Exponential-envelope tail-energy bound ($U_{\mathrm{tail}}\le A\tau$)

An exponential-envelope tail-energy bound limits unobserved energy by integrating a justified pointwise envelope on the true optical power. If $0\le P(t)\le A{e}^{-(t-b)/\tau }$ for every $t\ge b$, with $A\ge 0$ and $\tau >0$, then $U_{\mathrm{tail}}\le A\tau$. The parent bounding unobserved optical energy with an exponential tail model derives this result and distinguishes it from a statistical uncertainty.

At a later acquisition endpoint $b+L$, the remaining allowance isFor an original envelope with $A=0.12\mathrm{W}$ and $\tau =30\mathrm{ns}$, the initial allowance is $3.6\mathrm{nJ}$. Reducing the allowed missing energy to $0.036\mathrm{nJ}$ requires $L\ge 30\log 100\approx 138.2\mathrm{ns}$. This acquisition duration from an exponential tail-energy tolerance reduces the bound by a factor100. It does not prove that99% of an arbitrary underlying pulse has been captured: the actual power may lie strictly below the envelope at early times.

Joint parameter constraints also matter. An original allowed set uses dimensionless $a=A/(0.05\mathrm{W})$ and $q=\tau /(20\mathrm{ns})$ with $a\ge 0$, $q>0$ and $a+q\le 6$. Since $aq\le (a+q{)}^2/4$, every allowed envelope satisfiesattained at $a=q=3$. Taking separate maxima $A\le 0.30\mathrm{W}$ and $\tau \le 120\mathrm{ns}$ would instead give the looser $36\mathrm{nJ}$ bound; those two maxima cannot occur together in the stated joint set.

The guarantee concerns optical power, not an uncorrected detector tail. Optical tail and detector-response ambiguity explains that distinction, while NIST's primary response-calibration procedure characterizes the electrical measurement chain separately.^[ref] A delayed extra pulse or an unbounded constant background requires its own model; neither is excluded by a local exponential fit.

An acquisition duration from an exponential tail-energy tolerance chooses how long to continue recording when a validated optical-power envelope bounds the remaining energy. Suppose after time $b$ the true power satisfies $P(t)\le A{e}^{-(t-b)/\tau }$, with $A\ge 0$ and $\tau >0$. The exponential-envelope tail-energy bound gives a remaining allowance $A\tau e^{-L/\tau }$ after extension $L\ge 0$. For tolerance $\epsilon >0$,This is a conditional upper-bound calculation, not an inference from the last sample merely falling below a detector threshold.

An original uncertainty example has guaranteed parameter limits $A\le 0.10\mathrm{W}$ and $\tau \le 40\mathrm{ns}$. The remaining allowance increases with either parameter, so a $0.040\mathrm{nJ}$ tolerance requires $L\ge 40\log 100\approx 184.2\mathrm{ns}$. Using only best-fit values $A=0.08\mathrm{W}$ and $\tau =30\mathrm{ns}$ would give about $122.8\mathrm{ns}$, which does not guarantee the same tolerance over the allowed set. Joint parameter constraints can tighten a bound, but their coverage must be justified.

In the positive-duration branch, the local sensitivities areThis acquisition-time sensitivity to tail-model parameters explains why an uncertain slow decay can matter more than a small amplitude correction. The derivatives change interpretation at the zero-duration boundary.

The envelope must apply to optical power throughout the unrecorded region. A later pulse or an unremoved persistent background can invalidate it. NIST's separate measurement of detector response illustrates why an electrical waveform's decay alone is insufficient.^[ref] State the chosen time origin, tolerance units, valid envelope parameters and any other excluded intervals when reporting the acquisition requirement.

Acquisition-time sensitivity to tail-model parameters measures how an inferred recording extension changes when the assumed optical tail changes. In the positive-duration branch of acquisition duration from an exponential tail-energy tolerance, writeHere $A$ bounds the initial tail power, $\tau$ is its decay time and $\epsilon$ the tolerated missed energy. Their first-order relative effects areThese logarithmic condition numbers of a tail-acquisition requirement distinguish a fractional amplitude error from a fractional decay-time error. For a stringent tolerance with large $\ell$, decay-time sensitivity stays near unity while amplitude sensitivity decreases.

For an original example, take $A=0.20\mathrm{W}$, $\tau =20\mathrm{ns}$ and $\epsilon =0.040\mathrm{nJ}$. Then $A\tau /\epsilon =100$ and $L\approx 92.10\mathrm{ns}$. Let the independent standard uncertainties of $A$ and $\tau$ be $10\%$ and $5\%$, with the tolerance fixed. The amplitude contribution to $u(L)$ is $2.00\mathrm{ns}$; the decay-time contribution is $(1+\log 100)\times 1\mathrm{ns}\approx 5.605\mathrm{ns}$. Adding their variances gives $u(L)\approx 5.95\mathrm{ns}$, about $6.46\%$.

Correlated fitted parameters require covariance terms in this propagation.^[ref] A standard uncertainty is not a guaranteed recording margin: a worst-case requirement needs a justified allowed parameter set, as in the parent article. The zero-duration boundary is also outside this smooth positive-branch calculation.

Physically, these parameters must describe the unrecorded optical power, not merely an electrical detector tail. NIST separately measured the detector/cable/oscilloscope response before correcting pulse measurements.^[ref] A precise fit to the wrong response component does not justify the optical envelope or its acquisition-time bound.

Logarithmic condition numbers of a tail-acquisition requirement quantify the fractional response of a positive recording extension to fractional parameter changes. For $L=\tau \ell$ with $\ell =\log (A\tau /\epsilon )>0$, define the signed sensitivity ${\kappa }_x=\partial \log L/\partial \log x$. ThenTheir absolute values are componentwise relative condition numbers. Acquisition-time sensitivity to tail-model parameters gives the corresponding dimensional derivatives and uncertainty propagation.

These numbers depend on which coordinates are held fixed. If $M=A\tau$ denotes the initial tail-energy allowance, then $L=\tau \log (M/\epsilon )$. In coordinates $(M,\tau ,\epsilon )$, the decay-time sensitivity at fixed $M$ is exactly one, rather than $1+1/\ell$. This is the same physical formula: holding $M$ fixed while changing $\tau$ also changes $A$. Transforming the parameter covariance consistently leaves the propagated output uncertainty unchanged.^[ref]

An original boundary example takes $A\tau /\epsilon =1.01$ and $\tau =20\mathrm{ns}$. The required extension is only $L\approx 0.1990\mathrm{ns}$, but ${\kappa }_A\approx 100.50$. Increasing $A$ by $1\%$ multiplies the ratio by $1.01$, giving $1.0201=(1.01{)}^2$ and doubling the required extension exactly. Its relative change is large because the original extension was very small.

This does not mean the absolute requirement is singular there. In the active branch $\partial L/\partial \log (A\tau /\epsilon )=\tau$ remains finite. At and below the threshold, the physical extension is zero, so $\log L$ is undefined. A logarithmic tail-duration change at the zero-extension boundary must instead use the piecewise requirement or absolute changes. Large relative conditioning alone is not evidence of a large absolute recording-time risk.

Finally, conditioning describes the stated envelope model, not its experimental validity. The optical power tail must be distinguished from detector response, which NIST measured independently in its pulse calibration procedure.^[ref] A well-conditioned calculation based on an invalid optical envelope still gives an unjustified acquisition guarantee.

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.

A differential inequality for an exponential tail envelope supplies a sufficient decay condition for a global power bound. Let nonnegative $P$ be absolutely continuous on every finite interval after $b$, with $P(b)\le A$ andalmost everywhere, for constant $\tau >0$. The integrating factor argument in bounding unobserved optical energy with an exponential tail model gives $P(t)\le A{e}^{-(t-b)/\tau }$. Corners in the waveform are allowed under absolute continuity; an unmodeled upward jump is not.

The inequality is stronger than merely fitting a decay constant at a few sampled times. For example, $P(t)=A{e}^{-2(t-b)/\tau }$ obeys it and carries only $A\tau /2$ after $b$. The exponential-envelope tail-energy bound $A\tau$ remains valid but need not be tight. Conversely, a later pulse can violate the differential condition while leaving all earlier samples unchanged.

A useful extension is a time-dependent decay-rate envelope. For locally integrable $k(t)\ge 0$, the condition $P^{\prime }\le -kP$ givesThis follows from the same integrating factor and is a scalar comparison bound. If $k(t)\ge k_{\min }>0$ throughout the tail, its integrated energy is at most $A/k_{\min }$.

For an original piecewise rate, choose $A=0.20\mathrm{W}$, $k=1/(10\mathrm{ns})$ for the first $10\mathrm{ns}$ and $k=1/(5\mathrm{ns})$ thereafter. Integrating the resulting continuous envelope yieldsUsing only the minimum rate would give $2.0\mathrm{nJ}$. Faster guaranteed late decay tightens the result.

This is a conditional mathematical proof. A measured electrical waveform obeying a similar slope rule does not establish the optical condition without accounting for the detector impulse response. The primary NIST measurement procedure supplies an example of characterizing that response independently.^[ref]

A time-dependent decay-rate envelope bounds a nonnegative signal using a lower bound on its instantaneous decay rate. If $P$ is absolutely continuous on finite intervals, $P(b)\le A$ and $P^{\prime }\le -k(t)P$ almost everywhere for locally integrable $k(t)\ge 0$, thenThe differential inequality for an exponential tail envelope article proves this comparison using an integrating factor and gives a piecewise-constant example. A strictly positive constant lower bound on $k$ gives exponential decay, but a positive rate at every instant need not supply such a lower bound.

An original example takes $k(t)=\alpha /(t-b+t_0)$ with $t_0>0$ and $\alpha >0$. Its integrated rate is logarithmic, soThis power-law tail from a decaying rate envelope tends to zero for every positive $\alpha$. Its envelope energy is nevertheless finite only when $\alpha >1$, in which caseFor original chosen $A=0.20\mathrm{W}$, $t_0=5\mathrm{ns}$ and $\alpha =2$, the allowance is $1.0\mathrm{nJ}$. After a further $45\mathrm{ns}$ it is at most $0.10\mathrm{nJ}$. The remaining energy decreases algebraically rather than exponentially.

At $\alpha =1$ the integrated envelope diverges logarithmically despite its power tending to zero. This does not prove that every signal below it has infinite energy; it means these assumptions alone supply no finite all-time energy bound. Taking the envelope itself gives a mathematical counterexample to the claim that pointwise decay implies integrable energy.

An experimental rate law therefore needs both temporal validity and an integrability check before it bounds missed optical energy. A measured detector tail is not automatically the true power profile; NIST's primary calibration procedure measures the receiver response separately.^[ref] The power-law example is an independent comparison model, not a reported tail law of that apparatus.

A power-law tail from a decaying rate envelope arises when a positive instantaneous decay rate weakens with time. The time-dependent decay-rate envelope with $k(t)=\alpha /(t-b+t_0)$ givesfor $A\ge 0$, $t_0>0$ and $\alpha >0$. The parent article obtains this envelope by integrating the rate. Its use for recording design additionally requires integrating over the entire unrecorded tail.

For $\alpha >1$, integration after an extension $L\ge 0$ givesHence the acquisition duration from a power-law tail-energy tolerance $\epsilon >0$ is zero when $A{t}_0/(\alpha -1)\le \epsilon$; otherwise it is sufficient to choose

In an original example, $A=0.10\mathrm{W}$, $t_0=10\mathrm{ns}$ and $\alpha =2$ give an initial energy allowance $1.0\mathrm{nJ}$. Reducing the remaining allowance to $0.010\mathrm{nJ}$ requires $L\ge 990\mathrm{ns}$. An exponential envelope with the same initial allowance and a $10\mathrm{ns}$ decay time would require only $10\log 100\approx 46.05\mathrm{ns}$. Equal initial tail-energy allowances therefore do not imply equal acquisition durations.

For $0<\alpha \le 1$, the envelope tends to zero but its integral after every finite extension diverges. These assumptions then supply no finite missed-energy guarantee. They do not imply that every signal lying below the envelope has infinite energy.

This is an original comparison model, not a measured decay law attributed to a particular detector. In optical pulse metrology, NIST's response measurement separates receiver dynamics from the incident pulse before applying bandwidth corrections.^[ref] The same physical distinction is necessary before interpreting any measured slow tail as a global bound on optical power.

An acquisition duration from a power-law tail-energy tolerance extends a measurement until the validated envelope's unrecorded energy is below a chosen tolerance. For the model in power-law tail from a decaying rate envelope, with $\alpha >1$,The parent article integrates this envelope and inverts the tolerance. Writing $R=A{t}_0/[(\alpha -1)\epsilon ]$, its positive-duration requirement is $L=t_0(R^{1/(\alpha -1)}-1)$ when $R>1$; otherwise no additional duration is required.

In applying the rule, uncertainty in the exponent can dominate the result. With fixed $A,t_0$ and $\epsilon$, differentiating the active branch givesThe required duration decreases with a larger exponent. For a guaranteed parameter range, a validated lower bound on a power-law tail exponent is therefore more useful than its best-fit value alone. Since $(1+u/t_0{)}^{-\alpha }$ decreases with $\alpha$ for every $u\ge 0$, using the smallest allowed exponent bounds the entire tail, not just one selected sample.

For an original design example, let $A=0.10\mathrm{W}$, $t_0=10\mathrm{ns}$ and $\epsilon =0.005\mathrm{nJ}$. A nominal exponent $\alpha =3$ gives initial allowance $0.5\mathrm{nJ}$, $R=100$ and $L=90\mathrm{ns}$. If the validated exponent range is instead $2\le \alpha \le 3$, its slowest permitted tail has initial allowance $1.0\mathrm{nJ}$ and requires $L=1990\mathrm{ns}$. The nominal duration does not guarantee the tolerance over that range. Upper bounds on $A$ and $t_0$ can likewise be used conservatively because increasing either raises the envelope pointwise.

If the allowed exponents reach one or below, these envelope assumptions alone no longer supply a finite all-time energy guarantee. This is a limitation of the bound, not proof that the actual pulse has infinite energy. The model must also hold throughout the unobserved interval: later pulses, persistent backgrounds or an electrical detector tail misidentified as optical power can invalidate it. NIST's independent receiver-response measurement illustrates this last experimental distinction.^[ref] The power-law examples are original conditional models, not observed decay laws attributed to that apparatus.

A validated lower bound on a power-law tail exponent is an exponent constraint that is justified over the future interval where it is used to bound unrecorded energy. A fitted slope over a finite observation window is not by itself such a guarantee. Acquisition duration from a power-law tail-energy tolerance needs a global optical-power envelope, including a lower exponent limit above one for its finite-energy guarantee.

One sufficient condition, for positive absolutely continuous power, isIntegrating the logarithmic derivative gives $P(b+u)\le A(1+u/t_0{)}^{-{\alpha }_{\min }}$. The time-dependent decay-rate envelope provides the equivalent integrating-factor argument when zeros of the power need handling. This is a sufficient condition, not a necessary characterization of every bounded tail.

For an original conditional estimation example, assume an exact power-law family with known $t_0=10\mathrm{ns}$ and initial power $A=0.10\mathrm{W}$. At $u=90\mathrm{ns}$, a measured ratio range $0.008\le P(b+u)/P(b)\le 0.012$ yieldsapproximately $1.9208\le \alpha \le 2.0969$. Within that assumed family, the lower endpoint gives a conservative extension about $123.32\mathrm{ns}$ for a $0.10\mathrm{nJ}$ remaining-energy tolerance. A nominal exponent two would instead require $90\mathrm{ns}$.

The ratio interval constrains the exponent only conditional on the exact family, known time scale and suitable measurement uncertainty. It cannot exclude a later secondary pulse or a transition to a slower law after the observed window. Future-domain validity of a fitted power-law optical envelope therefore requires independent physical justification or explicitly limited claims. If the justified exponent set reaches one, this envelope alone supplies no finite all-time energy guarantee.

Finally, the measured trace must represent optical power rather than an uncorrected receiver tail. NIST measured the detector, cable and oscilloscope together using a short characterized optical input before correcting pulse measurements.^[ref] The exponent example is an original model calculation, not an observed law attributed to that apparatus.

Future-domain validity of a fitted power-law optical envelope means that the bound applies throughout the unobserved future interval where its integral is used, rather than merely matching a recorded waveform. The validated lower bound on a power-law tail exponent supplies a sufficient differential condition for such a bound. Agreement on a finite observation window alone does not establish it.

A concrete counterexample uses a fitted profile $P_0(u)=A(1+u/t_0{)}^{-2}$ observed only for $0\le u\le T$. Choose a later time $S>T$ and define an additional pulseBoth $P_0$ and $P_1=P_0+q$ are nonnegative and continuously differentiable, and they agree exactly on the recorded interval. Yet the unobserved extra energy isThus nonnegativity, this degree of smoothness and even a perfect finite-window fit cannot exclude an arbitrarily large later contribution.

For an original example, take $A=0.10\mathrm{W}$, $t_0=10\mathrm{ns}$ and $T=100\mathrm{ns}$. The fitted profile predicts remaining energy $A{t}_0/(1+T/t_0)=1/11\mathrm{nJ}\approx 0.0909\mathrm{nJ}$, below a $0.10\mathrm{nJ}$ tolerance. Put the later pulse at $S=300\mathrm{ns}$ with width $w=200\mathrm{ns}$ and peak $B=0.010\mathrm{W}$. It adds $1.0\mathrm{nJ}$, although all recorded data are unchanged. This unobserved delayed-pulse counterexample to a fitted tail bound violates the claimed missed-energy tolerance without altering the fit quality.

To use the envelope as a guarantee, state the physical assumptions that exclude such later input, justify the rate bound over the required future interval, or limit the claim to the interval actually validated. Recording a longer finite window improves evidence but does not alone supply an all-future theorem under only the assumptions used in this counterexample.

The waveform must also represent optical power rather than detector response. NIST's primary pulse calibration separately characterized the detector/cable/oscilloscope chain with a short optical input.^[ref] Correcting the receiver is necessary for interpreting the trace, but it does not rule out the additional source pulse constructed here. The example is a mathematical comparison, not a reported failure of that apparatus.

An unobserved delayed-pulse counterexample to a fitted tail bound constructs two admissible optical inputs giving the same finite measurement record but different missed energies. The future-domain validity of a fitted power-law optical envelope article gives a nonnegative, continuously differentiable delayed pulse with adjustable energy. The counterexample persists through a calibrated causal detector, not just an ideal instantaneous power readout.

Let $P_1=P_0+q$, where the additional pulse has support $[S,S+w]$ after the record ends at $T<S$. A linear receiver with impulse response $h$ and responsivity $g$ measures $y_i=g(P_i\ast h)$. Causality means $h(t)=0$ for $t<0$. Therefore, for every recorded time $t\le T$,since every argument $t-u$ is negative. This causal indistinguishability of delayed optical inputs on a finite record holds even when the impulse response is known exactly and the recorded data are noiseless.

For an original example, record through $T=80\mathrm{ns}$ and place the smooth sine-squared pulse of the parent article at $S=150\mathrm{ns}$, with width $w=40\mathrm{ns}$ and peak $B=0.020\mathrm{W}$. Its extra optical energy is $Bw/2=0.40\mathrm{nJ}$. A chosen causal exponential receiver with decay time $5\mathrm{ns}$ produces no difference in the initial record, although it will respond when the later pulse arrives. If its impulse response has unit area, the complete later output-area difference is $g\times 0.40\mathrm{nJ}$ in the receiver's output-area units.

Any rule using only the common initial record must return the same proposed bound for both inputs. Without a justified restriction on later illumination, the delayed pulse energy can be chosen larger than any proposed finite guarantee. More accurate calibration of that same initial record cannot resolve this information gap.

NIST characterized its detector, cable and oscilloscope together with a short known optical input before correcting pulse measurements.^[ref] Such response calibration addresses receiver distortion. Source shutoff, exclusion of later pulses or a future-valid rate envelope supplies the additional physical information needed for a missed-energy guarantee. The delayed pulse here is a mathematical construction, not a reported defect of that apparatus.

Causal indistinguishability of delayed optical inputs on a finite record means that an observation ending at a fixed time cannot discriminate inputs whose differences have not yet reached its receivers. The unobserved delayed-pulse counterexample to a fitted tail bound proves this for one calibrated convolution channel. The same information limit holds for any finite set of causal linear channels.

Let channel $j$ have response $h_j$ with $h_j(t)=0$ for $t<0$, and record $y_j=P\ast h_j$ only for $t\le T$. Every perturbation $q$ supported strictly after $T$ satisfiesfor every channel. Thus the finite-record observation operator has a whole family of invisible input directions. This is compatible with nonnegative optical power: choose any nonnegative delayed pulse and compare $P$ with $P+q$. No cancellation between unphysical negative powers is required.

Known propagation or detector latency can hide part of the input even before the record ends. If $h_j(t)=0$ for $t<d_j$, an input starting at $S$ produces no response before $S+d_j$. The input-time horizon of a delayed optical receiver record therefore depends on the channel's time reference, not just the last stored sample time.

For an original example, suppose two ideal delay channels measure $y_1(t)=P(t-2\mathrm{ns})$ and $y_2(t)=P(t-7\mathrm{ns})$, both recorded through $T=100\mathrm{ns}$. A pulse starting at input time $99\mathrm{ns}$ first appears at output times $101$ and $106\mathrm{ns}$. Both stored records are unchanged by that pulse, even though it began before the nominal record endpoint. Additional causal channels with smaller delays can extend the input horizon; none can observe an input that begins after their recording ends.

Offline processing of identical stored records still gives identical results, even if the processing filter itself uses samples on both sides of each stored time. To distinguish the inputs, obtain additional observations or supply independently justified restrictions on later illumination. Exact knowledge of the receiver response alone supplies neither.

NIST's pulse calibration characterized the detector, cable and oscilloscope as a combined response chain using a short optical input.^[ref] Such characterization is important for aligning the physical input and recorded output times. The two pure delays above are chosen models, not measured delays of that apparatus. A future-valid optical envelope or documented source cutoff remains additional information needed for an all-future missed-energy bound.

The input-time horizon of a delayed optical receiver record is the latest physical input time that could influence the recorded output, given the receiver's causal onset delay. If $h(t)=0$ for $t<d$ and the last output time is $T$, convolution depends only on input timesThus $T-d$ is a necessary timing limit on information in that record. Recoverability within the limit additionally depends on bandwidth, noise, sampling and the response's invertibility. Causal indistinguishability of delayed optical inputs on a finite record shows why later input changes cannot be inferred from the existing output samples.

For several channels with last recorded times $T_j$ and known onset delays $d_j$, the latest possible input horizon isAny input perturbation starting strictly after $H$ is invisible in every stored channel. A channel's latest output timestamp alone is insufficient to compare its input horizon with another channel's.

For an original example, channel 1 records through $120\mathrm{ns}$ with delay $15\mathrm{ns}$, while channel 2 records through $100\mathrm{ns}$ with delay $2\mathrm{ns}$. Their input limits are $105$ and $98\mathrm{ns}$; the combined limit is $105\mathrm{ns}$. Knowing both response functions exactly still supplies no observation of an input pulse beginning later than that limit.

A receiver-latency uncertainty in the inferred input-time horizon also changes the timing claim. If channel 1's delay is known only to lie between $14$ and $16\mathrm{ns}$, its possible horizon ranges from $104$ to $106\mathrm{ns}$. An input beginning at $104.5\mathrm{ns}$ could affect the record for one allowed delay but arrive too late for another. Using the conservative endpoint $104\mathrm{ns}$ avoids claiming timing access that depends on an unresolved latency. It still does not guarantee successful reconstruction of every earlier input.

NIST characterized detector, cable and oscilloscope as a combined response chain using a short known optical pulse.^[ref] Such a calibration must relate output timing to the physical input time origin. The delays above are illustrative, not quoted apparatus values. De-embedding known latency aligns the record's time reference; it cannot extend observations into an unrecorded future interval.

Receiver-latency uncertainty in the inferred input-time horizon propagates uncertainty in timing calibration into the latest input time that could have influenced a stored output record. For one channel, the horizon in input-time horizon of a delayed optical receiver record is $H=T-d$, with last output time $T$ and effective input-to-output timing offset $d$. HenceThis linear variance identity retains shared timing influences.^[ref]

A common calibration offset can cancel when it enters both quantities. In an original model, let errors be $\delta T=c+e_T$ and $\delta d=c+e_d$, with independent components of standard deviations $0.8$, $0.2$ and $0.6\mathrm{ns}$. Then $\delta H=e_T-e_d$ and $u(H)=\sqrt{0.2^2+0.6^2}\approx 0.6325\mathrm{ns}$. Discarding their common-offset covariance would instead give $\sqrt{0.8^2+0.2^2+0.8^2+0.6^2}\approx 1.296\mathrm{ns}$.

The shared component must come from the actual calibration model. Here $d$ is the effective timing offset relating the input and output coordinates, so an output-clock alignment error can enter both $T$ and $d$. A physical transport-delay error is not automatically correlated with the last-sample timestamp.

For hard timing limits, propagate validated bounds rather than reinterpret standard uncertainty as a guarantee. If $T=150\mathrm{ns}$ is known and $8\le d\le 12\mathrm{ns}$, then $138\le H\le 142\mathrm{ns}$. Claiming the optimistic endpoint as though it held for every allowed delay would overstate the input-time reach.

With several channels, $H={\max }_j(T_j-d_j)$. Channel switching under uncertainty in a combined receiver horizon makes this a nonsmooth function when the maximizing channel is uncertain. Propagate the justified joint timing set or distribution; a derivative of only one nominally best channel can miss such switching.

NIST's primary pulse calibration measured the detector, cable and oscilloscope together using a short known optical input.^[ref] The numerical timing errors above are chosen examples, not values from that apparatus. A timing horizon remains a necessary information limit; it does not guarantee invertibility of the receiver or bound future optical illumination.

Channel switching under uncertainty in a combined receiver horizon occurs when uncertainty changes which receiver has the latest accessible input time. For channel $j$, let $H_j=T_j-d_j$ be its last output time minus its effective input-to-output offset. The combined horizon is $H={\max }_j{H}_j$, as in receiver-latency uncertainty in the inferred input-time horizon. Away from ties a small perturbation may preserve the maximizing channel; near a tie it need not. Propagating only that channel's variance omits the switching operation.

An original two-channel model makes the effect exact. Suppose $H_1$ and $H_2$ are independent normal random variables with common mean $\mu$ and standard deviation $s>0$. Independence giveswhere $\Phi$ is the standard normal distribution cumulative distribution function. This square is already enough to show that $H$ is not normal: $\mathrm{Pr}(H\le \mu )=1/4$.

To calculate the resulting bias relative to the common nominal horizon, write $H=(H_1+H_2+|H_1-H_2|)/2$. The difference has normal distribution with variance $2s^2$, hence $E|H_1-H_2|=2s/\sqrt{\pi }$. The sum and difference are independent in this model, givingFor $\mu =100\mathrm{ns}$ and $s=1\mathrm{ns}$, the combined mean is about $100.564\mathrm{ns}$ and its standard deviation is about $0.826\mathrm{ns}$. The formulas concern the uncertainty distribution of the horizon, not a promise of extra observable future illumination. The normal cdf/ppf routines provide direct evaluation of the probabilities.^[ref]

Shared clock or calibration errors invalidate the independence product. The distribution of the maximum of correlated receiver horizons must retain the joint timing model; NIST's general uncertainty guidance retains input covariance and requires distributional justification for coverage.^[ref] For hard bounds instead, $H_1\in [99,101]\mathrm{ns}$ and $H_2\in [100,102]\mathrm{ns}$ imply $H\in [100,102]\mathrm{ns}$ without selecting a fixed winning channel. Neither calculation establishes invertibility of the physical receiver.

The distribution of the maximum of correlated receiver horizons must use the joint timing law. Let $H=\max (H_1,H_2)$. Then $\mathrm{Pr}(H\le h)=\mathrm{Pr}(H_1\le h,H_2\le h)$, which generally is not the product of the marginal probabilities. The independent special case is derived in channel switching under uncertainty in a combined receiver horizon.

For an explicit correlated model, suppose $(H_1,H_2)$ has a bivariate normal distribution with common mean $\mu$, common standard deviation $s>0$, and correlation coefficient $-1<\rho <1$. With $z=(h-\mu )/s$, the cumulative distribution function and density of the maximum areHere $\varphi ,\Phi$ are the standard normal distribution density and cdf, and ${\Phi }_2$ is the bivariate cdf. The density follows by differentiating along both boundaries and using the conditional normal law. The official SciPy reference supplies a cdf method parameterized by the mean vector and covariance matrix; it is a numerical evaluation, not an assumption of channel independence.^[ref]

The equal-variance model also gives exact moments. Write $H=(S+|D|)/2$, with $S=H_1+H_2$ and $D=H_1-H_2$. They are independent normal variables, with $\mathrm{Var}(D)=2s^2(1-\rho )$. The absolute-normal moment then givesFor an original shared-clock example, let $\mu =100\mathrm{ns}$, $s=1\mathrm{ns}$, and $\rho =1/2$. Then $EH\approx 100.399\mathrm{ns}$ and the standard deviation is about $0.917\mathrm{ns}$. Positive correlation reduces the expected gain from selecting the larger horizon, while increasing its variance relative to the independent equal-channel case.

A common-clock contribution to a maximum receiver horizon can be modeled for nonnegative $\rho$ by $H_j=\mu +s(\sqrt{\rho }{Z}_0+\sqrt{1-\rho }{Z}_j)$ with independent standard normal inputs. As $\rho$ tends to one, both channels share the same uncertain horizon: selection does not remove the common error. Actual receiver timing correlations must be justified by calibration and reference reuse; the example neither measures them nor guarantees that the receiver is invertible.