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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asked Nov 14, 2009 at 16:17
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3 Answers
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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 $$E(X^{s}) = (-1)^n\frac{1}{\Gamma(\alpha)} \int_0^\infty F^{(n)} (\theta) \theta^{\alpha-1} d \theta$$ with $\Gamma$ the usual Gamma function, $$\Gamma(\alpha) = \int_0^\infty \theta^{\alpha -1} e^{-\theta} d \theta.$$
Indeed by Tonelli's theorem and then a change of variable, $$\int_0^\infty F^{(n)}(\theta) \theta^{\alpha-1} d \theta = (-1)^n E \Bigl(\int_0^\infty X^n e^{-\theta X} \theta^{\alpha-1} d \theta \Bigr)= (-1)^n E(X^{n-\alpha}) \int_0^\infty \theta^{\alpha-1} e^{-\theta} d \theta,$$ and so long as $\alpha >0$ the integral on the right is convergent.
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$\begingroup$ How do I choose the parameter 'Alpha' ? In the question 's'=1/2. Is 'n'=1, 'Alpha'=1/2 the only choice? Or can I use as well 'n'=2, 'Alpha'=3/2 with no difference ? $\endgroup$ Commented Jan 8, 2014 at 10:16
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$\begingroup$ Yes. You can use $alpha =3/2$ and $n=2$. This amounts to integrating by parts once in the integral over $\theta$. $\endgroup$ Commented Jan 8, 2014 at 18:22
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$\begingroup$ @JeffSchenker Does the above work for $s<0$?? $\endgroup$– BobyCommented Jan 12, 2017 at 20:13
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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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$\begingroup$ 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 $E|X|$? $\endgroup$ Commented Nov 14, 2009 at 18:07
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$\begingroup$ 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|. $\endgroup$ Commented Nov 16, 2009 at 6:18
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$\begingroup$ 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 ). $\endgroup$ Commented Nov 17, 2009 at 14:31
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A version of the Laplace transform is the moment generating function. A paper answer to this question is The moment generating function has its moments by Noel Cressie and Marinus Borkent.
An example is at https://stats.stackexchange.com/a/317475
answered Feb 19, 2022 at 2:52