4.3 Further Properties of Poisson Process
4.4.1. Consider a Poisson process \(\{N(t), \, t\geq 0\}\) having rate \(\lambda \) and suppose that each time an event occurs, it is
classified as either a type \(I\) or type \(II\) event. Suppose further that each event is classified as
type \(I\) with probability \(p\) and as type \(II\) event with probability \(1 - p\), independently of all other
events.
Let \(N_1(t)\) and \(N_2(t)\) be the number of type \(I\) and type \(II\) events occuring in \([0,t]\).
Proposition. \(\{N_1(t),\, t\geq 0\}\) and \(\{N_2(t), \, t\geq 0\}\) are both Poisson process having rates \(\lambda p\) and \(\lambda (1 - p)\) respectively. Furthermore the two process are independent.
Proof. Clearly \(N(t) = N_1(t) + N_2(t)\). Let \(m\) be a non negative integer \begin {align*} P(N_1(t) = m) & = \sum ^{\infty }_{n =m} P\left (N_1(t) = m\, \big |\, N(t) = n\right )P(N(t) = n)\\ & = \sum ^{\infty }_{n = m}\binom {n}{m}\, p^m\, (1 - p)^{n - m}\, \frac {e^{-\lambda t}\, (\lambda t)^n}{n!}\\ & = p^m\, e^{-\lambda t}\, \sum ^{\infty }_{n = m} \frac {n!}{(n - m)!\, m!}\, (1 - p)^{n - m}\, \frac {(\lambda t)^n}{n!}\\ & = \frac {p^m\, e^{-\lambda t}}{m!}\, \sum ^{\infty }_{n =m} \, \frac {(1 - P)^{n - m}}{(n - m)!}\, (\lambda t)^n\\ & = p^m\, \frac {e^{-\lambda t}}{m!}\, (\lambda t)^m\, \sum ^{\infty }_{n = m}\frac {(1 - P)^{n - m}}{(n - m)!}\, (\lambda t)^{n -m}\\ & = (p\, \lambda t)^m\, \frac {e^{-\lambda t}}{m!}\, \left (1 + \frac {(1 - p)}{1!}\, \lambda t + \frac {(1 - p)^2\, (\lambda t)^2}{2!} + \cdots \cdots \right )\\ & = (p\, \lambda t)^m\, \frac {e^{-\lambda t}}{m!}\, e^{(1 - p)\lambda t}\\ & = \frac {e^{-p\lambda t}\, (p\lambda t)^m}{m!}. \end {align*}
Hence \(N_1(t)\) has a Poisson process of rate \(p\lambda \).
Similarly \(N_2(t)\) has a Poisson process of rate \((1 - p)\lambda \). \begin {align*} P\left (N_1(t) = n_1\, , \, N_2(t) = n_2\right ) & = P(N_1(t) = n_1\, ,\, N(t) = n_1 + n_2)\\ & = P(N_1(t) = n_1\, |\, N(t) = n_1 + n_2)\, P(N(t) = n_1 + n_2)\\ & = \binom {n_1 + n_2}{n_1}\, p^{n_1}\, (1 - p)^{n_2}\, e^{-\lambda t}\, \frac {(\lambda t)^{n_1 + n_2}}{(n_1 + n_2)!}\\ & = \frac {(n_1 + n_2)!}{n_1!\, n_2!}\, p^{n_1}\, (1 - p)^{n_2}\, \frac {e^{-\lambda t}\, (\lambda t)^{n_1 + n_2}}{(n_1 + n_2)!}\\ & = \frac {p^{n_1}\, (1 - p)^{n_2}}{n_1!\, n_2!}\, e^{-\lambda t}\, (\lambda t)^{n_1}\, (\lambda t)^{n_2}\\ & = \frac {e^{-\lambda t + p\lambda t - p\lambda t}}{n_1!\, n_2!}\, p^{n_1}\, (1 - p)^{n_2}\, (\lambda t)^{n_1}\, (\lambda t)^{n_2}\\ & = \frac {e^{-\lambda t + p\lambda t - p\lambda t}}{n_1!\, n_2!}\, (p(\lambda t))^{n_1}\, ((1-p)\lambda t)^{n_2}\\ & = \frac {e^{-p\lambda t} \, (p\lambda t)^{n_1}}{n_1!}\, \cdot \, \frac {e^{-\lambda t(1 - P)}\, ((1 - p)\lambda t)^{n_2}}{n_2!}\\ & = P(N_1(t) = n_1)\cdot P(N_2(t) = n_2). \end {align*} □
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