Monotonicity conjecture for the optimal grouped MSE constant

The unrestricted optimization has zero infimum and generally no minimum; the meaningful fixed-Gaussian monotonicity statement is false.

Use the definitions in Neri Merhav, Grouped Reverse Importance Sampling for the Partition Function, arXiv:2606.26748, Remark 2, equations (22), (24), and (30), and Conjecture 1. Let $\beta>0$, let $U\geq0$, and let $\nu_k$ be the group-energy counting measure or density of states. Assume

$$C_k=\int_0^\infty e^{-\beta u}\,d\nu_k(u) =Z(\beta)^k\in(0,\infty).$$

The generalized-Gaussian family and its objective are

$$\begin{aligned} m_{\alpha,s}(u) &=\exp\!\left(-\frac{u^\alpha}{2s}\right), &&\alpha>1,\quad s>0,\\ Q_k(m) &=\frac{ C_k\displaystyle\int m(u)^2e^{\beta u}\,d\nu_k(u) }{ \left(\displaystyle\int m(u)\,d\nu_k(u)\right)^2 },\\ V_k(m)&=\frac{Q_k(m)-1}{k}. \end{aligned}$$

1. Universal collapse of the unrestricted optimum

Cauchy–Schwarz gives $Q_k(m)\geq1$, with equality precisely when $m(u)=c e^{-\beta u}$ almost everywhere. Fix $s=(2\beta)^{-1}$ and let $\alpha\downarrow1$. Set

$$I_1(\alpha)=\int e^{-\beta u^\alpha}\,d\nu_k(u), \qquad I_2(\alpha)=\int e^{-2\beta u^\alpha+\beta u}\,d\nu_k(u).$$

Since $u-u^\alpha\leq1$ for $u\geq0$ and $\alpha>1$,

$$e^{-\beta u^\alpha}\leq e^\beta e^{-\beta u}, \qquad e^{-2\beta u^\alpha+\beta u} \leq e^{2\beta}e^{-\beta u}.$$

Dominated convergence consequently yields

$$I_1(\alpha),I_2(\alpha)\longrightarrow C_k, \qquad Q_k(m_{\alpha,(2\beta)^{-1}}) =\frac{C_kI_2(\alpha)}{I_1(\alpha)^2} \longrightarrow1.$$

Therefore, for every group size,

$$\boxed{ \inf_{\alpha>1,\,s>0}V_k(m_{\alpha,s})=0 \qquad(k\geq1). }$$

Moreover, if $\nu_k$ is supported on at least three distinct energies, equality cannot occur: it would require

$$\frac{u^\alpha}{2s}-\beta u=\text{constant}$$

on the support, although the left-hand side is strictly convex and can take any fixed value at no more than two points. Hence the minimum in the source's equation (24) does not exist. This includes its own example $U(x)=|x|$ on $\mathbb R$.

2. An explicit three-state boundary certificate

Take the source-admissible discrete system

$$\mathcal X=\{0,1,2\}, \qquad U(j)=j, \qquad\beta=\log2, \qquad Z(\beta)=\frac74.$$

For $n\geq1$, choose

$$\alpha_n =\log_2\!\left(2+\log_2\!\left(1+\frac1n\right)\right)>1, \qquad s=\frac1{2\log2}.$$

Then the one-sample weights are

$$\left(m_n(0),m_n(1),m_n(2)\right) =\left(1,\frac12,\frac{n}{4(n+1)}\right),$$

and exact rational arithmetic gives

$$V_1(m_n)=\frac{6}{(7n+6)^2}>0, \qquad V_1(m_n)\longrightarrow0.$$

Equality would require weights proportional to $(1,1/2,1/4)$. The values at energies $0$ and $1$ force $(2s)^{-1}=\log2$, while the value at energy $2$ then forces $\alpha=1$, which is excluded. Thus the infimum is zero but the minimum is absent.

3. Strict reversal for the fixed Gaussian family

The source's numerical Table 1 fixes $\alpha=2$ and optimizes only $s$. For this distinct substantive interpretation, write

$$\Delta_k^{(2)}=\inf_{s>0}V_k(m_{2,s}), \qquad t=e^{-1/(2s)}\in(0,1).$$

For the same three-state system,

$$\begin{aligned} M_1(t)&=1+t+t^4,\\ N_1(t)&=1+2t^2+4t^8,\\ V_1(t)&=\frac74\frac{N_1(t)}{M_1(t)^2}-1. \end{aligned}$$

At $t=7/10$,

$$\Delta_1^{(2)} \leq\frac{10454806}{376398801} <\frac7{250}.$$

For two samples, the energy multiplicities are $(1,2,3,2,1)$, so

$$\begin{aligned} M_2(t)&=1+2t+3t^4+2t^9+t^{16},\\ N_2(t)&=1+4t^2+12t^8+16t^{18}+16t^{32},\\ V_2(t)&=\frac12\left( \frac{49}{16}\frac{N_2(t)}{M_2(t)^2}-1 \right). \end{aligned}$$

In fact,

$$V_2(t)-\frac7{250} =\frac{F(t)}{4000M_2(t)^2},$$

where

$$\begin{aligned} F(t) ={}&95888t^{32}-8448t^{25}-12672t^{20}+89552t^{18} -8448t^{17}-4224t^{16}\\ &-25344t^{13}-16896t^{10}-8448t^9+54492t^8 -25344t^5-12672t^4\\ &+16052t^2-8448t+4013. \end{aligned}$$

To certify positivity globally, write $F(t)=\sum_{i=0}^{32}a_it^i$. On each interval $[j/8,(j+1)/8]$, use the Bernstein expansion

$$F\!\left(\frac{j+y}{8}\right) =\sum_{\ell=0}^{32} b_{j,\ell}\binom{32}{\ell}y^\ell(1-y)^{32-\ell}, \qquad 0\leq y\leq1,$$

whose coefficients are explicitly

$$b_{j,\ell} =\sum_{r=0}^{\ell} \frac{\binom{\ell}{r}}{\binom{32}{r}} \sum_{i=r}^{32}a_i\binom{i}{r}\frac{j^{i-r}}{8^i}.$$

Exact rational evaluation gives

$$\begin{array}{c|rrrrrrrr} j&0&1&2&3&4&5&6&7\\ \hline \min_{\ell}b_{j,\ell} &>3203&>2830&>2682&>2394&>1601&>312&>12&>2639. \end{array}$$

Since the Bernstein basis is nonnegative and sums to $1$, it follows that $F(t)>12$ throughout $[0,1]$. Also $M_2(t)\leq9$. Therefore

$$\boxed{ \Delta_2^{(2)} \geq\frac7{250}+\frac1{27000} >\frac7{250} >\Delta_1^{(2)}. }$$

Thus the unrestricted infimum formulation is identically zero, the literal minimum formulation is generally undefined, and the practically motivated fixed-Gaussian monotonicity formulation is false.