Integral of a Homogeneous Poisson Process Given that $X(t)$ is a homogeneous poisson process with arrival rate $\lambda$, how do I perform the transformation:
$$ Y(t) =\int ^t _{t-T} X(\zeta) d\zeta$$
to determine say $P(Y(t) < n)$?
 A: Separating the contributions of the events of the Poisson process before $t-T$ and between $t-T$ and $t$, one gets $Y(t)=TX(t-T)+Z_t(T)$ where $Z_t(T)$ is independent of $X(t-T)$ and distributed as $Z(T)=\sum\limits_{n\geqslant0}T_n\mathbf 1_{T_n\leqslant T}$, where $(T_n)_{n\geqslant1}$ is the sequence of jump times of a Poisson process with intensity $\lambda$. Equivalently, let $(U_n)_{n\geqslant1}$ denote some i.i.d. random variables uniform on $(0,1)$ and $N_T$ an independent Poisson random variable with mean $\lambda T$, then
$$
Z(T)=T\sum_{k=1}^{N_T}U_k.
$$
Summing up, considering a second Poisson random variable $\bar N_{t-T}$ with mean $\lambda(t-T)$ and independent of the rest, one gets
$$
Y(t)=T\bar N_{t-T}+T\sum_{k=1}^{N_T}U_k.
$$
Turning to distributions, the Laplace transform reads, for every nonnegative $s$,
$$
\mathbb E(\mathrm e^{-sY(t)})=\mathbb E(\mathrm e^{-sTN_{t-T}})\cdot\mathbb E(u_T(s)^{N_T}),
$$
where, for every $s$ and $t$,
$$
u_T(s)=\mathbb E(\mathrm e^{-sTU})=\frac{1-\mathrm e^{-sT}}{sT},\qquad \mathbb E(\mathrm e^{-sTN_t})=\mathrm e^{-\lambda t(1-\mathrm e^{-sT})}.
$$
Thus,
$$
\mathbb E(\mathrm e^{-sY(t)})=\exp\left(-\lambda (t-T)(1-\mathrm e^{-sT})-\lambda T\left(1-\frac{1-\mathrm e^{-sT}}{sT}\right)\right),
$$
which can be somewhat simplified, but not much. Moments, on the other hand, are easier, for example,
$$
\mathbb E(Y(t))=T\lambda(t-T)+T\lambda T\tfrac12=\lambda T(t-\tfrac12T).
$$
