Let $U_1,\dots ,U_n$ be independent uniform random variables on $[0,1]$, let ${\theta }_i>0$, and set $S={\sum }_i{\theta }_i\in (0,1)$ and $X_t={\sum }_i{\theta }_i{1}_{\{U_i\le t\}}$. ThenIndeed, the complementary crossing is that of level one by the uniform-arrival residual mass martingale $M_t=(S-X_t)/(1-t)$, with $M_0=S$. Its jumps are downward, so its first crossing of one is continuous. An arrival coincides with a possible crossing time with probability zero: the possible times belong to the finite deterministic set obtained from subsets of the weights. Stop at that crossing or at $r<1$. The stopped values are in $[0,1]$, and as $r\uparrow 1$ they converge to the crossing indicator because all arrivals occur before one almost surely. The optional stopping theorem and dominated convergence theorem give crossing probability $S$. For $S=1$, the strict displayed inequality already fails at time zero.

A labelled unicyclic component has a unique cycle of length $k\ge 3$, with a rooted forest attached to its vertices. Choose its vertices in $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ ways and its unoriented cycle in $(k-1)!/2$ ways. The labelled forest count with prescribed roots gives $k{n}^{n-k-1}$ attachments. Summing over $k$ and putting $j=n-k$ yields the displayed identity.

A basic sequence is suppression-unconditional with constant $C$ if every coordinate deletion satisfies the displayed inequality. When $C=1$, deletion never increases the norm. Uniformly bounded finite-coordinate projections imply unconditional convergence of the basis expansions; for real scalars, the sign-change constant is at most $2C$.

Let $T_C(X)$ consist of finite normalized sequences for which every scalar combination and sign change satisfy $\Vert {\sum }_i{\epsilon }_i{a}_i{x}_i\Vert \le C\Vert {\sum }_i{a}_i{x}_i\Vert$. These inequalities give a closed tree on a Polish space when $X$ is separable. An infinite branch is an unconditional basic sequence. If there is no such sequence, each fixed-constant tree is well-founded and has countable rank of a well-founded tree. Consequently, finite unconditional configurations with one fixed constant and arbitrarily high countable ranks force an infinite unconditional sequence.

An ultraviolet subdivergence arises when loop momenta internal to a proper subgraph tend to infinity while the momenta entering that subgraph are held fixed. It can occur even if simultaneous scaling of all momenta of the full Feynman diagram is convergent. Recursive local counterterms subtract these subgraph divergences before the remaining overall divergence is analyzed. A tadpole diagram inserted into an otherwise convergent six-point phi-fourth theory Feynman diagram is a simple massive example.

A topological space is both a compact space and a Hausdorff space exactly when every ultrafilter on its underlying set converges to exactly one point. Existence of all ultrafilter limits characterizes compactness. If a space is not Hausdorff, two distinct points have neighbourhood systems whose combined finite intersections are nonempty; the ultrafilter lemma extends these to an ultrafilter converging to both points. Conversely, disjoint neighbourhoods prevent such a common limit.

For an iid sample and symmetric square-integrable kernel $h$ of fixed order $m$, put $\eta =\mathbb{E}h(X_1,\dots ,X_m)$ and $h_1(x)=\mathbb{E}[h(x,X_2,\dots ,X_m)]-\eta$. If ${\zeta }_1=\mathrm{Var}(h_1(X_1))>0$, the displayed limit holds. The first Hoeffding projection is $(m/n){\sum }_i{h}_1(X_i)$; its orthogonal remainder is $O_{L^2}(n^{-1})$, giving the limit by the ordinary central limit theorem. Degenerate kernels require a different limiting theory.

Exchangeability of an independent identically distributed sample makes every permuted kernel have the same expectation. Averaging over permutations therefore leaves the target functional unchanged and gives a symmetric kernel. The U-statistic then averages this kernel over the unordered r-element sample subsets.

The symmetric kernel $h(x,y)=(x-y{)}^2/2$ has expectation equal to the population variance for independent identically distributed observations with finite second moment. Its U-statistic reduces to the displayed unbiased sample variance, using ${\sum }_{i<j}(X_i-X_j{)}^2=n{\sum }_i(X_i-\overline{X}{)}^2$.

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.