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

From papers

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.

Progress summary

Open

The conjecture that equal time scales make the two competing waiting times equally likely remains unproved, with only special cases and numerical support.

The conjecture asserts that independent Mittag-Leffler waiting times with equal scales satisfy $\mathbb{P}(X_{\alpha}<X_{\beta})=1/2$ for all indices in $(0,1]$ . The source presenting it states that this would make the queue’s recurrence classification depend only on the time scales, but gives no proof.

Known results

The equality holds when $\alpha=\beta$ , by symmetry.
It also holds when $\alpha=1$ or $\beta=1$ .
High-precision numerical calculations support the conjecture.
The associated recurrence conclusions are established in the relevant special cases.

2026 scaling-limit update

A newer paper classifies queue scaling behavior according to whether $p_{\alpha,\beta}^{\lambda,\mu}$ is above, equal to, or below $1/2$ , but does not prove or refute the equal-time-scale assertion.

Current status (as of August 2026): The conjecture is open; the cases $\alpha=\beta$ , $\alpha=1$ , or $\beta=1$ and numerical evidence are known, while the general case remains unresolved.

Sources

Sources & referencesView supporting material

Primary source

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

1 week ago

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.

0 endorsements

Shivam Patel · 1 day ago