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A U-statistic kernel is an integrable function of a fixed number r of observations whose expectation is the parameter functional to be estimated. Exchangeability permits symmetrization without changing that expectation. This is a function of r sample values, rather than the one-displacement kernel for density estimation.

For a symmetric degree-$m$ U-statistic kernel, the population functional is $T(F)=\int hd{F}^m$. Contaminating each of the $m$ identical distribution arguments and differentiating gives the displayed influence function. It equals $m$ times the first Hoeffding projection and gives leading asymptotic variance $m^2\mathrm{Var}(h_1(X))/n$. If that projection vanishes, the usual root-$n$ normal limit degenerates and higher-order projections determine the limiting law.

For a quantum state ensemble of pure vectors and an orthogonal projection $P$, measure $\{P,I-P\}$, preserve the successful subspace, and decode failure to a fixed normalized vector $\phi$. The resulting quantum channel is $\Lambda (X)=PXP+\mathrm{Tr}[(I-P)X]|\phi ⟩⟨\phi |$. For ${\alpha }_k=\Vert P{\psi }_k\Vert$, its squared quantum fidelity on the $k$th source vector is ${\alpha }_k^4+(1-{\alpha }_k^2)|⟨{\psi }_k|\phi ⟩{|}^2$. The inequality $a^2\ge 2a-1$ on $[0,1]$ proves the displayed average bound. It depends only on the source density operator's weight in $P$ and does not require the emitted vectors to be orthogonal.

For a finite-alphabet IID source and fixed $0<\epsilon <1$, the smallest set of length-$n$ strings carrying probability at least $1-\epsilon$ has size $2^{n(H(U)+o(1))}$. The typical set bounds give the upper estimate. For the lower estimate, any such set intersects the typical set in probability bounded away from zero, while each typical string has probability at most $2^{-n(H(U)-\delta )}$.

// Target: probability-and-statistics.bigb

In a model of ZFC, choose an external automorphism $j$ and an ordinal $\alpha$ with $j(\alpha )>\alpha$. Set $D_i=V_{j^i(\alpha )}$ for integer $i$, and declare the sets of type $i+1$ to be its objects that are subsets of $D_i$. Ordinary membership between successive types satisfies typed comprehension and set-extensionality. The automorphism shifts all types, yielding NFU by the type-zero collapse. A Skolem hull of integer-indexed increasing ordinal indiscernibles supplies such a $j$ by shifting the indices. The model is externally ill-founded; a transitive well-founded universe cannot have such an automorphism.

Define the head-count random variable $X:\Omega \to \{0,1,2\}$ byThe four elementary outcomes are inputs to this function; its three numerical values are outputs. Each fiber of a function collects all outcomes giving one value. Under the fair independent model, adding the masses in each fiber gives the probability mass functionConsequently the output values are not equally likely, although the inputs are. Using the reduced sample space $\{0,1,2\}$ with these masses gives the same head-count probability distribution; assigning $1/3$ to each value would describe a different model.

A question about a random variable's value becomes an event on the original sample space by taking an inverse image:For a real-valued random variable, this is an event whenever $D$ is a Borel set, by measurability. In this finite example every subset of $\{0,1,2\}$ is allowed. In particular,This is the same “at least one head” event counted above, expressed through an observable.^[ref]

Every event determined by $X$ is a union of complete fibers. There are $2^3=8$ such events, whereas the full power set of $\Omega$ has $2^4=16$. In particular, the first-head event $A=\{HH,HT\}$ is not determined by $X$: it contains $HT$ but excludes $TH$, although both give $X=1$. Choosing an observable makes calculations simpler precisely by forgetting distinctions; the useful choice is one that retains the distinctions the question needs.

Here $s(m)=\frac{1+m}{2}\ln \frac{1+m}{2}+\frac{1-m}{2}\ln \frac{1-m}{2}$. This free-energy density is the energy minus temperature times entropy of independent sublattice Ising spins. Since $s^{\prime }(m)=\mathrm{atanh}m$, stationarity yields the two-sublattice Ising mean-field approximation. Writing $M_{\pm }=(M_B\pm M_W)/2$, the quadratic coefficients are $(k_{B}T+qJ)/2$ for $M_{+}^2$ and $(k_{B}T-qJ)/2$ for $M_{-}^2$. The positive quartic coefficient $k_{B}T/12$ in the staggered direction stabilizes the continuous transition. Unlike a log-cosh auxiliary potential, this variational expression can be minimized away from self-consistency.

On the stable zero-field branch $M_B=-M_W=M$, put $a=\beta (1-M^2)$. Differentiating the two-sublattice Ising mean-field approximation with respect to the uniform field gives ${\chi }_B=a(1-qJ{\chi }_W)$ and ${\chi }_W=a(1-qJ{\chi }_B)$. Thus ${\chi }_{+}=a/(1+qJa)$ and $\partial M_{-}/\partial h=0$. The true staggered magnetic susceptibility, $\partial M_{-}/\partial g=a/(1-qJa)$, is a different response and diverges at the continuous transition. The factor $1-M^2$ is essential: the uniform response vanishes when the spins saturate at zero temperature.

In a two-generated two-step nilpotent group, use $c=[x,y]=x^{-1}{y}^{-1}xy$. Centrality of $c$ gives $yx=xy{c}^{-1}$, so every word collects to $x^a{y}^b{c}^r$. The multiplication of these collected expressions is $(a,b,r)(a^{\prime },b^{\prime },r^{\prime })=(a+a^{\prime },b+b^{\prime },r+r^{\prime }-a^{\prime }b)$. Relations on the central commutator can reduce $r$ modulo its order. To turn collection into a word problem for a group algorithm, one must also prove the independence of the remaining coordinates, for example by constructing a group with that cocycle multiplication.

In a steady inviscid planar incompressible flow, vorticity is constant along streamlines. A regular stream function supplies a local transverse coordinate, so locally $\Delta \psi$ depends only on $\psi$. A global single-valued function needs the same value on every component of each level set. This can fail: $\psi =x^3-x$ gives the steady Euler shear $(u,v)=(0,1-3x^2)$ with constant pressure, but $\psi (0)=\psi (1)=0$ while $\Delta \psi$ is respectively $0$ and $6$.

For two distinct massless final particles, $d{\Phi }_2/d\Omega =1/(32{\pi }^2)$ in the centre-of-mass frame. The invariant initial flux is $4p_1\cdot p_2=2s$. Their ratio gives the displayed differential cross-section. Equivalently the denominator is $128{\pi }^2{p}_1\cdot p_2$. Initial spins are averaged and final spins and colors summed when defining an inclusive unpolarized cross-section; averaging final colors would change the observable.

The clock hypothesis says that an ideal clock following a timelike curve measures its proper time: its instantaneous rate depends on velocity through $d\tau =dt\sqrt{1-v^2/c^2}$, with no separate acceleration term. Acceleration still changes the world line and hence the accumulated proper time. A physical clock may suffer disturbances under forces or environmental changes, so experiments must distinguish its mechanism's errors from the geometric prediction.

An atomic clock uses an atomic transition as a frequency reference for an oscillator; counting its calibrated cycles realizes a time interval. In a passive beam implementation, the chemical atoms are repeatedly prepared, interrogated and detected, while feedback control corrects the continuously running oscillator. The Ramsey separated oscillatory fields experiment explains how internal quantum phase becomes a measurable population signal.

The SI second fixes the unperturbed ground state hyperfine structure transition frequency of caesium-133 to exactly $9192631770\mathrm{Hz}$. This is a defining value for the ideal transition, rather than a guarantee that an operating apparatus has no shifts.^[ref]

A compact smooth embedding has a tubular neighborhood identified with its normal bundle. Collapse the complement of its open disk neighborhood to the basepoint and identify the remaining quotient with the Thom space of that normal bundle. Pulling back its Thom class constructs a Gysin class. For a graph embedding into a product with Euclidean space, desuspending the auxiliary coordinates gives the corresponding wrong-way map.

A smooth embedded submanifold has a neighborhood diffeomorphic to a neighborhood of the zero section of its normal bundle, with the zero section identified with the submanifold. One may prescribe the normal-bundle splitting of the derivative along that section. Choose a Riemannian metric making the splitting orthogonal and use its normal exponential map: its derivative there is the required isomorphism, so the inverse function theorem supplies local diffeomorphisms. A sufficiently small, possibly varying, fibre radius makes these fit into a tubular neighborhood.

For $m$ tribes of size $w$ with $m\sim (\log 2)2^w$ and $n=mw$, the success probability under the p-biased product measure is $1-(1-p^w{)}^m$. Its $x$-quantile is $p_x=[1-(1-x{)}^{1/m}{]}^{1/w}$. For fixed $0<x<1$,Thus the fixed-error window has order $1/\log n$. For small fixed error the leading coefficient grows with $\log (1/\epsilon )$, so the general Friedgut-Kalai sharp threshold theorem also has the correct error dependence up to absolute constants.

Choose $w\ge 1$ maximal such that $wm\le n$ for the displayed $m$, and take the OR of $m$ AND blocks of size $w$, ignoring unused coordinates. Its failure probability is $(1-2^{-w}{)}^m$, between one quarter and three quarters. Active coordinates have influence $2^{1-w}(1-2^{-w}{)}^{m-1}$ and unused ones have zero influence. Maximality gives $2^{1-w}<4(\log 2)(w+1)/n$; for $n\ge 2$, $w\le {\log }_{2}n$, so every influence is below $8\log n/n$. The construction is therefore a quite fair Boolean function with small influences for every $n\ge 2$. The logarithmic bound cannot hold at $n=1$, where every quite fair function has influence one.

Forward substitution solves a linear system $Ly=c$ with a lower triangular matrix $L$ and nonzero diagonal, from the first row downward:The first sum is empty, so $y_1=c_1/l_{11}$. Each subsequent row uses only components already found. For a unit triangular matrix $L$, omit the division because $l_{ii}=1$. The complementary procedure, back substitution, solves an upper triangular matrix system from the last row upward.^[ref]

If $np\to 0$, the triangle count in a binomial random graph has expectation at most $(np{)}^3/6\to 0$, so the Markov inequality shows there are no triangles in a graph with probability tending to one. If $np\to \infty$, the triangle variance in a binomial random graph divided by the squared mean is $O((np{)}^{-3})+O((n^{2}p{)}^{-1})\to 0$. The second moment method then gives at least one triangle in a graph with probability tending to one. At the intermediate scale $np\to c\in (0,\infty )$, a nontrivial Poisson limit for triangles in a binomial random graph describes the count.

Choose $n=d^7$ and a binomial random graph of edge probability $d/n$. With high probability its maximum degree is at most $2d$, it has $o(n)$ triangles, and every independent set has size at most $4n\log d/d$. Delete one vertex per triangle. The resulting triangle-free graph has chromatic number at least $d/(8\log d)$. Appending a disjoint star of degree $2d$, if necessary, makes the maximum degree exactly $2d$ without lowering the chromatic number. Thus the order $\Delta /\log \Delta$ is unavoidable.