I have this function:
$$\begin{alignat}{2}% & P(0) = (1 - \alpha) \notag \\ & P(1) = (1 - \alpha){(e^{\alpha} - 1)}\notag \\ & P(n) = (1-\alpha) \left( e^{n \alpha} +\sum_{k=1}^{n-1} e^{k \alpha}(-1)^{n-k} \right. \notag\\ & \quad \quad \quad \quad \quad \quad \quad \times \left. \left[ \frac{(k \alpha)^{n-k}}{(n-k)!} + \frac{(k\alpha)^{n-k-1}}{(n-k-1)!}\right]\ \right)\ n \geq 2\label{md1ssprob0} \end{alignat} %\end{fleqn}$$
that I want to create a probability density function (pdf) via the use of Laplace transforms and then derive the cumulative distribution (cdf) from the pdf.
Here is my attempt...
Let $L^n$ be the number of customers in the system and let $S_k$ be the service time of the $k^{th}$ customer. The random variables $S_k$ are independent and distributed according to a deterministic distribution with mean $\tau$. The quantity $W$ is the waiting time for the customer that arrives given $L^n$ customers are in the queue, that is, \begin{equation} \label{wsum} W = \sum_{k=1}^{L^n} S_k. \end{equation} From the above it follows, by conditioning on $L^n$, that
\begin{eqnarray} \widetilde{W}(s) & = & E(e^{-s W}) \\ \nonumber & = & \sum_{x=1}^{\infty} P(x = L_n) E(e^{-s(S_1 + \cdots S_{x})}) \\ \nonumber & = & \sum _{x=1}^{\infty } \left(\frac{P(x-1)}{\sum _{j=0}^{x-1} P(j)}-\frac{P(x)}{\sum _{j=0}^{x} P(j)}\right) E(e^{-s\ S_1}) \cdots E(e^{-s\ S_x}), \end{eqnarray} where $$ E(e^{-s S_k}) = e^{-s \tau}, $$ then \begin{eqnarray} \label{Ws} \widetilde{W}(s) = \sum _{x=1}^{\infty } \left(\frac{P(x-1)}{\sum _{j=0}^{x-1} P(j)}-\frac{P(x)}{\sum _{j=0}^{x} P(j)}\right) (e^{-s \tau})^x. \end{eqnarray}
To get the pdf, we need to invert the Laplace transform $\widetilde{W}(s)$
Inverting $\widetilde{W}(s)$, we get the (pdf)
\begin{align} \label{wt} w(t) & = {\cal{L}}^{-1} \{ \widetilde{W}(s)\}(t) \\ \nonumber & =\frac{1}{2 \pi i} \lim_{T \rightarrow \infty} \int_{\gamma - i T}^{\gamma + i T} e^{s t} \sum _{x=1}^{\infty } \left( \frac{P(x-1)}{\sum _{j=0}^{x-1} P(j)} -\frac{P(x)}{\sum _{j=0}^{x} P(j)} \right) (e^{-s \tau})^x dt \\ \nonumber & = \sum _{x=1}^{\infty }\left(\frac{P(x-1)}{\sum _{j=0}^{x-1} P(j)}-\frac{P(x)}{\sum _{j=0}^{x} P(j)}\right) \delta (t-\tau\ x), \end{align} that has (cdf) \begin{eqnarray} W(t) & = & \int_0^T \sum _{x=1}^{\infty }\left(\frac{P(x-1)}{\sum _{j=0}^{x-1} P(j)}-\frac{P(x)}{\sum _{j=0}^{x} P(j)}\right) \delta (t-\tau\ x) dt \\ \nonumber & = & \sum _{x=1}^{\infty } \left(\frac{P(x-1)}{\sum _{j=0}^{x-1} P(j)}-\frac{P(x)}{\sum _{j=0}^{x} P(j)}\right) \theta (T-\tau\ x),\ \ t \geq 0, \\ \nonumber & = & \sum _{x=1}^{T/\tau} \left(\frac{P(x-1)}{\sum _{j=0}^{x-1} P(j)}-\frac{P(x)}{\sum _{j=0}^{x} P(j)}\right) \\ \nonumber \end{eqnarray} where $\theta (t-\tau\ x)$ is the Heaviside step function.
Is the above correct for the pdf and cdf?
Thanks for your help...