Tetrahedron risk comparison at every degree

Under the uniform-simplex design, evaluation and noise assumptions of Theory of the skeleton risk, the new positive vertex sum closes the comparison on a tetrahedron at every degree. Here \(M=4\) is the number of vertices. The common positive noise and output factors cancel, so the two normalized asymptotic risks are

\[R_{\mathrm{ind}}(D)=\left(\sum_{m=0}^3\sqrt{a_m(4,D)}\right)^2, \qquad R_{\mathrm{all}}(D)=\binom{D+3}3.\]

Inactive coefficients are zero. This fixed-vertex-count result is distinct from Fixed-degree risk comparisons in every dimension, which fixes the degree and ranges over every vertex count.

Theorem 44 (Complete tetrahedron comparison)

For every integer \(D\geq1\), optimal inductive skeleton fitting on a tetrahedron has strictly smaller asymptotic risk than all-at-once fitting exactly when

\[D\in\{1,2\}\quad\text{or}\quad D\geq8.\]

All-at-once fitting has strictly smaller risk for \(D\in\{3,4,5,6,7\}\). Equality never occurs. The finite prefix in this theorem is computer-assisted.

Proof. We prove the comparison for \(D\geq625\) by inequalities and certify every remaining degree with exact rational square-root enclosures. No monotonicity of the risk difference is assumed.

Vertex bound. In Theorem 31, \(G_0=1/6\). For \(j\geq1\), use \(H_j\leq j\), \(H_j^{(2)}\leq j\) and drop the negative term to obtain

\[0<G_j\leq\frac2{j+3}+\frac16<1.\]

Vandermonde’s identity gives \(Z_D=\sum_j\binom{D-1}j\binom{D+1}j=\binom{2D}{D-1}\). Since \((D-1)!(D+1)!Z_D=(2D)!\), the vertex formula implies

\[a_0(4,D)\leq \frac{576}{(2D+1)(2D+2)(2D+3)} \leq\frac{72}{D^3}.\]

Edge bound. Fix one source-edge angular mode. Let \(p\) be its tetrahedron propagation and \(g\) its triangle propagation with the outside coordinate replaced by the sum of the two outside coordinates, as in the proof of Theorem 20. Their traces agree on every facet, so \(p-g\) is a full-simplex bubble and propagation orthogonality gives \(\langle p,p-g\rangle=0\). Hence \(\lVert p\rVert^2=\langle p,g\rangle\leq \lVert p\rVert\lVert g\rVert\), and \(\lVert p\rVert\leq\lVert g\rVert\).

If \(r\) is the sum of the two source-edge coordinates, its density is \(6r(1-r)\) on the tetrahedron and \(2r\) on the triangle. The conditional source-edge angular law is the same, and \(g\) depends on the outside coordinates only through their sum. Therefore \(\lVert g\rVert^2\) on the tetrahedron is at most three times its triangle norm. Dividing by the same source-edge norm and summing the modes, with the allocation factors \(\binom42^2=36\) and \(\binom32^2=9\), gives

\[a_1(4,D)\leq12a_1(3,D).\]

For \(D\geq2\), the triangle edge formula gives

\[a_1(3,D)=\frac{18}{2D+3} \left(H_{D-1}+H_{2D+2}-H_{D+3}\right)<9.\]

Indeed \(H_{D-1}\leq D-1\), while the difference of the other two harmonic numbers is the sum of \(D-1\) terms, each at most \(1/(D+4)\), and is smaller than one. Thus \(a_1(4,D)<108\).

Facet bound. Reindexing the codimension-one norm sum with \(j=D-2-k\) gives, for \(D\geq3\),

\[a_2(4,D)=48\sum_{j=1}^{D-2} \frac{D-1-j}{j(2D+4-j)} \leq24H_{D-2}\leq24(1+\log D)\leq48\sqrt D.\]

Here \((D-1-j)/(2D+4-j)\leq1/2\) and \(\log D\leq\sqrt D\) for \(D\geq1\).

Infinite tail. Put \(z=D^{1/4}\) and \(X=\sqrt{a_0}+\sqrt{a_1}+\sqrt{a_2}\). If \(D\geq625\), then \(z\geq5\) and the three bounds imply

\[X<7z+11+\frac9{z^6}\leq\frac{19}2z.\]

For the last inequality, \(11+9/z^6\leq11+9/5^6<25/2\leq(5/2)z\). The top coefficient is \(b=a_3(4,D)=\binom{D-1}3\), so \(\sqrt b\leq D^{3/2}/2\) and

\[\begin{split}\frac{R_{\mathrm{ind}}-b}{D^2} &=\frac{2\sqrt b\,X+X^2}{D^2}\\ &<\frac{19}{2z}+\frac{361}{4z^6} \leq\frac{19}{10}+\frac{361}{62500}<2.\end{split}\]

In contrast,

\[R_{\mathrm{all}}-b =\binom{D+3}3-\binom{D-1}3=2D^2+2.\]

Therefore inductive fitting strictly wins for every \(D\geq625\).

Finite prefix. For each \(1\leq D\leq624\), evaluate the vertex single sum, the proved tetrahedron edge recurrence representation, the harmonic facet formula, and the top coefficient over the rationals. Inactive coefficients are set to zero. For any nonnegative coefficient \(a=u/v\), let

\[\ell=\operatorname{isqrt} \left(\left\lfloor\frac{u2^{2p}}v\right\rfloor\right).\]

Then \(\ell/2^p\leq\sqrt a\leq(\ell+1)/2^p\), with equal endpoints when \(\ell^2v=u2^{2p}\). Sum the lower and upper endpoints separately and square them. Comparing these rational endpoints to the integer \(\binom{D+3}3\) certifies a strict sign; an overlapping interval is never accepted as a result. All 624 comparisons are separated already at \(p=32\). The loss set is exactly \(\{3,4,5,6,7\}\), completing the proof.

Reproducing the finite certificate

The optional standard-library implementation emits every exact coefficient, square-root enclosure and strict sign, together with the analytic tail bound. From the repository root, run

python -m tools.risk_research.tetrahedron_comparison \
  --output .research/skeleton-risk/tetrahedron-comparison.json

The finite prefix uses Fraction and integer square roots. It does not depend on floating-point sign decisions, a numerical large-degree fit, or an unproved crossover pattern. The theorem concerns the normalized asymptotic risks under the stated design assumptions; it makes no finite-sample risk-ordering claim.