Jobs arrive to a computer facility according to a Poisson process with rate $\lambda$ jobs / hour. Each job requires a service time $X$ which is uniformly distributed between $0$ and $T$ hours independently of all other jobs.
Let $Y$ denote the service time for all jobs which arrive during a one hour period. How should I find the Laplace transform of $Y$?
If $n$ jobs arrive during the period, $Y$ is distributed like the sum of $n$ i.i.d. random variables uniform on $(0,T)$. The number of jobs arriving during the period is Poisson $\lambda$. Putting these two remarks together and decomposing the universe along the number $n$ of jobs during the period, one gets, for every nonnegative $s$, $$ \mathrm E(\mathrm e^{-sY})=\sum_{n=0}^{+\infty}\mathrm e^{-\lambda}\frac{\lambda^n}{n!}\mathrm E(\mathrm e^{-sU})^n=\exp\left(-\lambda[1-\mathrm E(\mathrm e^{-sU})]\right), $$ where $U$ is uniform on $(0,T)$. To make this formula completely explicit, note that $$ \mathrm E(\mathrm e^{-sU})=\int_0^T\mathrm e^{-su}\frac{\mathrm du}T=\frac{1-\mathrm e^{-sT}}{sT}. $$
The general form of a Poisson process (or a Levy process) can be defined as; the number of events in time interval $(t, t + T]$ follows a Poisson distribution with associated parameter λT. This relation is given as
\begin{equation} P[(N(t+T)-N(t))=k] = \frac{(\lambda T)^k e^{- \lambda T}}{k!} \end{equation} Where $N(t + T) − N(t) = k$ is the number of events in time interval $(t, t + T]$. It will be obvious to you that $\lambda$ is the rate parameter. Assume in the simplest case for some $T$ where $k=1$ then
\begin{equation} f(\lambda) = \lambda e^{-\lambda T} \end{equation}
Then taking Laplace transforms yields
\begin{equation} \widehat{f(t)} = \frac{\lambda}{\lambda + s} \end{equation}
I'll leave it to you to fill in the more specific details, I've only dropped the textbook conclusions from a Poisson process and the Laplace transform of such a process.
