Rare-event discovery versus inverse-moment precision

Rare-event discovery versus inverse-moment precision separates seeing an influential thin region from estimating its contribution to inferred density. In a two-point thickness model, let a rare thickness occur with probability $p$, with reciprocal thickness $a$, while common regions have reciprocal thickness $b$. For $n$ independent observations, the rare count has a binomial distribution. The chance of at least one rare observation isTo make this at least $1-\alpha$, choose $n\ge \log (\alpha )/\log (1-p)$. For stipulated $p=10^{-4}$ and $\alpha =0.05$, the smallest integer is 29,956. This planning statement assumes the specified rare-event probability; it does not estimate that probability from a sample that missed the event.^[ref]

The sampling distribution of the estimated inverse moment answers another question. With $N$ the observed rare count,Thus a target relative standard uncertainty $\eta$ for the inverse moment requiresThis is a contribution-weighted precision target for inverse moments: the required count depends on both rarity and the leverage of the rare value.

Use the original model $a=1000\mu {\mathrm{m}}^{-1}$, $b=1/10.001\mu {\mathrm{m}}^{-1}$ and $p=10^{-4}$. Here $\mu \approx 0.19998\mu {\mathrm{m}}^{-1}$ and $(a-b)p/\mu =1/2$. ConsequentlyAt the 95%-discovery count, this relative standard uncertainty is still about 28.9%. Reaching 10% for the inverse moment needs 249,975 observations in this model, while reaching 10% for the rare fraction itself needs 999,900. The factor of four arises because the rare excess accounts for half the total inverse moment, so its relative fluctuation is diluted in the total.

These are exact variance calculations under a stipulated distribution, not confidence-coverage guarantees or measured material properties. Spatially clustered thin regions reduce the effective independent sample count. A credible physical lower-thickness bound or better resolved imaging can change the model more substantially than merely extending a simulation with an unsupported tail.

A contribution-weighted precision target for inverse moments chooses sample size according to how strongly rare observations affect the desired inverse moment, rather than according to the probability of seeing any rare observation. For positive thickness $H$, an inverse moment such as $\mu =\mathbb{E}(1/H)$ can be sensitive to very small thicknesses.

In an original two-point model, let reciprocal thickness be $a$ with rare probability $p$, and $b$ otherwise, with $a>b>0$. Then $\mu =b+(a-b)p$. For $n$ independent observations and rare count $N$, the estimate is $\hat{\mu }=b+(a-b)N/n$. Since $N$ has a binomial distribution,Define the fraction of the mean supplied by the rare excess over the baseline as $\theta =(a-b)p/\mu$. Its relative standard uncertainty isfor target relative uncertainty $\eta$. Here $0<\theta <1$. Rarity alone does not specify the required precision because the rare value's magnitude also enters.

For stipulated $p=10^{-4}$ and $\eta =0.10$, a rare-excess contribution $\theta =0.05$ requires at least 2,500 observations, whereas $\theta =0.50$ requires 249,975. A tenfold increase in contribution raises this variance-based count a hundredfold. The first case still has approximately $(1-10^{-4}{)}^{2500}=0.779$ probability of no rare observation. That is compatible with its standard-uncertainty target because missing the rare excess produces only a 5% relative error in that particular model.

This is not a 95% probability guarantee of a specified relative error. Discovery probability $1-(1-p{)}^n$, estimator variance and interval coverage are different criteria. Exact binomial distribution tail calculations address count probabilities; they do not turn a standard deviation into a coverage statement automatically.^[ref]

The calculation also treats $a$, $b$ and $p$ as specified planning quantities. Uncertainty in rare-event probability, imaging resolution or the reciprocal values requires further modeling; spatial dependence reduces the usefulness of an independent-sample count. In particular, increasing $n$ cannot establish an assumed physical lower-thickness bound that the measurement system never resolves.