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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.

Let $k=m/2$. After offset subtraction, each summed image has mean $N$ analogue-to-digital units per detector pixel. For a constant detector conversion gain $g$ and independent Poisson distribution detected counts, $\mathrm{Var}(A)=\mathrm{Var}(B)=N/g$: summing $k$ independent frames adds both their means and variances. Here $N$ is the mean of the sum, not one frame.

Write $A=N+a$, $B=N+b$. At high counts, the delta method gives $A/B\simeq 1+a/N-b/N$. Independence therefore givesThus the formula is a high-signal gain estimation from flat-field ratios, not an exact identity for a ratio of random variables. A Poisson distribution denominator can even be zero at low counts. The exposures must be linear and unsaturated, offsets removed, and read noise negligible. Being close to saturation increases the signal but does not establish linearity; clipping or charge-induced correlations biases a variance estimate. A large ensemble of suitable detector pixels supplies the measured standard deviation. In normal photodetector terminology $g$ is collected Electrons per ADU, consistent with detected photons here only under the stated unit-yield convention; see the instrument gain definition.

If spatial illumination is stable between the two sums, it acts just like the sensitivity factor: the mean ratio remains one but each detector pixel has a different shot-noise variance. A whole-image standard deviation is disproportionately influenced by faint detector pixels, so substituting a global arithmetic mean signal generally underestimates detector conversion gain. If the illumination pattern differs between the sums, their noise-free ratio is not constant; its structure adds apparent variance and causes a further systematic error.

Use locally uniform signal bins or subregions, remove offsets, mask defective or saturated detector pixels, and propagate the local shot and read variances. In the stable-pattern case a useful alternative to pooling raw ratios is the normalized differenceAt high signal and negligible read noise, treating the denominator at its mean gives $\mathrm{Var}(U_i)\simeq (2N_i/g)/(2N_i)=1/g$, independent of the illumination. Thus a variance fit or $g\simeq 1/\mathrm{Var}(U)$ can use a spatially varying flat without the arithmetic/harmonic mismatch. Measured noisy denominators cause higher-order corrections, so independent or smoothed signal estimates can be preferable. Fit and subtract a genuine changing illumination pattern before noise analysis, using sufficient smoothing not to fit away the random noise. Arbitrary flat-field normalization without propagating its noise does not by itself recover the original gain.

Use a two-sided top-hat line plume with full width $w(z)$, upward speed $U(z)$, and reduced gravity $b(z)$ uniform across the plume. Fluxes are measured per unit span along the line source: the volume flux is $V=wU$, the kinematic momentum flux is $M=w{U}^2$, and the buoyancy flux is $B=wUb$. If the inward edge speed is $eU$, where $e>0$ is the entrainment coefficient, the two exposed edges giveThese are the volume conservation, momentum conservation, and buoyancy flux balances for a Boussinesq approximation plume in an unstratified ambient. Entrained ambient fluid supplies neither vertical momentum nor reference buoyancy.

A pure plume has no persistent source length scale. Since a line-source $B$ has dimensions $L^3{T}^{-3}$, dimensional analysis gives constant $U\propto B^{1/3}$, $w\propto z$, and $V\propto z$. Substituting constant $U$ into $M=UV$ and the momentum balance determines the coefficients:Thus $\lambda =(2e{)}^{2/3}$ and $\mu =2e$ when width is full width and velocity and buoyancy have top-hat profiles. If width means half-width, $\mu =e$ instead. Other prescribed profile shapes change these numerical factors; the linear growth laws remain the same. The point-source $B$ in Question 1 has different dimensions, so its $z^{5/3}$ law must not be used here.

Let $C(x,z,t)$ be local contaminant concentration and let $c(z,t)=\int C(x,z,t)dx$ be horizontally integrated plume concentration. Its conserved amount per unit span is ${\int }_0^{\infty }cdz=V_c$. The dilute contaminant is a passive scalar; the maintained plume is unaffected by its impulsive release. A one-dimensional effective transport closure takes the integrated scalar flux to beThe constant advective speed and the eddy diffusivity scaling $K\sim Uw\propto B^{1/3}z$ follow from the line-plume scales. The entrainment coefficient alone does not determine the scalar dispersion coefficient: this Fickian closure for the integrated variable is an additional modelling assumption. Mass conservation $c_t+J_z=0$ then gives the displayed transport equation in the PDF. With this effective closure and advective speed chosen as $U$, $\alpha =(2e{)}^{-1/3}$; if $K=\kappa Uw$, then $\beta =\kappa (2e{)}^{2/3}$, with independent dimensionless mixing coefficient $\kappa$.

There is a second possible closure convention. If one instead applies local diffusive flux to horizontally averaged plume concentration $\overline{C}=c/w$, its integrated diffusive flux is $-Kw{\overline{C}}_z=-K(c_z-c/z)$. For $K=Dz$, the total scalar flux is then $(U+D)c-Dz{c}_z$. The same printed equation is obtained with $a=U+D$, rather than $a=U$. Thus the printed $\alpha$ and $\beta$ should be regarded as effective coefficients unless the averaging and closure conventions are specified; no equality between $\alpha$ and the velocity prefactor is universal.

Assume $a,D>0$, $z>0$, an initial impulse $V_c{\delta }_0$ at the origin, no further contaminant input, and zero endpoint scalar flux for $t>0$. For a similarity solution, putSubstituting into the advection-diffusion equation givesDecay at infinity makes the integrated constant zero. Hence $f^{\prime }/f=r/\eta -1$, and normalization with the gamma function gives $f(\eta )={\eta }^r{e}^{-\eta }/\Gamma (1+r)$. The resulting gamma impulse solution for linearly increasing diffusivity isIts integral over $z>0$ is $V_c$. Its flux is $J=(z/t)c$, which vanishes at both endpoints for $t>0$. The scale $Dt$ shrinks to zero as $t\downarrow 0$, so the normalized solution has weak convergence of probability measures to the required unit source impulse. The mass at the boundary is a full unit impulse on the half-line, not half of a whole-line impulse.

For the integrated quantity, ${\partial }_z\log c=r/z-1/(Dt)$, and consequentlyThis verifies the printed location for horizontally integrated plume concentration. It is a maximum because the derivative changes from positive to negative there. The normalized profile is a gamma distribution with shape $1+r$: its expected value is $(a+D)t$ and its variance is $(a+D)Dt$, so neither the mean position nor the spread should be mistaken for the modal position.

The PDF changes from “integral” to “averaged” concentration in its final request. A genuine horizontally averaged plume concentration divides by the growing width $w=\mu z$, and is thereforeIt is normalized by ${\int }_0^{\infty }w\overline{C}dz=V_c$, not by $\int \overline{C}dz$. When $a>D$, its interior maximum isWhen $a=D$ it decreases from a finite boundary supremum; when $0<a<D$ it is singular at the ideal point source and has no positive interior maximum. A finite source regularizes that singularity. Thus the final printed maximum is correct for the integrated variable $c$, or for an “average” using a fixed reference width, but is not generally the maximum of the local mean $c/w(z)$. Both quantities have been given rather than silently identifying them. If $D=0$, the smooth similarity formula is replaced by the advected impulse $V_c\delta (z-at)$; positive eddy diffusivity is essential to the gamma profile.

Condition on $X=k$. For $0<k<n$, the posterior expected loss is a strictly convex quadratic in the decision $d$. Its minimizer isThe beta density gives $\mathbb{E}[1/p\mid k]=(n+1)/k$ and $\mathbb{E}[1/(1-p)\mid k]=(n+1)/(n-k)$, so $d=k/n$. If $k=0$, every $d\ne 0$ has infinite posterior expected loss because of the nonintegrable $d^2/p$ term near zero, whereas $d=0$ has finite loss expectation. The analogous argument at $k=n$ forces $d=1$. Hence the Bayes estimator isIndeed $X/n$ is unbiased and has variance $p(1-p)/n$, which cancels the loss denominator. Its uniform-prior Bayes risk is $1/n$.

The printed loss is undefined at $p=0,1$, despite the stated closed parameter interval. A convention is necessary there. The natural extended loss assigns zero to the correct endpoint decision and infinity to an incorrect one; it gives endpoint risks zero for $X/n$. Alternatively a continuous extension of this estimator's risk gives endpoint values $1/n$. The interior result and the minimax value below are unchanged by either convention, and a uniform prior gives the endpoints measure zero. We do not identify an undefined $0/0$ with a value without specifying this choice.

The dashed curves are contours of constant Cook's distance, labelled 0.5 and 1, in the plot of standardized residual against regression leverage. Writing $h_{ii}$ for the diagonal of the hat matrix, $s^2$ for the residual variance estimate, and $p=8$ for the number of fitted coefficients,Thus a contour $D_i=d$ has $r_i=\pm \sqrt{dp(1-h_{ii})/h_{ii}}$. On the actual PDF plot, observation 579 has leverage about 0.084 and standardized residual about $-8$, lying beyond the $D=0.5$ contour and below $D=1$ in influence magnitude, approximately $D\simeq 0.7$ to $0.8$ from the plotted coordinates. It is unusually light relative to its fitted log-weight and potentially influential because deleting it can materially change fitted coefficients. It warrants checking, not automatic deletion. The solid smooth red curve is distinct from the dashed Cook's distance contours.

A residual-versus-fitted plot for the untransformed fit places $e_i=y_i-{\hat{y}}_i$ on the vertical axis and ${\hat{y}}_i$ on the horizontal axis. A fan-shaped cloud, with larger residual spread at larger fitted weight, suggests heteroscedasticity and approximately multiplicative errors. A logarithmic transformation may stabilize this spread when the conditional standard deviation is roughly proportional to the mean.

One can also use the scale-location plot of $\sqrt{|r_i|}$ against fitted values, where $r_i$ is a standardized residual. A systematic rise suggests nonconstant error variance. Neither plot proves that taking logarithms is sufficient: diagnostics of the transformed fit should be checked as well, and dependence between repeated chicken measurements is a separate issue.

The observations have finite mean and variance. The strong law of large numbers gives ${\overline{X}}_n\to \mathbb{E}{X}_1$ almost surely, so ${\tilde{\theta }}_n\to \theta$ almost surely and therefore is a consistent estimator. The central limit theorem gives

A standard Brownian motion is a real process $(B_t{)}_{t\ge 0}$ with $B_0=0$ almost surely, continuous paths, independent increments, and $B_t-B_s\sim N(0,t-s)$ for $0\le s<t$. If a filtration is specified, require adaptedness and independence of each increment from the past sigma-algebra. Thus the increments are centered Gaussian with variance equal to elapsed time.

Use the usual normal linear model $Y=X\beta +\epsilon$, with fixed full-column-rank design $X\in {\mathbb{R}}^{n\times p}$, $n>p$, and $\epsilon \sim N(0,{\sigma }^2{I}_n)$. These Gaussian distribution and constant-variance hypotheses are necessary for an exact finite-sample Student's t-test. Put $V=(X^{T}X{)}^{-1}$ and $\nu =n-p$. The vector of ordinary least squares estimators satisfiesWith $s^2=\Vert Y-X\hat{\beta }{\Vert }^2/\nu$, the test statistic isUnder the null hypothesis, $T$ has Student's t-distribution with $\nu$ degrees of freedom. A two-sided level-$5\%$ Student's t-test rejects when $|T|>t_{\nu ,0.975}$. If the intended test is one-sided against ${\beta }_0>0$, it rejects when $T>t_{\nu ,0.95}$; the statistical power conclusion below holds for either convention.

A maximum-likelihood estimate ${\hat{\theta }}_n$ is a measurable maximizer of ${\ell }_n(\theta )={\sum }_i\log f(X_i,\theta )$ over $\Theta$. Under identifiability, consistency conditions, an interior true parameter, sufficient differentiability and finite positive Fisher information, the standard limits areThe variance in the normal distribution is the reciprocal of the per-observation Fisher information, not of $nI$. Strong consistency needs additional hypotheses if it is to replace convergence in probability; the regular asymptotic result stated here does not require claiming it.

In the birth-death master equation, The upward jump has rate $\lambda$ and the downward jump rate $\beta x$, so the first two jump moments per unit time are $A(x)=\lambda -\beta x$ and $B(x)=\lambda +\beta x$. The second-order Kramers-Moyal expansion gives the diffusion approximation of a birth-death processFor the associated Fokker-Planck equation, integrating by parts with vanishing moment boundary terms gives $d⟨f⟩/dt=⟨A{f}^{\prime }+\frac{1}{2}B{f}^{\prime \prime }⟩$. Taking $f=x,x^2$ yieldsFor example, if $m_0=⟨x(0)⟩$, then $m(t)=\lambda /\beta +(m_0-\lambda /\beta )e^{-\beta t}$, while the variance satisfies $v^{\prime }=-2\beta v+\lambda +\beta m$. The two boxed moment equations also follow exactly from the discrete Markov jump-process generator, since quadratic functions have no higher jump terms. A reflecting diffusion confined to $x\ge 0$ can have extra boundary contributions; the formal approximation near extinction is not an exact replacement for the discrete process.

Under fit.2, observations in each period are independent Poisson random variables with a common mean, so each sample variance should fluctuate around its sample mean. The plot instead has variances roughly 130–300 for means roughly 80–120, well above the line variance equals mean. This is overdispersion.

Day-to-day traffic variation can produce a Poisson mixture: a random daily intensity increases the marginal variance and can correlate observations from the same day. A useful replacement is a negative binomial regression, with variance $\mu +{\mu }^2/\kappa$, or a Poisson generalized linear mixed model with a day random intercept. A day-factor effect can also be included if inference is restricted to the observed days. A Quasi-Poisson regression estimates a dispersion multiplier for uncertainty, but does not itself explain the shared-day dependence and has no ordinary likelihood-based AIC. The means can retain the period-factor or selected jump structure.