# Laplace transform and fractional moments.

Is there any "easy" way to calculate fractional moments from Laplace transform. To be more specyfic let us consider the following example. Let $X$ be a positive random variable and $L(\theta) := E \exp (-\theta X)$ be its Laplace transform. Of course it is easy to calculate $E X^n$ where $n$ is a natural number but what with e.g. $E X^{1/2}$.

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For your example of X^{1/2} you can evaluate the fractional half derivative of the Laplace transform (see for example the Wikipedia article on fractional calculus) at theta = 0.

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I thought about this also. I have even tried to use it but it does not very handy to use. I was hoping for something more "natural". Anyway, I have another question. Let us now drop the assumption that X is positive but assume that the Laplace transform exists. How can one calculate $E|X|$? –  Piotr Miłoś Nov 14 '09 at 18:07
If you write the Laplace transform of the probability measure of X as a sum of two pieces, the first corresponds to the integral from -inf to 0 and the second from 0 to +inf, i.e., F(s) = F+(s) + F-(s), then the Laplace transform of the probability measure of |X| will be F+(s)+F-(-s), from which you can derive the moments of |X|. –  David Bar Moshe Nov 16 '09 at 6:18
Yes, but the problem is that I have the Laplace transform of "whole" X that is F(s). How can I calculate F+ and F- using F only? May be I missing something elementary, but I can not see how to do it in an easy way (of course in principle one can apply the inverse Laplace transform and so on but this is not easy usualy ). –  Piotr Miłoś Nov 17 '09 at 14:31
If $X$ is positive the following works. Let $F(\theta)=E(e^{-\theta X})$ be the Laplace transform. Given $s\in \mathbb{R}$ write $s=n-\alpha$ with $n$ an integer and $\alpha >0$. Then $$E(X^{s}) = (-1)^n\frac{1}{\Gamma(\alpha)} \int_0^\infty F^{(n)} (\theta) \theta^{\alpha} d \theta$$ with $\Gamma$ the usual Gamma function, $$\Gamma(\alpha) = \int_0^\infty \theta^{\alpha -1} e^{-\theta} d \theta.$$
Indeed by Fubini, $$\int_0^\infty F^{(n)}(\theta) \theta^{\alpha-1} d \theta = (-1)^n E (\int_0^\infty X^n e^{-\theta X} \theta^{\alpha-1} d \theta )= (-1)^n E(X^{n-\alpha}) \int_0^\infty \theta^{\alpha-1} e^{-\theta} d \theta,$$ and so long as $\alpha >0$ the integral on the right is convergent.
Yes. You can use $alpha =3/2$ and $n=2$. This amounts to integrating by parts once in the integral over $\theta$. –  Jeff Schenker Jan 8 at 18:22