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Corrected exponent and added a few words of explanation.
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edit approved Apr 25, 2018 at 22:25
Russ Lyons
edit approved Apr 25, 2018 at 22:25
Russ Lyons
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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 $$E(X^{s}) = (-1)^n\frac{1}{\Gamma(\alpha)} \int_0^\infty F^{(n)} (\theta) \theta^{\alpha} d \theta$$$$E(X^{s}) = (-1)^n\frac{1}{\Gamma(\alpha)} \int_0^\infty F^{(n)} (\theta) \theta^{\alpha-1} d \theta$$ with $\Gamma$ the usual Gamma function, $$\Gamma(\alpha) = \int_0^\infty \theta^{\alpha -1} e^{-\theta} d \theta.$$

Indeed by FubiniTonelli's theorem and then a change of variable, $$\int_0^\infty F^{(n)}(\theta) \theta^{\alpha-1} d \theta = (-1)^n E (\int_0^\infty X^n e^{-\theta X} \theta^{\alpha-1} d \theta )= (-1)^n E(X^{n-\alpha}) \int_0^\infty \theta^{\alpha-1} e^{-\theta} d \theta,$$$$\int_0^\infty F^{(n)}(\theta) \theta^{\alpha-1} d \theta = (-1)^n E \Bigl(\int_0^\infty X^n e^{-\theta X} \theta^{\alpha-1} d \theta \Bigr)= (-1)^n E(X^{n-\alpha}) \int_0^\infty \theta^{\alpha-1} e^{-\theta} d \theta,$$ and so long as $\alpha >0$ the integral on the right is convergent.

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.

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.

added 13 characters in body
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Jeff Schenker
Jeff Schenker
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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 $$E(X^{s}) = \frac{1}{\Gamma(\alpha)} \int_0^\infty F^{(n)} (\theta) \theta^{\alpha} d \theta$$$$E(X^{s}) = (-1)^n\frac{1}{\Gamma(\alpha)} \int_0^\infty F^{(n)} (\theta) \theta^{\alpha} d \theta$$ with $\Gamma$ the usual Gamma function, $$\Gamma(\alpha) = \int_0^\infty \theta^{\alpha -1} e^{-\theta} d \theta.$$

Indeed by Fubini, $$\int_0^\infty F^{(n)}(\theta) \theta^{\alpha-1} d \theta = (-1)^n E (\int_0^\infty X^n e^{-\theta X} \theta^{\alpha-1} d \theta )= E(X^{n-\alpha}) \int_0^\infty \theta^{\alpha-1} e^{-\theta} d \theta,$$$$\int_0^\infty F^{(n)}(\theta) \theta^{\alpha-1} d \theta = (-1)^n E (\int_0^\infty X^n e^{-\theta X} \theta^{\alpha-1} d \theta )= (-1)^n E(X^{n-\alpha}) \int_0^\infty \theta^{\alpha-1} e^{-\theta} d \theta,$$ and so long as $\alpha >0$ the integral on the right is convergent.

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 $$E(X^{s}) = \frac{1}{\Gamma(\alpha)} \int_0^\infty F^{(n)} (\theta) \theta^{\alpha} d \theta$$ with $\Gamma$ the usual Gamma function, $$\Gamma(\alpha) = \int_0^\infty \theta^{\alpha -1} e^{-\theta} d \theta.$$

Indeed by Fubini, $$\int_0^\infty F^{(n)}(\theta) \theta^{\alpha-1} d \theta = (-1)^n E (\int_0^\infty X^n e^{-\theta X} \theta^{\alpha-1} d \theta )= E(X^{n-\alpha}) \int_0^\infty \theta^{\alpha-1} e^{-\theta} d \theta,$$ and so long as $\alpha >0$ the integral on the right is convergent.

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.

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Jeff Schenker
Jeff Schenker
  • 1.5k
  • 10
  • 16

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 $$E(X^{s}) = \frac{1}{\Gamma(\alpha)} \int_0^\infty F^{(n)} (\theta) \theta^{\alpha} d \theta$$ with $\Gamma$ the usual Gamma function, $$\Gamma(\alpha) = \int_0^\infty \theta^{\alpha -1} e^{-\theta} d \theta.$$

Indeed by Fubini, $$\int_0^\infty F^{(n)}(\theta) \theta^{\alpha-1} d \theta = (-1)^n E (\int_0^\infty X^n e^{-\theta X} \theta^{\alpha-1} d \theta )= E(X^{n-\alpha}) \int_0^\infty \theta^{\alpha-1} e^{-\theta} d \theta,$$ and so long as $\alpha >0$ the integral on the right is convergent.