Column orthogonality of the CKM matrix gives three complex numbersDrawing these as successive vectors in the complex plane produces a closed triangle. Its area is $\frac{1}{2}|\mathrm{Im}(a{b}^{\ast })|=|J|/2$, because unitarity identifies this imaginary quartet with a signed Jarlskog invariant. A nonzero area therefore detects physical CP violation for nondegenerate quark masses.

For a worked example, choose the exact parametrization of the CKM matrix with $s_{12}=1/4$, $s_{23}=1/10$, $s_{13}=1/100$ and $\delta =\pi /3$. These are illustrative choices, not fitted quark mixing parameters. They give $J\simeq 2.085597\times 10^{-4}$. The unscaled triangle has area approximately $1.042799\times 10^{-4}$.

Divide the closure relation by $-b$, assuming $b\ne 0$, and define the complex numberThe normalized triangle has vertices $0,1,z$, with $z\simeq 0.195376+0.335788i$ for these choices. Dividing each side by $|b|$ divides area by $|b{|}^2$: its area is $|J|/(2|V_{cd}{V}_{cb}^{\ast }{|}^2)=|\bar{\eta }|/2$, approximately $0.167894$. The familiar normalized picture therefore does not itself have area $|J|/2$.

Under quantum field rephasings, $a,b,c$ all acquire the same phase, so the original triangle merely rotates and $z$ stays fixed. Ratios and angles can thus compare quark mixing measurements without dependence on field conventions. If the Jarlskog invariant vanishes, the triangle collapses; additional quark mass degeneracies require extra care in interpreting the converse.

For an infinite-dimensional complex Hilbert space, the Calkin algebra is the unital quotient of the bounded operators by the closed two-sided ideal of compact operators. Its spectrum records invertibility up to compact errors. An operator is Fredholm exactly when its image in this quotient is invertible.

For a nonunital C-star algebra, adjoining an identity gives multiplication $(a,\lambda )(b,\mu )=(ab+\lambda b+\mu a,\lambda \mu )$ and involution $(a,\lambda {)}^{\ast }=(a^{\ast },\overline{\lambda })$. Its standard C-star norm extends that of $A$, with the new identity of norm one. It can be constructed by adjoining the identity operator to a faithful nondegenerate operator representation and taking the operator norm. The original algebra is a closed ideal. The usual spectrum of an element in a nonunital algebra is taken in this unitization, permitting unital spectral arguments without requiring the original algebra to contain an identity.

For a general nonautonomous Runge-Kutta method, Taylor expansion of stage $i$ gives $k_i=f+h[c_i{f}_t+(Ae{)}_i{f}_{y}f]+O(h^2)$. Matching the exact increment $hf+h^2(f_t+f_{y}f)/2$ gives the three displayed conditions. When internal consistency $c=Ae$ is imposed, the last two coincide. Internal consistency is a usual tableau convention, but is not logically necessary for order two of a generalized tableau.

In an M/G/1 queue, arrivals during the first service generate independent subtrees of work. Conditional on service time $S=s$, the number of descendants is Poisson with mean $\lambda s$. Thus $B\stackrel{d}{=}S+{\sum }_{j=1}^{N(S)}{B}_j$, with independent descendant busy periods. Conditioning and the Poisson generating function give the displayed transform equation wherever its moment-generating functions are finite. Taking expectations yields $\mathbb{E}B=\mathbb{E}S+\lambda \mathbb{E}S\mathbb{E}B$ in the stable regime.

Let $p$ be the smallest prime dividing the order of a finite group, and suppose its Sylow subgroup $P$ is cyclic of order $p^a$. Conjugation embeds $N_G(P)/C_G(P)$ into $\mathrm{Aut}(P)$, of order $p^{a-1}(p-1)$. Since $P\subseteq C_G(P)$, this quotient has order prime to $p$. Its remaining possible prime divisors are smaller than $p$ and cannot divide $|G|$. Thus $N_G(P)=C_G(P)$, and the Burnside transfer theorem gives a normal p-complement.

Ignoring end boundaries, the buoyancy-driven turbulent density diffusion model admits $\bar{\rho }={\rho }_0[1+A(t)C{z}^5]$ for $C>0$. Since $({\partial }_z\bar{\rho }{)}^{3/2}\propto A^{3/2}{z}^6$, the amplitude obeys $A^{\prime }=65^{3/2}K{W}^2\sqrt{gC}{A}^{3/2}$. With $A(0)=1$, $A=(1-t/t_{\ast }{)}^{-2}$ and $t_{\ast }=[35^{3/2}K{W}^2\sqrt{gC}{]}^{-1}$. Its formal growth is not a physical finite-time singularity of a closed mixing tube: ignored end fluxes, loss of the Boussinesq approximation and the finite energy budget invalidate the continuation.

In the Gamma prior for an inverse-gamma scale model, assume $k>2$. The variance of hypothetical means is $a=\alpha /[{\lambda }^2(k-1{)}^2]$, and the expected process variance is $v=\alpha (\alpha +1)/[{\lambda }^2(k-1{)}^2(k-2)]$. The Bühlmann credibility factor is $Z=n/(n+v/a)$, giving the displayed expression. The optimal affine credibility estimate is $(1-Z)\alpha /[\lambda (k-1)]+Z\overline{X}$. Its fixed-weight arithmetic-mean form differs from the reciprocal-based posterior mean.

Let a fixed policy intensity have prior density $3{\theta }^{-4}$ on $\theta >1$, and let $n$ annual claim counts be conditionally independent with Poisson distribution mean $\theta$. The expected process variance is $3/2$ and the variance of hypothetical means is $3/4$, so the Bühlmann credibility factor is $n/(n+2)$. A total count $k$ gives the displayed credibility premium. By contrast, the Bayesian posterior probability density function is proportional to ${\theta }^{k-4}{e}^{-n\theta }$ on $\theta >1$. Its mean is the ratio ${\int }_1^{\infty }{\theta }^{k-3}{e}^{-n\theta }d\theta /{\int }_1^{\infty }{\theta }^{k-4}{e}^{-n\theta }d\theta$. For $n=2,k=5$, that Bayesian credibility estimate is $5/3$, whereas the affine estimate is $2$.

For independent comparable risks observed over $n\ge 2$ years, estimate the expected process variance $v$ by the average within-risk sample variance. The sample variance $B$ of the risk means estimates $a+v/n$, where $a$ is the variance of hypothetical means. Thus $B-\hat{v}/n$ is unbiased for $a$; truncating at zero produces a practical nonnegative estimate but introduces bias. Use the grand mean for $m$ and plug these estimates into the Bühlmann credibility premium $\hat{Z}{\overline{X}}_s+(1-\hat{Z})\hat{m}$. At least two risks are needed for $B$; a single year cannot estimate within-risk variance this way.

Reflecting a Brownian path after its first hit of a level $b>0$ preserves Wiener measure and maps an endpoint $x\le b$ to $2b-x\ge b$.

Lévy area is the antisymmetric second-level integral of multidimensional Brownian motion. For distinct independent components the integrals are also Itô integrals. Combined with the symmetric tensor $\frac{1}{2}{B}_{s,t}^{\otimes 2}$, it gives the second level of enhanced Brownian motion. It is not continuously determined by the uniform norm of the Brownian path.

Planar Brownian motion started away from zero does not hit zero. Stopping the harmonic function $\log |x|$ on an annulus gives probability $(\log R-\log |x|)/(\log R-\log \epsilon )$ of reaching its inner circle before its outer circle. This tends to zero as $\epsilon \downarrow 0$. Exhausting the plane by outer circles proves point avoidance at finite times.

// Target: probability-and-statistics.bigb

For standard one-dimensional Brownian motion, the Itô formula for $B^k$ gives this recursion. Stop at exits from $[-n,n]$ to make the stochastic integral a true martingale. Gaussian tail integrability and the Doob Lp maximal inequality dominate all stopped powers on a fixed horizon, justifying the limit. Odd moments vanish and $\mathbb{E}{B}_t^{2k}=(2k)!t^k/(2^{k}k!)$.

// Target: probability-and-statistics.bigb

For standard Brownian motion started at zero in ${\mathbb{R}}^n$, every fixed orthogonal transformation preserves the process law. Its transformed increments remain independent centred Gaussian random vectors with covariance $(t-s)I$, and its paths remain continuous. Consequently its exit law from any centred sphere is normalized surface measure. Translations give the same assertion for spheres centred at its starting point.

The process $M_t=B_{t^2}$ is a martingale in the filtration ${\mathcal{G}}_t={\mathcal{F}}_{t^2}$ and has quadratic variation $t^2$. The integral $W_t={\int }_0^t(2s{)}^{-1/2}d{M}_s$ is well defined because its bracket is ${\int }_0^t(2s{)}^{-1}d(s^2)=t$. The Lévy characterization of Brownian motion makes $W$ Brownian in this filtration, and stochastic-integral associativity gives the displayed representation. The resulting integrand is continuous at zero despite the singular inverse integrand.

On a real Hilbert space $H$, this centered jointly Gaussian process is linear in its test vector $h$ and has the displayed covariance, with independent increments in time. Equivalently choose an orthonormal basis and independent real Brownian motions $W^j$, and define stochastic integration by $\int ⟨v_s,d{W}_s⟩={\sum }_j\int ⟨v_s,e_j⟩d{W}_s^j$ whenever $\int \Vert v_s{\Vert }^{2}ds<\infty$. In infinite dimension it need not be an $H$-valued random variable; the integral is the meaningful object.

The sign transform of standard Brownian motion is itself standard Brownian motion. The integrand is predictable because it is a Borel function of a continuous adapted process. Brownian motion spends zero Lebesgue time at a specified level: integrate $\mathbb{P}(B_s=0)=0$ over positive times. Consequently the transform's quadratic variation is $t$, regardless of the chosen sign value at zero. The Lévy characterization of Brownian motion applies.

For $t>0$, let $M_t={\sup }_{s\le t}{B}_s$ and $A_t={\sup }_{s\le t}|B_s|$. The Brownian reflection principle gives $M_t\stackrel{d}{=}|B_t|$, hence $\Vert M_t{\Vert }_2=\sqrt{t}$. Also $A_t\ge M_t$, with strict inequality on the positive-probability event $\{B_t<-\lambda ,M_t<\lambda \}$: reflection gives its probability as $\mathbb{P}(B_t>\lambda )-\mathbb{P}(B_t>3\lambda )>0$. The Doob L2 maximal inequality bounds $\Vert A_t{\Vert }_2\le 2\sqrt{t}$, so $\sqrt{t}<\Vert A_t{\Vert }_2\le 2\sqrt{t}$. At $t=0$ all quantities vanish and the lower inequality is equality.

// Target: probability-theory.bigb

Here $d_H$ counts the observations replaced. This curve measures the largest change after at most $m$ arbitrary replacements. Under the first-unbounded convention, the replacement breakdown point is $n^{-1}\min \{m:b_n(m;y)=\infty \}$. The largest fraction still guaranteed bounded is one grid step smaller; those conventions must not be confused. A sample mean breaks after one replacement, while a scalar sample median with a fixed ordinary tie convention needs about half the sample replaced.