2

While working on something, I have stumbled across the following expression $$\frac{\Gamma \left({\frac{k}{2\left(1-H\right)}} + \frac{1}{2H}\right)}{\Gamma \left(\frac{1}{2H}\right)}$$ where $0 < H \le 1$ is the Hurst Constant and $k \in Z^{+}$.

I am wondering, since $H$ can be irrational can I still use the first identity of the ratio of two Gamma functions given here. If I understand prerequisite condition correctly then in order to be able to use this identity, I need to establish $${\frac{k}{2\left(1-H\right)}} + \frac{1}{2H} + \frac{1}{2H} \\= {\frac{k}{2\left(1-H\right)}} + \frac{1}{H}\in Z$$ I am not quite sure how to proceed here, any help or advice would be much appreciated.

Edit 1 Actually on second thought this is not well posed question as $H=3/4$ would make it not work.

Edit 2 The real task I am undertaking is to derive a characteristic function of a distribution which I suspect is a transformed Gamma Distribution.

I have managed to reduce this derivation to $$\left(\frac{2\eta}{\zeta}\right)^{{\frac{1}{2\left(1-H\right)}}}\sum\limits_{k=0}^{\infty}\frac{1}{k!}\left(\frac{2\eta it}{\zeta}\right)^k\frac{\Gamma \left({\frac{k}{2\left(1-H\right)}} + \frac{\zeta}{2}\right)}{\Gamma \left(\frac{\zeta}{2}\right)}$$

where $\zeta = \frac{1}{H}$ and $\eta = \frac{H^{\left(2H-1\right)}}{\mu^{2H}\left(1-H\right)^{2\left(H-1\right)}}$. I am not sure how to manipulate it further so it appears more like a Gamma distribution's characteristic function.

Comic Book Guy
  • 5,748

1 Answers1

1

In general, I would say the answer to your question is no.

If $H$ can be any irrational number, it could be \emph{very} irrational in some sense.

Take $H=\frac{\pi}{4}$, then if your expression was an integer, you would get a too nice expression for $\pi$

Octania
  • 404
  • Can you be more precise about "very irrational", or "too nice" ? – Dietrich Burde Dec 23 '14 at 10:46
  • I think "transcendental" is more what I meant. In your case, taking $H=\sqrt{2}$, one could still hope getting something integer because you get a $H^2$ term if you write the common denominator – Octania Dec 23 '14 at 12:08
  • $\sqrt{2}$ is not too irrational in the sense that $\sqrt{2}^2$ is no longer irrational, but for $\pi$ there's nothing similar – Octania Dec 23 '14 at 12:10