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Let $X$ have rows $(1,y_i)$ and let ${\Sigma }_{\vartheta }$ be the spatial covariance matrix with $\vartheta =({\tau }^2,{\sigma }^2,a)$. Conditional on chosen covariance parameters, the best linear unbiased estimator of the drift coefficients is the generalized least squares estimatorThe printed procedure first estimates the linear drift from depth ~ y, constructs a binned residual empirical semivariogram, fits a Gaussian semivariogram to it, and then uses the fitted covariance in generalized least squares trend estimation and universal kriging. The missing fit could be supplied aswhere those arguments are initial values, not the reported final estimates. Such semivariogram fitting typically minimizes a weighted sum of squared discrepancies between binned empirical values and model values; it is not automatically a joint Gaussian likelihood maximization. One can iterate: form residuals using the current drift, refit the covariance model, recompute the generalized least squares drift, and repeat until changes are small. The displayed commands themselves show a single sequence, not evidence that such iteration occurred.

A likelihood-based alternative estimates the two blocks coherently fromSubstitute $\hat{\beta }(\vartheta )$ to obtain a profile log-likelihood, maximize over ${\tau }^2,{\sigma }^2\ge 0$ and $a>0$ with a positive-definite matrix ${\Sigma }_{\vartheta }$, and recover the drift coefficients from the optimizing covariance. Restricted maximum likelihood adds the design determinant term $\log \det (X^T{\Sigma }_{\vartheta }^{-1}X)$ and uses $n-2$ in place of $n$ in the normalizing term, up to a fixed-design constant. It can reduce bias from estimating the drift before the covariance. A zero fitted nugget is a legitimate boundary estimate, not proof that measurement error is absent. Estimate drift using covariance-weighted regression, and spatial dependence from drift-adjusted variation, with the two stages or likelihood explicitly distinguished.

Let ${\hat{\delta }}_D$ and ${\hat{\delta }}_I$ be independent, approximately unbiased estimates of one consistent treatment effect, with variances $v_D,v_I>0$. The inverse-variance weighted mean givesIndeed, weights $w,1-w$ give variance $w^2{v}_D+(1-w{)}^2{v}_I$; differentiating gives $w=v_I/(v_D+v_I)$, the weight on the direct estimate. Here the three displayed trials give ${\hat{\delta }}_D=-1.33$, $v_D=0.04$, and hence $\hat{\delta }\simeq -1.3203$ with variance $0.03448$. This numerical pooling uses the displayed evidence, not the unavailable data from all twelve trials. A shared study or shared control would introduce covariance, requiring the corresponding correlated-estimator formula; inconsistency would invalidate the common-effect interpretation.

Under the intended additional assumption of mutually independent, continuously calibrated null p-values, each test fails to reject with probability $0.95$. The familywise error rate is thereforeThere is a qualification to the printed use of “not correlated”: zero pairwise correlation coefficients do not generally imply mutual independence. A multivariate normal distribution with zero off-diagonal covariances does give independence, but an arbitrary biomarker distribution need not. The formula above requires independent rejection events, not just uncorrelated measurements.

The usual intuition for positively dependent tests is that they reject together, reducing the effective number of opportunities for a false positive. A rigorous sufficient assumption is positive association of random variables for the rejection indicators $I_1,\dots ,I_{10}$. Products of their non-rejection indicators are decreasing functions; association and induction giveThus under this stronger positive-dependence assumption the probability is no larger, and perfect dependence gives $0.05$ instead of $0.4013$. Strict inequality is not forced by every form of positive dependence.

Pairwise positive correlation coefficients alone, as printed, is insufficient. For a counterexample choose a vector of ten exchangeable random variables that are rejection indicators as follows: with probabilities $0.545,0.450,0.005$, respectively, reject none, reject one uniformly selected test, or reject all ten. Thenbut the familywise error rate is $0.455>1-0.95^{10}$. This can be realized with valid null p-values: conditional on the indicators, independently draw $p_i$ uniformly on $(0,0.05)$ if $I_i=1$ and on $(0.05,1)$ otherwise. Every $p_i$ is uniform on $(0,1)$, and $\mathrm{Cov}(p_i,p_j)=0.25\mathrm{Cov}(I_i,I_j)>0$. Taking biomarker statistics $X_i=1-p_i$ gives pairwise positively correlated null statistics with exactly these one-sided tests. Therefore the unconditional larger-or-smaller claim needs a specified dependence model.

Use one joint hypothesis test for $H_0:{\beta }_{\lambda }={\beta }_{\mu }=0$. Fit the unrestricted model with statistical parameters $({\lambda }_0,{\mu }_0,{\beta }_{\lambda },{\beta }_{\mu })$, where ${\lambda }_i={\lambda }_0{e}^{{\beta }_{\lambda }{z}_i}$ and ${\mu }_i={\mu }_0{e}^{{\beta }_{\mu }{z}_i}$ for treatment indicator $z_i$. Refit under $H_0$, estimating the two common baseline intensities; do not fix them at their unrestricted estimates. The likelihood-ratio test statistic isUnder regular identifiable positive-rate models, sufficient independent patient trajectories, and the null hypothesis, Wilks theorem gives two degrees of freedom because two coefficients are constrained. Reject if $D$ exceeds the chosen upper chi-squared distribution quantile. Alternatively a two-dimensional Wald test uses ${\hat{\beta }}^T\widehat{\mathrm{Cov}}(\hat{\beta }{)}^{-1}\hat{\beta }$, including the off-diagonal covariance. Two separate tests do not provide this single calibrated assessment. The point estimates alone cannot produce a numerical p-value without fitted likelihoods or a coefficient covariance matrix.

The most plausible cause is strong dependence between the latent variables and covariance hyperparameters in their joint posterior distribution. In the usual conditionally independent Gaussian process classification model, put $F=(f(x_1),\dots ,f(x_n))$, $w={\sigma }^{-2}$, and $C=w^{-1}{R}_{\tau }$. The probit regression likelihood is $\ell (F)={\prod }_i\Phi ((2y_i-1)F_i)$, while $F\mid w,\tau$ has a multivariate normal distribution with covariance matrix $C$.

For distinct inputs and positive length scales, $R_{\tau }$ is positive definite. Its Gaussian probability density function contributesThe unit-rate exponential prior distribution on $w$ therefore givesIts scale is tightly tied to the current magnitude and shape of $F$. Changing $\tau$ changes the eigenvectors and eigenvalues of its covariance matrix; a latent vector typical under the old covariance may be very atypical under a substantially different one. Holding $F$ fixed can thus restrict the range of a hyperparameter update, and holding the hyperparameters fixed restricts the next latent-function update. Even exact sampling of each full conditional distribution can move slowly along a narrow ridge of the joint posterior distribution.

Binary observations provide limited information about the absolute size of large correctly signed latent values. This can leave substantial scale uncertainty and strengthen the dependence. Long correlation lengths can also make the covariance matrix nearly singular and the latent coordinates highly dependent. These are plausible explanations for a large mixing time, not proofs that every data set causes slow convergence. A blocked Gibbs sampler removes dependence within its chosen blocks, but not dependence between them. Repeated identical inputs should be represented by one shared latent value; otherwise the nominal Gaussian matrix is singular and the displayed inverse formula must be replaced by a representation on its support.

Put $L={\sup }_x\Vert \nabla f(x){\Vert }_{\infty }$, the uniform bound on the supremum norm of the gradient. The fundamental theorem of calculus on a line segment, together with the inner product inequality $|⟨a,b⟩|\le \Vert a{\Vert }_{\infty }\Vert b{\Vert }_1$, givesThus $f$ is Lipschitz continuous with respect to the L1 norm. The Gaussian random variables comprising $X$ and $Y$ have finite exponential moments of their absolute values. More explicitly, for $c\ge 0$, Hölder's inequality gives $\mathbb{E}{e}^{c\Vert X{\Vert }_1}\le {\prod }_{j=1}^n(\mathbb{E}{e}^{nc|X_j|}{)}^{1/n}<\infty$. Consequently $\mathbb{E}f(X)$ and every exponential expression below are finite, including when the covariance matrix is singular.

Use the convex function $\Psi (v)=e^{\lambda |v|}$. Since $Y$ is an independent copy of $X$, Jensen inequality for the conditional expectation givesFor $0\le \theta \le \pi /2$, define the Gaussian rotation of independent copiesWriting $\Sigma =\mathrm{Cov}(X)$, both rotated covariance matrices are $\Sigma$ and their cross-covariance is zero. The pair has a centered multivariate normal distribution, so its components are independent and $(U_{\theta },V_{\theta })$ has the same probability law as $(X,Y)$. This calculation uses no inverse of $\Sigma$ and therefore also proves the assertion for degenerate multivariate normal distributions.

The path $\theta \mapsto U_{\theta }$ runs from $Y$ to $X$. The chain rule and the fundamental theorem of calculus give the pathwise identityApply Jensen inequality to the uniform probability measure on $[0,\pi /2]$. Then use Tonelli theorem and the Gaussian rotation of independent copies to obtainThis proves the Gaussian rotation interpolation inequality in the required form:The stated almost-sure domination now gives an exponential moment of an absolute standard normal variable bound. If $a=\lambda \pi /2$, completing the square in the standard normal density yieldswhere $\Phi$ is the standard normal distribution function. Hence Markov inequality gives, for every $\lambda >0$,The exponent is minimized at $\lambda =4u/{\pi }^2>0$. Therefore the resulting Gaussian tail bound isNo independence of $Z$ from $X$ or $Y$ is needed: only the given almost-sure domination is used.

Construct a mean-square derivative of a Gaussian process on the original probability space before choosing its continuous modification. The real covariance function satisfies $K(h)=K(-h)$, because covariance is symmetric. Three-times differentiability implies that $K^{\prime \prime }$ is continuous everywhere and differentiable at zero. Since $K^{\prime \prime }$ is an even function,This uses differentiability of $K^{\prime \prime }$ at zero, not a bound on $K^{\prime \prime \prime }$ throughout a neighborhood.

Let $D_{h}X(t)=(X(t+h)-X(t))/h$ for $h\ne 0$. Twice applying the fundamental theorem of calculus to the covariance function givesOriented integrals make this identity valid for negative as well as positive $h,k$. For a fixed $t$, the right side tends to $-K^{\prime \prime }(0)$ as $h,k\to 0$. In particular,By completeness of L2 space, the mean-square derivative of a Gaussian processexists. Finite vectors consisting of values of $X$ and $Y$ have a centered multivariate normal distribution: approximate the $Y$ coordinates by difference quotients and use mean-square convergence and characteristic functions. Thus $Y$ is a centered Gaussian process jointly constructed with $X$, withIn particular $-K^{\prime \prime }(0)\ge 0$, and $-K^{\prime \prime }$ is a positive-semidefinite kernel because it is an actual covariance function. Merely constructing an unrelated Gaussian process with that covariance function would not yet construct a modification of a stochastic process of $X$.

The increment variance of $Y$ isFor small $|t-s|$, differentiability of $K^{\prime \prime }$ at zero bounds this by $C|t-s|$. For the remaining distances in $[0,1]$, continuity of $K^{\prime \prime }$ enlarges $C$ to give the same bound. The fourth moment of a centered normal distribution is three times its variance squared, soThe one-dimensional Kolmogorov continuity theorem states that $\mathbb{E}|U(t)-U(s){|}^p\le C|t-s{|}^{1+\eta }$ with $p,\eta >0$ gives a continuous modification, with sample-path Hölder continuity of every order less than $\eta /p$. Apply it here with $p=4,\eta =1$. We obtain a continuous modification $\tilde{Y}$ on $[0,1]$. On its null exceptional event, set the entire path equal to zero.

Define, on this same probability space,The path integral exists because its integrand is a continuous function. It agrees in L2 space with the integral of $Y$: Riemann sums converge in L2 space because $Y$ is mean-square continuous; sums of $\tilde{Y}$ have the same values almost surely at each finite collection of mesh points and converge pathwise to the integral. These two limits therefore agree almost surely for each fixed $t$. Equivalently, the integral is the Bochner integral of the L2 space-valued map $s\mapsto Y(s)$.

To verify the modification of a stochastic process property directly, put $A_t=X(t)-X(0)$ and $I_t={\int }_0^{t}Y(s)ds$. The preceding covariance identities and the fundamental theorem of calculus giveTherefore $\mathbb{E}(A_t-I_t{)}^2=0$ and $\tilde{X}(t)=X(t)$ almost surely for every fixed $t$. This is also the mean-square fundamental theorem of calculus in this stationary setting. If a version on all of $\mathbb{R}$ is desired, retain the original $X(t)$ outside $[0,1]$; equality at every fixed parameter still holds.

The ordinary fundamental theorem of calculus now gives the simultaneous pathwise conclusionThus the requested differentiable modification of a stationary Gaussian process exists. Integrability of $K$ is compatible with the hypotheses but is unnecessary for this direct argument. It also avoids any assumption that a third absolute moment of a spectral density follows from three-times differentiability.

Compute the coupling flow at zero magnetic field, or assume the source has support only in retained Fourier modes, so that the shell Gaussian measure is centered. First justify the shell propagator. For a positive quadratic kernel on the shell, the Gaussian functional integral gives covariance equal to the inverse kernel,with both momenta restricted to $\Lambda /b<|\mathbf{p}|\le \Lambda$. Inserting the Fourier transforms and using the Dirac delta function yields the Gaussian shell covarianceChanging $\mathbf{p}$ to $-\mathbf{p}$ gives the negative-exponent convention as well. The dependence is on the separation, as required by translation invariance. The printed numerator uses $\mathbf{x}$ alone: it is correct only if that symbol means the separation or if $\mathbf{y}=0$. For example at $\mathbf{x}=\mathbf{y}\ne 0$, the covariance must equal the constant coincident value $G_{>}(0)$, while the literal printed integral generally depends on $\mathbf{x}$. Thus the missing separation is a real source qualification, not a change of Fourier-sign convention.

Integrating out the shell gives $H_{\mathrm{eff}}[{\varphi }_{<}]=H_0[{\varphi }_{<}]-\log ⟨e^{-V[{\varphi }_{<}+{\varphi }_{>}]}{⟩}_0^{\mathrm{shell}}$, up to a field-independent constant. At first order in $u_0$, the cumulant expansion of a coarse-grained free energy retains $⟨V⟩$. Odd moments of the centered Gaussian measure vanish and Wick theorem gives $⟨{\varphi }_{>}^4⟩=3G_{>}(0{)}^2$. Writing $I_1=G_{>}(0)$,The last term affects only the constant; the second is a momentum-independent tadpole diagram correction. Before rescaling it changes $r_0$ to $r_0+u_0{I}_1/2$, leaves the quartic coefficient at $u_0$, and produces no gradient correction. Combining with part (a) gives the leading coupling recursionThis is one-loop shell mass renormalization in scalar quartic theory. Here $I_1={\int }_{\mathrm{shell}}{d}^{D}p(2\pi {)}^{-D}\alpha /(p^2+\alpha r_0)$. The shell Gaussian is well defined if $p^2+\alpha r_0>0$ there; near $r_0=0$ this follows from $\alpha >0$. A negative slow-mode mass may be stabilized by the retained quartic term and does not require pretending that the full unconstrained quadratic measure at negative mass is normalizable.

Linearizing about $(r_0,u_0)=(0,0)$ givesIts mass and quartic eigenvalues are $b^2$ and $b^{4-D}$. The off-diagonal term is an additive critical-mass shift: setting the bare $r_0=0$ is not generally the critical tuning when $u_0>0$. The mass remains relevant for every $D$; at $D=2$ the two linear eigenvalues coincide and the matrix can have a Jordan block, but both perturbations are still relevant.

For $D>4$, the quartic interaction is an irrelevant operator, and the Gaussian fixed point attracts weak quartic perturbations after the mass and magnetic field have been tuned. The quartic can nevertheless be a dangerously irrelevant coupling: its positive value stabilizes the ordered phase. For $D<4$, it is a relevant operator, so the Gaussian description is unstable toward interactions. At $D=4$ it is a marginal operator; first-order perturbation theory alone does not decide its fate.

To settle that borderline case, retain the leading quartic contribution from the second cumulant, at zero external momentum in the local expansion. The two-fast-field part of $V$ is $(u_0/4)\int {\varphi }_{<}^2{\varphi }_{>}^2$. Since $⟨{\varphi }_{>}^2(\mathbf{x}){\varphi }_{>}^2(\mathbf{y}){⟩}_c=2G_{>}(\mathbf{x}-\mathbf{y}{)}^2$, its connected second cumulant contributesKeeping the local quartic term and using Parseval identity gives $\int d^{D}y{G}_{>}(\mathbf{x}-\mathbf{y}{)}^2=I_2$, where $I_2={\int }_{\mathrm{shell}}{d}^{D}p(2\pi {)}^{-D}[\alpha /(p^2+\alpha r_0){]}^2$. Thus the one-loop shell quartic renormalization isThe coefficient follows from multiplying the quartic density correction $-u_0^2{I}_2/16$ by $4!=24$. This local, zero-external-momentum coupling extraction is not a claim that the exact finite-shell effective action retains only its original polynomial; further operators are generated.

More explicitly, for a thin shell $b=e^{d\ell }$, let $c_D=\mathrm{area}(S^{D-1})/(2\pi {)}^D$, $R=\alpha r_0/{\Lambda }^2$ and $g=c_D{\alpha }^2{\Lambda }^{D-4}{u}_0$. With increasing length scale $\ell$, the leading flow isThis convention has the opposite scale direction to a renormalization-group beta function defined using increasing momentum. On the tuned critical surface at $D=4$, $R=O(g)$ and $dg/d\ell =-3g^2/2+O(g^3)$, so weak positive $g$ is marginally irrelevant, tending to zero with logarithmic corrections rather than remaining an exactly marginal parameter. Just below four dimensions, $D=4-\epsilon$ with $0<\epsilon \ll 1$, the same calculation yields the Wilson-Fisher fixed point $g_{\ast }=2\epsilon /3+O({\epsilon }^2)$, $R_{\ast }=-\epsilon /6+O({\epsilon }^2)$. It does not justify extrapolating a small-$\epsilon$ expansion to every lower dimension. The ordinary scalar quartic upper critical dimension is therefore

Write $K=k_1+k_2+k_3>0$ and $C=ϵ+1-c_s^2$. The inflationary scalar Fourier mode has real late-time value $A_k=H/\sqrt{4ϵM_{\mathrm{Pl}}^2{c}_s{k}^3}$ and $u_k^{\prime \ast }(\tau )=A_k{c}_s^2{k}^2\tau e^{ik{c}_s\tau }$. Thus the time integrand, before the factor $-2i$, isThe Bunch-Davies vacuum prescription supplies damping in the past. Equivalently, use a positive regulator $d$ and then take the limit:The real part of $-2i$ times this integral is $4/(c_s^3{K}^3)$. Including the six Wick contractions yields the cubic time-derivative curvature bispectrumThe resulting primordial bispectrum is a homogeneous function of degree $-6$ in the wavenumbers, as expected for scale-independent background parameters.

The printed limit $c_s\ll 0$ is not a physical sound-speed limit for these modes. The normalization and positive-frequency choice assume $c_s>0$; a negative value is outside the positive sound-speed convention used by these modes. The likely intended limit is $0<c_s\ll 1$, which is analyzed explicitly here rather than silently substituted for the PDF.

To compare primordial non-Gaussianity, normalize by the power spectrum, not only by the raw three-point amplitude. HereAt an equilateral bispectrum configuration, the exact normalized amplitude isThus $c_s\to 1$ gives an amplitude of order $ϵ$, consistent with weak non-Gaussianity in single-field slow-roll inflation. For $0<c_s\ll 1$, this vertex gives an amplitude of order one, enhanced relative to the slow-roll limit but not diverging as $c_s^{-2}$ after power-spectrum normalization. At fixed $H,ϵ,M_{\mathrm{Pl}}$ the raw bispectrum does scale as $c_s^{-2}$, but $P_{\zeta }^2$ has the same scaling. The often-quoted larger small-sound-speed amplitudes of more general inflationary actions require their actual interaction coefficients and other vertices; they do not follow from the lone supplied vertex. These conclusions concern its tree contribution within the stated weak-perturbation approximation.

The bispectrum shape function isAt fixed perimeter, the arithmetic-geometric mean inequality bounds this by $1/27$, attained only at an equilateral bispectrum configuration. In a squeezed bispectrum configuration with $k_1=q\ll k_2\simeq k_3=k$, it behaves as $q/(8k)\to 0$. At a flattened bispectrum configuration $(k,k,2k)$ it is finite, $1/32$, with no boundary singularity. Thus the shape prefers equilateral triangles and suppresses squeezed triangles, with a broad finite flattened contribution. This is a shape preference, not a complete signal-to-noise calculation: actual observational weighting also depends on cosmological transfer functions, covariance and accessible triangles. If $C=0$, this vertex produces no bispectrum at any configuration; if $C<0$, the signed shape reverses while its absolute triangle preference is unchanged.

Use the comoving curvature perturbation mode normalization. Take $H,ϵ,c_s$ to be effectively constant at leading slow-roll approximation order, with $H,ϵ,c_s>0$, and use the Bunch-Davies vacuum selected by the early-time $i\epsilon$ contour. The regulator $\epsilon$ is distinct from the Hubble slow-roll parameter $ϵ$.

Let $C=ϵ+1-c_s^2$ and use the Fourier transform convention $\zeta (\mathbf{x})=\int d^{3}k\zeta (\mathbf{k})e^{i\mathbf{k}\cdot \mathbf{x}}/(2\pi {)}^3$. Conversion to conformal time givesThe last sign follows from substituting $a=-1/(H\tau )$, which is positive for $\tau <0$; dropping the explicit minus sign in that substitution would reverse the final primordial bispectrum.

At the order needed by the in-in formalism, the unequal-time Wick contraction isIt uses the Gaussian quantum vacuum two-point function with the displayed operator ordering; an arbitrary classical unequal-time covariance cannot replace it. For the connected three-point function, each external field contracts with one of the three fields at the vertex. There are $3!=6$ such Wick contractions, all equal. The spatial integral conserves total momentum and leaves one factor $(2\pi {)}^3{\delta }^{(3)}({\mathbf{k}}_1+{\mathbf{k}}_2+{\mathbf{k}}_3)$. The other nine pairings are disconnected tadpole terms supported on a zero external momentum; they are excluded from the connected primordial bispectrum, equivalently by subtracting the one-point function or normal ordering the vertex.

The free Gaussian odd correlator is zero. Applying the stated first-order in-in formalism therefore givesBy definition, the full connected correlator is $(2\pi {)}^3{\delta }^{(3)}({\sum }_r{\mathbf{k}}_r)B$. The assumptions are a Gaussian Bunch-Davies vacuum, the given leading cubic vertex, tree order, the late-time limit $\tau \to 0^{-}$, effectively constant background parameters, and connected nonzero external modes. No symmetry factor $1/3!$ is inserted into the given Hamiltonian.

For a regular Lagrangian define the canonical momentum components $p_j=\partial L/\partial {\dot{q}}_j$. Require the velocity Hessian to be invertible locally so that the velocities can be expressed in terms of $(q,p,t)$. The Legendre transform in mechanics definesThe explicit time argument cannot generally be omitted when the given Lagrangian depends on time. Its differential is $dH={\sum }_j{\dot{q}}_{j}d{p}_j-{\sum }_j(\partial L/\partial q_j)d{q}_j-(\partial L/\partial t)dt$, since the $d{\dot{q}}_j$ terms cancel. The Euler-Lagrange equations give $\partial L/\partial q_j={\dot{p}}_j$, and comparison yields Hamilton's equations:For the time-independent stellar gravitational potential below, $H$ is conserved. Its sign convention differs from the binding energy in question 2: here $E=H=v^2/2-\psi$ is the ordinary specific orbital energy.

The prescribed nondegenerate prolate spheroidal coordinates require $0<\alpha <\beta$; equality collapses the coordinate family. Their physical velocity components are $v_j=h_j{\dot{q}}_j$, and $h_{\varphi }=R$. Per unit stellar mass the Lagrangian is $L=\frac{1}{2}{\sum }_j{h}_j^2{\dot{q}}_j^2+\psi$, so $p_j=h_j^2{\dot{q}}_j=h_j{v}_j$. Its Hamiltonian isThis uses acceleration $+\nabla \psi$, as specified, rather than importing the opposite gravitational potential convention.

For the Stäckel potential, use the Hamilton-Jacobi equation with principal function $S=W_{\lambda }(\lambda )+W_{\mu }(\mu )+l\varphi -Et$. Axisymmetry makes $p_{\varphi }=l=R{v}_{\varphi }$ constant. Write $d=\lambda -\mu$ and $A(\tau )=(\tau +\alpha )(\tau +\beta )$. Multiplying $H=E$ by $d$ and usingseparates it as $F(\lambda )-E\lambda =F(\mu )-E\mu$, where $F(\tau )=2A(\tau )(W_{\tau }^{\prime }{)}^2-(\alpha -\beta )l^2/[2(\tau +\alpha )]-G(\tau )$. Each side depends on its own coordinate only and therefore equals a constant, chosen as $-I$. Here $G(\tau )$ is the generating function of the gravitational potential, not the gravitational constant. HenceThis proves the conserved axisymmetric Stäckel third integral. Eliminating $E$ between the two expressions and using $p_j=h_j{v}_j$ givesAt regular generic phase-space points the three separation constants $E,l,I$ are independent; special circular or degenerate orbits can reduce their local independence. Thus the model is an integrable Hamiltonian system.

For the intended regular, phase-mixed steady model, Jeans theorem permits a galactic distribution function $f=F(E,l,I)$. At any fixed point, $E$ and $I$ depend on $v_{\lambda }$ and $v_{\mu }$ only through their squares, and $l$ depends only on $v_{\varphi }$. Thus $f$ is even separately in $v_{\lambda }$ and $v_{\mu }$, whether or not it is even in $v_{\varphi }$. Odd integration proves velocity-ellipsoid alignment in a Stäckel potential:The argument allows azimuthal streaming: it does not require zero mean $v_{\varphi }$. The means in the two meridional directions vanish, so the off-diagonal velocity covariance also vanishes. Second moments must exist for this conclusion.

There is a hypothesis behind the printed universal wording. On a generic nonresonant regular invariant torus, a continuous steady distribution is constant along a dense orbit and hence depends only on the torus integrals; this justifies the phase-mixed form. Bare stationarity alone is not sufficient in a degenerate resonant gravitational potential with additional integrals. For example $H=\frac{1}{2}(v^2+{\omega }^2{r}^2)$ is a Stäckel Hamiltonian: choose $G(\tau )=-\frac{1}{2}{\omega }^2[{\tau }^2+(\alpha +\beta )\tau ]$, up to a constant vacuum shift. The additional conserved quantity $K_{xy}=v_x{v}_y+{\omega }^{2}xy$ gives a positive normalizable steady tracer $f\propto \exp (-aH+ϵK_{xy})$ for $a>|ϵ|>0$. Its local $⟨v_x{v}_y⟩=ϵ/(a^2-{ϵ}^2)\ne 0$. At azimuth zero and a regular point where the $\lambda$ direction has a radial component, this gives nonzero $⟨v_{\lambda }{v}_{\varphi }⟩$. This tracer need not itself source the harmonic gravitational potential. It exhibits why the stronger, ordinary phase-mixed three-integral assumption is necessary instead of claiming that every imaginable steady distribution has the asserted alignment.

The geometry follows by multiplying the coordinate-defining equation and evaluating at $\tau =-\alpha ,-\beta$:Constant $\lambda$ surfaces are prolate spheroids, with equatorial semiaxis $\sqrt{\lambda +\alpha }$ and polar semiaxis $\sqrt{\lambda +\beta }$. Constant interior $\mu$ surfaces are two-sheeted hyperboloids. Their common foci lie at $z=\pm \sqrt{\beta -\alpha }$. The orbit volume in an axisymmetric Stäckel potential is determined byIts allowed coordinate intervals have turning boundaries or coordinate endpoints. A typical bound orbit with $l\ne 0$ oscillates between inner and outer spheroids, oscillates vertically between hyperboloidal limits, and circulates about the symmetry axis with $\dot{\varphi }=l/R^2$. The equatorial fold $\mu =-\beta$ joins the positive and negative $z$ branches; it is not necessarily a physical turning point in $z$.

For three incommensurate frequencies, the regular trajectory densely samples its projected allowed volume, with nonuniform occupation determined by its speeds. It does not follow one coordinate surface, and it need not close. Resonant orbits may be closed or lie in a smaller projected set; $l=0$ confines azimuthal motion to a meridional plane (with the usual axis coordinate qualification), and circular or equatorial limits are also lower-dimensional. Smooth Stäckel separation removes the generic chaotic behavior of nonintegrable potentials; it does not assert that every orbit fills a three-dimensional volume.

The common-age conditional probabilities giveHere $GW$ includes both possible component orders. Hence the binary star fractions have the concise ratioAs a check on the role of shared age, $p_{GG}=4p_G^2/3$ rather than $p_G^2$. This is covariance induced by a shared latent variable: the red giant indicator random variables have covariance equal to the variance of $g(Y)$, namely $1/97200$.

For a starburst galaxy with age $t$, put $y_b=t/(10\mathrm{Gyr})$. There is no age averaging. The component masses remain independent, so the fraction of systems with two red giants isThe last comparison is ambiguous in the printed question. Literally, the fraction of individual red giants in the first galaxy is $1/180$. Since $p_{GG}^{\mathrm{burst}}<1/8100<1/180$, no allowed starburst age satisfies that literal comparison. If the intended comparison is instead between systems containing two red giants in the two galaxies, compare with $p_{GG}=1/24300$; the answer isFor completeness, the individual red giant fraction in the burst is $g(y_b)=t/(900\mathrm{Gyr})$, which exceeds $1/180$ precisely for $5\mathrm{Gyr}<t<10\mathrm{Gyr}$. This is a third, different comparison; it does not change the computed fraction of two-giant systems.

The speckle contrast of a sinusoidal wavefront error gives a peak modal wave amplitude $h_0=(\lambda /\pi )\sqrt{C}$. If all $M$ residual modes have that wave amplitude, their coefficient norm is $H_{\mathrm{coeff}}=\sqrt{M}{h}_0$. Substituting the printed quadrant count gives the intended expressionThere is a normalization issue if this quantity is called physical spatial root mean square wavefront error. For $h(\mathbf{r})={\sum }_k{h}_{0k}\cos (2\pi {\mathbf{u}}_k\cdot \mathbf{r}/D+{\psi }_k)$, the actual pupil-averaged value isOn an orthogonal full-period basis, $⟨s_k^2⟩=1/2$ and cross terms vanish. Alternatively, independent uniform random wave phases give this result after wave phase averaging, even on a circular optical pupil. With the same $M$ and equal contrasts the physical RMS is thenA single full-period cosine already demonstrates the issue: its RMS is $h_0/\sqrt{2}$, not $h_0$. On the actual circular optical pupil, any nonconstant unit-peak cosine also has mean square strictly less than one; averaging over its wave phase gives exactly $1/2$. On a finite circular optical pupil, phase-dependent mode overlaps can additionally matter. If a constant piston is removed before computing RMS, replace the mode Gram matrix by its centred optical pupil covariance. The printed value can be recovered as a modal coefficient norm, or by counting $2M$ orthogonal peak-amplitude quadratures in the physical RMS; neither convention is specified by the quoted mode definition. This is the distinction between modal amplitude norm versus wavefront RMS, not a correction silently made to the PDF. Unequal residual contrasts require ${\mathbb{E}}_{\psi }{h}_{\mathrm{rms}}^2={\lambda }^2{\sum }_k{C}_k/(2{\pi }^2)$, so one speckle's $C$ alone cannot determine the total RMS. The weak-aberration approximation also requires the total wave phase error to stay small, not merely each coefficient separately.