Birth process

 

Defining the birth process 

Consider a colony of bacteria that never dies. We study the following process known as the birth process, also known as the Yule process.

1. The colony starts with $n_0$ cells at time $t=0$.
2. Assume that the probability that any individual cell divides in the time interval $(t,t+\delta t)$ is proportional to $\delta t$ for small $\delta t$.
3. Further assume that each cell division is independent of others.
4. Let $\lambda$ be the birth rate. The probability of a cell division for a population of $n$ cells during $\delta t$ is $\lambda n\delta t$.
5. We assume that the probability that two or more births take place in the time interval $\delta t$ is $o(\delta t)$. That is, it can be ignored.
6. Consequently, the probability that no cell divides during $\delta t$ is $1-\lambda n\delta t-o(\delta t)$.

Note that this process is an example of the Markov chain with states $n_0,n_0+1,n_0+2,\cdots$ where each integer $n$ represents the state with $n$ individuals. 

The population size (i.e., the number of individuals) at time $t$ is a random variable, $N(t)$. We denote by $p_n(t)$ the probability that $N(t)=n$. That is, $$p_n(t)=Pr(N(t)=n).$$ Our goal here is to find this probability distribution $p_n(t)$.

A sample path of a birth process is shown below:

Differential-difference equations

Based on the above assumptions, let us derive a set of differential-difference equations for $p_n(t)$. 

First of all, the initial condition is given by $$\begin{array}{}\text{(Eq:init)}& p_n(0)={\delta }_{n,n_0}\end{array}$$ where ${\delta }_{n,n_0}$ is Kronecker's delta.

Next, let $\delta t>0$ be a sufficiently small time interval. For $N(t+\delta t)=n$ to hold, we should have one of the following possibilities:

• $N(t)=n-1$ and a cell division has taken place during $\delta t$, or 
• $N(t)=n$ and no cell divisions have occurred during $\delta t$.

Thus, $$p_n(t+\delta t)=p_{n-1}(t)[\lambda (n-1)\delta t+o(\delta t)]+p_n(t)[1-\lambda n\delta t+o(\delta t)]$$ for $n\ge n_0+1$, and

$$p_{n_0}(t+\delta t)=p_{n_0}(t)[1-\lambda n_0\delta t+o(\delta t)]$$ for $n=n_0$. By rearranging these equations and taking the limit $\delta t\to 0$, we have the following differential-difference equations:

$$\begin{array}{}\text{(Eq:Birth0)}& \frac{d{p}_{n_0}(t)}{dt}& =& -\lambda n_0{p}_{n_0}(t),\text{(Eq:Birthn)}& \frac{d{p}_n(t)}{dt}& =& \lambda (n-1)p_{n-1}(t)-\lambda n{p}_n(t),(n\ge n_0+1)\end{array}$$

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

$$p_n(t)=\{\begin{array}{ll}0,& (n<n_0),\\ (\genfrac{}{}{0ex}{}{n-1}{n_0-1})e^{-\lambda n_{0}t}(1-e^{-\lambda t}{)}^{n-n_0},& (n\ge n_0).\end{array}$$

That is, $N(t)$ follows the Pascal or negative binomial distribution, $\mathrm{NB}(n_0,e^{-\lambda t})$.

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Generating function equation

Let's solve these equations. We use the method of the probability generating function (PGF). In the present case, the PGF with the dummy variable $s$ is defined as

$$\begin{array}{}\text{(Eq:PGF)}& G(s,t)=\sum _{n=n_0}^{\mathrm{\infty }}{p}_n(t)s^n.\end{array}$$

The initial condition (Eq:init) reads as $$\begin{array}{}\text{(Eq:InitG)}& G(s,0)=s^{n_0}.\end{array}$$

Now, multiply $s^n$ to the both sides of (Eq:Birthn), and sum over $n=n_0,n_0+1,\cdots$. We have

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

The left-hand side: $$\sum _{n=n_0}^{\mathrm{\infty }}\frac{d{p}_n(t)}{dt}{s}^n=\frac{\partial G(s,t)}{\partial t}.$$

The first term on the right-hand side:$$\lambda \sum _{n=n_0+1}^{\mathrm{\infty }}(n-1)p_{n-1}(t)s^n=\lambda s^2\frac{\partial G(s,t)}{\partial s}.$$

The second term on the right-hand side:$$\lambda \sum _{n=n_0}^{\mathrm{\infty }}n{p}_n(t)s^n=\lambda s\frac{\partial G(s,t)}{\partial s}.$$

Combining these together, we obtain the following PDE: $$\begin{array}{}\text{(Eq:PDE)}& \frac{\partial G(s,t)}{\partial t}=\lambda s(s-1)\frac{\partial G(s,t)}{\partial s}.\end{array}$$

Solving the PDE for $G(s,t)$

The PDE (Eq:PDE) can be greatly simplified if we can get rid of the factor $s(s-1)$ on the right-hand side. This can be achieved by the change of variables $s\to z$ where the new variable $z$ is defined by the ODE $$\frac{ds}{dz}=\lambda s(s-1).$$

Here, we are implicitly assuming that $0<s<1$. Solving this ODE (exercise!), we have

$$s=\frac{1}{1+e^{\lambda z}}.$$

Now, change the variable $s$ in $G(s,t)$ to $z$ and define

$$Q(z,t)=G(1/(1+e^{\lambda z}),t).$$

In terms of $Q(z,t)$, (Eq:PDE) becomes $$\begin{array}{}\text{(Eq:PDEQ)}& \frac{\partial Q(z,t)}{\partial t}=\frac{\partial Q(z,t)}{\partial z}.\end{array}$$

The general solution of (Eq:PDEQ) is known to be any differentiable function of the form $w(z+t)$. In fact, we have 

$$\frac{\partial w(z+t)}{\partial z}=\frac{\partial w(z+t)}{\partial t}$$

for any differentiable function $w(z+t)$.

Let $Q(z,t)=w(z+t)$. The functional form of $w$ is determined by the initial condition (Eq:InitG):

$$G(s,0)=s^{n_0}=\frac{1}{(1+e^{\lambda z}{)}^{n_0}}=Q(z,0)=w(z).$$

Thus,

$$\begin{array}{rcl}G(s,t)& =& Q(z,t)=w(z+t)=\frac{1}{(1+e^{\lambda (z+t)}{)}^{n_0}}\\ & =& \frac{1}{(1+\frac{1-s}{s}{e}^{\lambda t}{)}^{n_0}}=\frac{s^{n_0}{e}^{-\lambda n_{0}t}}{[1-(1-e^{-\lambda t})s{]}^{n_0}}.\end{array}$$

By applying the negative binomial theorem to the denominator, we have

$$\begin{array}{rcl}G(s,t)& =& \frac{s^{n_0}{e}^{-\lambda n_{0}t}}{[1-(1-e^{-\lambda t})s{]}^{n_0}}\\ & =& s^{n_0}{e}^{-\lambda n_{0}t}\sum _{m=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{m+n_0-1}{n_0-1})(1-e^{-\lambda t}{)}^m{s}^m\\ & =& e^{-\lambda n_{0}t}\sum _{n=n_0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{n-1}{n_0-1})(1-e^{-\lambda t}{)}^{n-n_0}{s}^n.\end{array}$$

Comparing the coefficients of $s^n$ with (Eq:PGF), we have $$p_n(t)=\{\begin{array}{ll}0,& (n<n_0),\\ (\genfrac{}{}{0ex}{}{n-1}{n_0-1})e^{-\lambda n_{0}t}(1-e^{-\lambda t}{)}^{n-n_0},& (n\ge n_0).\end{array}$$

Thus, $N(t)$ follows the Pascal or negative binomial distribution, $\mathrm{NB}(n_0,e^{-\lambda t})$. In particular, the mean population size is given by 

$$\mathbb{E}[N(t)]=\frac{\partial G}{\partial s}(1,t)=n_0{e}^{\lambda t}.$$

We can see that the population size increases exponentially with time.

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