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For Brownian motion with drift $dX=dB+\mu dt$ in $(-b,b)$, the displayed probability of exiting at $-b$ follows by stopping the harmonic function $e^{-2\mu X}$. A unit-time increment larger than $2b$ has a uniform positive probability and forces exit, so the exit time has a geometric tail bound from a uniform escape probability and all polynomial moments. At zero drift the limiting probability is $(b-x)/(2b)$.

The process $B_t^k-\mathbb{E}{B}_t^k$ is a martingale only for $k=1,2$. For higher powers its finite-variation drift is $k(k-1)(B_t^{k-2}-\mathbb{E}{B}_t^{k-2})/2$, which is not identically zero. Subtracting the random compensator $k(k-1){\int }_0^t{B}_s^{k-2}ds/2$ instead produces a true martingale on each finite horizon.

// Target: probability-and-statistics.bigb

For standard Brownian motion started at zero, choose an independent radius $R$ with $\mathbb{P}(R\le s)=(s/r{)}^n$, $0\le s\le r$, and stop on first reaching that radius. Rotational invariance of Brownian motion makes the conditional exit position uniform on each sphere. Mixing with radius density $n{s}^{n-1}/r^n$ gives the uniform probability distribution on the ball. In dimension one, the conditional sign has probabilities $1/2,1/2$. The stopped point samples volume measure even though each conditional exit law is supported on a boundary.

For independent coordinates of standard Brownian motion and $x_i>0$, the Brownian reflection principle gives ${\mathbb{P}}_x(T>t)={\prod }_i[2\Phi (x_i/\sqrt{t})-1]$. This is asymptotic to $(2/\pi {)}^{n/2}{\prod }_i{x}_i{t}^{-n/2}$. The tail integral formula for expectation therefore proves the stated criterion, with logarithmic divergence in dimension two.

For $d\ge 3$ and $|x|>\epsilon$, stop the harmonic function $u(x)=|x{|}^{2-d}$ at the two boundaries of $\{\epsilon <|x|<R\}$. Bounded optional stopping gives the probability of hitting the inner sphere first as $(u(x)-u(R))/(u(\epsilon )-u(R))$. Letting $R\to \infty$ gives the displayed formula; a finite hitting path is contained in some finite-radius ball.

For a cumulative distribution function $F$ and a standard Brownian bridge $\mathbb{G}$, the F-Brownian bridge is the centred Gaussian process ${\mathbb{G}}_F=\mathbb{G}\circ F$. Its covariance is $F(\min (s,t))-F(s)F(t)$, by substituting $F(s),F(t)$ into the bridge covariance. It can jump when $F$ jumps. Its law is tight in the supremum norm, since composition with $F$ is a continuous map from the separable space of continuous bridge paths.

For $t<1$ this stochastic integral solves $d{X}_t=d{B}_t-X_t(1-t{)}^{-1}dt$. The Itô product rule also gives $X_t=B_t-(1-t){\int }_0^t{B}_s(1-s{)}^{-2}ds$. The process is centered Gaussian, with covariance $\min (s,t)-st$, and has a continuous extension $X_1=0$. Time reversal of a Brownian bridge proves endpoint continuity without assuming it at the start.

The Brownian bridge covariance kernel is invariant under reversing both times, and its mean is zero. The two centered Gaussian processes therefore have equal finite-dimensional laws. Equality on countably many rational coordinates transfers the event of continuity at zero to the reversed endpoint. Interior continuity then extends the rational-time limit to all times. This proves the terminal continuity of the stochastic integral representation of a Brownian bridge.

The cubic crystal system has cubic point symmetry. Its three cubic Bravais lattices are primitive, body-centered, and face-centered. The body-centered cubic lattice is one member of that classification.

A tetragonal conventional cell with lattice points at its vertices and face centers describes a Bravais lattice. It is equivalent, after a rotation and a change of conventional cell, to a body-centered tetragonal description. The reciprocal lattice of a body-centered tetragonal lattice with conventional sides $a,a,b$ consists of $2\pi (h/a,k/a,l/b)$ with $h+k+l$ even; these points form a face-centered tetragonal conventional cell with sides $4\pi /a,4\pi /a,4\pi /b$.

A region whose translates tile a lattice's ambient space, together with the lattice points represented in it. A primitive unit cell represents one lattice point; a conventional unit cell may represent several to display symmetry.

Associate to each prime $p\equiv 3(\mathrm{mod}4)$ the class of $(-1,p{)}_{\mathbb{Q}}$. These classes have order two, since quaternion conjugation identifies the algebra with its opposite. The product rule $[(-1,a)][(-1,b)]=[(-1,ab)]$ follows by factoring their tensor product into the commuting subalgebras generated by $i_1,j_1{j}_2$ and by $i_1{i}_2,j_2$; the second is the split algebra $(1,b)$. Every nonempty finite product of distinct such primes has an odd valuation at one of them. The quaternion norm obstruction at primes congruent to three modulo four therefore makes its class nonzero, proving independence. Euclid's construction $4p_1\cdots p_m-1$ proves there are infinitely many such primes.

For a number field $K$, localization of the Brauer group is injective. Its image is exactly the finite-support tuples of local classes whose local Brauer invariants sum to zero. The non-Archimedean summands have invariant group $\mathbb{Q}/\mathbb{Z}$, real summands have $\{0,1/2\}$, and complex summands are zero. For a cyclic character $\chi$ and parameter $a$, the invariant of the local cyclic algebra is $\chi ({\mathrm{Art}}_{K_v}(a))$. Thus the zero-sum condition on principal parameters is the additive form of Artin reciprocity.

The Brauer class of a central simple algebra is its equivalence class under adjoining matrix-algebra factors. Two algebras have the same class exactly when their central division-algebra representatives are isomorphic. The zero class consists of split matrix algebras; the inverse of $[A]$ is the class of the opposite algebra. For a cyclic algebra $(L/K,\sigma ,a)$, the class is zero exactly when $a$ is a field norm from $L$.

Assume four spacetime dimensions, $\Phi \ne 0$, constant $\omega$, no scalar potential and matter independent of $\Phi$. In the stated action convention, scalar variation gives $\Phi R+\omega {\Phi }^{-1}(\nabla \Phi {)}^2-2\omega \square \Phi =0$. The trace of the metric tensor equation gives $\Phi R+\omega {\Phi }^{-1}(\nabla \Phi {)}^2+3\square \Phi =8\pi GT$. Their subtraction proves the displayed wave equation with $T=g^{ab}{T}_{ab}$. The degenerate coupling of Brans-Dicke theory needs separate treatment before dividing by its coefficient.

Take two copies of each direction $(1,0),(0,1),(1,1),(1,-1)$. The resulting box spline has total polynomial degree six and $C^4$ continuity: there are eight directions and a rank-deficient remaining set has at most two, so the removal number is six. With a centred origin its support is the zonotopeIts centred binary symbol isThis symbol factors as $A(z,w)A(zw,z/w)$ with $A=(2+z+z^{-1})(2+w+w^{-1})/8$, giving an efficient quincunx subdivision implementation. The box spline criterion is also described in the primary research paper 4–8 Subdivision by Velho and Zorin.

This box spline is the density of the image of the uniform unit four-cube under the indicated direction matrix. Its polynomial pieces have total degree $4-2=2$. Removing three directions is necessary to leave a nonspanning set, giving $C^1$ continuity. Its knot arrangement is a criss-cross triangulation of a square lattice. Its refinement mask is $\frac{1}{4}(1+z)(1+w)(1+zw)(1+z/w)$, up to the monomial that fixes the origin.

The unit-direction univariate box spline is the density of the sum of $m$ independent uniform unit-interval coordinates. It is the order-$m$ Cardinal B-spline: its closed support is $[0,m]$, it is strictly positive on the interior, and its polynomial degree is $m-1$. For $m\ge 2$ it is $C^{m-2}$, generally not $C^{m-1}$. The recurrence $B_m(x)={\int }_0^1{B}_{m-1}(x-u)du$ proves its support and positivity. Starting with a half-open unit interval, integrating the translate partition recursively proves ${\sum }_{j\in \mathbb{Z}}{B}_m(x-j)=1$. Orthogonal projection onto the unit diagonal rescales support width from $m$ to $\sqrt{m}$.

A transformed-response regression estimates the conditional mean $m$ on the transformed scale. Inverting that mean does not generally estimate the original-scale mean, because nonlinear transformations do not commute with expected values. For a smooth inverse $g$ and centered transformed error of variance ${\sigma }^2$, a second-order Taylor expansion gives the displayed local correction. For $g(z)=z^q$ at positive $m$, it is $m^q+\frac{1}{2}q(q-1)m^{q-2}{\sigma }^2$. The variance is the response's residual variance, not the sampling variance of the estimated mean. This approximation requires errors small enough that the inverse is meaningful over the relevant range; it is not an exact moment formula for arbitrary transformed noise.

A bounding volume hierarchy traversal visits only nodes whose ancestors passed a conservative enclosure test. It can use depth-first search; its time complexity depends on how many nodes pass, not just the height of a binary tree.

// Target: optics.bigb