# Laplace Stieltjes transform of production time

I've a question about the laplace stieltjes transform of a production time.

The production time of a product consist of two phases. The first phase is the same for every customer and is equal to 1 hour. The second phase is not the same for every customer and takes an exponentially distributed time with mean=1 hour.

Now i'm looking for the Laplace stieltjes transform of this distribution (in hours). I know for phase 2 it's equal to $\frac{1}{1+s}$, but I don't know how to do this with the first part.

Is it correct to say: if Z=X+Y, then $\tilde{Z}(s)=\tilde{X}(s)\tilde{Y}(s)$ So the Laplace stieltjes transform is $\tilde{X}(s) \cdot \frac{1}{1+s}$

and $\tilde{X}(s)$ is equal to $\int_0^\infty e^{-st}\cdot 1 dt=-\frac{1}{s}$?

I never had learn Laplace stieltjes transforms, so i'm not sure what to do with the first phase. Hope somebody can help me and tell me if this is correct.

The production time is $W = 1 + T$ where $T\sim\mathsf{Exp}(\lambda)$, so for $t\geqslant 1$ we have \begin{align} \mathbb P(W\leqslant t) &= \mathbb P(1+T\leqslant t)\\ &= \mathbb P(T\leqslant t-1)\\ &= 1 - e^{-\lambda(t-1)}. \end{align} The density of $W$ is then $$f_W(t) = \lambda e^{-\lambda(t-1)}\cdot\mathsf 1_{[1,\infty)}(t),$$ and hence the Laplace-Stieltjes transform is \begin{align} \widetilde W(s) &= \mathbb E\left[e^{-sW}\right]\\ &= \int_1^\infty e^{-st}\lambda e^{-\lambda(t-1)}\ \mathsf dt\\ &= \frac\lambda{\lambda+s}e^{-s},\quad s>-\lambda. \end{align} If $\lambda=1$ then this reduces to $$\widetilde W(s) = \frac1{1+s}e^{-s}.$$