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Cubic intensity-noise bias in pulse correlation is the difference between the average of a third-order signal and the signal predicted from the average intensity when a fluctuating common intensity multiplier is cubed. Assume every shot has the same pulse shape, $I_j(t)=K_j{I}_0(t)$, with nonnegative scalar $K_j$. The ideal third-order integral givesThe three intensity factors follow from the second-harmonic generation and sum-frequency generation gate.^[ref]

Let $K=\mu +\delta$, with $\mathbb{E}[\delta ]=0$, variance ${\sigma }^2$ and third central moment $m_3$. Expanding the cube givesFor $\mu >0$, write the coefficient of variation as $v=\sigma /\mu$ and, when $\sigma >0$, skewness as $\gamma =m_3/{\sigma }^3$. The relative multiplier is $1+3v^2+\gamma v^3$. For zero variance the multiplier is one without needing to define skewness.

Equally likely $K=0.9$ and $1.1$ have mean one and give mean cubic multiplier $1.03$. Rare bright shots can matter much more: stipulated probabilities 0.9 and 0.1 at $K=0.8$ and $2.8$ also give mean one, butThe extra signal need not mean that the mean input intensity increased. These are original probability examples, not measurements reported in the correlator paper.

For unchanged pulse shape, common fluctuations multiply every delay by the same expected factor. Normalizing the averaged trace by its peak cancels that factor. Shape changes correlated with brightness destroy this simplification. For example, two equally frequent shot classes with central signals 1 and 8 and side/center ratios 0.01 and 0.10 give ratio of averaged signals $(0.01+0.8)/(1+8)=0.09$, whereas averaging the two ratios gives 0.055. They answer different questions.

A controlled measurement records a synchronized input-intensity monitor and detector background alongside each trace. Compare intensity-binned signals in an unsaturated range and test whether normalized shapes agree. Common attenuation in third-order pulse correlation predicts the cubic scaling only while the nonlinear gate, pulse shape and readout retain the assumed response. A mean power reading alone does not determine the required third moment.

The expected value premium principle charges the expected value of the covered loss plus a proportional margin. For a nonnegative random variable $L$ with finite expected value, and a fixed dimensionless loading $\theta \ge 0$, it setsHere $\theta =0$ gives the expected claim cost alone; $\theta >0$ gives a positive margin when $\mathbb{E}[L]>0$. The loading is a fraction of the expected value, not of the largest possible claim size. Specify the covered period and use a consistent monetary valuation date. The rule itself does not determine $\theta$. ^[ref].

Consider an explicitly toy one-year policy: it pays £1,000 with probability $p=0.1$, and zero otherwise. Write $L=1000B$, where $B$ has the Bernoulli distribution with parameter $p$. The expected value is $1000(0.1)=\pounds 100$. A loading $\theta =0.2$ adds £20, making the premium £120. Neither the £100 expected value nor the £120 premium is a possible realized claim: the actual payment is £0 or £1,000. The first diagram separates these quantities. All numerical examples and both diagrams here are exact consequences of invented assumptions, not measured insurance data.

Why pay £120 for a loss whose expected value is only £100? Conditional on the insurer meeting its obligation, the policyholder replaces an uncertain £1,000 bill with a known £120 payment. The insurer can combine many such policies, so a few claims are funded by premiums from the whole pool. Winkel develops this motivation and the importance of independence in §15.2 of ^[ref].

To see the pooling mechanism precisely, suppose there are $n$ toy policies with mutually independent Bernoulli random variables $B_1,\dots ,B_n$, each of parameter $0.1$. Every claim size is £1,000, with at most one claim per policy. The aggregate claims model isUsing linearity of expectation and addition of variances for independent random variables, the expected value and standard deviation of this binomial distribution aggregate areThus 100 policies have expected aggregate claims £10,000 and aggregate standard deviation £3,000; 10,000 policies have expected aggregate claims £1,000,000 and aggregate standard deviation £30,000. Total uncertainty grows, but much more slowly than the expected value. Per policy, $S_n/n$ has expected value £100 and standard deviation $\pounds 300/\sqrt{n}$. Equivalently, the aggregate coefficient of variation, its standard deviation divided by its positive expected value, isIt falls from $300\%$ for one policy to $30\%$ for 100 policies and $3\%$ for 10,000. This is the useful beauty of independent pooling: an erratic individual cost becomes a comparatively stable average cost. The law of large numbers describes the corresponding convergence of the average toward its expected value.

At the same fixed loading $\theta =0.2$, total premium is $\Pi (S_n)=120n$, with expected surplus before other outgoings $120n-100n=20n$. The loading margin divided by the aggregate standard deviation is $20n/(300\sqrt{n})=\sqrt{n}/15$. This ratio rises with $n$ under our independence assumption; it is not a specified loss probability. For $n=100$, premiums total £12,000, but 13 claims cost £13,000. The one-year probability that claims exceed collected premiums is exactlyA positive loading therefore leaves about a $19.8\%$ chance of a one-year claims deficit in this toy pool. A claims deficit need not mean insolvency if capital is available. This calculation assumes premiums are collected before claims and ignores expenses and investment income; it is not an ultimate ruin probability.

The expected value premium principle also has a revealing limitation. With a common fixed $\theta$, linearity of expectation gives $\Pi (L_1+L_2)=\Pi (L_1)+\Pi (L_2)$ even without independence. Pooling therefore does not automatically reduce the premium per policy under this rule, and two losses with the same expected value receive the same premium regardless of their variance or large-loss probabilities. Any decision to change the loading after pooling requires information beyond the formula.

Dependence can remove the stability benefit. For arbitrary square-integrable losses, the aggregate variance includes their covariances:In the extreme toy common-shock model, every policy has $L_i=1000B$ for the same Bernoulli random variable $B$. Each policy still has expected value £100, so total premium is still $120n$, but $S_n=1000nB$ has standard deviation $300n$ and coefficient of variation $3$ for every $n$. With probability $0.1$, all policies claim together and losses exceed premiums by $880n$. The second diagram compares this model with independent pooling. Real shared events can invalidate independence; Winkel uses flood claims as an example in §15.2 of Actuarial Science II. The common-shock curve is an illustrative limiting model, not a fit to flood data.

The earlier $n^{-1/2}$ relative-variability calculation needs finite variance and the stated model of independent and identically distributed random variables. A finite expected value alone suffices to define the premium but does not imply finite variance. The rule makes no promise of solvency: losses, available capital and the timing of payments must be modelled separately. In the classical risk model, claim arrival rate $\lambda$ and mean claim size $\mu$ give premium rate $c=(1+\theta )\lambda \mu$; relative safety loading is then $\theta$. That model tracks capital through time and asks for ultimate ruin probability, a different question from calculating an expected margin. ^[ref]. Under quota share reinsurance, the reinsurer can apply the expected value premium principle to its ceded fraction $(1-\alpha )S$ using its own loading, which need not equal the direct insurer's loading.

Finite inverse moments in thickness models determine whether a positive but uncertain layer thickness permits a finite mean or variance after conversion to density. With fixed positive mass and area, write $\rho =K/\delta$, where $K=M/A$. A finite thickness mean does not ensure that $E({\delta }^{-1})$ or $E({\delta }^{-2})$ exists.

Suppose the thickness probability density function behaves near zero as $p(\delta )\sim C{\delta }^{a-1}$ with $C>0$. The small-thickness part of the inverse moment is proportional toFor $q>0$, this converges exactly when $a>q$. Thus a finite mean density requires $a>1$, while finite density variance requires $a>2$, assuming the rest of the distribution is well behaved.

An exact example uses a gamma distribution with shape $a$ and scale $\theta$. Substitution into its density and the gamma function integral givesThe distribution and parameter conventions are specified by NIST; the inverse-moment calculation here follows directly from the integral.^[ref] For $a=2$ and $\theta =d/2$, mean thickness is $d$ and mean density is $2K/d$, yet its second moment diverges. A simulation can return finite sample variances without establishing a finite population variance.

A lognormal distribution provides a different positive-tail model. If $\log \delta \sim N(\mu ,s^2)$, then $E({\delta }^{-q})=\exp (-q\mu +q^2{s}^2/2)$, finite for every finite positive $q$. NIST defines this logarithmic-normal parameterization and its coefficient of variation.^[ref] With mean thickness $d$ and coefficient of variation $c$, mean density is $K(1+c^2)/d$. At $c=0.20$, it exceeds $K/d$ by 4%.

Tail-model dependence of inferred density uncertainty is therefore experimentally consequential. Choosing a convenient positive distribution changes the implied inverse moments. Justify near-zero behaviour from dimensional evidence or physical bounds rather than treating positivity or a stable finite simulation as sufficient validation.

The mean geometric chord, also called the standard mean chord, is the span-average airfoil chord of an aircraft wing:where $S$ is the wing area and $b$ the wingspan. The area is the projected wing planform, not the sum of the upper and lower skin areas.^[ref]

A useful exact identity explains its difference from the mean aerodynamic chord. Represent uniform spanwise averaging by a coordinate $Y$ with a continuous uniform distribution on $[-b/2,b/2]$, and put $C=c(Y)$. This is a mathematical averaging device, not a claim that the wing geometry is random. With $\mu =\mathbb{E}[C]=S/b>0$,Thus the relative excess is the squared coefficient of variation of the local chords. The mean aerodynamic chord can also be interpreted as the expected value under the corresponding size-biased distribution, since larger chords receive proportionally larger weights.

For the original linear taper from $2$ to $1\mathrm{m}$, uniform spanwise sampling makes $C$ uniform on $[1,2]\mathrm{m}$. Its expected value is $1.5\mathrm{m}$ and variance is $1/12{\mathrm{m}}^2$. Hence ${\overline{c}}_a=1.5+(1/12)/1.5=1.5556\mathrm{m}$, a relative excess of $1/27\simeq 3.70\%$. A constant-chord wing has zero variance, so its two averages coincide.

Equal $S/b$ does not determine how local chord varies or where the wing lies relative to a reference axis. Those additional data enter wing sweep and the mean aerodynamic chord reference position.

Measurement precision describes agreement among repeated indications or measured values obtained under stated conditions. It is commonly characterized using a standard deviation, variance or coefficient of variation, which describe the observed spread. Measurement repeatability is precision under a particular restricted set of conditions; intermediate measurement precision and measurement reproducibility concern other specified conditions.^[ref]

For an original balance example, take three repeat readings of a reference object: $100.100$, $100.080$ and $100.120\mathrm{g}$. Their arithmetic mean and sample standard deviation areSuppose the reference object's assigned mass is $100.000\mathrm{g}$ with standard uncertainty $0.002\mathrm{g}$. The observed mean differs from that reference by $0.100\mathrm{g}$, much more than the within-series scatter. These chosen readings therefore illustrate why close repeat readings do not establish agreement with a reference. They do not determine the true systematic offset from only three trials.

Measurement accuracy and measurement trueness answer different questions from precision. A common additive offset leaves the deviations $x_j-\bar{x}$, and hence $s$, unchanged. Averaging more independent readings can reduce random uncertainty in the mean but does not by itself remove that common offset.^[ref]

A useful physical study repeatedly removes and replaces the reference object while recording load procedure, environmental conditions and elapsed time. Specify which parts of the measurement were repeated: repeated display readings alone may omit placement variation or instrumental drift. The statistic describes the stated experiment, not an unconditional property of the instrument.

With zero initial messenger RNA and protein concentration, solving the two linear ordinary differential equations in sequence givesThe last expression applies after the short messenger RNA transient when ${\alpha }_R\gg {\alpha }_P$. All three mean curves have this shape. At a fixed steady state, a larger translational burst size requires fewer transcription events: ${\beta }_R={\alpha }_P{X}_{P,\ast }/b$. Thus $b=400$ gives infrequent large upward jumps followed by protein degradation, whereas $b=3$ gives many smaller jumps and a smoother trace.

Noise depends on molecule counts. If the fixed cell volume is $V$, write $\bar{N}=V{X}_{P,\ast }$. In the instantaneous-burst limit, independent Poisson process arrivals of geometrically distributed bursts, with count arrival rate $\lambda =V{\beta }_R$, and unit protein degradation giveIndeed, the moment equations are $d\mathbb{E}N/dt=\lambda b-{\alpha }_P\mathbb{E}N$ and $d\mathrm{var}N/dt=\lambda b(1+2b)+{\alpha }_P\mathbb{E}N-2{\alpha }_P\mathrm{var}N$. With zero initial counts the variance becomes $\bar{N}[(1+b)-e^{-{\alpha }_{P}t}-b{e}^{-2{\alpha }_{P}t}]$. These equations explain both the approach to the steady state and the scatter about it.

For illustration only, $V=1\mu {\mathrm{m}}^3$ gives $\bar{N}=10^4$ at the higher concentration: the two values of the coefficient of variation are $0.020$ and $0.200$. At the lower concentration, $\bar{N}=100$ and $b=3$ gives $0.200$. Decreasing the concentration by a factor of 100 therefore increases relative fluctuations tenfold, despite keeping small bursts. Absolute fluctuation sizes cannot be inferred without a cell volume. The original sketches below stipulate that volume; they are model realizations, not source data.^[ref]

Convert number density into molecule count: $\bar{n}=X_{P,\mathrm{st}}V$. For the explicitly assumed representative cell volume $V=1\mu {\mathrm{m}}^3$, the two densities correspond to $\bar{n}=10^4$ and $10^2$. The concentrations alone do not fix fluctuation sizes without a volume.

For activation from zero messenger RNA and protein, solving the two mean rate equations givesThus all three normalized mean curves are almost identical. Sample paths differ. For Poisson transcription with biological translation bursts, the stationary Fano factor and squared coefficient of variation areOne way to obtain the fast-messenger RNA limit is to use burst arrivals at rate $a$, a geometrically distributed burst $B$ with $⟨B⟩=b$ and $⟨B^2⟩=b+2b^2$, and independent unit protein degradation at rate ${\alpha }_{P}n$. The first stationary moment gives $ab={\alpha }_P\bar{n}$; the second gives $\mathrm{Var}n=\bar{n}(1+b)$.

At $\bar{n}=10^4,b=3$, typical relative fluctuations are $\sqrt{4/10^4}=0.02$ and many small bursts make the trajectory smooth. At $\bar{n}=10^4,b=400$, they are $\sqrt{401/10^4}\simeq 0.20$; individual bursts have mean normalized height $0.04$, with only about $25$ bursts per protein lifetime. At $\bar{n}=100,b=3$, the relative fluctuations are again $0.20$, but the typical burst height is $0.03$ and about $33$ bursts occur per lifetime. The latter two examples can look similarly noisy after normalization even though their mean molecule counts differ by two orders of magnitude. Burst arrival times and sizes fluctuate: a high-burst trace is not a periodic sawtooth.

The upper panels are independently simulated burst-arrival and unit-removal processes, shown alongside the common mean. The lower panels illustrate d; their point clouds are schematic, not experimental data.

An open ion channel carries the electric currentThe macroscopic mean is $I=N{P}_{o}i$, so $I/i\simeq 204$ is the mean number open, not the total number present. Reading $P_o(-20\mathrm{mV})\simeq 0.40$ from the original graph givesThe graph reading is approximate, so a precision of hundreds is appropriate. Only a few hundred active ion channels generate a nanoampere-scale cellular electric current; molecular discreteness and stochastic gating therefore matter. For independent identical ion channels, the open-count standard deviation is $\sqrt{N{P}_o(1-P_o)}\simeq 11$, corresponding to a coefficient of variation in electric current of about $5.4\%$. This is an instantaneous ideal independent-gating estimate, before temporal filtering and other noise sources, and requires the single-channel recording $P_o$ to apply to the cellular recording conditions.

A stable multiplicative sensitivity factor $s_i$ at detector pixel $i$ changes both means to $N_i=s_i{N}_0$. The noise-free ratio is still one, so flat-field correction is not required merely to remove that fixed pattern from the mean ratio. This is the principal advantage of using two matched flats rather than the variance of a single image.

However, the photon shot noise is not cancelled. At high counts the conditional ratio variance is $2/(g{N}_i)$. Pooling detector pixels gives $E^2\simeq (2/g)⟨1/N_i⟩$, where brackets denote a spatial average. If the formula uses $\overline{N}=⟨N_i⟩$, it returnsThe inequality follows from the Jensen inequality for $1/x$ on positive signals. For small fractional response variation, the relative bias is approximately minus its squared coefficient of variation. A locally nearly uniform region or a fit using detector pixel-dependent variances avoids this bias from mixing the harmonic mean with the arithmetic mean. Stable response cancellation also presumes the conversion gain itself is uniform; gain variations, offset errors or changing sensitivity do not obey that simple cancellation. The fixed response pattern cancels in the ratio mean; unequal shot-noise levels can still bias a globally pooled gain estimate.

For arbitrary nonnegative raw weights $W_i$, let $\bar{W}=K^{-1}{\sum }_i{W}_i$. The empirical squared coefficient of variation, using variance divisor $K$, isSubstitution into the stated definition gives the usual effective sample size of importance samplingFor normalized weights $w_i=W_i/{\sum }_j{W}_j$, this reduces to $\boxed{\widehat{\mathrm{ESS}}=1/\sum _i{w}_i^2}$.

Pulse-to-pulse energy fluctuation is variation in the laser pulse energy of successive emitted pulses. For a stationary population with finite variance, its standard deviation is ${\sigma }_U=\sqrt{\mathbb{E}[(U-\mathbb{E}U{)}^2]}$. A finite record supplies an estimate, whose convention should be stated rather than being confused with the uncertainty of the record's mean.

For an original four-pulse record, take measured energies $9,11,11,9\mu \mathrm{J}$. Their mean is $10\mu \mathrm{J}$. The sample standard deviation isIts coefficient of variation is 11.55 percent. The root-mean-square deviation of this particular list, using divisor four rather than three, is instead $1\mu \mathrm{J}$ or 10 percent. The definition matters when comparing reported stability figures. The familiar $s_U/\sqrt{N}$ estimate of uncertainty in a mean additionally assumes independent, identically distributed observations; an ordered record alone does not establish that assumption.

A historical NIST stability assessment compared an internal excimer-laser pulse monitor with a pyroelectric laser-energy meter. Its calorimeter integrated a pulse train, so averaging and individual-pulse variation entered different parts of the analysis.^[ref]

Observed variability also includes readout noise. Separating detector noise from pulse-energy variance requires an appropriate measurement model and independent information about that noise. Detector saturation, drift and correlated pulse errors are not removed merely by collecting a longer record. An average power of a laser pulse train measurement by itself does not resolve this sequence.

Reciprocal moments of an estimated variance determine whether a precision estimate such as $1/S^2$ has a finite expected value or variance. Small positive values of a sample variance become large values after inversion. An unbiased estimator of a variance therefore need not give an unbiased or even integrable estimate of its reciprocal.

For $n\ge 2$, let $X_1,\dots ,X_n$ be independent normal random variables with common mean $\mu$ and variance ${\sigma }^2>0$. For the unbiased sample variance $S^2=(n-1{)}^{-1}{\sum }_i(X_i-\overline{X}{)}^2$, put $\nu =n-1$. Then $Z=\nu S^2/{\sigma }^2$ has a chi-squared distribution. Its probability density function is $z^{\nu /2-1}{e}^{-z/2}/[2^{\nu /2}\Gamma (\nu /2)]$ for $z>0$.^[ref]

For any real $p>0$, direct integration and the Gamma function giveThe convergence restriction is essential: near zero the integrand has exponent $\nu /2-p-1$, so the integral diverges when $\nu \le 2p$. Substitution of a negative Gamma function argument into the displayed quotient does not define a nonexistent moment.

Rescaling gives the useful general resultIn particular, the mean reciprocal exists for $n>3$, while its variance requires $n>5$:For $n=6$ the mean is $5/(3{\sigma }^2)$, the variance is $50/(9{\sigma }^4)$ and the coefficient of variation is $\sqrt{2}$. For $n=4$ the mean is finite but the variance is infinite. For $n=3$ even the mean diverges. Increasing the number of groups does not repair a nonexistent moment of each raw reciprocal estimate.

Inverse-variance weighting commonly normalizes these reciprocals. Those normalized weights lie between zero and one, so their moments can exist even when the raw reciprocal moments fail. This distinction matters in uncertainty in estimated inverse-variance weights: neither the moment calculation nor a plug-in precision formula alone establishes the accuracy of the final weighted estimator. The results here use the exact normal-sample law; a nonnormal sample requires a different analysis.

For a random variable $X$ with finite variance and nonzero mean, its relative variance is $R_X=\mathrm{Var}(X)/(\mathbb{E}X{)}^2$. It equals the square of the coefficient of variation under the usual positive-mean convention. Multiplying $X$ by a nonzero constant leaves $R_X$ unchanged; adding a constant generally does not.

For an average of $n$ independent identically distributed observations with mean $\mu \ne 0$ and variance ${\sigma }^2$, the relative variance is ${\sigma }^2/(n{\mu }^2)$. Chebyshev's inequality gives $\mathbb{P}(|X/\mathbb{E}X-1|\ge a)\le R_X/a^2$, explaining its role in self-averaging. At zero mean the ratio is undefined, so another reference scale is needed.

Tail-model dependence of inferred density uncertainty is the sensitivity of a density estimate's distribution to the assumed probability of very small positive thicknesses. For $\rho =K/\delta$ with fixed $K=M/A>0$, a tiny thickness has disproportionate influence through $1/\delta$. The main body of a thickness histogram can look stable while the inferred mean or variance remains model-dependent.

A gamma distribution and a lognormal distribution provide useful explicit models. NIST specifies their density functions and parameter conventions.^[ref] At mean thickness $d$ and coefficient of variation $c$, the gamma model has shape $a=1/c^2$ and givesThe lognormal model gives $(1+c^2)/d$. The two agree closely for small $c$, but that agreement concerns those model families, not every positive distribution with a similar spread.

An original rare-event example makes the limitation explicit. Let thickness be $0.001$ micrometres with probability $0.0001$ and $10.001$ micrometres otherwise. Mean thickness is exactly 10 micrometres and its standard deviation is approximately 0.10 micrometres, about 1% of the mean. Nevertheless,nearly twice the reciprocal of mean thickness. The rare component supplies about half the inverse moment. These are deliberately constructed model values, not measured polymer-layer dimensions.

Rare-thickness leverage on inverse moments can escape an apparently reassuring sample. With 1,000 independent thickness observations, the chance of seeing none of that rare component is $(0.9999{)}^{1000}\approx 0.905$. A finite simulation or a smooth fit to common observations need not validate the unobserved lower tail.

A useful experiment investigates whether dimensional resolution, specimen geometry or physical constraints rule out such near-zero layers. Report sensitivity to defensible alternative lower-tail models or measured lower bounds. Choosing a positive distribution guarantees positive thickness, but does not establish that its inverse moments describe the measurement process.