Is there any way to calculate the restricted Laplace transform of the random variable $X$, i.e., $ \int_{0}^{u}e^{-sx}dF(x)\ $ $(u<\infty)$, based on its Laplace transform?
Computing the Restricted Laplace Transform of a Random Variable
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probability
laplace-transform
1 Answers
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Let's suppose for simplicity $X$ has a continuous density $f(x)$ with $|f(x)|
$R(s) = \int_0^u e^{-xs} f(x)\ dx = \frac{1}{2\pi i} \int_L \int_0^u e^{(z-s)x} Y(z)\ dx dz = \frac{1}{2\pi i} \int_L \frac{e^{(z-s)u}-1}{z-s}\ Y(z)\ dz$