Existence of an independent factor for fractional operators

From papers

Let $W$ be a random variable satisfying the support, normalization, and limiting conditions denoted by,, and. Let $U$ be the random variable appearing in. The relation can be written as

U=WV,

where $V$ is required to be independent of $W$ and supported in $[0,1]$ . Equivalently, with $\widetilde{W}=-\log W$ ,

-\log W-\log V=-\log U\sim\operatorname{Exponential}(1).

Existence conjecture. Under certain conditions on $W$ , there exists a random variable $V$ , independent of $W$ and with support in $[0,1]$ , such that holds. In terms of $\widetilde{W}$ , this amounts to decomposing an $\operatorname{Exponential}(1)$ random variable as a sum of two non-negative independent random variables, with $\widetilde{W}$ fixed and satisfying

\mathbb{P}[\widetilde{W}\in[0,\epsilon)]>0\quad\text{for all }\epsilon>0,

and

\lim_{n\to\infty}n\mathbb{E}[\mathrm{e}^{-n\widetilde{W}}]=0.

The problem asks for conditions on the prescribed factor $W$ that guarantee the existence of an independent factor $V$ . The preceding discussion shows that a beta-distributed choice for $V$ generally fails, although it works in the L-fractional-calculus case $\alpha=1$ ; the general existence question remains open.

Progress summary

Partially solved

The general factorization question remains open: one paper gives examples where the independent factor exists and examples where it cannot exist, but no theorem supplies the desired conditions.

The problem asks when a prescribed random variable $W$ can be multiplied by an independent $V\in[0,1]$ so that $WV$ is uniform on $[0,1]$ , equivalently so that $-\log W$ and $-\log V$ sum to an exponential random variable. The general existence question is explicitly left open.

Known results

Gamma factors give existence when $-\log W\sim\operatorname{Gamma}(\alpha,1)$ and $-\log V\sim\operatorname{Gamma}(1-\alpha,1)$ , with $0<\alpha<1$ .
A second discrete-mixture construction gives existence using a half-weighted exponential factorization.
For $-\log W\sim2\,\operatorname{Bernoulli}(1/2)\cdot\operatorname{Exponential}(1)$ , the required quotient fails to be a characteristic function, so no such $V$ exists.

November 2024 restatement and counterexample

The published article formulates the question as Conjecture 1, records the positive constructions above, and gives the negative example above; it also notes that analogous absolutely continuous counterexamples can be obtained by approximation. No proof of general sufficient conditions or claimed settlement was found.

Current status (as of August 2026): Specific positive and negative examples are known, but the general conditions guaranteeing an independent factor $V$ remain open.

Sources

mdpi.com

Sources & referencesView supporting material

Primary source

Marc Jornet, “Theory on new fractional operators using normalization and probability tools”, arXiv:2403.06198 (2024).

1 week ago

There is first a limiting-condition inconsistency: equation (4.4) assumes

\lim_{n\to\infty}n\mathbb E[W^n]=+\infty,

whereas the later reformulation replaces this by $0$ . Under the displayed zero-limit condition, no factor can exist: independence and $WV\sim\operatorname{Uniform}(0,1)$ would give

\mathbb E[V^n] = \frac1{(n+1)\mathbb E[W^n]} \longrightarrow+\infty,

contradicting $0\le V\le1$ .

For the corrected problem there is a necessary-and-sufficient criterion. Put

\mu_n=\mathbb E[W^n]>0, \qquad b_n=\frac1{(n+1)\mu_n}.

An independent $V\in[0,1]$ with $WV\sim\operatorname{Uniform}(0,1)$ exists if and only if

\boxed{ \sum_{j=0}^k \frac{(-1)^j\binom{k}{j}} {(n+j+1)\mathbb E[W^{n+j}]} \ge0 \qquad(n,k\ge0). }

Necessity follows because this sum equals

\mathbb E[V^n(1-V)^k].

Conversely, the Hausdorff moment theorem gives a probability law on $[0,1]$ with moments $b_n$ . Take $V$ with this law independently of $W$ . Then

\mathbb E[(WV)^n]=\frac1{n+1}

for every $n$ , and compact moment determinacy implies that $WV$ is uniform.

For beta distributions there is a complete explicit classification:

\boxed{ W\sim\operatorname{Beta}(a,b) \text{ admits such a factor} \iff a\ge1,\quad 0<b\le1. }

Indeed, $WV\le W$ forces

\mathbb P(W\le t)\le t.

For $a<1$ ,

\mathbb P(W\le t) \sim\frac{t^a}{a\,\mathrm B(a,b)}>t \qquad(t\downarrow0),

which is impossible. Moreover, the forced moments are

\mathbb E[V^n] = \frac{(a+b)_n}{(n+1)(a)_n} \sim \frac{\Gamma(a)}{\Gamma(a+b)}n^{b-1},

ruling out $b>1$ .

Conversely, let $a\ge1$ and $0<b\le1$ . If $a+b\le2$ , take independent

X\sim\operatorname{Beta}(1,a-1), \qquad Y\sim\operatorname{Beta}(a+b,2-a-b), \qquad V=XY,

interpreting a zero second parameter as a point mass at $1$ . Then

\mathbb E[V^n] = \frac{(1)_n}{(a)_n} \frac{(a+b)_n}{(2)_n} = \frac{(a+b)_n}{(n+1)(a)_n}.

If $a+b\ge2$ , set $c=a+b-1$ , let

Z\sim\operatorname{Beta}(c,1-b),

and independently take

T\sim \frac1c\delta_1+ \left(1-\frac1c\right) \operatorname{Uniform}(0,1).

For $V=ZT$ ,

\mathbb E[V^n] = \frac{(c)_n}{(a)_n} \frac{n+c}{c(n+1)} = \frac{(a+b)_n}{(n+1)(a)_n}.

Both constructions give $WV\sim\operatorname{Uniform}(0,1)$ .

In particular, the entire proposed family

W\sim\operatorname{Beta}(a,b), \qquad0<a,b<1,

admits no such factor, even without independence. Thus the limiting typo, the exact general existence condition, and the complete beta-family classification are all resolved.

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Shivam Patel · 22 hours ago