Death process

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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 \[p_{n}(t) = \delta_{n,n_0}. \tag{Eq:Init}\]

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:
\[p_n(t+\delta t) = [\mu(n+1)\delta t + o(\delta t)]p_{n+1}(t) + [1 - \mu n\delta t + o(\delta t)]p_n(t), ~~ (n \leq n_0 - 1).\]
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:
\[\begin{eqnarray} \frac{{d}p_n(t)}{{d}t} &=& \mu(n+1) p_{n+1}(t) - \mu np_n(t),~~(0 \leq n \leq n_0 - 1)\tag{Eq:DeathN}\\ \frac{{d}p_{n_0}(t)}{{d}t} &=& - \mu n_0p_{n_0}(t)\tag{Eq:DeathN0}. \end{eqnarray}\]

In the next post, we will solve these equations and show that \(\{p_n(t)\}\) is given by
\[p_n(t) = \binom{n_0}{n}e^{-n\mu t}(1 - e^{-\mu t})^{n_0 - n}, ~~ (0 \leq n \leq n_0).\]
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 \[G(s,t) = \sum_{n=0}^{n_0}p_n(t)s^n.\tag{Eq:PGF}\] 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
\[\sum_{n=0}^{n_0}\frac{{d}p_n(t)}{{d}t}s^n =\mu\sum_{n=0}^{n_0-1}(n+1) p_{n+1}(t)s^{n} - \mu\sum_{n=1}^{n_0}np_n(t)s^{n}.\]
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
\[G(s, 0) = s^{n_0}.\tag{Eq:InitG}\]
By employing the same technique as in the birth process, we can show (exercise!) that
\[\begin{eqnarray} G(s,t) &=& (1 - e^{-\mu t} - se^{-\mu t})^{n_0}\\ &=& \sum_{n=0}^{n_0}\binom{n_0}{n}e^{-n\mu t}(1 - e^{-\mu t})^{n_0 - n}s^{n}. \end{eqnarray}\]
Hence, we have
\[p_n(t) = \binom{n_0}{n}e^{-n\mu t}(1 - e^{-\mu t})^{n_0 - n}, ~~ (0 \leq n \leq n_0).\]
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:
\[\mathbb{E}(N(t)) = n_0e^{-\mu t}.\]
In fact, the probability of ultimate extinction is
\[\lim_{t\to\infty}p_0(t) = \lim_{t\to\infty}(1 - e^{-\mu t})^{n_0} = 1\]
so that the population will become extinct almost surely.

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