0
$\begingroup$

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.

$\endgroup$

1 Answer 1

0
$\begingroup$

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 $$ \widetilde W(s) = \frac1{1+s}e^{-s}. $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .