Minimality conjecture for the dense mPH subclass

From papers

Let R+d\mathbb{R}_+^d denote the positive orthant, and let MPH^\ast distributions have general phase-type (PH) margins. The mPH minimality conjecture. The mPH class is the smallest possible subclass of the MPH^\ast distributions with general PH margins that is dense in the set of distributions supported on R+d\mathbb{R}_+^d. The preceding denseness theorem shows that the mPH class is dense; the claim that no proper subclass with the stated properties is dense remains unsubstantiated here.

Progress summary

Open

The density result is known, but the claim that this class is the smallest dense one remains unproved.

The conjecture asserts that mPH\mathrm{mPH} is the smallest subclass of MPH\mathrm{MPH}^\ast distributions with general phase-type margins that is dense on the positive orthant. The source states this as Conjecture 3.6 and gives no proof or resolution.

Known results

  • The mPH\mathrm{mPH} class is dense in distributions supported on R+d\mathbb{R}_+^d.
  • Approximants can be finite mixtures of dd-dimensional vectors with independent Erlang components, with the number of mixture components diverging.
  • Keeping the number of mixture components fixed does not give density in fixed finite dimension.

Current status (as of August 2026): Density of mPH\mathrm{mPH} is settled, but its minimality among subclasses with general phase-type margins remains open, with no public proof, counterexample, or claimed resolution found.

Sources
Sources & referencesView supporting material

Primary source

Martin Bladt, “A tractable class of multivariate phase-type distributions for loss modeling”, arXiv:2110.05179 (2022).

Solutions 1

Counterexample

The source explicitly orders subclasses by set inclusion and does not require closure under finite mixtures. Under that stated meaning, the conjecture is false, even if “general PH margins” requires realization of every possible tuple of phase-type marginal laws.

Fix d≥2 and let M_d denote the mPH class. Define E_d={finite mixtures of independent products of Erlang laws}, I_d={independent products of arbitrary univariate PH laws}, C_d=E_d∪I_d. The source's Theorem 3.5 proves E_d⊆M_d and weak density of E_d in all probability laws on the positive orthant. Also I_d⊆M_d: given PH marginals PH(α_i,T_i), use the product initial state space ∏_i[p_i], product initial distribution ∏_iα_i, and coordinate generators acting separately. The resulting absorption times are independent and have the prescribed marginals. Hence C_d⊆M_d is dense and realizes every tuple of arbitrary PH marginals.

To prove strict containment, take d=2 and the mPH law with common initial distribution π=(1/3,1/3,1/3) and transient generators T₁=[[-2,1,0],[0,-2,1],[1,0,-3]], T₂=diag(−1,−2,−3). Both have nonnegative off-diagonals and strictly negative row sums. Conditional mean vectors are −T₁^{-1}1=(10,9,7)/11, −T₂^{-1}1=(1,1/2,1/3). Conditional independence and the law of total covariance give Cov(X₁,X₂)=17/594>0, so this law is not in I_2.

Its first marginal has Laplace transform L₁(s)=(4s²+22s+33)/[3(s³+7s²+16s+11)]. The cubic denominator has discriminant −23, and its resultant with the numerator is 11. Consequently L₁ has genuine nonreal poles. Every finite Erlang mixture has only negative real poles, since its Laplace transform is a finite sum of terms (λ/(s+λ))^r. Therefore the displayed first marginal is not a finite Erlang mixture, and the joint law is not in E_2. Thus C_2⊊M_2, although C_2 remains dense and realizes every PH marginal pair. Appending independent exponential coordinates gives the same contradiction for every d≥2.

The precise repair is M_d=Mix_fin(I_d). Indeed, conditioning any mPH law on its finite common initial state expresses it as a finite mixture of independent PH products; conversely I_d⊆M_d and M_d is closed under finite mixtures. Hence mPH becomes the unique smallest class only after adding the unstated requirement of finite-mixture closure together with inclusion of all independent PH products. The conjecture as written does not impose that requirement.

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Shivam Patel ·