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Let $X$ be a positive random variable with density function $f$ and $\mathbb{E}X<\infty$, let $a>0$ such that $a\cdot\mathbb{E}X<1$. Denote the Laplace transform of $f$ by $$\hat{f}(s)=\int_0^{\infty}e^{-sx}f(x)dx.$$ I don't understand why the following inequaltiy holds: $$a\left(\frac{1}{s}-\frac{\hat{f}(s)}{s}\right)<1~~\forall s>0$$

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    $\begingroup$ $\frac{1-\hat{f}(s)}{s}$ is the Laplace transform of $1_{x > 0}-\int_0^x f(t)dt $ $\endgroup$
    – reuns
    Apr 12, 2017 at 21:39
  • $\begingroup$ @user1952009 So I have $$a\left(\frac{1-\hat{f}(s)}{s}\right)=a\left(\int_0^{\infty}e^{-sx}(1-F(x))dx\right)\leq a\int_0^{\infty}1-F(x)dx=a\mathbb{E}X<1$$ for all $s>0$ $\endgroup$
    – Vroko
    Apr 13, 2017 at 8:54
  • $\begingroup$ Positive random variable means $P(X < 0) = 0$ ? then yes, $\int_0^\infty (1-F(x))dx = \mathbb{E}[X]$. Can you write an answer, proving everything ? $\endgroup$
    – reuns
    Apr 13, 2017 at 17:56
  • $\begingroup$ @user1952009 This one is all over the site already... $\endgroup$
    – Did
    Apr 15, 2017 at 20:01

1 Answer 1

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Since $\hat{f}(s)=\int_0^{\infty}e^{-sx}f(x)dx=s\int_0^{\infty}e^{-sx}F(x)dx$ we get $$\int_0^{\infty}e^{-sx}(1-F(x))dx=\frac{1-\hat{f}(s)}{s}$$ and therefore $$a\left(\frac{1-\hat{f}(s)}{s}\right)=a\left(\int_0^{\infty}e^{-sx}(1-F(x))dx\right)\leq a\int_0^{\infty}1-F(x)dx=a\mathbb{E}X<1.$$

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