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If I have a positive random variable $X$, how can I show that its Laplace transform $$L_X(t)=E[e^{-tX}] = \int_{\mathbb{R}} e^{-tx}f(x)dx, \quad t>0$$ where $f(x)$ is the density of the r.v. $X$, is uniformly continuous ?

I just know the $\epsilon$ - $\delta$ definition of uniform continuity.

I would start with $$|L(t+h) - L(t)|=||\int_{\mathbb{R}} f_X(x) e^{-tx} \bigl( e^{-hx} - 1 \bigr)dx|| \leq \int_\mathbb{R}|| f_X(x)|| e^{-tx} |e^{-hx} - 1 |dx$$ but then I don't know how to go on.

Actually, I wanted to use the same trick used for the fact that the Fourier transform is uniformly continuous as explained in this answer, but I get stuck.

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  • $\begingroup$ How is $E$ defined? $\endgroup$ Jan 21, 2020 at 23:23
  • $\begingroup$ It's the expected value , editing $\endgroup$
    – andereBen
    Jan 21, 2020 at 23:23
  • $\begingroup$ How is it defined as an integral? $\endgroup$ Jan 21, 2020 at 23:24
  • $\begingroup$ Just edited my question. It's the usual laplace transform, where $f$ is the density function of a random variable $\endgroup$
    – andereBen
    Jan 21, 2020 at 23:25
  • $\begingroup$ You seem to have started your proof in a way that's different to the answer for the Fourier transform. $\endgroup$ Jan 21, 2020 at 23:27

1 Answer 1

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If $t_n \to t$ then $Ee^{-t_nX} \to Ee^{-tX}$ by DCT, so $L_X$ is continuous. Also $L_X(t) \to 0$ as $ t\to \infty$ by DCT and any continuous function on $[0,\infty)$ which vanishes at $\infty$ is uniformly continuous.

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  • $\begingroup$ Thanks @Kavi Rama Murthy, but I am not allowed to use the tool you had in your answer. I need some explicit computation as the one I tried to do above $\endgroup$
    – andereBen
    Jan 21, 2020 at 23:32
  • $\begingroup$ By the way, the DCT applies because for $t>0$ I have $e^{-t X(\omega)}| \leq 1 $ since $X(\omega) \leq 0$ for every $\omega \in \Omega$, right? $\endgroup$
    – andereBen
    Jan 22, 2020 at 8:11
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    $\begingroup$ @andereBen Yes. DCT applies because $0 <e^{-tX} \leq 1$. $\endgroup$ Jan 22, 2020 at 8:18

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