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.

One-sided derivatives at an optimal Raman frame-count switch describe the change in the best acquisition variance when several integer designs tie. Let $V_{\ast }(W)={\min }_n{V}_n(W)$ as in frame-count switching in Raman acquisition-budget sensitivity, and let $A$ be the active designs with $V_n(W_{\ast })=V_{\ast }(W_{\ast })$. Suppose the relevant branches are differentiable near the tie and nonactive designs have a positive value gap.

Writing their slopes as $s_n=V_n^{\prime }(W_{\ast })$ givesIndeed, $V_n(W_{\ast }+h)=V_{\ast }(W_{\ast })+s_{n}h+o(h)$. Positive $h$ selects the smallest slope; negative $h$ selects the largest because multiplication reverses their ordering. A single derivative exists only when the selected left and right slopes agree. If an active branch itself has a kink, use its appropriate one-sided slope on each side.

For an original balanced-noise example, take $\alpha =\beta =a=10^{-6}\mathrm{s}$, exposure cap $c=0.080\mathrm{s}$ and per-frame readout $d=0.015\mathrm{s}$. Near the three-pair/four-pair switch,The branches tie at $W_{\ast }=6c+8d=0.60\mathrm{s}$, with variance about $8.3333\times 10^{-6}$. The three-pair branch has slope zero; the four-pair branch has slope $-4a/(6c{)}^2\approx -17.3611\times 10^{-6}{\mathrm{s}}^{-1}$. Thus these are the left and right optimum slopes. Smaller pair counts have less capacity and larger counts have more readout near this tie, so they do not change the local optimum.

Joint time-and-readout perturbations at a Raman design tie require following the direction in both resource coordinates; the scalar slopes above hold when only $W$ changes. They also quantify a local sensitivity, not a universal prescription to change exposure count whenever a small budget fluctuation occurs.

The primary stimulated-Raman apparatus limited exposures for saturation and switched pump state during shuttered readout.^[ref] The rounded timings here are a separate mathematical example. Read noise, drift or switching costs can change the competing branches and their derivatives.

Felidae is the taxonomic family of mammals that includes the domestic cat and its wild relatives, within the order Carnivora. It is broader than the taxonomic genus Felis: the lion and tiger belong to Panthera, another taxonomic genus within Felidae.^[ref]

The distinction is easy to see by tracing two branches. A domestic cat belongs to Felis and then Felidae; a lion belongs to Panthera and then Felidae. Sharing a taxonomic family does not make them the same species or give them identical ecology. Conversely, being a stray cat or feral cat does not move a domestic cat outside Felidae: those categories concern its relationship with humans.

Classification is a testable account of relationships. The IUCN Cat Specialist Group's 2017 revision combined evidence from morphology, molecular studies and geographical variation, and recorded uncertainty where evidence or expert judgments differed. Such work explains why names below the family level can change as evidence improves, without making the distinction between Felidae and Felis arbitrary.^[ref]

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.

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.

Scalar rescaling invariance of a tie-line weight matrix means that multiplying every fit weight by the same positive number leaves the fitted phase fraction unchanged. Let endpoint compositions $\mathbf{a},\mathbf{b}$ and overall composition $\mathbf{y}$ be expressed in independent mole fraction coordinates. With $\mathbf{d}=\mathbf{b}-\mathbf{a}$, a fixed symmetric positive-definite matrix $W$ defines the tie-line phase-fraction fit by minimizingFor $c>0$, $Q_{cW}(\lambda )=c{Q}_W(\lambda )$. Multiplication by a positive constant preserves every comparison between objective values, so the sets of minimizers agree. The proof also applies when the same constraint $0\le \lambda \le 1$ is imposed in both fits.

For distinct endpoints, the unconstrained formula makes the cancellation visible:The differential in sensitivity of a tie-line fit to its weight matrix agrees: for a perturbation $\delta W=\epsilon W$, its numerator is $\epsilon {\mathbf{d}}^{T}W\mathbf{r}=0$ by the fit's normal equation. Thus the scalar direction is insensitive even when the residual is nonzero.

In an original example, take $\mathbf{a}=(0.65,0.10)$, $\mathbf{b}=(0.25,0.50)$, $\mathbf{y}=(0.42,0.36)$ and $W=\mathrm{diag}(1,4)$. The remaining composition coordinate is fixed by closure. These values giveReplacing $W$ by $7W$ gives the same fraction, although the minimum objective increases from0.00072to0.00504. Changing only the second diagonal weight from4to7 instead changes the fitted fraction to0.640625. Relative weights determine the geometry; their common multiplier does not.

Point-estimate invariance does not by itself establish invariance of physical measurement uncertainty. Merely rewriting the same objective with $cW$ leaves propagation through the same measured-input covariance matrix unchanged. If instead $W={\Sigma }^{-1}$ represents absolute precision and $cW$ asserts a genuinely different covariance $\Sigma /c$, that is a different noise model. With exact endpoints, the linear-model variance $1/({\mathbf{d}}^{T}W\mathbf{d})$ then decreases by a factor $c$. The measurement model and covariance interpretation must be stated.^[ref]

Reference lineage in covariance models records which primitive observations and calibration quantities enter each derived measurement, with what coefficients. Two inputs can share the same reference yet respond to it differently. A shared name alone therefore does not establish either equal errors or exact cancellation.

For a linear measurement model, collect primitive inputs in $\mathbf{z}$, derived inputs in $\mathbf{x}=A\mathbf{z}$ and their covariance matrices in $U_z,U_x$. ThensoThe matrix records the data lineage as well as the algebra. This implements the GUM approach of expressing correlated quantities in terms of their common inputs.^[ref]

An original example uses independent inputs $T,B,C,{ϵ}_1,{ϵ}_2$ with variances $1,9,9,4,4\mathrm{m}{\mathrm{g}}^2$, respectively. Let two reference readings be $A_1=C+{ϵ}_1$ and $A_2=2C+{ϵ}_2$. Correct a specimen by $Y=T-B+A_1-A_2$. Substitution givesThe shared $C$ does not cancel, because its sensitivities are one and two. Equivalently, $\mathrm{Var}(A_1)=13$, $\mathrm{Var}(A_2)=40$ and $\mathrm{Cov}(A_1,A_2)=18$ in squared milligrams. Their difference has variance $13+40-2(18)=17$, giving the same total 27. Treating the reference readings as independent would incorrectly give 63 for the total variance.

If both readings instead contained $C$ with coefficient one, that shared contribution would cancel and the total would be 18 in the same units. These are different acquisition models, not two uncertainty conventions for the same data.

A usable lineage record therefore includes acquisition identifiers, calibration versions, reuse of numerical values and sensitivity coefficients. For nonlinear data reduction, a local sensitivity matrix replaces the linear $A$ when first-order propagation is justified. Covariance must follow the actual reuse and dependence structure; assigning independence because derived columns have different labels can discard the information needed for correct cancellation.

Finite ratio moments under a positive denominator cutoff follow from a simple pointwise bound: dividing by a denominator known to be at least $c>0$ cannot amplify an accepted numerator by more than $1/c$. The conclusion still needs a numerator-moment assumption and a nonzero acceptance probability. It applies to the pulse-contrast denominator acceptance rule, including dependent numerator and denominator noise.

Let $A=\{C\ge c\}$, $p_A=\mathbb{P}(A)>0$, and let $p>0$. If $\mathbb{E}|S{|}^p<\infty$, thenTo prove it, on $A$ one has $|S/C{|}^p\le |S{|}^p/c^p$. Integrate over $A$ and divide by $p_A$:Here $1_A$ is an indicator function. The accepted ratio is defined on $A$; any indicator-weighted expression is assigned zero outside $A$, so this proof does not divide by an unaccepted zero denominator. No independence assumption was used.

For $p=2$, the accepted ratio has finite second moment and hence finite variance. The Cauchy-Schwarz inequality also gives a finite first absolute moment:This makes the accepted expected value well-defined. Standard sampling conclusions, such as a law of large numbers for independent accepted ratios with the same conditional distribution, still require their own sampling assumptions.

For an original example matching the two-state model in selection bias from a noisy denominator threshold, take $S=0.02$, $C\in \{4,6\}$ with equal probabilities and $c=5$. Then $p_A=1/2$ and $\mathbb{E}{S}^2=0.0004$, so the displayed second-moment bound is0.000032. The actual accepted ratio is constantly1/300, whose second moment is $1/90000\approx 0.00001111$. The bound is conservative. Using the conditional numerator moment directly gives the sharper bound $0.0004/25=0.000016$.

A cutoff alone is insufficient if the numerator has no required moment. If $C=2$ always and $S$ has a standard Cauchy distribution, accepting $C\ge 1$ retains every shot, but $S/2$ still has infinite first absolute moment and infinite second moment. Conversely, if $|S|\le M$ on accepted shots, the stronger deterministic bound $|S/C|\le M/c$ gives all positive moments without the factor $1/p_A$.

The cutoff and probability must remain explicit. A positive cutoff for each analysis does not supply a uniform bound if cutoffs approach zero or accepted numerator distributions change without control. Nor does moment finiteness identify an unbiased measurement: the accepted conditional expectation can differ from the desired pulse-population quantity, as selection bias from a noisy denominator threshold demonstrates.

Selection bias from a noisy denominator threshold occurs when accepting measurements according to the same noisy central signal used for normalization changes the distribution of the reported ratios. For side signal $S$, measured central signal $C$, positive cutoff $c$ and acceptance event $A=\{C\ge c\}$, the sample mean over accepted shots targets $\mathbb{E}[S/C\mid A]$. It need not target the true-signal ratio or the unfiltered mean ratio. This is a consequence of conditioning, even when every physical pulse has the same true brightness.

For an original two-state readout model, let the true central signal be5, let measured $C$ equal4 or6 with equal probabilities, and let $S=0.02$ be constant. Both denominator errors have zero mean. The noiseless ratio is0.004, whereas the unfiltered mean measured ratio isA cutoff $c=5$ accepts only $C=6$, giving $\mathbb{E}[S/C\mid A]=1/300\approx 0.003333$ and acceptance probability1/2. This separates the unfiltered nonlinear ratio bias from the further change introduced by selection. Both denominators were already safely positive; the selection effect is not merely a division-by-zero problem.

The direction is not universal when numerator and denominator errors are related. Keep $C=4$ or6 but let $S=0.02+k(C-5)$. ThenFor $k=0.01$, the two observed side signals are0.01 and0.03, with ratios0.0025 and0.005. The same cutoff now gives0.005, above the same true-signal ratio0.004. For $k=0.004$, the observed signals are exactly proportional and every ratio is0.004, so selection leaves it unchanged. The noise dependence and measurement model determine whether thresholding shifts the accepted ratios upward, downward or not at all.

When a joint density and integrable accepted ratio exist, the dependence is explicit:The conditional numerator response cannot generally be replaced by its unconditional mean. Report the acceptance rule and accepted fraction alongside the shot-resolved pulse-contrast estimator; increasing the number of accepted shots reduces sampling scatter but does not undo the conditioning.

A physical motivation comes from a single-shot correlator that spatially encoded the nonlinear mixing signal and imaged it onto a filtered CCD detector.^[ref] If central pixels decide whether a trace is retained, later normalization uses data already conditioned by that choice. The two-state examples are original readout models, not measurements attributed to that instrument. Finite ratio moments under a positive denominator cutoff establish mathematical moment control under stated assumptions; they do not establish absence of selection bias.

A numerical feasibility residual for baseline constraints measures how closely computed pulse-estimator weights satisfy their requested responses. For $F^T\mathbf{a}=\mathbf{b}$, evaluateThis is not the same diagnostic as the projected unattainable response in null-space feasibility test for baseline-estimator constraints. That test asks whether a solution exists at the chosen matrix rank; the residual checks the weights actually returned.

Small numerical components need physical scaling. If the measurement model is $\mathbf{x}=\theta \mathbf{s}+{\sum }_j{\beta }_j{\mathbf{h}}_j+\mathbf{ϵ}$, an estimator intended to preserve the signal and reject the backgrounds instead has biasThus an unweighted Euclidean residual norm is not a universal error bound. Coefficient magnitudes and the units of each constraint matter.

For an original example, suppose the constant-background residual is $r_0=10^{-6}$ and the linear-time residual is $r_1=10^{-12}\mathrm{s}$. A background offset $1\mathrm{mV}$ contributes $1\mathrm{nV}$ through the first residual. A drift rate $1mV/\mu \mathrm{s}=10^{3}V/s$ also contributes $1\mathrm{nV}$ through the second. The much smaller raw numerical value is not a much smaller physical leakage. Expressing time in microseconds changes $r_1$ to $10^{-6}\mu \mathrm{s}$ without changing that leakage.

A response-normalized tolerance for baseline-feasibility residuals can specify allowed response errors ${\delta }_j>0$ in matching units and check $\eta ={\max }_j|r_j|/{\delta }_j\le 1$. When coefficient bounds $|\theta |\le B_s$, $|{\beta }_j|\le B_j$ are justified, the corresponding bias bound is $B_s|r_s|+{\sum }_j{B}_j|r_j|$. A noise-minimizing estimator still needs a separate variance calculation from the actual covariance.^[ref]

NIST's before/after pulse baselines provide the physical setting for rejecting background and drift.^[ref] Report the chosen constraint scaling and tolerated leakage. A small solver residual alone establishes neither low variance nor negligible bias for every possible background.

Frame-count switching in Raman acquisition-budget sensitivity occurs when a different integer number of exposure pairs becomes optimal as the available time changes. For each feasible pair count $n$, let $V_n(W)$ be the minimum predicted ratio variance after optimizing its live exposures. The complete design isThe marginal value of unused time below a Raman frame cap gives slopes for a fixed $n$. At a branch switch, those slopes need not be the derivative of the lower envelope on both sides.

A simple original example uses equal shot-noise coefficients $\alpha =\beta =a$, readout duration $d$ per frame and cap $c$ per exposure. Equal on/off totals are optimal, givingTake $a=10^{-6}\mathrm{s}$, $c=0.10\mathrm{s}$ and $d=0.010\mathrm{s}$. Two pairs saturate their $0.40\mathrm{s}$ live-time capacity once $W\ge 0.44\mathrm{s}$. Three pairs have live time $W-0.06\mathrm{s}$ near this budget. Their variances agree atJust below this value the two-pair variance is the constant $10\times 10^{-6}$. Just above it the three-pair variance is $4\times 10^{-6}/(W-0.06\mathrm{s})$ and begins decreasing. One pair has lower exposure capacity, while four or more pairs incur more readout and have less useful live time near the switch, so these two branches determine the global optimum there.

The one-sided derivatives at an optimal Raman frame-count switch are zero on the left and $-25\times 10^{-6}{\mathrm{s}}^{-1}$ on the right. The optimal variance is continuous at this switch, but has a kink. At $W=0.45\mathrm{s}$ two pairs give $10\times 10^{-6}$; at $0.47\mathrm{s}$ three pairs give about $9.7561\times 10^{-6}$. Extra time becomes useful again after the discrete design changes, despite the previous fixed-count plateau.

The primary Raman experiment's saturation-limited exposures and shuttered on/off readout motivate these competing resources.^[ref] The example uses chosen rounded timings. Read noise, drift and switching requirements can move the branch crossings and must be included when choosing an actual acquisition protocol.

Sign-sensitive measurements that resolve a covariance-axis ambiguity add an observable whose distribution changes when the fitted axis direction is reversed. In the global orientation ambiguity in a rank-one covariance fit, $\mathbf{u}$ and $-\mathbf{u}$ give the same covariance. An independently calibrated signed projection can distinguish them.

Consider two candidates with an additional observation $z=\pm m+\eta$, where $m=a{\mathbf{w}}^T\mathbf{u}\ne 0$ is known, $a$ has known positive sign and $\eta \sim N(0,s^2)$. Assume this calibration reading is independent of the covariance data conditional on the candidate. The Gaussian densities giveThe normal density and cumulative distribution function used here are documented in the pinned numerical reference.^[ref] The likelihood-ratio expression follows by expanding the two squared residuals.

For an original example, let $m=0.20\mathrm{V}$, $s=0.050\mathrm{V}$ and observe $z=0.10\mathrm{V}$. The log likelihood ratio is 16, strongly favoring the positive candidate under this stated model. With equal prior probabilities and equal decision costs, the sign-threshold decision has repeated-measurement error probability $\Phi (-|m|/s)=\Phi (-4)\approx 3.17\times 10^{-5}$. This decision-error probability is different from assigning a posterior error probability to that one observed value.

A practical calibration applies a known signed perturbation and records the signed detector-response component. Reversing the applied perturbation checks readout polarity. A calibrated phase-sensitive readout can provide such a projection; measuring only noise covariance does not supply its sign.

The failure of sign recovery from magnitude-only calibration data is important. For symmetric Gaussian noise, $|z|$ has the same distribution under means $+m$ and $-m$. If $\mathbf{w}$ is orthogonal to the axis, then $m=0$ and even the signed reading adds no separation. If the amplitude sign itself is unknown, reversing both amplitude and axis preserves the mean. The additional measurement resolves the ambiguity only when its reference sign and response model are independently established.

An ellipsoidal admissibility test for overlap data checks whether every point of a specified joint calibration ellipsoid permits finite Gaussian radius inversion. For the inversion of a symmetric Gaussian overlap log-ratio, write $x=(D,h{)}^{\mathsf{T}}$ with $D$ the measured logarithmic ratio and $h$ the positive radius-setting parameter. The admissible set is $D-4h<0$ and $-D-4h<0$. Together these strict inequalities already imply $h>0$.

Suppose the joint set isFor nonsingular $\Sigma$ this is equivalently $(x-\mu {)}^{\mathsf{T}}{\Sigma }^{-1}(x-\mu )\le k^2$. The parameterized form also accommodates a singular covariance matrix or a lower-dimensional deterministic set. Its support function in direction $l$ isIndeed, $l^{\mathsf{T}}Ay=(A^{\mathsf{T}}l{)}^{\mathsf{T}}y$, so the Cauchy-Schwarz inequality gives the bound, attained by a parallel $y$ when $A^{\mathsf{T}}l\ne 0$.

Let $\mu =(D_0,h_0{)}^{\mathsf{T}}$, marginal scales $s_D,s_h$, and off-diagonal entry $c_{Dh}$. Substituting $l_{+}=(1,-4{)}^{\mathsf{T}}$ and $l_{-}=(-1,-4{)}^{\mathsf{T}}$ gives two exact tests:Both must hold. Equality means that the closed joint set touches a boundary where no finite positive radius ratio exists; testing only the center would miss that point.

For an original example, take $D_0=2.4$, $h_0=0.7$, $s_D=0.1$, $s_h=0.03$ and $k=2$. Zero covariance gives upper-boundary excursion $2\sqrt{0.0244}\approx 0.3124$, smaller than the central margin $0.4$. The lower-boundary test also passes. With correlation coefficient $-0.8$, however, $c_{Dh}=-0.0024$ and the upper excursion becomes $2\sqrt{0.0436}\approx 0.4176$: the joint set crosses the forbidden region despite identical marginal scales.

The offending point need not be found by grid search. When $l^{\mathsf{T}}\Sigma l>0$, the maximizing point isFor the correlated example and $l=l_{+}$, this gives approximately $(D,h)=(2.5877,0.6425)$, with $D-4h\approx 0.0176>0$. It is a concrete incompatible calibration pair.

The ellipsoid here is a stated joint set. Interpreting it as a confidence region requires a separate sampling or uncertainty model; choosing $k=2$ alone supplies no universal coverage probability. JCGM 101 distinguishes assignment of joint input distributions from propagation and summaries.^[ref] If the set fails, report the incompatible region and reconsider the input model or measurement. Clipping it before inversion changes the set being propagated.

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.

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.

Noninvertible spectral compression versus a coordinate change distinguishes discarding spectral coordinates from expressing the same measurements in different units. The covariance-consistent rescaling of whitened spectral coordinates article assumes an invertible transformation. A compression $\mathbf{z}=B\mathbf{y}$ with fewer independent rows than input coordinates instead has covariance $C_z=BC{B}^T$ and may lose information even when this covariance is propagated correctly.^[ref]

For positive-definite $C$ and a known signal shape $\mathbf{s}$, compare the squared noise-weighted signal sizesassuming $B$ has independent rows. Put $\mathbf{t}=C^{-1/2}\mathbf{s}$ and $M=B{C}^{1/2}$. ThenThe matrix $P$ is an orthogonal projector. This proves that compression cannot increase this signal measure under the stated covariance model. The condition for preserving a spectral signal under linear compression is that its whitened shape lie entirely in the retained subspace.

For an original example, let $C=I_2$, desired shape $\mathbf{s}=(1,1{)}^T$ and nuisance shape $\mathbf{x}=(1,-1{)}^T$. Keeping only the first channel, $B=(1,0)$, reduces $J$ from two to one and makes both shapes the same scalar. An unknown nuisance amplitude can then imitate the desired signal exactly.

Summing instead, $B=(1,1)$, gives desired response two, nuisance response zero and noise variance two. It retains $J_B=2$ for this particular signal while rejecting that nuisance. Thus a noninvertible map can preserve a chosen estimation task, but cannot in general be reversed to recover every original spectral component.

The primary Raman apparatus records array spectra before subsequent spectral processing.^[ref] Choose compression against the scientific signal and nuisance model, and propagate the full covariance. Correct uncertainty accounting does not restore a spectral distinction removed by the map.

The marginal value of unused time below a Raman frame cap measures how the optimum predicted Raman ratio variance changes with the allowed acquisition budget. Hold the pair count $n$, readout time $d$ and per-frame exposure cap $c$ fixed. Put $H=nc$ and $T=W-2nd>0$. The wall-clock slack in saturation-limited Raman frame allocation becomes positive once $T>2H$; additional budget then cannot increase exposure at that pair count.

For shot-noise coefficients $\alpha >\beta >0$, define $K=(\sqrt{\alpha }+\sqrt{\beta }{)}^2$. The unconstrained on fraction is $p=\sqrt{\alpha }/(\sqrt{\alpha }+\sqrt{\beta })$. It reaches its cap when $T=H/p$. The minimized variance is thereforeThis follows by using the unconstrained ratio, then holding the weaker on state at its cap while extending the off state, and finally holding both at their caps.

Away from the transition points, the positive marginal improvement per extra wall-clock second isThe slope agrees on both sides of the first transition. At the final cap it changes to zero, because new time is unused.

For an original example, choose $n=2$, $c=0.10\mathrm{s}$, $d=0.010\mathrm{s}$, $\alpha =4\times 10^{-6}\mathrm{s}$ and $\beta =10^{-6}\mathrm{s}$. The wall-clock transitions are $0.34$ and $0.44\mathrm{s}$:At the last setting, $0.05\mathrm{s}$ is slack. Optimizing the pair count as well introduces frame-count switching in Raman acquisition-budget sensitivity; the fixed-count slopes above need not describe that full integer optimization.

The primary Raman experiment capped exposures to avoid saturation and used shuttered readout between pump states.^[ref] The numerical timings here are illustrative. These marginal values concern the stated noise objective, without additional drift, read-noise or switching-cadence costs.

The eastern phoebe, Sayornis phoebe, is a North American flycatcher belonging to Tyranni, the suboscine branch of Passeriformes. It often watches for flying insects from a low perch, then makes a short flight to capture prey. Sheltered ledges, including those on bridges and buildings, provide nesting sites. Feeding behavior and nest placement help identify the living species alongside its appearance and voice.^[ref]

Its vocal development provides a controlled test of learning. In a 1991 study, four nestlings collected at 10–12 days were hand-reared and deafened at about 35 days, before they began singing. Their later songs were judged normal in comparison with wild males and intact laboratory birds. The tested birds could therefore develop the species-typical song without auditory feedback from their own voices.^[ref]

The result is narrower than saying that the brain or experience has no role in singing. It tests whether intact hearing during this developmental interval is necessary for the measured song pattern. Comparing it with a songbird experiment in which removing auditory experience disrupts song makes the experimental question clear: similar adult behavior can arise through different developmental requirements.

The reset dwell time below threshold in an RC pulse train is the continuous time for which the filtered electric potential difference lies below an electronic comparator's threshold between successive accepted pulses. A positive threshold-reset margin for a periodic RC pulse train ensures a crossing in the ideal model, but does not ensure a long enough low interval for the intended electronics.

Consider the periodic RC low-pass filter response to a high level $V_0>0$ lasting $T$, followed by zero input for a quiet time $D$. Let $\tau >0$ be the time constant, $V_b$ the periodic baseline at the start of the high interval and $V_p$ the peak at its end. Assume a nonhysteretic ideal electronic comparator and $V_b<V_{\mathrm{th}}<V_p$. Solving the two exponential segments gives the rising crossing time $t_{\uparrow }$, measured from the high interval's start, and the falling crossing delay $s_{\downarrow }$, measured from its end:The below-threshold interval runs from the falling crossing to the next rising crossing. ThereforeIncluding $t_{\uparrow }$ matters: the low interval usually extends into the next input high interval. The natural exponential decay underlying these segments is directly testable with the switched-RC apparatus in the MIT laboratory.^[ref]

For an original exact example, take $V_0=1\mathrm{V}$, $\tau =1\mu \mathrm{s}$ and $T=D=\tau \log 2$. The periodic solution in missing comparator reset in an RC pulse train gives $V_b=1/3\mathrm{V}$ and $V_p=2/3\mathrm{V}$. At $V_{\mathrm{th}}=0.5\mathrm{V}$, both crossing delays equal $\tau \log (4/3)$, so $t_{\mathrm{low}}=D\approx 0.693147\mu \mathrm{s}$. The reset margin is a positive $1/6\mathrm{V}$, yet a stipulated requirement of at least $0.8\mu \mathrm{s}$ low time is not met.

The strict crossing assumptions are essential. At $V_{\mathrm{th}}=V_b$, the waveform only touches the threshold and the low interval vanishes. At or above the peak there is no finite high interval to separate. A Schmitt trigger uses distinct switching thresholds, so its output-state interval requires the corresponding falling and rising crossings rather than substituting one threshold indiscriminately.

A proposed measurement records the pulse generator, capacitor and electronic comparator outputs on an oscilloscope. Measure the low interval between actual output edges while also locating both capacitor-level crossings. Their difference can expose switching delays; the capacitor formula alone does not specify a real device's minimum pulse duration.

Reference identity in algebraic uncertainty cancellation asks whether two appearances of a reference are the same measured quantity or merely readings given the same descriptive name. If a transfer is $D=B-A_1$ but a correction subtracts $A_2$, thenThe last difference vanishes identically only when the same reference value is reused. Separate acquisitions may be correlated, independent, or affected by drift; sharing a label does not decide their covariance.

For independent $T$ and $B$, also independent of the reference pair, propagation givesWith equal reference standard uncertainties $s$ and correlation coefficient $\rho$, the residual reference contribution is $2s^2(1-\rho )$. It is zero for identical reused values, $2s^2$ for independent readings, and is not generally zero just because the displayed numbers agree. The GUM requires significant correlations from common inputs to be included.^[ref]

An original acquisition model separates what cancels:Here $C$ is a common uncertain calibration contribution, while ${ϵ}_1,{ϵ}_2$ are independent repeat errors. Their difference removes both $A_0$ and $C$, but leaves ${ϵ}_1-{ϵ}_2$. If $u(C)=3$ mg and each repeat error has standard deviation 2 mg, each reading has variance $13\mathrm{m}{\mathrm{g}}^2$ and their covariance is $9\mathrm{m}{\mathrm{g}}^2$. The difference therefore has variance $13+13-18=8\mathrm{m}{\mathrm{g}}^2$, not zero.

Adding independent $u(T)=1$ mg and $u(B)=3$ mg gives $u^2(Y)=18\mathrm{m}{\mathrm{g}}^2$. Reusing exactly one recorded reference value in both places instead gives $u^2(Y)=10\mathrm{m}{\mathrm{g}}^2$. These constructed cases have different data-reduction histories despite similar-looking equations.

Reference lineage in covariance models records acquisition identifiers, shared calibration inputs and coefficients in each derived quantity. A calibration change or time-dependent drift must appear in that model if significant. Algebraic simplification is valuable only when it preserves the identity of the underlying inputs; renaming two acquisitions with the same symbol does not make their errors cancel.