Logarithmic tail-duration change at the zero-extension boundary ($L=\tau \max (0,\log R)$)

Logarithmic tail-duration change at the zero-extension boundary describes a recording requirement whose input ratio crosses the point where no further acquisition is needed. For fixed decay time $\tau >0$, put $R=A\tau /\epsilon$ and $x=\log R$. The physical extension isHere the logarithm applies to the input ratio, not to a zero output duration. Logarithmic condition numbers of a tail-acquisition requirement applies only where $L>0$.

At $x=0$, the left derivative is zero and the right derivative is $\tau$. The directional derivative of a clipped logarithmic acquisition time along a signed input change $h$ isThus no single ordinary derivative describes threshold-crossing perturbations. Nevertheless, the duration is stable in absolute units:This follows because the positive-part map has Lipschitz continuity with constant one.

For an original interval example, take $\tau =25\mathrm{ns}$ and a justified input-ratio range $0.99\le R\le 1.02$. The corresponding duration range is $0\le L\le 25\log 1.02\approx 0.4951\mathrm{ns}$. Its lower part is clipped at zero. Linearizing only on the zero-duration side would incorrectly predict no possible extension even though the allowed set crosses the threshold. A relative error divided by a nominal $L=0$ is undefined, while this absolute bound remains meaningful.

If $\tau$ is uncertain too, optimize the full expression over the justified joint parameter set rather than treating this fixed-$\tau$ inequality as a complete uncertainty analysis. The tail model must bound optical power throughout the missed interval. NIST's calibration apparatus measured detector/cable/oscilloscope response separately, illustrating why a recorded electrical decay cannot automatically certify such an optical envelope.^[ref] The clipping analysis is an original property of the stated conditional acquisition model.

The directional derivative of a clipped logarithmic acquisition time describes the first-order recording extension when several envelope parameters change together at the acquisition threshold. For positive $A,\tau ,\epsilon$, writeThe logarithmic tail-duration change at the zero-extension boundary article treats a fixed decay time and logarithmic input coordinate. Here consider a boundary point $A\tau =\epsilon$ and an additive parameter direction $\mathbf{h}=(a,t,e)$.

The one-sided directional derivative uses positive step sizes:The logarithmic chain rule gives the expression inside the maximum. Differentiating the prefactor contributes zero at the boundary because the clipped logarithm there is zero.

For an original example, let $A=0.10\mathrm{W}$, $\tau =10\mathrm{ns}$ and $\epsilon =1.0\mathrm{nJ}$. Increase amplitude in direction $a=0.002\mathrm{W}$ while decreasing decay time in direction $t=-0.10\mathrm{ns}$, with $e=0$. The net fractional direction is $0.02-0.01=0.01$, so $D^{+}L=0.10\mathrm{ns}$ per unit step. Reversing the direction gives derivative zero, not $-0.10\mathrm{ns}$.

The derivative is positively homogeneous in the direction but is not linear, so no single gradient represents all threshold-crossing perturbations. Above threshold the usual smooth derivative applies; strictly below threshold all sufficiently small changes leave the extension zero. Gaussian input uncertainty at a clipped acquisition-time threshold can consequently produce a mass of outcomes exactly at zero rather than a symmetric distribution around a nominal duration. The directional calculation alone does not assign a confidence level.

This is a conditional optical-envelope calculation. Separating optical power from the receiver's electrical response remains necessary before its parameters describe missed light. NIST independently characterized the detector/cable/oscilloscope response in its pulse calibration procedure.^[ref] A derivative can describe parameter sensitivity accurately while the envelope itself still needs experimental justification.

Gaussian input uncertainty at a clipped acquisition-time threshold produces a mixed output distribution rather than an ordinary Gaussian error bar. Hold the decay time $\tau >0$ fixed and assume the logarithmic input ratio $X=\log (A\tau /\epsilon )$ has distribution $N(\mu ,s^2)$, with $s>0$. The required extension isThe directional derivative of a clipped logarithmic acquisition time explains the kink; here the full input distribution is propagated through it.

Writing $\Phi$ and $\phi$ for the standard normal cumulative distribution function and density, respectively,For $\ell >0$ the continuous part has density $\phi ((\ell /\tau -\mu )/s)/(\tau s)$. It integrates to less than one because the remaining probability is an atom at zero. Treating this density alone as the entire distribution loses that atom.

At the nominal threshold $\mu =0$, half the outcomes are zero and the conditional positive part has a half-normal distribution. Direct integration of the normal density givesFor an original example, $\tau =20\mathrm{ns}$ and logarithmic standard deviation $s=0.02$ give mean extension about $0.1596\mathrm{ns}$ and standard deviation $0.2335\mathrm{ns}$, although inserting the nominal input into the clipped formula gives zero. The probability of an extension above $0.80\mathrm{ns}$ is $1-\Phi (2)\approx 0.0228$.

The quantiles of a clipped Gaussian acquisition duration must likewise include the zero atom. A symmetric interval around the nominal zero duration cannot represent these probabilities. NIST's uncertainty discussion stresses the need to know and combine input distributions when assigning an interval probability.^[ref]

This is a stated uncertainty model for a conditional recording requirement, not a worst-case optical-energy guarantee. Uncertain decay time, correlation among the original envelope parameters or a non-Gaussian logarithmic ratio requires the corresponding joint propagation instead. Establishing the optical envelope remains a separate experimental task.

Quantiles of a clipped Gaussian acquisition duration assign probability levels to a recording extension whose distribution has a point mass at zero. Under the model of Gaussian input uncertainty at a clipped acquisition-time threshold, let $X\sim N(\mu ,s^2)$, $s>0$, and $L=\tau \max (0,X)$ for fixed $\tau >0$. Its zero mass is $p_0=\Phi (-\mu /s)$.

Using the lower generalized quantile $q_p=\inf \{\ell :\mathrm{Pr}(L\le \ell )\ge p\}$ for $0<p<1$ givesEquivalently, $q_p=\tau \max (0,\mu +s{\Phi }^{-1}(p))$. The constant branch is essential: directly using a negative untruncated normal quantile would assign an impossible negative duration. SciPy's normal-distribution documentation identifies ppf as the inverse cumulative distribution function used for the Gaussian part of this calculation.^[ref]

For an original threshold example, take $\mu =0$, $\tau =30\mathrm{ns}$ and $s=0.01$. Half the probability is at zero, so the median is zero. The 90th and 95th percentiles are approximately $0.3845$ and $0.4935\mathrm{ns}$. The one-sided interval $[0,q_{0.95}]$ has probability $0.95$ under this stated model.

A nominal equal-tail 95 percent construction behaves differently: $[q_{0.025},q_{0.975}]=[0,0.5880]\mathrm{ns}$ actually contains probability $0.975$. Its lower endpoint includes the entire zero atom instead of excluding a 2.5 percent lower tail. This coverage excess from central quantiles at a zero-duration atom is a property of the mixed distribution, not a rounding error.

Quantiles describe uncertainty in the conditional acquisition requirement. They are not automatically a frequentist confidence set for an unknown fixed physical duration, nor a worst-case guarantee on unrecorded optical energy. NIST explains why an interval's probability interpretation requires the underlying distributions and assumptions.^[ref] Propagate any uncertainty in decay time or other envelope parameters jointly before applying this fixed-$\tau$ formula.

Coverage excess from central quantiles at a zero-duration atom is the additional distributional probability included when a closed quantile interval captures an entire point mass at zero. Here coverage means probability under the stated duration model. For $L=\tau \max (0,X)$ with $X\sim N(\mu ,s^2)$, the zero mass is $p_0=\Phi (-\mu /s)$ and the positive part is continuous. Let $a=\alpha /2$, $b=1-\alpha /2$, and use the lower quantiles from quantiles of a clipped Gaussian acquisition duration.

For the closed central interval $[q_a,q_b]$, its probability isTo verify this, write the probability as $F(q_b)-F(q_a^{-})$. When both endpoints are positive, the lower left limit is $a$ and the upper value is $b$. Once the lower endpoint is zero, its left limit is zero; if the upper endpoint is zero too, the interval contains exactly the zero atom. Endpoint inclusion is therefore part of the probability calculation.

For a nominal 95 percent central construction, zero masses $0.01$, $0.20$ and $0.99$ give actual probabilities $0.95$, $0.975$ and $0.99$ respectively. In the middle case, the excess is $0.025$, not the whole $0.20$ atom: the interval already includes the probability between zero and the upper quantile.

As an original dimensional example, set $\tau =20\mathrm{ns}$, $s=0.02$ and $\mu /s\approx 0.84162$, giving $p_0=0.20$. The nominal central 95 percent interval is approximately $[0,1.1206]\mathrm{ns}$ and contains probability $0.975$. SciPy documents the normal cdf and inverse-cdf ppf used to evaluate these Gaussian-component values.^[ref]

A one-sided upper interval or a separately specified probability construction can target a different included probability. Randomized endpoint inclusion at a zero-duration atom would require an explicit randomization rule, not a change made silently after seeing an endpoint equal zero. NIST explains why interval interpretation requires the input distributions and the chosen construction.^[ref] Distributional probability for this conditional acquisition model is separate from a worst-case optical-energy guarantee or a confidence procedure for an unknown fixed parameter.

Randomized endpoint inclusion at a zero-duration atom adds an independent random decision about whether an interval contains the atom, allowing an exact included probability under a specified distribution. Let the duration $L\ge 0$ have zero mass $p_0$ and a continuous positive part. For a nominal central level $1-\alpha$, put $a=\alpha /2$, $b=1-\alpha /2$ and assume $a\le p_0<b$. Choose upper endpoint $u$ with $\mathrm{Pr}(L\le u)=b$.

The coverage excess from central quantiles at a zero-duration atom shows that the closed set $[0,u]$ has probability $b$, while $(0,u]$ has probability $b-p_0$. Draw an independent $U\sim \mathrm{Uniform}(0,1)$ and reportThen the joint probability over both $L$ and the auxiliary random draw isThe independence assumption is essential to the multiplication $r{p}_0$.

For an original example, let $p_0=0.20$ and $\alpha =0.05$. The endpoint is included with probability $r=0.875$. The positive part contributes $0.775$ probability and randomized zero inclusion contributes $0.175$, totaling $0.95$. An individual reported set is still either closed, with duration-model probability $0.975$, or open at zero, with probability $0.775$. Exact $0.95$ applies to the randomized procedure jointly, not conditionally to each realized set.

The distribution sensitivity of randomized zero-endpoint probabilities matters if $p_0$ or the upper endpoint is estimated inaccurately. Plugging a guessed atom probability into $r$ does not establish the same equality for the true distribution. The pinned numerical reference documents a standard uniform distribution and its random-variates method, while its normal cdf/ppf methods evaluate the Gaussian component when the duration comes from a clipped normal input.^[ref]

This is a mathematical probability construction. It does not remove the atom physically, improve measurement precision, or create a worst-case guarantee on missed optical energy. Its randomization rule and probability interpretation must be specified as part of the procedure.

Distribution sensitivity of randomized zero-endpoint probabilities quantifies how the probability equality in randomized endpoint inclusion at a zero-duration atom changes when its assumed duration law is wrong. Let the nominal law have zero-atom probability ${\tilde{p}}_0$, choose a positive upper endpoint $\tilde{u}$ with nominal cumulative distribution function $\tilde{F}(\tilde{u})=b$, and put $r=1-a/{\tilde{p}}_0$, with $0<a\le {\tilde{p}}_0<b<1$. Use the parent's independently randomized decision to include zero.

If the actual zero probability is $p_0$ and the actual cumulative distribution function is $F$, the included probability isSubtracting the target $b-a$ separates the two model errors:The first term is upper-endpoint probability error; the second is atom-probability error in a randomized endpoint rule. This identity is an algebraic consequence of independence of the auxiliary random draw, not a normal-approximation result. The documented uniform distribution random-variates construction supplies that auxiliary draw; normal cdf/ppf routines can evaluate a specified clipped-normal duration model.^[ref]

For an original misspecification example, choose $a=0.025$, $b=0.975$, and ${\tilde{p}}_0=0.20$, so $r=0.875$. Suppose instead $p_0=0.30$ and $F(\tilde{u})=0.970$. Thenwhich is $0.0175$ below $0.95$: $0.005$ comes from the upper endpoint and $0.0125$ from the atom. These are stipulated distributions, not measured failure rates.

More generally, bounds $|F(\tilde{u})-b|\le {ϵ}_F$ and $|p_0-{\tilde{p}}_0|\le {ϵ}_p$ imply $|C-(b-a)|\le {ϵ}_F+(a/{\tilde{p}}_0){ϵ}_p$. This propagates validated model-error bounds only. Estimating the nominal law from the same observation used for the interval can add dependence; the unconditional fixed-rule identity then requires an appropriate joint or conditional argument.

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 ${\hat{p}}_0\ge a$, its independent endpoint-inclusion probability is $r=1-a/{\hat{p}}_0$. Conditional on that chosen value,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.^[ref]

If ${\hat{p}}_0$ 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 givesThis 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.

Reciprocal-estimation bias in randomized atom inclusion quantifies the nonlinear plug-in loss beyond the sign supplied by Jensen's inequality. Keep the independent-calibration model of atom-probability error in a randomized endpoint rule: true atom probability $p>0$, unbiased estimate $\hat{p}=p+\epsilon$, and included probability $C=b-ap/\hat{p}$, with the upper endpoint calibrated to actual probability $b$. Assume $0<a<b<1$, $\hat{p}\ge a$, and $E\epsilon =0$.

The exact algebraic identitygivesThus the loss depends on squared errors weighted by the reciprocal estimated probability. The second-order delta method approximation is $-a\mathrm{Var}(\hat{p})/p^2$, but the exact remainder shows why small estimated probabilities deserve particular attention.

If $|\epsilon |\le \eta p$ almost surely for $0<\eta <1$, a nonasymptotic bound isFor an original small-error example, let $p=0.4$, let measurement calibration return $0.36$ or $0.44$ equally often, and take $a=0.025$, $b=0.975$. Here $\eta =0.1$ and the variance is $0.0016$. The exact included probability is about $0.9497475$; the second-order approximation is $0.94975$, while the bounded-error calculation puts the coverage loss between about $0.0002273$ and $0.0002778$.

The bounded-relative-error assumption can fail for a finite binomial measurement calibration: an empirical atom fraction can be zero with positive probability. Such near-zero probability estimates in reciprocal plug-in rules require an explicitly defined guard or different measurement calibration model, not division by zero. The auxiliary endpoint decision must remain independent of the future duration and measurement calibration as stipulated. The documented uniform random-variates construction supplies that auxiliary randomness; it does not validate the atom estimate.^[ref]

Near-zero probability estimates in reciprocal plug-in rules can make a nominal randomization undefined even when the underlying probability estimate is unbiased. For an independent measurement calibration count $J\sim \mathrm{Binomial}(M,p)$, the empirical atom fraction $\hat{p}=J/M$ has $E\hat{p}=p$, butThus its reciprocal is not a finite statistic on every measurement calibration outcome. The binomial probability mass function supplies this zero-count probability directly.^[ref]

In reciprocal-estimation bias in randomized atom inclusion, a fixed upper endpoint contributes actual probability $b$ and the auxiliary rule uses $r=1-a/\hat{p}$. Valid probabilities require a separate rule when $\hat{p}<a$, including zero. One possible numerical guard is ${\hat{p}}_{ϵ}=\max (J/M,ϵ)$ with $a\le ϵ\le 1$. It makes $r=1-a/{\hat{p}}_{ϵ}$ defined and between zero and one. It does not preserve the unbiasedness or the coverage properties of the unguarded formula.

For an original finite measurement calibration example, take $M=20$, $p=0.02$, $a=0.005$, $b=0.995$, and $ϵ=0.01$. A zero count occurs with probability about $0.6676$. On that outcome the guarded rule includes zero with probability $0.5$, and its included duration-model probability is $b-ap/ϵ=0.985$, below the nominal $b-a=0.99$. Averaging over every measurement calibration outcome givesThe calculation assumes the future duration, measurement calibration sample, and auxiliary random draw have the stipulated independence, and the upper-endpoint probability is exactly $b$.

Such finite-calibration guards for reciprocal endpoint randomization solve a domain problem, not a coverage proof. A validated upper probability bound can provide a different conservative construction under the parent's assumptions, but an arbitrary floor is not automatically such a bound. Calibration size, lower-tail events and the actual guarded rule must all enter the final joint-probability calculation.

Finite-calibration guards for reciprocal endpoint randomization must specify both a valid endpoint decision and how calibration uncertainty enters its probability guarantee. An arbitrary positive floor avoids division by zero but does not establish nominal inclusion, as the example in near-zero probability estimates in reciprocal plug-in rules demonstrates.

One finite-sample construction uses an upper confidence bound $U(J)$ for the atom probability $p$, based on independent calibration count $J\sim \mathrm{Binomial}(M,p)$. Supposewhere $0<\gamma <1$ is a calibration failure budget. An exact one-sided binomial bound can be obtained by inversion: for $j<M$, choose $U(j)$ satisfying ${\mathrm{Pr}}_{U(j)}(J\le j)=\gamma$, and put $U(M)=1$. This is a confidence statement over calibration samples, not a posterior probability for a fixed $p$ after observing one count. The pinned reference documents the exact binomial proportion-interval method.^[ref]

For a fixed upper duration endpoint with actual cumulative probability $b$, setThe rule is defined and $0\le r\le 1$. On the event $p\le U(J)$, its conditional included probability is $b-ap/q(J)\ge b-a$. This calculation also requires independence of the calibration, future duration and auxiliary endpoint draw as stipulated by the parent model.

For an original zero-count example, take $M=20$ and $\gamma =0.01$. Since ${\mathrm{Pr}}_p(J=0)=(1-p{)}^{20}$, inversion gives $U(0)=1-0.01^{1/20}\approx 0.20567$. With $a=0.005$, $b=0.995$, and stipulated true $p=0.02$, the zero-count conditional included probability is approximately $0.99451$. The bound is much larger than the raw estimate zero, reflecting the uncertainty of a small calibration sample rather than inventing a measured atom fraction.

The calibration confidence level is not free. Since included probabilities are nonnegative even on the failure event, a general joint guarantee from this argument is onlyAt conditional target $0.99$ and calibration confidence $0.99$, this lower guarantee is $0.9801$, not $0.99$. To certify an unconditional target $t$ using this argument, the conditional target must be at least $t/(1-\gamma )$, provided it can fit below $b$, or the failure budget must be reduced. Any upper-endpoint model error also needs separate accounting.