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A nonprincipal ultrafilter on the natural numbers gives a Banach limit by taking the ultralimit of the Cesàro means of each bounded sequence. Compactness of a closed bounded interval, or disk for complex sequences, supplies the limit. Finite intersections in the filter give linearity; nonnegative averages give positivity; constants give norm one. The Cesàro theorem for convergent sequences gives agreement with ordinary limits because the ultrafilter contains every cofinite set. For the left shift on bounded sequences, the two means differ by $(x_{n+1}-x_1)/n$, which tends to zero, proving shift invariance.

The displayed sublinear functional on the real l-infinity sequence space agrees with the ordinary limit on convergent sequences. A dominated extension by the Hahn-Banach theorem therefore defines a generalized limit bounded between the lower and upper limits of the Cesàro means. For the left shift on bounded sequences $S$, the averages of $Sx-x$ equal $(x_{N+1}-x_1)/N$, which tend to zero. Domination applied to both signs gives $L(Sx-x)=0$, hence shift invariance. Positivity, agreement on constant sequences and the operator norm one follow from the bounds, so $L$ is a Banach limit.

For a nonempty ordered coordinate set $A$, write $A^{-}=[\min A-1]$ and $A^{+}=[\max A-1]\setminus A$. An exact $k$-cover satisfies ${\sum }_{A}H(X_A\mid X_{A^{+}})\le kH(X)\le {\sum }_{A}H(X_A\mid X_{A^{-}})$. A short proof uses the chain rule for information entropy contributions $h_i=H(X_i\mid X_{[i-1]})$: each conditional subset entropy bounds ${\sum }_{i\in A}{h}_i$ on the appropriate side, by conditioning reduces entropy. Weighting gives the lower inequality for a fractional packing and the upper inequality for a fractional cover. See Madiman and Tetali, "Information Inequalities for Joint Distributions".

The proportionally fair allocation on a four-cycle has aggregate service $P/(P+Q)$ on each active odd route and $Q/(P+Q)$ on each active even route. With Poisson arrival rates ${\nu }_r$ and exponential document-size parameters ${\mu }_r$, the departure intensity is ${\mu }_r{n}_r{x}_r$. The adjacent ratios of the displayed binomial coefficient weight cancel those intensities in detailed balance, with ${\rho }_r={\nu }_r/{\mu }_r$.

For linear smoothers, let $C=I-11^T/n$ and $T_j=C{S}_{j}C$. The centered fixed-point equations of the backfitting algorithm have diagonal identity blocks and off-diagonal blocks $T_j$. On the centered component space, a consistent system has a unique answer exactly when its homogeneous system has only the zero solution. For two components the reduced equation is $(I-T_1{T}_2)f_1=T_1(I-T_2)Cy$. A common centered vector reproduced by both smoothers yields opposite perturbations of the components without changing their sum.

// Target: probability-and-statistics.bigb

For increasing sites and an ordered nonnegative normalized B-spline basis, the matrix $A_{ij}=N_j(x_i)$ is a totally nonnegative matrix. One way to see the minors' signs is knot insertion. Each inserted knot changes coefficients by $b_j={\alpha }_j{a}_j+(1-{\alpha }_j)a_{j-1}$ with $0\le {\alpha }_j\le 1$, using the standard constant endpoint pieces. This is a nonnegative rectangular bidiagonal map, whose minors are nonnegative. Products preserve that property by the Cauchy–Binet formula. Insert each site to full knot multiplicity; a suitable refined coefficient is then the value $s(x_i)$. Thus collocation is an ordered row submatrix of the refinement map, and has nonnegative minors. The strict support conditions $t_i<x_i<t_{i+k}$ give invertibility by the Schoenberg–Whitney theorem; its inverse has the checkerboard inverse of a totally nonnegative matrix sign pattern.

For $|y|=a$ and $n>-1$, this surface integral is finite: near $x=y$ its integrand has size $O(|x-y{|}^{2n})$ and the two-dimensional area element supplies one further power of distance. Rotational covariance and symmetry make it an axisymmetric symmetric rank-two tensor. Put $K=\pi (2a{)}^{2n+2}$. Polar integration about the $y$ axis gives $\mathrm{tr}T=K/(n+1)$ and $y^{T}Ty=a^{2}K/(n+2)$. Solving $3\alpha +\beta a^2=\mathrm{tr}T$ and $\alpha +\beta a^2=y^{T}Ty/a^2$ gives $\alpha =K/[2(n+1)(n+2)]$ and $\beta =\alpha (2n+1)/a^2$. Thus the weighted chord tensor becomes an isotropic tensor exactly at $n=-1/2$.

With $A=\mathcal{V}/\pi$, $g_0^{\prime }=g_0{c}_0$ and initial radius $R_0$, eliminating time from the axisymmetric settling box model gives $\sqrt{c/c_0}=(R_{\infty }^4-R^4)/(R_{\infty }^4-R_0^4)$. Put $s_0=R_0^2/R_{\infty }^2$ and $\tau =4A/(w_s{R}_{\infty }^2)$. Then $R^2/R_{\infty }^2=\tanh [t/\tau +\mathrm{artanh}{s}_0]$. Thus the finite limiting runout is approached at infinite time; treating it as an exactly attained finite-time stopping point contradicts these balances.

In a mixed particle-laden gravity current, a size class with settling speed $w$ has concentration $c(w,t)=c_0(w)e^{-wJ(t)}$. The shared reduced gravity is $g^{\prime }(J)=\int g_0(w)c_0(w)e^{-wJ}dw$, and $d{R}^4/dJ=4\mathrm{Fr}(\mathcal{V}/\pi {)}^{3/2}\sqrt{g^{\prime }(J)}$. Finite runout therefore requires ${\int }_0^{\infty }\sqrt{g^{\prime }(J)}dJ<\infty$. A positive minimum settling speed guarantees this; a distribution with arbitrarily small settling speeds need not. For $g_0{c}_0(w)\sim C{w}^p$ at zero, with $p>-1$, the integral converges precisely when $p>1$.

For a slender centreline $\mathbf{X}(\theta )=(A\cos \theta ,B\sin \theta ,b\theta )$, $-\pi \le \theta \le \pi$, let $s^{\prime }(\theta )=(A^2{\sin }^2\theta +B^2{\cos }^2\theta +b^2{)}^{1/2}$ and use local force density $\mathbf{f}=c(2I-\mathbf{tt})\mathbf{v}$, $\mathbf{t}=d\mathbf{X}/ds$, $c=2\pi \mu /\ln (L/R)$. Rotation with angular speed $\Omega$ gives $\mathbf{t}\cdot \mathbf{v}=\Omega AB/s^{\prime }$, henceAxial translation with speed $W$ gives exactly the same coefficient for the torque: $G_z=-cWABb\int d\theta /s^{\prime }$. Indeed $(\mathbf{X}\times \mathbf{t}{)}_z=AB/s^{\prime }$. Both forces and torques here are exerted on the fluid. In the circular case $A=B=a$, this reduces to $F_z/\Omega =G_z/W=-2\pi c{a}^{2}b/\sqrt{a^2+b^2}$. The equality is also a consequence of the Lorentz reciprocal theorem, since the local resistance tensor is symmetric.

For $x^{\prime \prime }+\epsilon (x^2-1)x^{\prime }+x=0$, the averaged amplitude for position-dependent damping has the displayed logistic-type equation. With $A(0)=1$, its positive solution is $A(T)=2/\sqrt{1+3e^{-T}}$. Small amplitudes grow and large ones decay toward the stable value two; the leading oscillation is $A(\epsilon t)\cos t$.

For $x^{\prime \prime }+\epsilon |x|x^{\prime }+x=0$ and unit initial amplitude, averaging uses $⟨|\cos \psi |{\sin }^2\psi ⟩=2/(3\pi )$. Thus $A_T=-2A^2/(3\pi )$ and the displayed algebraic envelope follows. The leading phase has no first-order drift. The dissipation is amplitude dependent, so an exponential envelope with a constant decay rate would give the wrong long-time scale.

A correlation time measures how rapidly a stationary stochastic process loses temporal dependence. Specify whether it means an exponential decay time, an integral of the normalized autocorrelation, or another width. For $R(t)=R(0)e^{-|t|/\tau }$, the exponential time and the one-sided integral ${\int }_0^{\infty }R(t)dt/R(0)$ both equal $\tau$.

An augmenting path maximizing the possible increase in flow, namely its smallest residual network edge capacity. If the remaining maximum flow is $F>0$ and the original undirected network has $m$ edges, a path has bottleneck at least $F/m$. Otherwise the vertices reachable through arcs of capacity at least $F/m$ define a cut of capacity strictly less than $F$, contradicting the max-flow min-cut theorem. Only one directed arc per original crossing edge contributes to that cut.

Give $N$ the subspace filtration and $M/N$ the quotient filtration. If the degree-$i$ class of $m\in F_{i}M$ maps to zero, then $m=n+m^{\prime }$ with $n\in N$ and $m^{\prime }\in F_{i-1}M$. Thus $n\in N\cap F_{i}M$ and the class comes from $\mathrm{gr}N$. Inclusion is injective in every degree and the quotient map is surjective, proving the displayed short exact sequence of associated graded modules. Exactness can fail for arbitrarily chosen incompatible filtrations.

An Asian option whose averaging uses the geometric mean. In the constant-coefficient Black-Scholes model, the logarithm of the continuous geometric average is Gaussian with mean $\log S_0+(r-{\sigma }^2/2)T/2$ and variance ${\sigma }^{2}T/3$, because $\mathrm{Var}({\int }_0^T{W}_{t}dt)=T^3/3$. Its forward mean is $S_0{e}^{rT/2-{\sigma }^{2}T/12}$. The discounted lognormal call formula gives its price. It is strictly cheaper than the terminal European call when $\sigma ,K,T>0$ and $r\ge -{\sigma }^2/6$, since both its forward mean and log-variance are no larger, with strictly smaller variance. No unconditional comparison holds for arbitrary negative interest rates: for $r<-{\sigma }^2/6$, sufficiently small positive strikes reverse the inequality.

If a traded payoff matrix $A$ has initial price vector $q=A^{T}p$ with $p\ge 0$, there cannot be a negative-cost portfolio with nonnegative terminal payoff. Nevertheless a nonzero nonnegative attainable payoff supported only in zero-price states can cost zero and give an arbitrage. When every state has positive physical probability, excluding this possibility requires strictly positive Arrow state prices, not merely a nonnegative supporting vector.

Write a presheaf on $0\to 1$ as an arrow $r:B\to A$. Evaluation $\Gamma (r)=B$ has the displayed chain, where $\Pi (r)=A$, $\Delta (S)=(S\stackrel{1}{\to }S)$, $\nabla (S)=(S\to 1)$ and $\Lambda (S)=(\varnothing \to S)$. The Hom-set bijections follow by solving the commuting-square equation. The functor $\nabla$ has no right adjoint because it fails to preserve the initial object: $\nabla (\varnothing )=(\varnothing \to 1)$ is not $(\varnothing \to \varnothing )$.

The normalized information matrix for a homoscedastic linear regression model is the average outer product of its regressor vector. It is positive-definite precisely when the supported regressors span the parameter space. The full Fisher information matrix under normal errors is $NM(\xi )/{\sigma }^2$.

Use the north-pole stereographic projection and the cross-ratio convention $[a,b;c,d]=(a-c)(b-d)/[(a-d)(b-c)]$. Antipodes have coordinates $-1/\bar{u}$. Rotate the first point to the south pole, whose coordinate is zero: its antipode becomes infinity, while a point at spherical distance $d$ has projected modulus $\tan (d/2)$. Möbius invariance of the cross-ratio then gives the displayed identity. Antipodal endpoints require the extended-value interpretation.