# Generalised (nonhomogeneous) Poisson process

Define a generalised Poisson process as an arrival process that begins at time 0 and that satisfies:

• The independence property: the number of arrivals during two non-overlapping intervals are independent.
• Small interval probabilities:

\begin{align*} P(X_{t+\varepsilon} - X_t \ge 2) &= o(\varepsilon) \\ > P(X_{t+\varepsilon} - X_t = 1) &= \lambda(t)\varepsilon + > o(\varepsilon) \\ P(X_{t+\varepsilon} - X_t = 0) &= > 1-\lambda(t)\varepsilon + o(\varepsilon) \end{align*}

The function $\lambda(t)$ is called the intensity function.

Let ${X_t, t \ge 0}$ be a generalised Poisson process (and suppose $\lambda(t)$ is continuous for simplicity). Let $p_n(t) = P(X_t = n)$ and $m(t) = \int_0^t \lambda(u) du$

Question (a)

Obtain $p_0(t) = P(X_t = 0)$ by showing that $p_0(t)$ solves a first order differential equation.

My Working

Define $p_0(t) = P(X_t = 0)$, then for $h > 0$, \begin{align*} p_0(t+h) & = P(X_{t+h} = 0) \\ & = P(X_t = 0, X_{t+h} - X_t = 0) \\ & = P(X_t=0)P(X_{t+h} - X_t = 0) \ \ \ \ \ \ \text{By the independence property} \\ & = p_0(t)\left[1-\lambda(t)h + o(h) \right] \ \ \ \ \ \ \text{By the small interval probabilities} \\ & = p_0(t) - p_0(t)\lambda(t)h + p_0(t)o(h) \end{align*} By rearranging, \begin{align*} \nonumber \frac{p_0(t+h) - p_0(t)}{h} & = -p_0(t)\lambda(t) + p_0(t)\left(\frac{o(h)}{h} \right) \\ \nonumber \lim_{h \rightarrow 0} \left(\frac{p_0(t+h) - p_0(t)}{h} \right) & = -p_0(t)\lambda(t) + p_0(t)\lim_{h \rightarrow 0}\left(\frac{o(h)}{h} \right) \\ \nonumber p_0'(t) & = -p_0(t)\lambda(t) \\ p_0'(t) + \lambda(t) p_0(t) & = 0 \end{align*} where the above is a separable first order linear differential equation.

Solving this differential equation yields, \begin{align*} \frac{p_0'(t)}{p_0(t)} &= -\lambda(t) \\ \int \frac{p_0'(t)}{p_0(t)} dt &= -\int \lambda(t) dt \\ \ln\left(|p_0(t)|\right) + C_1 &= -\int_0^t \lambda(u) du + C_2 \end{align*} where $C_1$ and $C_2$ are the constants of integration.

Note that since $p_0(t)$ is a probability, then $p_0(t) > 0$ for all $t \ge 0$. Hence, \begin{align*} \ln\left(p_0(t)\right) = -m(t) + C_3 \end{align*} Since $p_0(0) = 1$ and $m(0) = 0$, then $C_3 = 0$, so the solution is given by, \begin{align*} p_0(t) = \exp\left(-m(t)\right) \ \ \ \ \forall \ \ 0 \le t < +\infty \end{align*}

Query

My main query is when removing the absolute values from $\ln\left(|p_0(t)|\right)$ by saying "...since $p_0(t)$ is a probability, then $p_0(t) > 0$ for all $t \ge 0$." How can I actually prove $p_0(t) > 0$ for all $t \ge 0$ based on the properties of a generalised Poisson process?

Question (b)

Let $T_1$ be the time of the first arrival. Find the density of $T_1$.

My working/Query

Is my working correct here?

From part (a), we know $p_0(t) = \exp\left(-m(t)\right)$. Assume that the process starts from time zero and let $T_1$ denote the time of the first arrival. The CDF of $T_1$ is given by, \begin{align*} F_{T_1}(t) = P(T_1 \le t) = 1- P(T_1 > t) = 1-P(X_t = 0) = 1-\exp(-m(t)) \end{align*} To find the PDF of $T_1$, differentiate the CDF with respect to $t$ yields, \begin{align*} f_{T_1}(t) & = m'(t)\exp(-m(t)) \\ & = \lambda(t)\exp(-m(t)) \ \ \ \ \ \ \text{By the Fundamental Theorem of Calculus} \end{align*} Thus, $f_{T_1}(t) = \lambda(t)\exp(-m(t))$ for all $0 \le t < +\infty$.