I just saw the following on wikipedia about Laplace transformations:


"In probability theory and applied probability, the Laplace transform is defined as an expected value. If $X$ is a random variable with probability density function $f$, then the Laplace transform of $f$ is given by the expectation: $L\lbrace f \rbrace(s) = \mathbb{E}\left[e^{-sX} \right].$


By abuse of notation, this is referred to as the Laplace transform of the random variable $X$ itself.


Of particular use is the ability to recover the cumulative distribution function of a continuous random variable $X$, by means of the Laplace transform as follows:
$F\_X(x) = \mathcal{L}^{-1}\! \left\{\frac{1}{s}\mathbb{E}\left[e^{-sX}\right]\right\}\! (x) = \mathcal{L}^{-1}\! \left\{\frac{1}{s}\mathcal{L}\{f\}(s)\right\}\! (x)$."


So I understand, how we can use the Inverse Laplace transformation of a random variable $X$ to get the CDF of $X$, but why is this possible? Is there some literature on this? Or even better an easy explanation, which I just don't realize?



Why is it possible to use the Inverse Laplace transform to get CDF?
Let $f$ be the density and $F$ the distribution function of $X \geq 0$. Then $F' = f$ (a.s.) and $\mathcal{L}\{F'\}(s) = \mathcal{L}\{f\}(s) = \int\_0^\infty f(x) e^{-sx} dx$ and $\mathcal{L}\{F\}(s) = \int\_0^\infty F(x) e^{-sx} dx$. Since $F(0) = 0$ and using partial integration we get
$$\int\_0^\infty F'(x) e^{-sx} dx = s \int\_0^\infty F(x) e^{-sx} dx,$$
i.e. $\frac{1}{s}\mathcal{L}\{f\}(s) = \mathcal{L}\{F\}(s)$. This is equivalent to the assertion.


