Hello and Happy holidays.

I am interested in generalizations of the following product formula for the gamma function $\Gamma(z)= \int_{0}^{\infty} t^{z-1}e^{-t}dt$ when $n \geq 2$:

\begin{align} \displaystyle\prod_{k = 1}^{n} \frac{\Gamma(\frac{z}{2^k}+\frac{1}{2})}{\Gamma(\frac{1}{2})} = & \frac{\Gamma(z+1)}{2^{2z(1-\frac{1}{2^n})} \Gamma(\frac{z}{2^n}+\frac{1}{2})} \end{align}

Let $H_1,H_2,...H_n \in (0,1)$ and $z \in \mathbb{R^+}$.
1) Then is it true that the following formula holds for $n \geq 2$?

\begin{align} \frac{\Gamma(zH_1 + \frac{1}{2})\Gamma(zH_1H_2 + \frac{1}{2}) \dotsb \Gamma(zH_1H_2 \dotsb H_n + \frac{1}{2})}{\prod_{k=1}^{n} \Gamma(\frac{1}{2})} = \end{align}

$\frac{\Gamma(z+1)}{2^{2z(1-H_1H_2 \dotsb H_n)} \Gamma( z H_1 H_2 \dotsb H_n + \frac{1}{2})}$

2) As $n$ tends to $\infty$ is the LHS of the last expression finite?

3) Does question 1) hold if $H_1 = 1$?

(In the context of my research the $H_i$'s are Hurst parameters from n+1 independent fractional Brownian motions)

  • $\begingroup$ Could you give some more context on why you believe this to be true? $\endgroup$ Dec 25, 2010 at 7:47
  • $\begingroup$ In brief we are studying moments of iterated stochastic processes and each product can be thought of as representing the moment of a random walk iterated $n$ times . $\endgroup$
    – jzadeh
    Dec 25, 2010 at 8:28
  • $\begingroup$ In particular a paper titled "Iterated Random Walk" L. Turban 2004 Europhys. Lett. 65 627 uses the product formula for all $H_i=1/2$ to show that successive iterations of a random walk converge to a fixed process as $n$ tends to $\infty$.We think something like the last expression should hold for general $H_i \in (0,1)$ because we where able to arrive at a type of generalization of Turban's result from a different perspective than his method of moments. $\endgroup$
    – jzadeh
    Dec 25, 2010 at 8:28
  • $\begingroup$ For both assignments: z,h1,h2=0.5,0.3,0.4 and z,h1,h2,h3=0.5,1/3,1/3,1/3 - sage says that equality does not hold. $\endgroup$ Dec 25, 2010 at 9:04
  • $\begingroup$ Out of curiosity, did you test any values for this conjecture before posting the question? $\endgroup$ Dec 26, 2010 at 6:55

1 Answer 1


Certainly not. Let all $H_i$ tend to zero. By continuity yu get $$ 1=\frac{\Gamma(z+1)}{2^{2z}}, $$ which is false.

  • $\begingroup$ Thank you very much for your time and nice one line answer. We also have been considering the case $H_1 = 1$. For fixed $z \in \mathbb{R}$ and $H_1 = 1$ is there a way to multiply by an appropriate constant to still make the equality hold? $\endgroup$
    – jzadeh
    Dec 26, 2010 at 0:06

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