4.1 Poisson Process

The counting process \(\{N(t),\, t\geq 0\}\) is said to be a Poisson process having rate \(\lambda , \hspace {0.2cm} \lambda > 0\) if

(i): \(N(0) = 0\)
(ii): \(\{N(t)\}\) has independent and stationary increments.
(iii): \(P\left (N(h) = 1\right ) = \lambda \, h + O(h)\)
(iv): \(P(N(h) > 1) = O(h)\)

\[P(N(h) = 0) = 1 - P(N(h) = 1) + P(N(h) \geq 1) = 1 - \lambda \, h.\]

Let \(P_n(t) = P(N(t) = n)\) \begin {align*} P_0(t + \Delta t) & = P(\text {no event in}\, (0,t)\, \text {and}\, \text {no event in}\, (t, t + \Delta t))\\ & = P(N(t + \Delta t) = 0)\\ & = P_0(t) \,(1 - \lambda \,\Delta t). \end {align*}

As \(\Delta t \longrightarrow 0\), we get \[\frac {P_0(t + \Delta t) - P_0(t)}{\Delta t} = -\lambda \]

\[P'_0(t) = -\lambda \, P_0(t)\] We’ll solve this differential equation using the initial condition \(P_0(0) = 1.\) \[\frac {P'_0(t)}{P_0(t)} =- \lambda \] \[\int \frac {P'_0(t)}{P_0(t)}\, dt =\int (-\lambda )\, dt\] \[\ln P_0(t) = -\lambda \, t + C\] \[P_0(t) = K\, e^{-\lambda t}\] where \(K = e^C\). \[P_0(0)= K\, e^0 = 1\, \implies \, K = 1\]

\[\therefore \,\hspace {0.3cm} P_0(t) = e^{-\lambda t}.\]

\[P_1(t + \Delta t) = P_0(t) (\lambda \, \Delta t) + P_1(t)(1 - \lambda \, \Delta t).\] As \(\, \Delta t \longrightarrow \, 0\), \[P'_1(t) = \lambda \, P_0(t) - \lambda \, P_1(t)\] \[P'_1(t) = \lambda \, e^{-\lambda t} - \lambda \, P_1(t)\] \[P_1'(t) + \lambda \, P_1(t) = \lambda \, e^{-\lambda t}\]

\[I.F = e^{\int \lambda \, dt} = e^{\lambda t}\]

\[e^{\lambda t}\, P_1(t) = \lambda \int e^{\lambda t}\, e^{-\lambda \, t}\, dt + K\] \[e^{\lambda \, t}\, P_1(t) = \lambda \, t + K\] Since \(N(0) = 0, \hspace {0.3cm} P_1(0) = 0\), substituting \(t = 0\), in the above equation. We get \( K = 0\). Hence \[P_1(t) = e^{-\lambda t}\, \lambda \, t.\]

Generalizing the above two results \[P_n(t) = \frac {e^{-\lambda t}\, (\lambda t)^n}{n!}\hspace {0.2cm} , \hspace {0.5cm} n = 0, \, 1, \, 2, \, \cdots \cdots \]

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