Talk:Laplace–Stieltjes transform

At the Probability distribution sections, the convolution is NOT done over the CDF-s but over the PDF-s, as it is correctly written in the LTS integral. Am I right? In every other source I have found that the convolution should be done on the PDF-s to get the PDF of the sum of the variables. e.g.https://en.wikipedia.org/wiki/Convolution_of_probability_distributions

Does the convolution theorem *actually* hold for this?
Assume that $$f, g : \mathbb R \to \mathbb R$$ are both differentiable, with their support only being in the positive real numbers. Then
 * $$\begin{align}\mathcal L^*(f * g) &= \mathcal L((f * g)') = \mathcal L\left(\frac{d}{dx}\int_0^\infty f(t) g(x - t)\, dt\right) = -\mathcal L\left(\int_0^\infty f(t) g'(x - t)\, dt\right) \\&= -\mathcal L(f * g') = -\mathcal L(f) \mathcal L(g') = -\mathcal L(f) \mathcal L^*(g),\end{align}$$

where I used differentiation under the integral sign. This seems to suggest the article is wrong about this, no? --Svennik (talk) 19:01, 15 February 2024 (UTC)