5.1 The Simple Birth Process

Let \(X(t)\) denote the size of population at time \(t\). Let \(P_n(t) = P\left (X(t) = n\right ).\)

Assume all individuals in the population are capable of giving birth to new individuals. Let the probability of a given individual producing a new one in time \(\Delta t\) be \(\lambda \Delta t\) and not producing in time \(\Delta t\) be \(1 - \lambda \Delta t\). Assume further that individuals produce independently.

Let \(P_n(t) = P\left (X(t) = n\right )\)

\(*\): \(\lambda \Delta t -\) individual produces
\(*\): \(1 - \lambda \Delta t-\) individual doesn’t produce.

\begin {align*} P_n\left (t + \Delta t\right ) = & P\left (X(t) = n\, \, \text {and no birth in time}\,\, \Delta t\right ) + P\left (X(t) = n - 1\,\,\text {and one birth in time}\, \, \Delta t\right )\\ = & P_n(t)\left [\lambda \Delta t (1 - \lambda t)(1 - \lambda \Delta t) \, \cdots \, (1 - \lambda \Delta t)\right ] + P_{n - 1}(t)[\lambda \Delta t (1 - \lambda \Delta t)\, \cdots \, (1 - \lambda \Delta t)\\ & + (1-\lambda \Delta t)\lambda \Delta t(1 - \lambda \Delta t)\, \cdots \, (1 - \lambda \Delta t) + \cdots + (1 - \lambda \Delta t) \cdots (1 - \lambda \Delta t)(\lambda \Delta t)]\\ & = P_n(t)\left (1 - n\lambda \Delta t\right ) + P_{n - 1}(t)(n - 1)\lambda \Delta t. \end {align*}

\[\frac {P_n(t + \Delta t) - P_n(t)}{\Delta t} = -n\, \lambda \, P_n(t) + (n - 1)\, \lambda \, P_{n - 1}(t)\] as \(\Delta t \longrightarrow 0\), \[P'_n(t) = -n\lambda P_n(t) + (n - 1)\, \lambda \, P_{n - 1}(t).\]

Assume \(X(0) = a.\) i.e \(P_a(0) = 1\).

Let \(M(\theta ,t)\) denote the mgf of \(X(t)\). Then \begin {align*} M(\theta ,t) & = E\left (e^{\theta \, X(t)}\right )\\ &= \sum ^{\infty }_{n = 0} e^{\theta n}\, P_n(t)\\ & = P_0(t) + P_1(t)\, e^{\theta } + P_2(t)\, e^{2\theta } + \cdots \end {align*}

Hence \(M(\theta ,0) = e^{\theta \, a}\). \[M(\theta ,t) = \sum ^{\infty }_{n = 0} e^{n\theta }\, P_n(t)\] \[\frac {\partial M}{\partial \theta } = \sum ^{\infty }_{n = 0} n\, e^{n\theta }\, P_n(t)\] \[\frac {\partial M}{\partial t} = \sum ^{\infty }_{n = 0} e^{n\theta }\, P_n'(t).\] Now coming back to the equation \[P'_n(t) = -n\, \lambda \, P_n(t) + (n- 1)\, \lambda \, P_{n - 1}(t)\] multiply across by \(e^{n\theta }\) and sum for \(n\) from 0 to \(\infty \). \[\sum ^{\infty }_{n = 0} P'_n(t)\, e^{n\theta } \, = \, \sum ^{\infty }_{n = 0}-n\, \lambda \, P_n(t)\, e^{n\theta } \, + \, \sum ^{\infty }_{n = 0}(n - 1)\, \lambda \, P_{n - 1}(t)\, e^{n\theta }.\]

\[\frac {\partial M}{\partial t} = -\lambda \, \frac {\partial M}{\partial \theta } + \lambda \, \sum ^{\infty }_{n = 0}(n -1)\, e^{\theta (n - 1)}\, P_{n -1}(t)\, e^{\theta }.\]

\[\frac {\partial M}{\partial t} = -\lambda \, \frac {\partial M}{\partial \theta } \, + \, \lambda \, e^{\theta }\, \frac {\partial M}{\partial \theta }.\] We’ll solve this PDF with the initial condition \(M(\theta ,0) = e^{a\theta }\). \[\frac {\partial M}{\partial t} \, +\, \lambda \left (1 - e^{\theta }\right )\, \frac {\partial M}{\partial \theta } = 0.\]

For solving the PDF, we find two independent general solution for the equations \[\frac {dt}{1} = \frac {d\theta }{\lambda \left (1 - e^{\theta }\right )} = \frac {dM}{0}\] \[\frac {dt}{1} = \frac {dM}{0}\, \implies \, dM = 0\hspace {0.5cm}\text {i.e}\hspace {0.2cm} M = \text {constant}\] Find a general solution of \[\frac {dt}{1} = \frac {d\theta }{\lambda \left (1 - e^{\theta }\right )}\]

\[\lambda \, dt = \frac {d\theta }{e^{\theta }\, \left (e^{\theta } - 1\right )}\]

\[-\lambda \, dt = \frac {- e^{-\theta }}{e^{-\theta } - 1}\, d\theta \] \[C - \lambda \, t = \ln \left (e^{-\theta } - 1\right )\] \[e^C\, e^{-\lambda \,t} = e^{-\theta } - 1\] \[e^{\lambda \,t}\, \left (e^{-\theta } - 1\right ) = \text {constant}.\] Hence the general solution of the PDF is \[M(\theta ,t) = \Phi \left (e^{\lambda t}\, \left (e^{-\theta } - 1 \right )\right )\]

\[M(\theta ,0) = \Phi \left (e^{-\theta } - 1\right )\] but \(M(\theta ,0) = e^{a\theta }\). Hence \[\Phi \left (e^{-\theta } - 1\right ) = e^{a\theta }.\] Let \(y = e^{-\theta } - 1\) \[e^{-\theta } = y + 1\] \[e^{\theta } = \frac {1}{y + 1}\] \[e^{a\theta } = \left (\frac {1}{y + 1}\right )^a.\]

Hence the mapping \(\Phi \) is \[\Phi (y) = \left (\frac {1}{y + 1}\right )^a.\] Hence \begin {align*} M(\theta , t) & = \left (\frac {1}{e^{\lambda t}\left (e^{-\theta } - 1\right ) + 1}\right )^a = \left (e^{\lambda t}\, \left (e^{-\theta } - 1\right ) + 1\right )^{-a} = \left [1 - e^{\lambda t}\, \left (1 - e^{-\theta }\right )\right ]^{-a}. \end {align*}

Now, the Negative Binomial \(-\) Number of failures before the \(r^{\text {th}}\) success. We find the MGF of the Negative Binomial \[P(Y=n) = \binom {n - 1}{r - 1}\, P^{r - 1}\, q^{n - r}\, P\, , \hspace {0.3cm} n\geq r.\]

PGF of \(y = E\left (\theta ^y\right )\)

\begin {align*} E\left (\theta ^y\right ) & = \sum ^{\infty }_{n = r}\theta ^n\, \binom {n - 1}{r - 1}\, p^r\, q^{n - r}\\ & = \left (\frac {p}{q}\right )^r\, \sum ^{\infty }_{n = r} \frac {(n - 1)!}{(n - r)!(r - 1)!}\, \left (\theta q\right )^n\\ & = \left (\frac {p}{q}\right )^r\,\left [\left (\theta q\right )^r + r\left (\theta q\right )^{r + 1} + \frac {r(r + 1)}{2!}\, \left (\theta q\right )^{r + 2} + \cdots \cdots \right ]\\ & = \left (\frac {p}{q}\right )^r\,\left (\theta q\right )^r\left [1 + r \left (\theta q\right ) + \frac {r(r+1)}{2!}\, \left (\theta q\right )^2 + \cdots \cdots \right ]\\ & = \left (p\theta \right )^r\, \left [1 + r\, \left (\theta q\right ) + \frac {r(r + 1)}{2!}\,\left (\theta q\right )^2 + \cdots \cdots \right ]\\ & = \left (p\theta \right )^r\, \left (1 - \theta q\right )^{-r}. \end {align*}

MGF of negative Binomial distribution is derived from the PGF of the distribution by replacing \(t\) with \(e^t\).
Hence MGF of Neg-distribution is \begin {align*} M(\theta ) & = \left (p\,e^{\theta }\right )^r\, \left (1 - e^{\theta }\, q\right )^{-r} = \left (\frac {1 - e^{\theta }\, q}{p\, e^{\theta }}\right )^{-r}\\ & = \left [\frac {1}{p\, e^{\theta }} - \frac {q}{p}\right ]^{-r}\\ & = \left [\frac {1}{p}\, e^{-\theta } - \frac {1 - p}{p}\right ]^{-r}\\ & =\left (p^{-1}\, e^{-\theta } - p^{-1} + 1\right )^{-r}\\ & = \left (1 - p^{-1}\left (1 - e^{-\theta }\right )\right )^{-r}. \end {align*}

Therefore \[M(t,\theta ) = \left (1 - e^{\lambda t}\, \left (1 - e^{-\theta }\right )\right )^{-a}.\] This is the mgf of Negative Binomial distribution with \( p = e^{-\lambda t}\) and \(r = a\).
Hence \(X(t)\) is a negative Binomial random variable with \[Prob\left (X(t) = n\right ) = \binom {n - 1}{a - 1}\, e^{-a\lambda t}\, \left (1 - e^{-\lambda t}\right )^{n - a}\, ,\hspace {0.3cm} n\geq a.\] Stochastic mean \[E(X(t)) = a\left (e^{\lambda t} - 1\right )\] \[\sigma ^2(t) = ae^{\lambda t}\, \left (e^{\lambda t} - 1\right )\]

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