A proof about a Poisson Process

Question

Let the $N(t)$ be a Poisson process with a rate $\lambda >0$. The sequence $T_1, T_2, ...$ is a sequence of interval times between events. The sums $S_k=\sum_{i=1}^{k}T_i$ are the moments in which the events occured. My problem is to show that for each natural number $k, t>0, u\in (0, t)$ and $v>0$ we have:

$P(t-u<S_k\leq t \land t<S_{k+1}\leq t+v) = \frac{(\lambda t)^k - (\lambda (t-u))^k}{k!} e^{-\lambda t}(1 - e^{- \lambda v})$

I have no idea even how to start, which properities of Poisson Process may be useful for that?

Thanks for any help.

Answer 1

The inequalities, $\;t-u\lt S_k\leq t\lt S_{k+1}\leq t+v,\;$ mean two things:

• Up to time $t$ we need exactly $k$ events but we cannot have it such that those $k$ events occured up to time $t-u$ and then none between $t-u$ and $t$.
• Between times $t$ and $t+v$ we have at least $1$ event.

Therefore,

\begin{eqnarray*} P(t-u\lt S_k\leq t &\cap& t\lt S_{k+1}\leq t+v) \\ &=& P((N(t) = k \setminus [N(t-u) = k \cap N(t-(t-u)) = 0] \cap N((t+v)-t) \geq 1) \\ && \\ &=& \left[P(N(t) = k) - P(N(t-u) = k)\cdot P(N(u) = 0)\right] \cdot P(N(v) \geq 1) \\ && \qquad\qquad\qquad\qquad\qquad\text{by independent and stationary increments} \\ && \\ &=& \left[\dfrac{(\lambda t)^k}{k!}e^{-\lambda t} - \dfrac{(\lambda (t-u))^k}{k!}e^{-\lambda (t-u)} \cdot e^{-\lambda u}\right] \cdot \left(1-e^{-\lambda v}\right) \\ && \\ &=& \dfrac{(\lambda t)^k - (\lambda (t-u))^k}{k!}e^{-\lambda t} \cdot \left(1-e^{-\lambda v}\right). \end{eqnarray*}