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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 $P(Y(t) < n)$?

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Got something from the answer below? –  Did Mar 28 '13 at 22:26
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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, $$ \mathbb E(Y(t))=T\lambda(t-T)+T\lambda T\tfrac12=\lambda T(t-\tfrac12T). $$

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How are your two dexcriptions for the process $Z(T)$ equivalent? The first is very intuitive, attained by constructing the "area" under a Poisson process by adding rectangles, but I can't see how the $Uni((0,1))$ distribution arises in this context. –  madison54 Mar 28 '13 at 17:16
    
In a homogenous Poisson process, conditionally on $N_T=n$, the set of jump times before time $T$ is distributed like the set of values of $n$ i.i.d. random variables uniformly distributed on $(0,T)$. And any uniform random variable on $(0,T)$ coincides with $T$ times a uniform random variable on $(0,1)$. –  Did Mar 28 '13 at 22:25
    
I thought the jump times are distributed as $T_n = t_1 + \dots + t_n$ with $t_i$ i.i.d. Exp($\lambda$) and thus follows a Gamma distribution. Where is my mistake? Does it have to do with conditioning on a fixed number of arrival times in a fixed interval? –  madison54 Mar 29 '13 at 9:04
    
Yes, the "uniform" description is valid conditionally on the number of jumps in (0,T). –  Did Mar 29 '13 at 9:45
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