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I am learning about Birth and Death (Stochastic) Processes (https://en.wikipedia.org/wiki/Birth%E2%80%93death_process):

When a birth occurs, the process goes from state $n$ to $n + 1$. When a death occurs, the process goes from state $n$ to state $n - 1$. The process is specified by positive birth rates$\{\lambda_i\}_{i=0\dots\infty}$ and > positive death rates $\{\mu_i\}_{i=1\dots\infty}$. The number of individuals in the process at time $t$ is denoted by $X(t)$. The process has the Markov property and $P_{i,j}(t)=\mathsf{P}\{X(t+s)=j|X(s)=i\}$ describes how $X(t)$ changes through time. For small $\triangle t>0$, the function, $P_{i,j}(\triangle t)$ is assumed to satisfy the following properties:

$$P_{i,i+1}(\triangle t) = \lambda_i \triangle t + o(\triangle t), \quad i \geq 0$$ $$P_{i,i-1}(\triangle t) = \mu_i \triangle t + o(\triangle t), \quad i \geq 1$$
$$P_{i,i}(\triangle t) = 1 - (\lambda_i + \mu_i) \triangle t + o(\triangle t), \quad i \geq 1$$

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The way I see it, in each infinitesimal time unit, the Birth and Death process describes the probability of net population change. Assuming a fixed birth rate and death rate, the dynamics of the process can be described as follows. The probabilities $p_n(t)$ of being in state $n$ at time $t$ satisfy the following differential equations (Kolmogorov):

$$ \begin{align*} \frac{dp_0(t)}{dt} &= -\lambda_0 p_0(t) + \mu_1 p_1(t), \\ \frac{dp_n(t)}{dt} &= \lambda_{n-1} p_{n-1}(t) - (\lambda_n + \mu_n) p_n(t) + \mu_{n+1} p_{n+1}(t) \quad \text{for } n \geq 1. \end{align*} $$

Here:

  • $\lambda_n$ for $n \geq 0$ and death rates $\mu_n$ for $n \geq 1$,
  • $\lambda_n$ is the rate of "births" or transitions from state $n$ to state $n+1$,
  • $\mu_n$ is the rate of "deaths" or transitions from state $n$ to state $n-1$,
  • $p_n(t)$ is the probability of being in state $n$ at time $t$.

Based on this, I am trying to understand the following questions:

  • How is the Birth and Death Process related to the Poisson Process?
  • When will the next Birth and Death happen? Is this somehow related to the Exponentially Distributed Holding/Sojourn time of a Continuous Markov Chain, or the Exponentially Distributed Interarrival Times of a Poisson Process?

I know that a Poisson Process can be defined as:

$$\mathsf{P}(X(t) = k) = \frac{(\lambda t)^k e^{-\lambda t}}{k!}, \quad k = 0, 1, 2, \ldots$$ $$ \mathsf{P}[N(t + \Delta t) - N(t) = 1] = \lambda \Delta t + o(\Delta t)$$

I am guessing that in a Birth Death process, the number of births and deaths each follow a Poisson Process? Using the standard properties of a Continuous Time Markov Chain, I am guessing that in a Birth Death process:

  • The time until the next event (either a birth or a death) when in state $n$ is exponentially distributed with parameter $\lambda_n + \mu_n$ (this is like a holding/sojourn time)
  • The probability that the next event is a birth (transition to state $n+1$) is given by $\frac{\lambda_n}{\lambda_n + \mu_n}$.
  • The probability that it is a death (transition to state $n-1$) is given by $\frac{\mu_n}{\lambda_n + \mu_n}$

Is my understanding correct? Is it possible to derive these formulas?

RobPratt
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konofoso
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1 Answers1

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Poisson process is a birth and death process with constant positive birth rate and zero death rates.

To work out what happens first, birth or death, look at the minimum of two independent exponential variables with rates $\lambda_i$ and $\mu_i$ respectively.

van der Wolf
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