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Let the positive undisturbed surface tension be $\bar{\sigma }={\sigma }_0-A{\Gamma }_0>0$. An insoluble surfactant perturbation $g$ produces ${\sigma }_1=-Ag$. The tangential Marangoni stress changes the surface condition to $\mu u_z(h)={\sigma }_x$. Solving the lubrication theory momentum equation now givesAt leading order, conservation of insoluble surfactant on a moving interface reduces to ${\Gamma }_t+(\Gamma u_s{)}_x=0$. Corrections from the changing circumference are smaller by $h_0/a$. Likewise the normal-pressure contribution $-{\sigma }_1/a$ gives a coupling smaller by $h_0/a$ than the retained tangential Marangoni stress, so it is omitted at the same order.

For thickness and concentration amplitudes $H,g$ proportional to $e^{\beta t+ikx}$, put $D=a^{-2}-k^2$ and use $p_1=-\bar{\sigma }DH$. The two conservation equations becomeSet $s=k^2{a}^2$, $\lambda =h_0^3\bar{\sigma }/(3\mu a^4)$ and $\alpha =h_{0}A{\Gamma }_0/(\mu a^2)$. The uncoupled capillary growth and elastic relaxation rates are $L=\lambda s(1-s)$ and $S=\alpha s$. The off-diagonal product is $3LS/4$, so the determinant is $(\beta -L)(\beta +S)+3LS/4$. Thus the surfactant stabilization of an annular liquid lining relation isThis proves the dispersion relation before its lettered deductions. The physical assumption $\bar{\sigma }>0$ is needed for the claimed capillary instability band.

For $B_0>0$, change to logarithmic magnitude and its correctly transformed probability density function:Using ${\partial }_B=B^{-1}{\partial }_x$, the radial equation becomesThis is diffusion with constant drift. Set $y=x-3Dt$ and $g(y,t)=f(y+3Dt,t)$; then $g_t=D{g}_{yy}$. The initial Dirac delta transforms to $f(x,0)=\delta (x)$, so the heat-kernel solution givesDividing by $B$ yields the lognormal magnetic-field amplification lawExpanding the square and using $B=B_0{e}^x$ gives the equivalent requested formThe logarithm is a Gaussian random variable with mean $3Dt$ and variance $2Dt$, which proves normalization and spreading directly. The exponential moments of that Gaussian distribution give $⟨B^q⟩=B_0^q{e}^{Dq(q+3)t}$, independently checking the energy growth. The limit at $t\downarrow 0$ is the initial Dirac delta distribution, not a pointwise finite function.

Write $ϵ=l/d_i\ll 1$ and $\delta =v_A\tau /l\ll 1$. For magnetic amplitudes of order $B_0$, ion acceleration is of order $v_A^2/l$. Thus its change over the fast time isThe ions are consequently fixed to leading order on this timescale. Their first nonzero response is driven by the momentum equation, with negligible self-advection for an initially resting ion flow; one need not assert that the dimensional Lorentz acceleration literally vanishes. In fast-time, Alfvén-velocity dimensionless variables, the ion equation reads ${\partial }_{\hat{t}}\hat{\mathbf{u}}=O(\delta )$, giving ${\partial }_{\hat{t}}\hat{\mathbf{u}}=0$ at leading order.

Meanwhile the electron-current velocity is $j/(en)\sim v_A{d}_i/l=v_A/ϵ$, larger than a bulk flow of order $v_A$. The ion-advection term in induction is smaller than Hall induction by $O(ϵ)$ even if that slow flow is retained. The leading magnetic equation is thereforewith stationary ions at leading order. This is the fast-time electron-MHD limit of Hall magnetohydrodynamics and is precisely electron magnetohydrodynamics.

The already obtained fast-wave eigenvector has $u/b=|q/\omega |\sim 1/(k{d}_i)\ll 1$, confirming that ion motion cannot affect its leading frequency. The closed magnetic equation therefore suffices to obtain that branch without using momentum to close it. Its timescale is ${\tau }_w\sim 1/(v_A{d}_i|k_{\parallel }|k)$; compatibility with ${\tau }_w\ll l/v_A$ requires $|k_{\parallel }|d_i\gg 1$ when $l\sim k^{-1}$. Arbitrarily small $k_{\parallel }$ can violate this fast-time assumption even though $k{d}_i\gg 1$. The slow branch does not obey the stated ordering and is not retained by the fixed-ion reduction.

When $k{d}_i\gg 1$, expand the square root in the exact frequencies. The fast branch haswhile the slow branch has ${\Omega }_{\mathrm{slow}}=|q|[(k{d}_i{)}^{-1}+O((k{d}_i{)}^{-3})]$. HenceThe fast dispersive wave is the mode sought in the question's subsequent fast-time limit. In a strongly oblique wave, $k\simeq k_{\perp }$, so its frequency scales as $|\omega |\simeq |k_{\parallel }|v_A{k}_{\perp }{d}_i$. The slow branch instead tends to $(|k_{\parallel }|/k){\Omega }_i$, where the ion cyclotron frequency is ${\Omega }_i=v_A/d_i=e{B}_0/(m_{i}c)$.

There is a terminology qualification: in the pressure-free incompressible Hall equations used here, the fast branch is conventionally a whistler wave. The question calls the requested small-scale waves kinetic Alfvén waves, but a general kinetic-Alfvén dispersion includes pressure/kinetic ordering absent from these equations. The boxed dispersion is the result for the printed model; no extra pressure response or kinetic term has been silently added.

Solving the polarization quadratic gives all four signed roots:A sign-independent description is the pair of positive frequency magnitudesFor wavelengths long compared with the ion skin depth, $k{d}_i\ll 1$, both magnitudes approach $|k_{\parallel }|v_A$. The Hall correction is then small and the usual Alfvén wave dispersion ${\omega }^2=k_{\parallel }^2{v}_A^2$ is recovered. In an incompressible system the two transverse polarizations include the shear and pseudo-Alfvén wave sectors; away from this limit Hall coupling selects circular helicity combinations. A small parallel wavenumber alone does not remove the relative Hall splitting: the expansion parameter is the full $k{d}_i$.

Linearize about zero bulk flow and the uniform guide field. Put $\mathbf{b}=\delta \mathbf{B}/\sqrt{4\pi \rho }$, and take plane waves proportional to $e^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}$. Let $k_{\parallel }=\mathbf{k}\cdot \hat{\mathbf{z}}$, $k=|\mathbf{k}|$, $q=k_{\parallel }{v}_A$, andBoth $\mathbf{u}$ and $\mathbf{b}$ are perpendicular to $\mathbf{k}$. Projecting momentum onto that plane eliminates total pressure. The remaining magnetic-tension term givesIn the induction equation, the Hall contribution is $-c(B_0{\partial }_z)(\nabla \times \delta \mathbf{B})/(4\pi en)$; its two derivative factors of $i$ in the Fourier mode produce a positive coefficient multiplying $\mathbf{k}\times \mathbf{b}$. ThusEliminating $\mathbf{u}$ givesOn the transverse plane, $(i\mathbf{k}\times {)}^2=k^2$, so applying that operator again proves the incompressible Hall-MHD wave dispersion:Equivalently, in a helicity decomposition of a transverse Fourier mode with $i\mathbf{k}\times {\mathbf{b}}_h=hk{\mathbf{b}}_h$, $h=\pm 1$, each polarization obeys ${\omega }^2-hqk{d}_i\omega -q^2=0$. For $k_{\parallel }=0$ the projected linear restoring terms vanish and these modes have zero frequency; the propagating calculation assumes $k_{\parallel }\ne 0$.

True as a major quantitative test of the hot Big Bang, with a significant lithium qualification. Big Bang nucleosynthesis predicts several abundances from one main baryon-density parameter using measured nuclear and weak rates, rather than fitting a separate primordial composition for every species. Comparisons involve distinct observing environments and an independent cosmic microwave background determination at a much later epoch. Their overall agreement is powerful; it is not a claim that every light-element measurement is mutually consistent without stellar or observational corrections. The cosmological lithium problem must be included in that assessment.

Use the relative potential $\Psi ={\Phi }_{\ast }-\Phi$ and integrate the stated, untruncated Maxwell-Boltzmann velocity distribution over ${\mathbb{R}}^3$. The normalization cancels its three Gaussian integrals:The gravitational potential sign reverses under the relative-potential convention, so the spherical Poisson equation for Newtonian gravity isAlso ${\Psi }^{\prime }=-Gm(r)/r^2$, with $m^{\prime }=4\pi r^2\rho$. Differentiating the density relation yieldsAn ideal gas with constant temperature has $P=\rho k_{B}T/m_0$. Its hydrostatic equilibrium equation is precisely the same whenTogether with the same enclosed-mass equation, central density and boundary conditions, this proves the isothermal stellar and gas density equivalence. In dimensionless terms, $y=-[\Psi -\Psi (0)]/{\sigma }^2$ and $x=r\sqrt{4\pi G\rho (0)}/\sigma$ obey $x^{-2}(x^2{y}^{\prime }{)}^{\prime }=e^{-y}$ with $y(0)=y^{\prime }(0)=0$ in both systems.

This calculation uses the distribution as printed for all velocities. Imposing an additional cutoff at positive relative energy changes the integral to ${\rho }_1[e^W\mathrm{erf}\sqrt{W}-2\sqrt{W}/\sqrt{\pi }]$, where $W=\Psi /{\sigma }^2\ge 0$, and no longer gives an exactly isothermal gas profile. An untruncated self-gravitating isothermal sphere also needs an outer modification for finite total mass. The function $\mathrm{erf}$ here is the error function. Finally, the half-line Gaussian integral printed in the note lacks a factor: the correct value is ${\int }_0^{\infty }{e}^{-\alpha x^2}dx=\frac{1}{2}\sqrt{\pi /\alpha }$.

The Gaussian integral over the full vertical velocity line is $\sqrt{2\pi }{\sigma }_z$. Thus the mass density of the self-gravitating isothermal slab isNeglecting radial derivatives makes the Poisson equation for Newtonian gravity ${\Phi }_{zz}=4\pi G\rho (z)$. Substituting $\Phi ={\sigma }_z^2\varphi$ and $z=z_0\zeta$, with $z_0^2={\sigma }_z^2/(8\pi G{\rho }_0)$, givesMultiply by ${\varphi }^{\prime }$ and integrate. The central value and symmetry under $z\mapsto -z$ fix the constant of integration:On the positive-$\zeta$ side, Newtonian gravity makes ${\varphi }^{\prime }\ge 0$. Put $y=e^{\varphi /2}$; then $4(y^{\prime }{)}^2=y^2-1$ with $y(0)=1$. Integration gives $2\mathrm{arcosh}y=\zeta$, and even reflection supplies the other side. ThereforeDirect differentiation verifies both the equation and central conditions. Uniqueness of the smooth initial value problem rules out other branches meeting those conditions. The surface density is $4{\rho }_0{z}_0$, and the conventional slab scale height is $2z_0$; the factor of two is only a definition of the length scale.

Let $d=k-1$ and let $e_m$ denote the elementary symmetric polynomial of degree $m$ in the $d$ interior knots $t_{i+1},\dots ,t_{i+d}$, with $e_0=1$. Expand the two sides of the Marsden identity as polynomials in $x$:andComparing the coefficient of $x^{d-m}$ and cancelling the common sign gives the monomial B-spline coefficientsEquivalently,In particular $a_i^{(0)}=1$, giving partition of unity on the interior interval. For $k\ge 2$, $a_i^{(1)}=(t_{i+1}+\cdots +t_{i+k-1})/(k-1)$, and the highest-degree coefficient is the product of those interior knots. The formulas also cover $k=1,m=0$ through empty products.

For fixed $t$, put $g_t(u)=(u-t{)}_{+}^{k-1}$, a truncated power function. Begin with distinct increasing knots. The two interpolation polynomials ${\ell }_i(\cdot ,t)$ and ${\ell }_{i+1}(\cdot ,t)$ agree at their $k-1$ shared knots. Their difference has degree at most $k-1$, soFor $k=1$ the empty product is one and the same assertion holds for constant interpolants. The leading coefficient of a degree-at-most-$k-1$ interpolant on $k$ nodes is its order-$k-1$ divided difference, by the Newton interpolation polynomial. ThereforeThis proves the Lee interpolation identityThe identity holds for every real $x,t$, with the usual convention for the order-zero truncated power. Admissible repeated knots are handled by the corresponding confluent divided differences and consistent one-sided limits at breakpoints; taking these limits preserves the Lee interpolation identity.

Sum over $i$ to telescope:If $t_k<t<t_{n+1}$, the first $k$ knot values of $g_t$ are zero, so ${\ell }_1=0$. At the last $k$ knots $t_{n+1},\dots ,t_{n+k}$, the function $g_t$ agrees with the polynomial $(u-t{)}^{k-1}$, so uniqueness of polynomial interpolation gives ${\ell }_{n+1}(x,t)=(x-t{)}^{k-1}$. Thus the Marsden identity is

Work with real-valued functions and put $\mathrm{sign}(0)=0$. Let $r=f-p_{\ast }$, and write an arbitrary competitor as $p=p_{\ast }+v$ with $v\in U$. Pointwise,For $r\ne 0$ this is $|r-v|\ge \mathrm{sign}(r)(r-v)$; for $r=0$ the right side is zero. Integrating and using the assumed orthogonality givesThus $p_{\ast }$ is a best $L^1$ approximation from $U$. This sign criterion for best L1 approximation is sufficient without any finite-dimensionality or uniqueness assumption; possible residual zeros do not invalidate the inequality.

Use the approximants just chosen. The weak type (1,1) bound givesThe measures of these exceptional sets are summable. Indeed, the measure of their union for $n\ge m$ is at most $C{\sum }_{n\ge m}{2}^{-n}\to 0$. Hence their limsup is null; this is the Borel-Cantelli first lemma. Outside it, $S(g-f_n)\to 0$.

Remove also the countable null set from part (b). At each remaining point and for every $n$, the estimate in part (a) yieldsLetting $n\to \infty$ makes the right side zero. ThereforeThis proves almost-everywhere convergence from weak-type domination. The argument uses countably many measurable exceptional sets and does not assume that a supremum over all real parameters is measurable.

The Metropolis–Hastings algorithm is useful when a target posterior can be evaluated up to a normalizing constant but direct independent sampling or simple conditional sampling is unavailable. It constructs a Markov chain with that posterior as invariant law. After adequate convergence, averages of functions of the chain estimate posterior expectations, and the retained draws approximate marginal distributions, quantiles and credible intervals.

For the random-walk Metropolis algorithm, at current $\theta$ propose ${\theta }^{\prime }=\theta +Z$, where $Z$ has a symmetric distribution, often $N(0,\Sigma )$. With a symmetric proposal the acceptance probability isAccept the proposal when an independent uniform is below this probability; otherwise retain the current value. For a nonsymmetric proposal $q({\theta }^{\prime }\mid \theta )$, include $q(\theta \mid {\theta }^{\prime })/q({\theta }^{\prime }\mid \theta )$ in the ratio. Detailed balance follows because the accepted flow is the minimum of the two proposed directional flows; rejection supplies the remaining self-transition probability. Irreducibility, appropriate recurrence and aperiodicity are also needed for convergence.

In the displayed trace plots, the first parameter moves in small increments with pronounced long-term wandering and a shift from its early region to a later region. This is consistent with an undersized proposal or a strongly correlated slow direction. Measure its acceptance rate: if it is very high, increase that coordinate's proposal variance. The second parameter moves rapidly within regions but changes its level around iteration 4500; that is evidence against treating the full run as a settled stationary sample, and it suggests investigating dependence with the first parameter or separated posterior regions. Its local scale may already be reasonable, so a blanket reduction of all proposal variances would be unwarranted. The third parameter has conspicuous long plateaus, indicating many repeated states: if the recorded acceptance is low, reduce its proposal variance; also check constraints and dependence before deciding that scale alone is the cause.

Rerun with coordinate-specific scale changes, inspect joint dependence, consider a covariance-based block proposal, and run longer chains from dispersed starting points. The MCMC trace diagnosis and proposal tuning should include actual acceptance rates and effective sample sizes; the three marginal trace plots do not by themselves identify correlations or prove convergence. Further practical details are in the three subparts.

The plug-in estimate isFor a concrete bootstrap construction, suppose the rows $(x_i,y_i)$ are independent and identically distributed, $\mathbb{E}(e_i\mid x_i)=0$, the design moment matrix is nonsingular, and the required moments exist. Use a paired bootstrap: sample $n$ rows with replacement, form $X^{\ast },Y^{\ast }$, refit by ordinary least squares, and compute ${\hat{\theta }}^{\ast }={\hat{\beta }}_1^{\ast }{\hat{\beta }}_2^{\ast }$. Repeat many times. Singular resampled designs require a suitable rank-preserving implementation or replacement replicate; with a regular continuous design their frequency vanishes asymptotically.

The bootstrap-t confidence interval method also needs a standard error in the original sample and every replicate. One asymptotically valid heteroscedasticity-robust covariance estimate isThe delta method for the coefficient product givesRecompute ${\hat{V}}^{\ast },d^{\ast },{\hat{s}}^{\ast }$ in each paired replicate. This is bootstrap inference for a product of regression coefficients.

With fixed rather than random design, resample centered residuals if errors are identically distributed with homoskedasticity, or use a properly specified wild bootstrap for heteroscedastic errors, keeping $X$ fixed and refitting after generating $Y^{\ast }$. The printed assumption $\mathbb{E}{e}_i=0$ alone does not establish independent observations or validity of a bootstrap confidence interval. The formulas below give asymptotic 95% coverage under the stated regularity conditions, not an exact finite-sample guarantee. In particular the ordinary first-order studentization needs $({\beta }_1,{\beta }_2)\ne (0,0)$ and a nonzero asymptotic variance; when both coefficients vanish the product has nonstandard behavior.

Assume the parents are unaffected, the disease has complete penetrance with a single recessive causal allele, and there are no phenocopies, new disease mutations or alternative causal loci. An affected first child has genotype $aa$, so each unaffected parent must be a heterozygote $Aa$. Conditional on these parental genotypes, Mendelian segregation in the second child's paternal and maternal meioses is independent of the transmissions to the first child. ThereforeThis is the same sibling recurrence risk as for unrelated, unaffected parents under the same recessive model. Parental kinship increases the prior probability that both parents carry the same rare disease allele; after both are identified as carriers, it does not alter their Mendelian transmission probabilities. The factor $63.4$ from the preceding part is therefore not an additional multiplier on the sibling recurrence risk.

If unaffected parents are not explicitly required, the rare-disease assumption is an approximation supporting the usual carrier-by-carrier calculation, rather than a logically exact consequence of an affected child alone. Under complete penetrance, the compatible parental matings are $Aa\times Aa$, $Aa\times aa$ and $aa\times aa$, with second-child risks $1/4,1/2,1$ respectively. Let their prior probabilities be $w_{11},w_{12},w_{22}$, including both orders in $w_{12}$. Conditioning on the affected first child givesWithout negligible or excluded affected-parent matings, this mixture need not equal $1/4$ and may depend on consanguinity. Likewise incomplete penetrance or disease heterogeneity invalidates the assertion that an affected child of unaffected parents necessarily identifies an $Aa\times Aa$ mating. The stated $1/4$ answer uses the unaffected-parent, fully penetrant single-locus assumptions.

Assume a single disease allele $a$, complete penetrance of $aa$, no disease in $Aa$ or $AA$, and Hardy-Weinberg equilibrium in the outbred population. If its population frequency is $q$, the disease prevalence is $q^2=10^{-6}$, so $q=10^{-3}$.

For a child with inbreeding coefficient $F$, the two inherited copies are identical by descent with probability $F$. In that component there is just one ancestral allele draw, and it is $a$ with probability $q$. In the remaining component there are two independent ancestral draws, with probability $q^2$ of $aa$. Thus the inbreeding genotype frequencies giveUsing the first-cousin inbreeding coefficient $F=1/16$, the disease-risk ratio isThe absolute risk in this model is $6.34375\times 10^{-5}$. This calculation averages over the parental genotypes in first-cousin marriages before observing an affected child. It is not the risk conditional on both parents being known genetic carriers, and assumes the specified cousin relationship is independent of disease status rather than conditioning on a selected disease pedigree.

Let the first cousins be $X,Y$, with parents $(P,U)$ and $(Q,V)$ respectively, where $P,Q$ are full siblings and $U,V$ have no ancestral relationship to one another or to that sibling family. To compute ${\varphi }_{XY}$, choose one allele copy from each cousin. They originate from $P$ and $Q$ with probability $1/4$. All other parental combinations have zero kinship coefficient under the otherwise outbred pedigree assumption. HenceThe two gene copies within a child of $X,Y$ are random transmissions from these parents. They are identical by descent precisely in the event defining parental kinship, so the child's inbreeding coefficient isFor a direct ancestral-path check, there are two shared noninbred grandparents. Through either grandparent, the two cousin-to-grandparent paths have two meioses each, and matching the ancestral grandparental copy supplies a further factor $1/2$. Each contributes $2^{-2-2-1}=1/32$, giving the same first-cousin inbreeding coefficient. Additional ancestral relationships or inbred grandparents would change the result.

The kinship coefficient ${\varphi }_{XY}$ is the probability that a randomly selected allele copy from individual $X$ and an independently selected copy from individual $Y$ at the same genetic locus are identical by descent. The sampling is of gene copies, not of whole genotypes. For a noninbred individual $X$, sampling its two copies with replacement gives ${\varphi }_{XX}=1/2$; in general ${\varphi }_{XX}=(1+F_X)/2$, where $F_X$ is its inbreeding coefficient.

For two full siblings in an outbred pedigree, the two sampled copies are both paternal with probability $1/4$, both maternal with probability $1/4$, and of different parental origin with probability $1/2$. Given that both are paternal, their independent Mendelian segregation transmissions select the same paternal ancestral copy with probability $1/2$; likewise for both maternal. The different-parent case cannot be identical by descent when the parents are unrelated. ThereforeEquivalently, conditional on an IBD-sharing count $J$, the chance that the two random copies match is $J/4$, so ${\varphi }_{\mathrm{sib}}=E[J]/4=1/4$. The expected shared proportion $E[J/2]=1/2$ and the usual additive relationship coefficient $2\varphi =1/2$ are twice this kinship coefficient; confusing these conventions would double the subsequent inbreeding coefficient.

Put $N=n_0+n_1+n_2$ and $s=2x=2\theta (1-\theta )$. The usual recombination fraction range $0\le \theta \le 1/2$ gives $0\le s\le 1/2$, and the preceding likelihood function becomesTo see its relation to nonparametric linkage analysis, use Bayes' theorem and the constant disease probability $1/16$ to obtain the affected-pair marker-sharing probabilitiesTheir multinomial likelihood has exactly the same $s$-dependent kernel, since its extra factor $2^{n_1}$ is constant. Set $A=2n_0+n_1$ and $B=2n_2+n_1$. Then $A+B=2N$, and $B$ is the total number of shared ancestral copies. At fixed $N$, the Fisher-Neyman factorization theorem makes $B$ a sufficient statistic. The observed mean shared proportion is $\hat{p}=B/(2N)$, and the binomial likelihood kernel givesThe unlinked null is $s=1/2$, whereas linkage corresponds to $s<1/2$, or expected shared proportion $1-s>1/2$. If $\hat{p}\le 1/2$, the constrained likelihood is maximized at the unlinked null. For $\hat{p}>1/2$, its log likelihood ratio iswhose derivative in $\hat{p}$ is $2N\log [\hat{p}/(1-\hat{p})]>0$. Hence it orders samples exactly by excess IBD sharing. Under this fully penetrant recessive model, parametric linkage testing and a one-sided test of the overall IBD-sharing proportion use the same information. This is allele-sharing sufficiency for recessive linkage, not a general equivalence for all disease models. Use limiting values for terms such as $0\log 0$, and assume $N>0$ for the estimators.