= Atom-probability error in a randomized endpoint rule
<Atom-probability error in a randomized endpoint rule> is the change in included duration-model probability caused by using an inaccurate zero-atom probability in the randomization. Let $0<a<b<1$, and hold the positive upper endpoint $u$ fixed at an actual <cumulative distribution function> value $F(u)=b$. If the true zero probability is $p_0$ but the rule uses a positive $\widehat p_0\ge a$, its independent endpoint-inclusion probability is $r=1-a/\widehat p_0$. Conditional on that chosen value,
$$
C=b-a\frac{p_0}{\widehat p_0},\qquad
C-(b-a)=a\left(1-\frac{p_0}{\widehat p_0}\right).
$$
This is the atom term isolated in <distribution sensitivity of randomized zero-endpoint probabilities>, with upper-endpoint error deliberately removed. Underestimating $p_0$ reduces inclusion; overestimating it increases inclusion. The independent auxiliary draw can be generated from a <uniform distribution>.<SciPy 1.16.2 documentation>[scipy.stats.uniform, support and rvs]{ref}
If $\widehat p_0$ comes from an independent calibration, conditional probabilities must also be averaged over that calibration. For an original example, take $p_0=0.20$, $a=0.025$, and $b=0.975$. Suppose the calibration returns $0.10$ or $0.30$ with equal probability. Its estimate is unbiased, yet the resulting included probabilities are $0.925$ and $0.95833$, averaging about $0.94167$ rather than $0.95$.
More generally, for an unbiased positive estimate in the valid rule range, <Jensen's inequality> gives
$$
E\!\left(\frac1{\widehat p_0}\right)\ge\frac1{p_0},\qquad
EC=b-ap_0E\!\left(\frac1{\widehat p_0}\right)\le b-a.
$$
This <reciprocal-estimation bias in randomized atom inclusion> comes from a nonlinear plug-in rule, even with independent calibration. It does not identify the size of the error without a calibration model. Values below $a$, including zero, require an explicitly defined clipped or guarded rule; the displayed inverse formula cannot simply be used there.
If a validated upper bound $p_U\ge p_0$ is available and $p_U\ge a$, choosing $r=1-a/p_U$ instead guarantees $C\ge b-a$ under the fixed-upper-endpoint and independence assumptions. It may be conservative rather than exact. Uncertainty in $F(u)$ remains a separate error term, and these are probabilities for a duration distribution, not a removal of the physical zero-duration atom.
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