Characterisation of a Poisson process

Question

Consider a non homogeneous Poisson process, $N(t)$ with rate $r(t)$. The probability distribution of the first event $T_1$, conditioned on the fact that this happens in a finite time (not assuming necessary that $r(t)$ sums up to infinity) is given by \begin{equation} f_{T_1}(t)=\frac{r(t)e^{-\int_0^tr(u)du}}{\mathbb{P}(T_1<\infty)} \end{equation} Under the assumption tha $f_{T_1}$ and $r(t)$ are continuous function, if I know $f_{T_1}$ can I be sure that $r(t)$ and the undelrying non homogeneous Poisson process are univocly determined?

Answer 1

Let consider the cumulative function $F_{T_1}$ related to $f_{T_1}$ ($F_{T_1}$ and $f_{T_1}$ are bijectively related) and let define $R(t)=\int_0^tr(u)du$. You have $R'(t)=r(t)$ and $R(0)=0 $

You can easily prove (by integrating $f_{T_1}$ and noticing that ($R'(t)e^{R(t)}=(e^{R(t)})'$ ) that

$F_{T_1}(t) = \frac{1-e^{-R(t)}}{1-e^{-L}}$ (1)

where L is the limit $R(\infty)$ and $1-e^{-L}=P(T_1<\infty)$

And therefore, $R(t)=-ln(1-F_{T_1}(t)(1-e^{-L})$

You can see that if you chose any $L>0$ the right part of this expression has L as limit when t approaches $\infty$

So that defines a function $R_L(t)$ that is valid for equation 1

Therefore, you have an infinity of functions $R_L(t)$ (and their derivatives $r_L(t)$) that provide your distribution $f_{T_1}$. You have one function for each $P(T_1<\infty)$ so only one of them such that $P(T_1<\infty)=1$.