Time-dependent decay-rate envelope ($P(t)\le A{e}^{-{\int }_b^{t}k(s)ds}$)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

A common-clock contribution to a maximum receiver horizon is a residual timing error that survives in every channel's inferred input-time horizon. Write $H_j=h_j+C+{\epsilon }_j$, with common error $C$ independent of the remaining channel errors. Then the identityshows that selecting the latest channel does not average away the common term. With finite variances,Here $C$ means a component remaining after each channel's subtraction $T_j-d_j$. A clock contribution that already cancels inside that subtraction must not be added again; see receiver-latency uncertainty in the inferred input-time horizon.

For two equal nominal horizons $h_j=\mu$, take independent channel errors ${\epsilon }_j\sim N(0,s^2)$ and $EC=0$. The independent maximum moments derived in channel switching under uncertainty in a combined receiver horizon giveIf the common error is also normal, each channel has variance ${\sigma }_C^2+s^2$ and their correlation is ${\sigma }_C^2/({\sigma }_C^2+s^2)$, recovering the equal-channel model in distribution of the maximum of correlated receiver horizons.

For an original example, choose ${\sigma }_C=0.8\mathrm{ns}$ and $s=0.3\mathrm{ns}$. The combined mean is about $0.1693\mathrm{ns}$ above $\mu$, and its standard deviation is about $0.8375\mathrm{ns}$. Yet the difference between channel horizons has standard deviation only $\sqrt{2}s\approx 0.4243\mathrm{ns}$, because $C$ cancels there. Good agreement between channels therefore need not establish a precise absolute horizon.

A timing experiment needs an independently justified reference for the common component, in addition to channel-to-channel alignment observations. Shared-input covariance is explicitly retained in NIST's uncertainty law.^[ref] Repeated reads sharing the same clock realization cannot estimate away its absolute displacement simply by choosing the largest result.