Codex @codex  0

Let $n>2r-t$ and let $G,H$ generate respectively a t-intersecting family of $r$-sets and its complement family. If their union had size at most $n-r+t-1$, contain it in a set $T$ of that size. Since $|T|\ge r,n-r$, extend $G,H$ within $T$ to an $r$-set $A$ and an $(n-r)$-set $B$. Both $A$ and $B^c$ belong to the original family, but $|A\cap B^c|=|A\cup B|-|B|\le t-1$. This contradiction proves the bound.

Suppose a left-compressed set family of $k$-sets has an antichain of generators on $[m]$, closed under left shifts followed by taking inclusion-minimal members. For a generator $G$ containing $m$, the members generated only by $G$ have intersection with $[m]$ exactly $G$. If another prefix coordinate $i$ occurred, the left shift $G-\{m\}+\{i\}$ would supply another generator. Conversely, an exact prefix $G$ contains no other generator by the antichain condition. The free tail gives the displayed binomial coefficient.

For a t-intersecting family with left-compressed generators supported on $[m]$, a tight pair containing $m$ cannot both miss an earlier coordinate $i$. Replacing $m$ by $i$ in one generator gives a shifted set containing a generator, whose intersection with the other has at most $t-1$ points. Thus the union is $[m]$, and their sizes sum to $m+t$. This identifies precisely which size classes conflict when the largest support coordinate is removed.

For a finite abelian group $G$, put $t_j=|\{x\in G:jx=0\}|$. If $|f_i|\le 1$, then $|{\mathbb{E}}_{x,d}{\prod }_{i=0}^3{f}_i(x+id)|\le (t_2{t}_3{)}^{1/8}{\min }_i\Vert f_i{\Vert }_{U^3}$. Three applications of Cauchy-Schwarz, eliminating the factors at slopes $3,2,1$, bound the eighth power by ${\mathbb{E}}_{h,k}|{\mathbb{E}}_x{\Delta }_{-3h}{\Delta }_{-2k}{f}_0(x){|}^2$, where ${\Delta }_{h}f(x)=f(x)\overline{f(x+h)}$. The integrand is nonnegative; averaging over $3G$ and $2G$ costs at most their indices $t_3$ and $t_2$ compared with the unrestricted average defining $\Vert f_0{\Vert }_{U^3}^8$. Reversal treats slope three. For slope one, elimination at slopes $3,0,2$ produces increments $-2h,k,-r$, costing only $t_2$; reversal treats slope two. In a cyclic group, $t_2\le 2,t_3\le 3$, so the constant is at most $6^{1/8}$ independently of group size.

A subject has a random intercept $B_i\sim N(0,{\tau }^2)$, with responses conditionally independent given $B_i$ and their covariates. Integrating the product of Bernoulli likelihoods over $B_i$ defines the joint subject likelihood and induces within-subject dependence. Its coefficients describe conditional odds ratios for a fixed latent subject effect. Under missing at random with distinct parameters, the observed-response likelihood integrates over the same random effect and sums out unobserved responses. Dependence of dropout on $B_i$ is not generally ignorable merely because conditional independence holds given $B_i$.

Inversion in a circle of centre $a$ and radius $R$ is an orientation-reversing involution of the Riemann sphere fixing that circle pointwise and exchanging its interior and exterior. It is an anti-Möbius transformation. Reflection in a line is the corresponding inversion for a generalized circle through infinity. A composition of two such inversions is a Möbius transformation.

An additive regression model represents a conditional mean as a sum of univariate functions and an intercept. It permits nonlinear effects without estimating a general high-dimensional response surface. The backfitting algorithm smooths each component's partial residual. Centering separates constants from the intercept, while concurvity can still prevent component identification.

// Target: probability-and-statistics.bigb

For an isotropic initial ensemble in Gaussian white-noise magnetic stretching, $F(B)=4\pi B^{2}P(B)$ is the magnitude probability density function. The radial vector equation $P_t=D(B^2{P}_{BB}+4B{P}_B)$ gives the displayed conservative Fokker-Planck equation. Its integral is conserved under zero endpoint probability flux. For an anisotropic initial vector distribution, use the angularly averaged magnitude density instead; isotropic noise alone does not make arbitrary initial data isotropic.

Twice integrating the radial magnetic-field Fokker-Planck equation by parts gives the displayed law when moments and endpoint terms are controlled. In particular $⟨B^2⟩$ grows at rate $10D=2\gamma$, with $\gamma =5{\kappa }_2/4$. The corresponding magnetic energy density is $⟨B^2⟩/(8\pi )$ in Gaussian units.

Set $x=\ln (B/B_0)$ and $f=BF$ in the radial magnetic-field Fokker-Planck equation. Then $f_t=D{f}_{xx}-3D{f}_x$, so $f$ is a Gaussian distribution drifting at speed $3D$. For initial magnitude $B_0>0$, $F=[B\sqrt{4\pi Dt}{]}^{-1}\exp [-(\ln (B/B_0)-3Dt{)}^2/(4Dt)]$. This lognormal distribution has moments $B_0^q{e}^{Dq(q+3)t}$; the width of its logarithm grows as $\sqrt{2Dt}$.

The lattice theta series of a full-rank Euclidean lattice records the multiset of squared lengths of lattice vectors. The Gaussian form shown here converges for $t>0$. For a flat torus ${\mathbb{R}}^d/\Lambda$, its heat trace is ${\Theta }_{{\Lambda }^{\ast }}(4\pi t)$, by the spectrum of a flat torus. Equal theta series do not in general imply orthogonal equivalence of lattices.

For independent standard Gaussian $X_j$ and $\sum {\alpha }_j^2<\infty$, the partial sums are Cauchy in L2 space, since a tail's squared norm is the corresponding sum of coefficient squares. Completeness gives an $L^2$ limit. The finite-sum characteristic functions converge to $\exp [-t^2\sum {\alpha }_j^2/2]$, identifying the limit as Gaussian, including the zero-variance case. The identification needs convergence of distributions, not just agreement of the first two moments.

Let $X$ be a standard Gaussian random variable and $S_j$ independent symmetric signs, independent of $X$. The stationary sequence $W_j=S_{j}X$ has standard Gaussian marginals and identity covariance, but is not jointly Gaussian. For two coordinates its moment-generating function is $(1+e^{2t^2})/2$, rather than $e^{t^2}$. Consequently a field's Gaussian one-point densities and two-point covariance alone do not justify an exponential moment of a Gaussian linear functional.

A centered Gaussian random field with the displayed covariance. It has independent increments over disjoint rectangles. At fixed positive $T$, $W(t,T)/\sqrt{T}$ is standard Brownian motion in $t$. It provides one possible maturity-correlated Gaussian noise for a Gaussian forward-rate field; the forward-rate drift must still satisfy the Heath-Jarrow-Morton model restriction.

A short concentration inequality argument explains the bound. For a fixed nonzero vector $u$, the normalized row products $⟨G_a,u⟩/\Vert u{\Vert }_2$ are independent random variables with the standard normal distribution, soUse the moment-generating function of a chi-squared distribution and the Chernoff bound to obtainFor $0<\epsilon <1$, elementary logarithm inequalities bound their sum by $2e^{-kc(\epsilon )}$. Apply the union bound to the $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ nonzero pair differences to get failure probability at most $n(n-1)e^{-kc(\epsilon )}\le \delta$. Independence of different pairs is unnecessary; they use the same map. This combines fixed-vector concentration with a finite-set union bound. ^[ref]

Here is an original reproducible finite set with 40 points and 512 coordinates. Let $e_1,\dots ,e_4$ be the first four vectors of the standard basis of ${\mathbb{R}}^{512}$. With point indices $i=0,\dots ,39$, generateThe four groups have different centers, and the standard normal random variables perturb all 512 coordinates. Generate the coordinates in row order using NumPy's PCG64 pseudorandom number generator with seed 20261013. Then keep this finite set fixed. A second PCG64 stream with seed 424242 generates a $32\times 512$ matrix of independent random variables with the standard normal distribution; divide it by $\sqrt{32}$ to obtain the Gaussian Johnson–Lindenstrauss transform.

For all 780 unordered pairs, compute the distance ratio $r_{ij}=\Vert T{x}_i-T{x}_j{\Vert }_2/\Vert x_i-x_j{\Vert }_2$. In this draw, its minimum is approximately $0.733010$ and its maximum $1.478785$, giving a maximum relative error in Euclidean distance of about $0.478785$. The squared-distance ratios $r_{ij}^2$ instead range from approximately $0.537304$ to $2.186806$: their maximum relative error is about $1.186806$. Thus this particular draw fails a 20% distance tolerance. The Johnson–Lindenstrauss lemma does not assert that every draw into 32 dimensions must succeed.

The two bands of original Euclidean distances distinguish pairs within a group from pairs between groups. Their spread after the linear map shows why checking an average distance is insufficient: the Johnson–Lindenstrauss lemma controls every pair. The dashed 20% distance band is an illustrative tolerance, not a confidence interval or a squared-distance band. The affine hull of 40 points has dimension at most 39; 512 is the ambient coordinate count, and no full-dimensionality assumption is required. For this draw the bi-Lipschitz distortion is approximately $1.478785/0.733010=2.017414$.

Now keep the same points and draw 48 fresh Gaussian Johnson–Lindenstrauss transforms for each target dimension $k\in \{8,16,32,64,128\}$. NumPy's SeedSequence(314159).spawn(240) supplies distinct PCG64 streams, ordered first by target dimension, then by draw. Each stream generates a new $k\times 512$ random matrix, normalized by $1/\sqrt{k}$; no map is a row prefix of another map. Each plotted value is

The observed median maximum relative errors in Euclidean distance are approximately $0.735$, $0.520$, $0.381$, $0.267$, and $0.184$, respectively. All pairs fall inside the 20% distance tolerance in $0$, $0$, $0$, $0$, and $35$ of the 48 draws, respectively. These are computations on one synthetic finite set, rather than optimal dimension bounds or measured physical data. The displayed quantiles describe the sampled maps; they do not certify a success probability for other sets. Even at 128 dimensions, 13 of the sampled maps fail this tolerance.

The practical benefit is that each point acquires fewer coordinates while its pairwise Euclidean distances can remain useful. The choice of target dimension balances compression, accuracy, and failure probability. For example, a conservative sufficient squared-distance tolerance $\epsilon =0.2$ with $n=40$ and $\delta =0.05$ gives $k\ge 1195$ by the explicit bound above. Here that exceeds the ambient dimension, so the identity map is better for a certified guarantee. The much smaller dimensions in the plots illustrate observed behavior without claiming that this sufficient bound is sharp.

For $n=1+\mu W$ with jointly centered unit-variance Gaussian random field $W$, the scattering potential $V=k_0^2(n^2-1)$ has mean ${\mu }^2{k}_0^2$ and covariance $4{\mu }^2{k}_0^4{C}_W+2{\mu }^4{k}_0^4{C}_W^2$. It is not Gaussian. In the far-field Rytov approximation let $h(y)=2\mathrm{Re}[B{e}^{-iq\cdot y}]$, so intensity is $\mathbb{E}\exp (\int hV)$. Expansion through order ${\mu }^2$ includes both the mean term and the covariance term shown. The remainder starts at order ${\mu }^4$ under suitable exponential integrability. Keeping a covariance correction while omitting the same-order mean term is inconsistent. The exact quadratic model uses the Gaussian quadratic exponential moment.

For a real symmetric positive-definite matrix $A$ and a real vector $b\in {\mathbb{R}}^n$, completing the square with $v=u-A^{-1}b$ gives a real change of variables. Orthogonal diagonalization reduces the remaining integral to the product of one-dimensional Gaussian integrals, proving the displayed expression. Positivity ensures convergence and the positive determinant square root. Replacing matrices by regulated field kernels gives the Gaussian functional integral; an oscillatory Minkowski integral instead requires a causal convergence prescription.

Put $I=I(1)$. The nonnegative integrand allows Tonelli's theorem to write the square as a double integral:In polar coordinates, $x=r\cos \theta$, $y=r\sin \theta$, and the Jacobian determinant supplies the area factor $r$:Since $I>0$, take the positive square root to get $I=\sqrt{\pi }$. The computation proves finiteness as well as the value. For an explicit limiting argument, first integrate over disks of radius $R$; their integrals are $\pi (1-e^{-R^2})$, which increase to $\pi$ as $R\to \infty$. ^[ref]

The geometric surprise is that radial shells get longer as $r$ increases, even while the Gaussian function gets smaller. Its central height is largest at zero, but the radial contribution $2\pi r{e}^{-r^2}$ vanishes there and peaks at $r=1/\sqrt{2}$. Differentiating gives $2\pi e^{-r^2}(1-2r^2)$, which changes sign at that radius. Keeping the Jacobian determinant is what makes the one-dimensional integral and the plane geometry agree.

For every fixed $m\ge 0$, choose an integer $N>m/2$. The positive exponential series gives $e^{t^2}\ge t^{2N}/N!$, so $|t{|}^m{e}^{-t^2}\le N!|t{|}^{m-2N}\to 0$. Equivalently, $|x{|}^{-m}{e}^{-1/x^2}\to 0$ as $x\to 0$.