Nassim Nicholas Taleb's Statistical Consequences of Fat Tails has a proof (Section 7.7) of the expected value of the the mean deviation for a stable distribution. In the midst of the proof, Taleb says:
The Hilbert transform $H$
$(H f)(t)=\frac{2}{\pi i}\int_0^\infty \chi_S(u+t)-\chi_S(u-t) dt$
can be rewritten as
$(H f)(t)=-i \frac{\partial}{\partial u}(1 + \chi_S(u)+\frac{1}{\pi}\int_0^\infty \chi_S(u+t)-\chi_S(u-t)-\chi_S(t)+\chi_S(-t)dt)$
Should these be equations for $(H f)(u)$, rather than $(H f)(t)$? In the first equation, shouldn't we divide the integrand by $t$, as this is a Hilbert transform? And what is the justification for the expansion in the second equation?
More broadly, does anyone know an alternative proof of the expected value of $|X|$ (given that $X$ is from a stable distribution) that I could review?
Taleb computes the expected value of the absolute value of a stable distribution with mean zero. The version below gives some of the details that are omitted in his book.
In what follows, let $\mathscr{F}(f)$ denote the Fourier transform of $f$, with the convention that $\mathscr{F}(f)(\omega)=\int_{-\infty}^{\infty}e^{i\omega t}f(t)dt$. Then, $$\partial_u \mathscr{F}(f)(u)\rvert_{u=0}=i\int_{-\infty}^{\infty}x f(x)dx$$
Let $f(x)$ denote the PDF of the stable distribution. So, $$E({\mid}X{\mid})=\int_{-\infty}^{\infty}{\mid}x{\mid}f(x)dx=\int_{-\infty}^{\infty}sgn(x) x f(x)dx=\frac{\partial_u \mathscr{F}(sgn(x) f(x))(u)\rvert_{u=0}}{i}$$
By the Convolution Theorem (with the Fourier transform convention given above), $$\mathscr{F}(sgn(x) f(x))(u)=\frac{1}{2\pi}(\mathscr{F}(sgn)*\mathscr{F}(f))(u)$$
As the characteristic function of the distribution $\chi(t)$ is the Fourier transform of the PDF $f(x)$, and as $\mathscr{F}(sgn)(u)=\frac{2i}{u}$,
$$E({\mid}X{\mid})=\partial_u \frac{1}{2{\pi}i} \int_{-\infty}^{\infty}\frac{2i}{u} \chi(u-x) dx\rvert_{u=0}=\partial_u\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{\chi(u-x)}{u}dx\rvert_{u=0}$$ $$=\partial_u\frac{1}{\pi}\int_{0}^{\infty}\frac{\chi(u-x)-\chi(u+x)}{u}dx\rvert_{u=0}$$
The characteristic function of a stable distribution is of the form
$$\chi(t)=e^{i \mu t - {\mid}t c{\mid}^\alpha(1- i \beta tan(\frac{\pi \alpha}{2})sgn(t))}$$
As we are assuming $\mu=0$, this simplifies to:
$$\chi(t)=e^{- {\mid}t c{\mid}^\alpha(1- i \beta tan(\frac{\pi \alpha}{2})sgn(t))}$$
We reverse the order of differentiation and integration in the computation of $E({\mid}X{\mid})$. The derivative at $u=0$ of the quantity inside the integral is (according to Mathematica):
$$\frac{2 \alpha e^{-(c t)^{\alpha }} (c t)^{\alpha } \left(\beta \tan \left(\frac{\pi \alpha }{2}\right) \sin \left(\beta \tan \left(\frac{\pi \alpha }{2}\right) (c t)^{\alpha }\right)+\cos \left(\beta \tan \left(\frac{\pi \alpha }{2}\right) (c t)^{\alpha }\right)\right)}{t^2}$$
Changing variables, letting $z=log(t)$ yields:
$$2 \alpha e^{-\left(c e^z\right)^{\alpha }-z} \left(c e^z\right)^{\alpha } \left(\beta \tan \left(\frac{\pi \alpha }{2}\right) \sin \left(\beta \tan \left(\frac{\pi \alpha }{2}\right) \left(c e^z\right)^{\alpha }\right)+\cos \left(\beta \tan \left(\frac{\pi \alpha }{2}\right) \left(c e^z\right)^{\alpha }\right)\right)$$
Replacing $sin(x)$ with $\frac{e^{i x}-e^{-i x}}{2i}$ and $cos(x)$ with $\frac{e^{ix}+e^{-ix}}{2}$ and integrating (using the principal value) gives the solution:
$$E({\mid}X{\mid})=\frac{c \Gamma \left(\frac{\alpha -1}{\alpha }\right) \left(\left(1+i \beta \tan \left(\frac{\pi \alpha }{2}\right)\right)^{1/\alpha }+\left(1-i \beta \tan \left(\frac{\pi \alpha }{2}\right)\right)^{1/\alpha }\right)}{\pi }$$