1
$\begingroup$

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?

$\endgroup$
2
  • 1
    $\begingroup$ Simply write it as $\mathcal{L}\{F_X\}(x) = \frac{1}{x}\mathcal{L}\{f\}(x)$ and use $F_X'(x) = f(x)$. $\endgroup$ Jun 9 at 15:08
  • $\begingroup$ @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! $\endgroup$
    – Linus
    Jun 9 at 16:58
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thank you Dieter!! This is the explanation I was loking for. Greetings Linus $\endgroup$
    – Linus
    Jun 15 at 6:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.