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. $E X^{1/2}$.
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
|
2
1
|
|
|
|
|
1
|
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. |
|||||||||
|
You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
|
0
|
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. |
|||
|
|
