# Why is it possible to use the Inverse Laplace transform to get CDF?

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?

• Simply write it as $\mathcal{L}\{F_X\}(x) = \frac{1}{x}\mathcal{L}\{f\}(x)$ and use $F_X'(x) = f(x)$. Jun 9 at 15:08
• @DieterKadelka thank you for the fast reply, but I am not quite sure, how this helps me. Why can we write $\mathcal{L}\lbrace F_X \rbrace (x)$ as $\frac{1}{x} \mathcal{L}\lbrace F_X \rbrace (x)$? If we use the Laplace transformation on the equation mentionend in my question, we are getting your equation, but I would like to understand why the equation in my question holds... I apologize for this basic questions, but I'm kinda new to probability theory. Thanks in advance! Jun 9 at 16:58

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.