Let alpha,beta∈(0,1], and let X_alpha and X_beta be independent Mittag-Leffler random variables with power indices alpha and beta, respectively, and equal time scales lambda=mu. The embedded queue is governed by the arrival probability p_alpha,beta^lambda,mu=PX_alpha lambda; its behaviour depends only on the time scales, not on the power-tail indices. The preceding special cases and numerical simulations support the claim, but no proof or resolution is provided here.

The equal-time-scale conjecture for Mittag-Leffler renewal queues

At least 3 years old · documented by 2022

Let $\alpha,\beta\in(0,1]$ , and let $X_{\alpha}$ and $X_{\beta}$ be independent Mittag-Leffler random variables with power indices $\alpha$ and $\beta$ , respectively, and equal time scales $\lambda=\mu$ . The embedded queue is governed by the arrival probability $p_{\alpha,\beta}^{\lambda,\mu}=\mathbb{P}\{X_{\alpha}<X_{\beta}\}$ . Equal-time-scale conjecture. One has

\mathbb{P}\{X_{\alpha}<X_{\beta}\}=p_{\alpha,\beta}^{\lambda,\lambda}=\frac{1}{2}.

Consequently, the embedded discrete queue is null recurrent if and only if $\lambda=\mu$ , and positive recurrent if and only if $\mu>\lambda$ ; its behaviour depends only on the time scales, not on the power-tail indices. The preceding special cases and numerical simulations support the claim, but no proof or resolution is provided here.

References

Primary source

Jacob Butt, Nicos Georgiou and Enrico Scalas, “Queuing models with Mittag-Leffler inter-event times”, arXiv:2211.13127 (2022).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A reader-written complete proof claims the conjecture and a stronger scale-threshold result, but neither has been independently verified.

The conjecture, posed by Jacob Butt, Nicos Georgiou, and Enrico Scalas in 2022, says that equal time scales make two independent Mittag-Leffler waiting times equally likely to occur first, regardless of their indices $\alpha,\beta\in(0,1]$ . The original paper explicitly states that no rigorous proof was obtained.

Known results

The equality is proved for $\alpha=\beta$ by symmetry (Butt, Georgiou, and Scalas, 2022).
It is known when $\alpha=1$ or $\beta=1$ ; the source gives an explicit comparison formula in that case (Butt, Georgiou, and Scalas, 2022).
For general indices, a critical scale ratio exists, but identifying it with $1$ is precisely the conjecture (Butt, Georgiou, and Scalas, 2022).

Posted attempt

A reader-written argument claims a complete proof: the Mellin transform makes $\log X_\alpha-\log X_\beta$ symmetric, yielding probability $1/2$ at equal scales and a universal threshold at scale ratio $1$ . The attempt has not been independently verified.

Current status (as of August 2026): A complete proof has been claimed, but the conjecture remains unverified; the classical special cases are settled and the general queue classification remains open.

Sources

arxiv.org

ProofThis solution needs a summarySee full solutionHide full solution

The conjecture holds for every pair of Mittag-Leffler indices. In fact, the complete arbitrary-scale comparison has a universal threshold independent of those indices.

Let $0<\gamma\le1$ , and let $X_\gamma$ have the standard Mittag-Leffler waiting-time distribution

\mathbb E e^{-sX_\gamma}=\frac1{1+s^\gamma}.

The source uses the scale convention

X_{\gamma,\lambda}=\frac{X_\gamma}{\lambda}.

For $0<q<\gamma$ , the fractional-moment identity and Euler's beta integral give

\begin{aligned} \mathbb EX_\gamma^q &=\frac{q}{\Gamma(1-q)} \int_0^\infty \frac{s^{\gamma-q-1}}{1+s^\gamma}\,ds\\ &= \frac{\Gamma(1+q/\gamma)\Gamma(1-q/\gamma)} {\Gamma(1-q)}. \end{aligned}

Similarly,

\mathbb EX_\gamma^{-q} = \frac1{\Gamma(q)} \int_0^\infty\frac{s^{q-1}}{1+s^\gamma}\,ds <\infty.

Hence the Mellin transform is holomorphic on

-\gamma<\Re z<\gamma,

and analytic continuation yields

\mathbb EX_\gamma^z = \frac{\Gamma(1+z/\gamma)\Gamma(1-z/\gamma)} {\Gamma(1-z)}.

For independent $X_\alpha,X_\beta$ , set

W=\log X_\alpha-\log X_\beta.

Its characteristic function is

\begin{aligned} \varphi_W(t) &=\mathbb EX_\alpha^{it}\,\mathbb EX_\beta^{-it}\\ &= \frac{ |\Gamma(1+it/\alpha)|^2 |\Gamma(1+it/\beta)|^2 }{ |\Gamma(1+it)|^2 }. \end{aligned}

This is real and even. Uniqueness of characteristic functions therefore gives

W\stackrel d=-W.

The waiting-time laws are continuous, so $\Pr(W=0)=0$ . Consequently

\Pr(X_\alpha<X_\beta)=\frac12

for every $\alpha,\beta\in(0,1]$ , and therefore

p_{\alpha,\beta}^{\lambda,\lambda} = \Pr(X_{\alpha,\lambda}<X_{\beta,\lambda}) =\frac12.

More generally,

p_{\alpha,\beta}^{\lambda,\mu} = \Pr\left(W<\log\frac{\lambda}{\mu}\right).

Both waiting-time densities are strictly positive on $(0,\infty)$ , so the distribution function of $W$ is strictly increasing. Its symmetry gives the complete classification

p_{\alpha,\beta}^{\lambda,\mu} \begin{cases} <1/2,&\lambda<\mu,\\ =1/2,&\lambda=\mu,\\ >1/2,&\lambda>\mu. \end{cases}

Thus the critical scale ratio is exactly $1$ for every pair of Mittag-Leffler indices.