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For independent $Z\sim N(0,1)$ and $\Lambda \sim {\chi }_{\nu }^2$, the displayed ratio has the Student t-distribution with $\nu >0$ statistical degrees of freedom. Equivalently, conditional on $\Lambda =\lambda >0$, $T$ has normal distribution $N(0,\nu /\lambda )$. This is a normal scale mixture representation of Student t: a small latent denominator occasionally produces a much larger spread. The result is a heavy-tailed distribution even though every conditional probability distribution is a normal distribution.

For a short derivation, independence lets us multiply the conditional normal density and the chi-squared distribution's probability density function, then integrate over the unobserved $\lambda$:The resulting gamma function integral has shape $(\nu +1)/2$ and rate $(1+t^2/\nu )/2$, yielding the normalized probability density function above. Also, for $\nu >2$, conditional expectation gives $\mathbb{E}[T^2]=\nu \mathbb{E}[{\Lambda }^{-1}]=\nu /(\nu -2)$. Thus unit scale does not mean unit variance. Finally, $\mathbb{E}[\Lambda /\nu ]=1$ and $\mathrm{Var}(\Lambda /\nu )=2/\nu$ imply convergence in probability to 1 by Chebyshev's inequality; Slutsky's theorem then gives the standard normal distribution limit.

Take four distinct objects $a,b,x,d$ and six nonidentity arrows $m:a\to b$, $h:x\to b$, $r,s:b\rightrightarrows d$, $k:a\to d$, $l:x\to d$, with $rm=sm=k$ and $rh=sh=l$. These and identities are all morphisms. The only nonidentity epimorphisms have target $d$, which admits no arrow to $b$, so every lifting square against $m$ either has an identity left side or does not exist. Thus $m$ is a strong monomorphism. It is not a regular monomorphism: $h$ equalizes $r,s$ but cannot factor through $m$, and an identical pair has $1_b$ as equalizer. Every strong nonregular monomorphism in any category occurs as the image of $m$ under a faithful functor from this category. Indeed, such a monomorphism is not epic, so choose distinct $r,s$ agreeing on it; failure of its equalizer property then supplies $h$.

Let $X$ switch between $\pm 1$ at rate $q$ and be killed into zero at rate $\lambda$. To exceed a new level it must cross it in state $+1$. Restarting there by the Strong Markov property gives the multiplicative identity ${\psi }_{+}(c+d)={\psi }_{+}(c){\psi }_{+}(d)$, hence an exponential tail. First-step conditioning gives ${\psi }_{-}(c)={\int }_0^{\infty }q{e}^{-(q+\lambda )u}{\psi }_{+}(c+u)du$ and ${\psi }_{+}(c)=e^{-(q+\lambda )c}+{\int }_0^{c}q{e}^{-(q+\lambda )u}{\psi }_{-}(c-u)du$. Substituting the exponential determines its rate as $\sqrt{\lambda (2q+\lambda )}$.

For a singular strong limit cardinal $\lambda$, let $\mu =\mathrm{cf}(\lambda )$ and choose cofinal cardinals ${\lambda }_i<\lambda$. Subset restrictions inject $\mathcal{P}(\lambda )$ into ${\prod }_{i<\mu }\mathcal{P}({\lambda }_i)$, whose cardinal is at most ${\lambda }^{\mu }$ by the strong-limit property. Conversely ${\lambda }^{\mu }\le (2^{\lambda }{)}^{\mu }=2^{\lambda }$. This equality alone does not imply the singular cardinals hypothesis.

The capacity is the exponential growth rate of the independence numbers of strong graph powers. Products of independent sets make these numbers supermultiplicative, and Fekete lemma gives the limit and its equality to the supremum. Passing to every second power gives $c(G⊠G)=c(G{)}^2$.

For a finite-dimensional convex optimization problem with affine equalities $Ax=b$ and convex inequalities $g_i(x)\le 0$, Slater's condition requires a point in the relative interior of the common effective domain that satisfies the equalities and makes every inequality strict. A refined version only requires strictness of the nonaffine inequalities; affine inequalities may hold with equality. If the primal optimal value is finite, these conditions imply strong duality and dual attainment.^[ref]

The relative-domain hypothesis matters. Minimize $-\sqrt{x}$ over its domain $x\ge 0$ subject to $x\le 0$. The only feasible point is zero, giving primal value zero. For multiplier $\lambda >0$, the dual function isIts supremum is zero but is not attained by a finite multiplier. The feasible point lies outside the relative interior $(0,\infty )$ of the objective domain, so the refined condition does not apply. Feasibility of affine constraints alone suffices only when the relevant domain hypothesis is also met.

For a closed string in a string sigma model in an isotropic plane wave, insert $e^{-i{\omega }_n\tau +2\pi in\sigma /L}$ into the transverse wave equation. Periodicity requires integer $n$, giving the displayed dispersion relation. The zero spatial mode is an oscillator of frequency $|m\kappa |$; when this mass vanishes it becomes a free particle mode and needs the position-plus-linear-time expansion.

Expand every transverse closed string coordinate in a real orthonormal Fourier basis of the spatial circle. The canonical momenta are $p_{Ia}=T{\dot{q}}_{Ia}$, and spatial orthogonality reduces the field Hamiltonian to the displayed independent harmonic oscillators. Canonical quantization gives ${\omega }_a(a_{Ia}^{†}{a}_{Ia}+1/2)$ for each nonzero-frequency mode; a zero-frequency mode retains the free particle term $p_{Ia}^2/(2T)$. The formal zero-point energy requires a regulator; it is not automatically the massless flat-background intercept. Opposite travelling-wave oscillator combinations diagonalize the translation generator and satisfy closed-string level matching.

The binary Brier score applies squared-error loss to a reported success probability $p$ and the observed indicator $y\in \{0,1\}$. It is a loss, so smaller is better. Its values lie in $[0,1]$: predicting the outcome with certainty gives zero, while assigning it zero probability gives one. Unlike the logarithmic scoring rule loss, the binary Brier score remains bounded even for a confidently wrong forecast.

For a Bernoulli distribution with true success probability $q$, averaging the two squared losses givesThe report controls only $(p-q{)}^2$; the remaining term is the variance of the Bernoulli random variable. Consequently the Brier score is a strictly proper scoring rule. At $q=0.7$, the expected loss is $(p-0.7{)}^2+0.21$, with unique minimum $0.21$ at $p=0.7$. Reporting $p=0.2$ instead gives expected loss $0.46$, despite its better loss on the particular outcome $Y=0$ above.

For several possible categories, the Brier score sums the squared differences between reported probabilities and observed category indicators. With two categories that sum is $2(p-y{)}^2$, so its numerical values are twice the binary convention used here, with the same optimizing report. Brier's original ^[ref] averages these category-sum losses over forecasts. Strictly Proper Scoring Rules, Prediction, and Estimation by Tilmann Gneiting and others (Section 3.1, Example 1, and Table 1) gives the category-sum and binary forms as negative rewards; negating them gives the losses described here.

Reflexification turns a strict partial order into a partial order. Reflexivity is immediate, transitivity follows by the four disjunction cases, and antisymmetry follows because two opposed strict comparisons imply a forbidden self-comparison. The construction uses only coherent logic, so is preserved by inverse image functors of geometric morphisms.

Let $K\subseteq {\mathbb{R}}^2$ be compact and convex, and assume that it contains a point with both coordinates positive. The product $xy$ has a unique maximizer over $K\cap [0,\infty {)}^2$, and that maximizer lies in the positive quadrant. Compactness gives existence; the strictly positive attainable product excludes points on the axes. On the positive quadrant, maximizing $xy$ is equivalent to maximizing $\log x+\log y$, a strictly concave function. A segment between two distinct maximizers would contradict strict concavity, proving uniqueness.

The quadrant restriction matters. The line segment from $(-10,-10)$ to $(1,1)$ is a compact convex set meeting the positive quadrant, but its global product maximum over the entire segment is $100$ at the negative endpoint. The result concerns the positive feasible portion, unless the original feasible set is itself confined to the nonnegative quadrant.

If the otherwise force-free second sphere is held against rotation, its angular velocity relative to the incident fluid is $-3a({\mathbf{U}}_0\times \mathbf{n})/(4R^2)$. The rotlet reflected back to the first sphere addsThis follows by evaluating the rotating sphere in Stokes flow at displacement $-R\mathbf{n}$ and using the vector triple-product identity. The holding couple does no work because the held sphere's angular velocity is zero.

Within each prespecified stratum, choose a block containing fixed equal counts of the two treatment assignments, then randomly permute their order. Every completed block gives exact treatment balance. Variable block sizes and allocation concealment protect against predicting the next assignment. Incomplete final blocks permit a small unavoidable imbalance; odd stratum sizes cannot have exact equal allocation.

For constant nonzero Coriolis parameter and positive constant buoyancy frequency, the anisotropic quasi-geostrophic potential vorticity operator becomes the ordinary three-dimensional Laplacian in the displayed coordinates. Its decaying whole-space inverse is $\psi (\mathbf{X})=-\int [Q({\mathbf{X}}^{\prime })-f]/(4\pi |\mathbf{X}-{\mathbf{X}}^{\prime }|)d^3{X}^{\prime }$. Boundary or far-field data are needed to exclude arbitrary harmonic additions.

The Prandtl ratio compares horizontal and vertical scales in rotating density stratification. In the three-dimensional quasi-geostrophic potential vorticity operator, horizontal curvature scales as $\psi /L^2$ and vertical stretching as $(f_0^2/N^2)\psi /H^2$. Comparable terms therefore require $H/L\sim |f_0|/N$. For constant $N$, the stretched coordinate $Z=(N/|f_0|)z$ makes the operator isotropic. This aspect ratio concerns rotational and stratification effects, not the viscous-to-thermal diffusivity Prandtl number.

The compact space $L$ in a stratified neighbourhood ${\mathbb{R}}^s\times cL$ is the link of the stratum. Its dimension is one less than the codimension of that stratum. A disconnected link records distinct local branches.

A rational Witt space is a stratified pseudomanifold such that the link $L$ of every stratum of odd codimension $2r+1$ satisfies $I{H}_r^{\bar{m}}(L;\mathbb{Q})=0$. This removes the only difference between lower and upper middle extensions and gives rational middle-perversity Poincare duality on closed oriented Witt spaces.

The link of an isolated stratum of a closed two-dimensional pseudomanifold is a finite disjoint union of circles. Separate these branches, replacing the cone on their union by separate disks. This produces a closed surface and a finite surjective map to the original space, a homeomorphism on the regular stratum. Orientations on the regular stratum extend over the disks.

Principal strain rates are the eigenvalues $a,b,c$ of the symmetric rate-of-strain tensor. Positive values are local extension and negative values local compression. In incompressible flow $a+b+c=0$, implying $a^3+b^3+c^3=3abc$. Their eigenvectors describe the principal deformation directions.

In three-dimensional isotropic linear elasticity, decompose the symmetric strain as $e=pI+d$ with $p=\mathrm{tr}(e)/3$ and $\mathrm{tr}d=0$. The stored energy is $E=\mu {\sum }_{ij}{d}_{ij}^2+\frac{3}{2}(3\lambda +2\mu )p^2$. Its nonnegativity for every strain is equivalent to the displayed inequalities: the traceless and purely volumetric test strains prove necessity, and the sum of squares proves sufficiency.