Bounding unobserved optical energy with an exponential tail model

A finite optical record can bound later radiant energy if an independently justified exponential decay envelope applies at every later time. Let the recorded window end at $b$, and suppose the true nonnegative radiant power satisfiesIntegration gives the exponential-envelope tail-energy boundEquality holds when the entire tail follows the envelope. This is a bound on an unobserved contribution, not automatically a standard uncertainty or a probabilistic interval.

For an original numerical example, suppose the energy inside the recorded window is known to lie in $[48,52]\mathrm{nJ}$, while a justified envelope has $A=0.20\mathrm{W}$ and $\tau =20\mathrm{ns}$. The missing contribution is at most $4\mathrm{nJ}$, so the total laser pulse energy lies in $[48,56]\mathrm{nJ}$. The optical energy outside a time gate fraction, considering only this late contribution, is at most $4/(48+4)\approx 7.69\%$. Any unobserved early energy requires its own bound.

One way to justify the envelope mathematically is a differential inequality for an exponential tail envelope. Suppose $P$ is absolutely continuous on each finite interval after $b$, $P(b)\le A$, and $P^{\prime }(t)\le -P(t)/\tau$ almost everywhere. Multiplying by the integrating factor $e^{(t-b)/\tau }$ givesso $e^{(t-b)/\tau }P(t)\le P(b)\le A$. The decay law must apply throughout the unobserved region; it is a physical assumption to validate, rather than a consequence of fitting the last few samples.

The distinction is testable by a counterexample. Append a later nonnegative pulse whose support lies entirely beyond the recorded interval. Its energy can be arbitrarily large while every recorded sample is unchanged. Such a pulse violates the global envelope but cannot be excluded by a good local exponential fit alone. Similarly, an envelope $A{e}^{-(t-b)/\tau }+P_0$ with a positive constant background $P_0$ has an infinite integral over an infinite time interval. A finite observation horizon or an adequately defined and validated background subtraction is then essential.

Uncertain envelope parameters require a bound valid for their allowed combinations. If the allowed set is the rectangle $0\le A\le A_{\max }$ and $0<\tau \le {\tau }_{\max }$, then $U_{\mathrm{tail}}\le A_{\max }{\tau }_{\max }$. For $A_{\max }=0.25\mathrm{W}$ and ${\tau }_{\max }=25\mathrm{ns}$, this is $6.25\mathrm{nJ}$ and the preceding total interval becomes $[48,58.25]\mathrm{nJ}$. More restrictive joint parameter constraints can tighten this result; inserting separate best-fit values does not establish a guaranteed bound.

Finally, an electrical tail need not be an optical tail. In an original idealized linear time-invariant system, let a photodetector have unit-area impulse response $h(t)={\tau }_d^{-1}{e}^{-t/{\tau }_d}$ for $t\ge 0$, and zero otherwise. An impulsive optical input $U\delta (t)$, with $\delta$ the Dirac delta function, gives the calibrated output $s(t)=Uh(t)$ by convolution. Its electrical tail after $L$ has area $U{e}^{-L/{\tau }_d}$ even though all incident optical energy arrived at time zero. This optical tail and detector-response ambiguity requires a separate response measurement. NIST's historical procedure characterized the photodetector, cable and oscilloscope together before convolving measured responses with trial waveforms. Appendix C describes typical optical excitation of about $120\mathrm{ps}$, a photodetector rise time near $5\mathrm{ns}$, and a $1\mathrm{GHz}$ oscilloscope with $0.4\mathrm{ns}$ rise time. The optical excitation was much shorter than the detector response, while the oscilloscope was fast enough that the detector dominated the measured waveform. Response data were collected at each gain setting and several signal levels.^[ref]

Optical tail and detector-response ambiguity arises because a measured temporal trace combines the optical waveform with the receiver's dynamics. In a calibrated linear time-invariant system,where $P$ is incident radiant power and $h$ the unit-area detector impulse response. One trace does not generally identify both unknown functions. The parent bounding unobserved optical energy with an exponential tail model gives an impulsive-input example; two finite-width profiles provide another exact ambiguity.

Define $h_{\tau }(t)={\tau }^{-1}{e}^{-t/\tau }$ for $t\ge 0$, and zero otherwise. Consider model A with optical power $P=U{h}_{2\mathrm{ns}}$ and detector response $h_{10\mathrm{ns}}$, and model B with these two time constants exchanged. By commutativity of convolution, both give exactlyTheir optical energies after $20\mathrm{ns}$ are nevertheless $U{e}^{-10}$ and $U{e}^{-2}$. For original chosen $U=100\mathrm{nJ}$, those are about $0.00454$ and $13.53\mathrm{nJ}$. Identical recorded tails therefore coexist with very different true late optical energy.

Known-input calibration of a pulse detector supplies information missing from this inverse problem. NIST measured the detector/cable/oscilloscope response with optical excitation much shorter than the response before calculating waveform distortions.^[ref] The example above is an independent mathematical construction. A fitted electrical decay constant alone cannot establish a global optical-tail envelope.

Known-input calibration of a pulse detector uses a characterized optical waveform to separate the incident signal from the receiver response. In a linear, baseline-corrected model,where $P_{\mathrm{ref}}$ is known input power, $h$ a normalized impulse response and $g$ the overall responsivity. Recording the complete response givesThus a known reference energy calibrates the area gain without requiring the input to be an ideal impulse. Shape calibration additionally uses the reference waveform.

For nonnegative normalized profiles with finite second moments, define $p=P_{\mathrm{ref}}/U_{\mathrm{ref}}$ and $q=y/\int y$. Since $q=p\ast h$, their temporal variances satisfy ${\sigma }_q^2={\sigma }_p^2+{\sigma }_h^2$. In an original chosen example, an input standard deviation $0.60\mathrm{ns}$ and recorded deviation $1.00\mathrm{ns}$ imply ${\sigma }_h=0.80\mathrm{ns}$. Treating that finite input as a delta impulse would incorrectly assign the full $1.00\mathrm{ns}$ to the detector. This finite-input correction in pulse-detector calibration concerns standard deviations; a universal subtraction rule for arbitrary FWHM values does not follow. Negative ringing also invalidates the nonnegative-profile interpretation used for the variance identity.

NIST's primary procedure illuminated the detector with a pulse much shorter than the detector response and characterized detector, cable and oscilloscope together at the relevant gain and signal levels.^[ref] Such a known input breaks the optical tail and detector-response ambiguity described by the parent article.

In the frequency domain, the same model is $Y=g{P}_{\mathrm{ref}}H$. Division can estimate $H$ where the input spectrum is known and sufficiently nonzero; small input components amplify measurement noise. A short reference pulse is useful because it excites a broad frequency range, but detector linearity, baseline, area coverage and reference uncertainty still need checking.

Finite-input correction in pulse-detector calibration removes the known input duration from an inferred receiver width, while propagating uncertainty in both widths. Under the nonnegative, normalized, finite-variance convolution model of known-input calibration of a pulse detector,where $p$ is the known input profile, $q$ the measured output and $h$ the receiver's impulse response. The subtraction concerns temporal variances, not arbitrary reported full widths.

Differentiating on the positive-width branch givesFirst-order propagation of standard uncertainty therefore yieldsThe covariance matters when both widths use the same timebase or fitted reference.^[ref]

An original near-resolution example has ${\sigma }_p=0.95\mathrm{ns}$ and ${\sigma }_q=1.00\mathrm{ns}$, each with independent standard uncertainty $0.01\mathrm{ns}$. The corrected receiver width is $0.31225\mathrm{ns}$ and its propagated standard uncertainty is about $0.04417\mathrm{ns}$, or $14.1\%$. Width measurements individually precise to about one percent can thus give a much less precise difference-derived width.

NIST's primary bandwidth-correction procedure used roughly 120 ps optical inputs and recorded the detector, cable and oscilloscope as one response chain at different gains and signal levels.^[ref] A sufficiently short characterized input reduces this resolution-limit conditioning in finite-input detector calibration. Near a zero variance difference, linear uncertainty propagation may fail; a nonpositive measured difference does not represent a negative physical detector width. Report an interval or limit under the stated model rather than silently replacing the difference by zero. Negative ringing also violates the nonnegative-profile assumption behind the variance subtraction.

Resolution-limit conditioning in finite-input detector calibration is the sensitivity of a small inferred detector width to uncertainty in two nearly equal measured widths. In finite-input correction in pulse-detector calibration, the nonnegative normalized convolution model givesThe signed relative sensitivities areThey diverge as the measured output width ${\sigma }_q$ approaches the known input width ${\sigma }_p$. Subtracting two well-measured squared widths can therefore be a poorly conditioned inference.

For an original example, ${\sigma }_q=1.00\mathrm{ns}$ and ${\sigma }_p=0.99\mathrm{ns}$ give $D=0.0199\mathrm{n}{\mathrm{s}}^2$ and ${\sigma }_h\approx 0.1411\mathrm{ns}$. The two relative sensitivity coefficients are about $50.25$ and $-49.25$. A small independent fractional width error can produce a much larger fractional response-width error; a common multiplicative timebase error instead partly cancels because the coefficients sum to one. Correlations must therefore be retained.^[ref]

Near the boundary, report uncertainty in the variance difference before transforming it. For a chosen interval $D\in [-0.010,0.050]\mathrm{n}{\mathrm{s}}^2$, imposing the physical constraint $D\ge 0$ gives a detector-width interval $[0,\sqrt{0.050}]\mathrm{ns}$, approximately $[0,0.2236]\mathrm{ns}$. A coverage-preserving detector-width bound from a variance interval uses this monotone square-root transformation on the physical intersection. It need not resemble a symmetric error bar around a point estimate. An interval wholly below zero contains no admissible width under this model; replacing it silently with a zero-width result hides the inconsistency.

NIST characterized its detector/cable/oscilloscope chain with an optical pulse much shorter than the receiver response.^[ref] That design avoids subtracting two nearly equal broad widths. The variance identity still requires nonnegative profiles and finite second moments; arbitrary full widths at half maximum and responses with negative ringing need a different model.

A coverage-preserving detector-width bound from a variance interval transforms uncertainty in a nonnegative response variance without imposing a symmetric width error bar. Let $D={\sigma }_h^2\ge 0$ and let a random interval $I(X)$ have stated frequentist coverage for $D$. Define the width confidence set byFor the true width ${\sigma }_h\ge 0$,The events are identical, so their coverage probabilities are identical. The transformation does not create coverage that the original interval lacked. Resolution-limit conditioning in finite-input detector calibration explains why working with the variance difference first is useful near a barely resolved detector response.

For an original explicit measurement model, suppose $\hat{D}$ has a normal distribution with mean $D$ and known standard deviation $s=0.025\mathrm{n}{\mathrm{s}}^2$. A reading $\hat{D}=0.040\mathrm{n}{\mathrm{s}}^2$ gives an approximately 95 percent interval $\hat{D}\pm 1.96s=[-0.009,0.089]\mathrm{n}{\mathrm{s}}^2$. Intersecting with the physical range and taking square roots gives $J=[0,0.2983]\mathrm{ns}$. This is an assumed normal model for measuring $D$, not a claim that any difference of fitted squared widths is automatically normal.

If the entire variance interval is negative, the result is an empty physical confidence set for a detector variance difference. Do not replace its negative upper endpoint by zero and report an apparently exact zero width: that changes the confidence-set construction. An empty realization signals disagreement between that interval and the physical parameter range; it can occur even in a valid procedure on an occasional sample that misses the true value.

NIST distinguishes a standard uncertainty from an interval with a justified confidence level and explains the distributional assumptions behind a coverage factor near two.^[ref] The normal interval above needs its stated model; the monotone transformation proof itself does not. Application to a detector additionally requires the nonnegative-profile convolution assumptions of finite-input correction in pulse-detector calibration; a correct statistical transformation cannot repair an invalid physical variance identity.

An empty physical confidence set for a detector variance difference occurs when the entire reported interval for the inferred response variance lies below its admissible range. In the nonnegative-profile detector model, $D={\sigma }_q^2-{\sigma }_p^2={\sigma }_h^2\ge 0$. An interval $I=[\ell ,u]$ gives a physical width setIt is empty exactly when $u<0$. The coverage-preserving detector-width bound from a variance interval proves why this intersection and transformation retain the original coverage event.

For an original reading, suppose an explicitly normal measurement model for $D$ gives $\hat{D}=-0.030\mathrm{n}{\mathrm{s}}^2$ with known standard deviation $s=0.010\mathrm{n}{\mathrm{s}}^2$. The approximately 95 percent interval $\hat{D}\pm 1.96s=[-0.0496,-0.0104]\mathrm{n}{\mathrm{s}}^2$ contains no nonnegative variance. This is disagreement between that interval and the physical model, not a negative physical detector width or an exact measurement of zero width.

Empty sets can occur occasionally even when the model is correct. If $\hat{D}\sim N(D,s^2)$ and $I=[\hat{D}-zs,\hat{D}+zs]$, thenwhere $\Phi$ is the standard normal cumulative distribution function. At the physical boundary $D=0$ and $z=1.96$, the probability is about $2.5\%$; at $D=s$ it is about $0.154\%$. The boundary probability of an empty detector-width confidence set explains why one empty realization is not a logical disproof of the detector model.

Replacing an empty set by the point zero changes the procedure. It must not be presented as though the original interval precisely determined that point. NIST likewise distinguishes an uncertainty interval from an automatically justified confidence level, emphasizing the distributional assumptions behind coverage factors.^[ref]

In practice, examine the width estimates, their correlations and the physical convolution assumptions, while retaining the actual negative interval in the measurement record. NIST's short-input receiver-response measurement illustrates how an independently characterized input helps resolve detector broadening.^[ref] Neither clipping the estimate nor declaring the response model false follows solely from the sign of one noisy variance difference.

The boundary probability of an empty detector-width confidence set is the chance that a valid uncertainty procedure yields no admissible nonnegative response width when the true variance difference is at its boundary. Under the explicit model $\hat{D}\sim N(D,s^2)$, $D\ge 0$, take the central interval $I=[\hat{D}-zs,\hat{D}+zs]$. The empty physical confidence set for a detector variance difference occurs when its upper endpoint is negative:This probability decreases with $D$, so the largest value over the physical parameter range occurs at $D=0$. If $z$ is the usual two-sided $1-\alpha$ normal critical value, the boundary probability is $\alpha /2$.

At that boundary the other way to miss zero is a strictly positive lower endpoint. The two tail events are disjoint and each has probability $\alpha /2$; the remaining probability $1-\alpha$ produces a confidence set containing zero. The joint probability of empty and strictly positive detector-width intervals therefore accounts for the full two-sided noncoverage, not just the empty cases.

For an original repeated-measurement calculation, use $z=1.96$ and 400 independent measurements of a stipulated zero response-variance difference. The expected number of empty physical sets is approximately ten, with standard deviation $\sqrt{400p(1-p)}\approx 3.12$, where $p=\Phi (-1.96)\approx 0.025$. The count follows a binomial distribution only under the stated independent, common-probability model. Roughly ten additional intervals are expected to lie wholly above zero. Their existence is compatible with approximately 95 percent coverage at the boundary.

For an interior example $D=s$, the empty-set probability falls to $\Phi (-2.96)\approx 0.00154$. Empty results are thus most common near the resolution boundary, where subtracting similar measured variances is already delicate.

NIST explains that interpreting expanded uncertainty as a confidence interval requires distributional assumptions.^[ref] These probabilities assume an actual normal model with known $s$; they are not inferred merely from an error bar labelled standard uncertainty. Correlation, systematic offsets or a misspecified detector response can change the repeated-measurement behavior. Retaining empty results in the record allows that behavior to be assessed rather than hiding it by replacing them with zero-width estimates.

The joint probability of empty and strictly positive detector-width intervals must distinguish outcomes of one measurement from counts across repeated measurements. For one normal variance-difference estimate $\hat{D}\sim N(D,s^2)$, let $I=[\hat{D}-zs,\hat{D}+zs]$. An empty physical width set requires $\hat{D}+zs<0$; a strictly positive interval requires $\hat{D}-zs>0$. Those events are disjoint for that same interval, so their intersection has probability zero.

Writing $d=D/s$ and $\Phi$ for the standard normal cumulative distribution function, the three category probabilities areThe last category is an interval containing zero. At $D=0$ and a central $1-\alpha$ interval, these become $(\alpha /2,\alpha /2,1-\alpha )$, as in boundary probability of an empty detector-width confidence set. For interior $D>0$, the categories describe the relation to zero; they are not themselves the coverage categories for the true positive value.

For $N$ independent repetitions with the same probabilities, the counts $(N_E,N_P,N_Z)$ have a multinomial distribution. For nonnegative integers $e,p,k$ summing to $N$,Although the measurement repetitions are independent, their category counts are not. An empty and a positive indicator cannot both equal one in the same repetition, givingThis negative covariance of empty and positive detector-interval counts follows from the fixed total count.

For an original example, take 20 independent boundary measurements with exact central 95 percent normal intervals. Then $p_E=p_P=0.025$. Each count has mean $0.5$, and their covariance is $-0.0125$. The probability of exactly one empty and one strictly positive interval is $20\times 19\times 0.025^2\times 0.95^{18}$, about $0.094$. It is not zero, unlike the intersection of those two outcomes for a single interval.

NIST's uncertainty guidance emphasizes the assumptions needed to assign confidence levels.^[ref] The joint-count calculation above additionally assumes independent repetitions with a common normal model; shared calibration errors can invalidate that independence even when each marginal interval is computed in the same way.

The negative covariance of empty and positive detector-interval counts arises because a fixed number of measurement intervals is divided into mutually exclusive categories. Let $E_i,P_i$ indicate that interval $i$ is respectively physically empty or strictly positive. The same interval cannot satisfy both, so $E_i{P}_i=0$ andFor $N$ independent repetitions with common probabilities, the counts $N_E={\sum }_i{E}_i$ and $N_P={\sum }_i{P}_i$ therefore haveThe joint probability of empty and strictly positive detector-width intervals gives the underlying multinomial model. The sign describes category-count dependence, not physical antagonism between two detectors.

For an original boundary example, take $N=100$ and $p_E=p_P=0.025$. Each count has variance $100(0.025)(0.975)=2.4375$, while their covariance is $-0.0625$. The total tail-category count has variance $2.4375+2.4375-0.125=4.75$. Treating the two counts as independent would give $4.875$ instead. The covariance is needed even though the individual measurement repetitions are independent.

Conditioning on the total is important. If the number of repetitions is random and independent of the category assignments, the law of total covariance givesAn independent total with a Poisson distribution has equal mean and variance, making this covariance zero; a more variable total can make it positive. This is a different sampling protocol, not a contradiction of the fixed-total formula.

Shared calibration drift is another change of model. Repetitions may then have dependent category indicators, and cross-repetition covariance terms must be added. Overdispersion of detector-interval counts under shared drift can make a simple multinomial analysis inadequate.

NIST's measurement guidance retains common influences in uncertainty propagation and requires assumptions for confidence-level interpretation.^[ref] State whether the repetition total is fixed, how the categories are defined and which calibration influences are shared before interpreting their empirical count covariance.

Overdispersion of detector-interval counts under shared drift is excess count variability caused by a common experimental condition changing the chance of an interval outcome. Let $M$ count empty physical detector-width intervals among $N$ readings. Conditional on a block's calibration state $H$, suppose the readings are independent with empty probability $p(H)$, so $M\mid H\sim \mathrm{Binomial}(N,p(H))$.

Writing $\overline{p}=\mathbb{E}[p(H)]$, the law of total variance givesThe first term is the binomial variance using the mean probability. The second comes from sharing one fluctuating condition across the block. Individual indicators have covariance $\mathrm{Var}(p(H))$ across different readings, even though they are independent after conditioning on $H$.

For an original mixture example, suppose equally likely block conditions have empty probabilities $0.01$ and $0.09$. Then $\overline{p}=0.05$ and $\mathrm{Var}(p(H))=0.0016$. With $N=100$, the expected count is five, but its variance is $4.75+15.84=20.59$, with standard deviation about $4.54$. An independent binomial model with probability $0.05$ would give standard deviation only about $2.18$.

The effective repetition count for detector intervals with shared drift can consequently be much smaller than the number of stored traces. Repeating measurements within one unchanged calibration block does not supply the same information about drift as acquiring independent blocks. Record acquisition grouping and reference reuse before pooling interval categories.

The negative covariance of empty and positive detector-interval counts applies to a fixed-probability multinomial model. Shared drift adds cross-repetition dependence and can change that count covariance as well as each count's variance. In the example, the two probability states are stipulated; a physical model would relate them to the receiver's drift, variance-difference estimator and uncertainty calibration.

NIST retains shared influences in uncertainty propagation and explains that confidence interpretations depend on the assumed distributions.^[ref] Observed overdispersion can motivate checking the block model, but does not by itself identify which physical influence caused it.

The effective repetition count for detector intervals with shared drift is the number of independent repetitions that would give the same variance for an average indicator as the actual dependent repetitions. In overdispersion of detector-interval counts under shared drift, let $X_i$ indicate an empty interval, with common probability $\overline{p}$ satisfying $0<\overline{p}<1$. If the covariance between distinct indicators is $v=\mathrm{Var}(p(H))$, their correlation coefficient isThe parent's law of total variance calculation therefore givesThis is an observable-specific variance equivalence, not a literal count of independent traces or a guarantee of confidence interval coverage. A different statistic or a different dependence pattern may require a different equivalence.

For an original feasible mixture, take $\overline{p}=0.1$ and $\rho =0.1$. These moments can arise when the block probability is $0$ with probability $9/19$ and $0.19$ with probability $10/19$: its variance is $0.009$. With $N=50$, the variance-inflation multiplier is $5.9$ andDoubling the repetitions to $100$ only raises $N_{\mathrm{eff}}$ to about $9.17$. For a fixed positive $\rho$, its large-$N$ limit is $1/\rho$, here $10$.

The experiment must therefore record which traces share a reference or drift state. Acquiring genuinely independent calibration blocks addresses a different limitation from oversampling one block; effective repetition count across independent calibration blocks depends on that grouping. A cluster bootstrap can preserve the observed within-block dependence when blocks are suitable independent resampling units. NIST's uncertainty law explicitly retains shared-input covariance, supporting the need to model reference reuse rather than discard it.^[ref]

Effective repetition count across independent calibration blocks accounts for dependence confined to each acquisition block rather than shared by every stored trace. Suppose block $j$ contains $n_j$ empty-interval indicator random variables, all with common probability $0<p<1$ and within-block correlation coefficient $\rho \ge 0$. Assume different blocks are independent. With $N={\sum }_j{n}_j$, summing within-block covariances givesMatching this to the variance of an average of independent indicators yieldsFor $K$ equal blocks of size $m$, this reduces to $N_{\mathrm{eff}}=Km/[1+(m-1)\rho ]$. The denominator involves the block size, not the total $Km$; the single-block expression in effective repetition count for detector intervals with shared drift must not be applied as though the same drift state affected every block.

For an original comparison, fix $N=120$ and $\rho =0.05$. One block gives $N_{\mathrm{eff}}=120/6.95\approx 17.27$. Six independent blocks of twenty give $N_{\mathrm{eff}}=120/1.95\approx 61.54$. Six very uneven blocks of sizes $(100,4,4,4,4,4)$ instead giveAll three designs store the same number of traces. The difference comes from how often the same fluctuating condition is reused. For fixed $N$, $K$, and positive $\rho$, minimizing ${\sum }_j{n}_j^2$ favors balanced blocks; this motivates optimal allocation of detector repetitions across calibration blocks when block-preparation costs also matter.

Independence must come from the physical acquisition model. Starting another file does not erase a shared clock, reference calibration, or slowly varying environmental influence. NIST's uncertainty law retains shared-input covariance explicitly.^[ref] If drift persists between nominal blocks, those cross-block terms must be added and the displayed formula changes. A cluster bootstrap likewise requires blocks that are suitable independent resampling units.

Optimal allocation of detector repetitions across calibration blocks chooses block sizes using the dependence and acquisition-cost model, not just the desired number of stored traces. For fixed total $N$, number of independent blocks $K$, common success probability, and positive within-block correlation $\rho$, effective repetition count across independent calibration blocks givesMaximizing it is equivalent to minimizing ${\sum }_j{n}_j^2$. If two integer block sizes differ by at least two, transferring one repetition from the larger to the smaller decreases that sum by $2(n_i-n_j-1)>0$. Therefore the optimal integer sizes differ by at most one.

For an original fixed-count example, $N=23$ and $K=5$ give balanced sizes $(5,5,5,4,4)$. At $\rho =0.1$, their effective count is $529/31.4\approx 16.85$. Sizes $(11,3,3,3,3)$ give only $529/36.4\approx 14.53$ under the same assumptions.

More independent blocks can help, but preparing one may consume time. In a continuous planning model let each block cost setup time $h>0$, each repetition cost $\tau >0$, and each block contain $n$ repetitions. With total time $T$, the effective count becomesFor $0<\rho <1$, differentiation gives the interior choice $n_{\ast }=\sqrt{h(1-\rho )/(\tau \rho )}$. Integer block counts, minimum sizes, and hardware restrictions must subsequently be imposed.

For chosen $h=9\mathrm{s}$, $\tau =1\mathrm{s}$, $\rho =0.1$ and $T=180\mathrm{s}$, the continuous optimum $n_{\ast }=9$ is feasible with ten blocks and gives $N_{\mathrm{eff}}=50$. Single-repetition blocks give 18, while two blocks of 81 repetitions also give 18. These are model-based comparisons, not measured apparatus timings.

Unequal block noise or different drift correlations require unequal-noise allocation across calibration blocks. Independence between prepared blocks must also be justified by the reference and environmental model; NIST explicitly retains shared-input covariance.^[ref]

Unequal-noise allocation across calibration blocks distributes repetitions using block-specific variability while keeping the intended estimator fixed. Suppose independent blocks supply means ${\overline{Y}}_j$ and the target estimate is $\hat{\theta }={\sum }_j{w}_j{\overline{Y}}_j$ with specified weights $w_j$. Let one observation in block $j$ have variance ${\sigma }_j^2$ and equicorrelation $0\le {\rho }_j<1$ within that block. ThenThe first sum is a block-specific floor; only the second decreases with repetition count. Changing the estimator weights merely to improve this calculation could change the estimand, so they remain fixed here.

For continuous positive allocations with positive costs $c_j>0$, budget ${\sum }_j{c}_j{n}_j=C$ and $a_j=w_j^2{\sigma }_j^2(1-{\rho }_j)>0$, a Lagrange-multiplier calculation givesFor equal observation costs, repetitions are proportional to $|w_j|{\sigma }_j\sqrt{1-{\rho }_j}$, not to the variance itself. Minimum counts and integer rounding must subsequently respect the actual acquisition constraints.

For an original two-block example, take $w_1=w_2=1/2$, equal costs, $C=100$, independent readings within each block, and standard deviations $1$ and $3$ in the output unit. The optimal counts are $25$ and $75$, giving variance $0.04$ instead of $0.05$ from fifty each. If both within-block correlations are instead $0.1$, the same proportions give variance $0.286$, while equal counts give $0.295$. The common-error floor $0.25$ now dominates and cannot be eliminated by taking more observations inside those same blocks.

This extends optimal allocation of detector repetitions across calibration blocks, whose balanced solution used homogeneous noise. For binary detector indicators, the per-reading variance is $p_j(1-p_j)$; differing probabilities require a clearly specified weighted target. The planning variances and correlations must come from an appropriate measurement calibration model, not merely from counting files as independent blocks. NIST's uncertainty propagation retains shared-input covariance for exactly this reason.^[ref]

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]