$p_{n_0}(t+\delta t)=p_{n_0}(t)(1-\mu n_0\delta t+o(\delta t)).$$

Next, for $N(t+\delta t)=n$ to hold, there are two possibilities:

• $N(t)=n+1$ and one individual dies during $\delta t$, or
• $N(t)=n$, and no individuals die during $\delta t$.

Hence, we have the following equation:

$$p_n(t+\delta t)=[\mu (n+1)\delta t+o(\delta t)]p_{n+1}(t)+[1-\mu n\delta t+o(\delta t)]p_n(t),(n\le n_0-1).$$

In particular, we have, for $n=0$,

$$p_0(t+\delta t)=[\mu \delta t+o(\delta t)]p_1(t)+[1+o(\delta t)]p_0(t).$$

Rearranging these, and taking the limit $\delta t\to 0$, we have the following differential-difference equations:

$$\begin{array}{}\text{(Eq:DeathN)}& \frac{d{p}_n(t)}{dt}& =& \mu (n+1)p_{n+1}(t)-\mu n{p}_n(t),(0\le n\le n_0-1)\text{(Eq:DeathN0)}& \frac{d{p}_{n_0}(t)}{dt}& =& -\mu n_0{p}_{n_0}(t).\end{array}$$

In the next post, we will solve these equations and show that $\{p_n(t)\}$ is given by

$$p_n(t)=(\genfrac{}{}{0ex}{}{n_0}{n})e^{-n\mu t}(1-e^{-\mu t}{)}^{n_0-n},(0\le n\le n_0).$$

That is, $N(t)$ follows the binomial distribution $\mathrm{B}(n_0,e^{-\mu t})$.

Generating function equation

We can similarly solve the above equations as the birth process. The probability generating function for this death process is defined as $$\begin{array}{}\text{(Eq:PGF)}& G(s,t)=\sum _{n=0}^{n_0}{p}_n(t)s^n.\end{array}$$ Note the sum is now from $n=0$ to $n=n_0$.

As usual (by now), multiply the both sides of (Eq:DeathN) by $s^n$ and sum over $n$ to have

$$\sum _{n=0}^{n_0}\frac{d{p}_n(t)}{dt}{s}^n=\mu \sum _{n=0}^{n_0-1}(n+1)p_{n+1}(t)s^n-\mu \sum _{n=1}^{n_0}n{p}_n(t)s^n.$$

By now, you should be able to see that this equation implies the following:

$$\frac{\partial G(s,t)}{\partial t}=\mu (1-s)\frac{\partial G(s,t)}{\partial s},$$

and the initial condition (Eq:Init) reads

$$\begin{array}{}\text{(Eq:InitG)}& G(s,0)=s^{n_0}.\end{array}$$

By employing the same technique as in the birth process, we can show (exercise!) that

$$\begin{array}{rcl}G(s,t)& =& (1-e^{-\mu t}-s{e}^{-\mu t}{)}^{n_0}\\ & =& \sum _{n=0}^{n_0}(\genfrac{}{}{0ex}{}{n_0}{n})e^{-n\mu t}(1-e^{-\mu t}{)}^{n_0-n}{s}^n.\end{array}$$

Hence, we have

$$p_n(t)=(\genfrac{}{}{0ex}{}{n_0}{n})e^{-n\mu t}(1-e^{-\mu t}{)}^{n_0-n},(0\le n\le n_0).$$

This is the binomial distribution $\mathrm{B}(n_0,e^{-\mu t})$.

Probability of extinction

The probability of extinction at time $t$ is 

$$p_0(t)=(1-e^{-\mu t}{)}^{n_0}.$$

The mean population size decreases exponentially:

$$\mathbb{E}(N(t))=n_0{e}^{-\mu t}.$$

In fact, the probability of ultimate extinction is

$$\underset{t\to \mathrm{\infty }}{lim}{p}_0(t)=\underset{t\to \mathrm{\infty }}{lim}(1-e^{-\mu t}{)}^{n_0}=1$$

so that the population will become extinct almost surely.

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