5.3 The Simple Birth and Death Process
Now each individual independently gives birth with probability \(\lambda h + o(h)\) and dies with probability \(\mu h + o(h)\) in \((t, t + h]\). With \(n\) individuals the population rises by one with probability \(n\lambda h + o(h)\) and falls by one with probability \(n\mu h + o(h)\), so \[P_n'(t) = -n(\lambda + \mu )P_n(t) + (n - 1)\lambda P_{n-1}(t) + (n + 1)\mu P_{n+1}(t),\qquad n\geq 0,\] with \(P_{-1}\equiv 0\), and the m.g.f. satisfies \[\frac {\partial M}{\partial t} = \left [\lambda (e^\theta - 1) + \mu (e^{-\theta } - 1)\right ]\frac {\partial M}{\partial \theta } .\]
Theorem 5.3.1. For the simple birth and death process with \(X(0) = a\), \(E\left [X(t)\right ] = ae^{(\lambda - \mu )t}\). The probability that the population eventually dies out is \(1\) if \(\lambda \leq \mu \) and \(\left (\frac \mu \lambda \right )^a\) if \(\lambda >\mu \).
Proof. Differentiating the equation for \(M\) with respect to \(\theta \) and putting \(\theta = 0\), where \(M = 1\) and \(\frac {\partial M}{\partial \theta } = m(t) = E[X(t)]\), gives \(m'(t) = (\lambda - \mu )m(t)\), so \(m(t) = ae^{(\lambda - \mu )t}\). For extinction, look only at the sequence of births and deaths: from a population of \(n\), the next event is a birth with probability \(\frac {n\lambda }{n\lambda + n\mu } = \frac {\lambda }{\lambda + \mu }\) and a death otherwise, so the sizes after successive events form a random walk with these step probabilities. By the gambler’s ruin calculation, this walk started at \(a\) ever reaches \(0\) with probability \(1\) if \(\lambda \leq \mu \) and \(\left (\frac {\mu }{\lambda }\right )^a\) if \(\lambda >\mu \). □
In the critical case \(\lambda = \mu \) with \(a = 1\) the distribution can be written down: \[P(X(t) = 0) = \frac {\lambda t}{1 + \lambda t},\qquad P(X(t) = n) = \frac {(\lambda t)^{n-1}}{(1 + \lambda t)^{n+1}},\quad n\geq 1 .\] The mean stays at \(1\) for all \(t\), yet the probability of extinction by time \(t\) tends to \(1\): a population that survives does so by becoming large.
Example 5.3.2. A colony starts with \(2\) cells. Each cell divides at rate \(\lambda = 0.5\) per hour and dies at rate \(\mu = 0.3\) per hour. Find the expected size after \(10\) hours and the probability that the colony eventually dies out.
Solution. \(E[X(10)] = 2e^{(0.5 - 0.3)10} = 2e^2 = 14.8\) cells. Since \(\lambda >\mu \), the extinction probability is \(\left (\frac {0.3}{0.5}\right )^2 = 0.36\). More than a third of such colonies vanish, even though the expected size grows without limit.
Questions on this section
Stuck on something here? Ask below and it stays attached to this topic.