Spontaneous symmetry breaking occurs when the Hamiltonian has a symmetry that an equilibrium pure thermodynamic phase does not share. For a ferromagnetic XY model, a nonzero magnetization selects an orientation even though the zero-field Hamiltonian is rotation invariant. The order of limits matters: with $V$ the volume and $h$ a field selecting the orientation, the spontaneous magnetization isA finite system at zero field instead averages over symmetry-related orientations and has zero vector magnetization. Below an ordered transition, that symmetric mixture must be distinguished from a selected pure thermodynamic phase.

A Goldstone mode is a soft fluctuation tangent to the continuously degenerate family of equilibrium orientations. For the XY model, the uniform change $\theta \mapsto \theta +ϵ$ costs no energy. Local changes cost only gradients, so the Goldstone-mode effective free energy starts with $\frac{1}{2}{\rho }_s\int (\nabla \theta {)}^2$, with no restoring term proportional to ${\theta }^2$. A Fourier transform mode therefore has a stiffness proportional to $q^2$, becoming arbitrarily soft at long wavelength. The dimensionless phase stiffness here is $\bar{K}=\beta {\rho }_s$. These are the spin waves of the angle field. Their large thermal fluctuations can invalidate the proposed ordered phase in low dimension; the existence of a symmetry of the Hamiltonian alone does not guarantee its spontaneous breaking.

An algebra involution is a conjugate-linear map $a\mapsto a^{\ast }$ satisfying $(ab{)}^{\ast }=b^{\ast }{a}^{\ast }$ and $(a^{\ast }{)}^{\ast }=a$; on a unital algebra it fixes $1$. In the Banach-star convention it is continuous, although the following spectral identity uses only its algebraic properties. An element $a$ is invertible precisely when $a^{\ast }$ is: its inverse is $(a^{-1}{)}^{\ast }$. Applying this to $a=\lambda 1-x$ provesFor a nonunital algebra, extend the involution to its unitization of an algebra by $(a,\lambda {)}^{\ast }=(a^{\ast },\overline{\lambda })$ and use the spectrum in that unitization; the same argument proves the identity.

A unital C-star algebra is a complex Banach algebra with an involution satisfying the C-star identity $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$. A Normal element of a C-star algebra satisfies $x^{\ast }x=x{x}^{\ast }$. The norm identity for the nonzero identity implies $\Vert 1\Vert =1$.

Use the common nonzero identity $1$. If $\Vert h\Vert <1$, completeness and submultiplicativity give the Neumann seriesFor $a\in G(A)$ and $\Vert a^{-1}\Vert \Vert b-a\Vert <1$, write $b=a(1+a^{-1}(b-a))$ and apply that series. This proves that the group of invertible elements of a Banach algebra is open. Locally the same series bounds $\Vert b^{-1}\Vert$, andproves continuity of inversion. Inversion is its own inverse, hence it is a homeomorphism of $G(A)$ onto itself.

The spectrum of an element is ${\sigma }_A(x)=\{\lambda :\lambda 1-x\notin G(A)\}$, and its complement is the algebra resolvent set ${\rho }_A(x)$. Openness of $G(A)$ makes the algebra spectrum closed. For $|\lambda |>\Vert x\Vert$,so it is bounded and thus compact. For nonemptiness, suppose $R$ existed everywhere. The local Neumann expansion around each point makes $R$ a Banach-space-valued entire function. For every bounded linear functional $\ell$ on $A$, $\ell (R(\lambda ))$ is entire, bounded on compact discs, and tends to zero at infinity by the displayed series. The Liouville theorem makes it identically zero. The Hahn-Banach theorem separates points, so $R(\lambda )=0$, contradicting $(\lambda 1-x)R(\lambda )=1$. Hence the algebra spectrum is nonempty and compact.

If $B$ is a closed unital subalgebra with the same identity, then ${\rho }_B(x)\subseteq {\rho }_A(x)$. It is relatively open by the Neumann series in $B$. If ${\lambda }_j\in {\rho }_B(x)$ tends to $\lambda \in {\rho }_A(x)$, continuity of inversion in $A$ gives $({\lambda }_{j}1-x{)}^{-1}\to (\lambda 1-x{)}^{-1}$. Closedness puts this last inverse in $B$, so $\lambda \in {\rho }_B(x)$. Therefore ${\rho }_B(x)$ is relatively open and closed in ${\rho }_A(x)$. This proves the required resolvent components in a closed unital subalgebra directly, with no spectral-boundary assumption.

The absorption question concerns the geometric-optics capture cross-section, assuming rays reaching the central singular boundary are absorbed. Null geodesic motion alone does not specify a wave boundary condition at a singularity.

By spherical symmetry choose an equatorial ray. Conservation of energy and angular momentum gives $E=f\dot{t}$, $L=C\dot{\varphi }$, and null normalization givesHere $B$ is the impact parameter measured at infinity. A returning ray has an exterior turning point; capture occurs when none exists. The critical impact parameter obeys $B_c^2=1/\sup H$, and the cross-section is the area $\pi B_c^2$ in the incident impact-parameter plane.

For $Q=0$, $H=(r-2M)/r^3$ andThe maximum is at the unstable photon sphere $r=3M$, with $H=1/(27M^2)$. Therefore

For $Q^2=2M^2$, cancellation for $r>2M$ gives $H=1/r^2$. There is no regular exterior circular photon orbit: the supremum occurs at the singular limiting surface, with $\sup H=1/(4M^2)$. Rays with $B>2M$ turn at $r=B$; those with $B<2M$ reach the central boundary. Thus the extremal dilaton metric photon capture result isThe critical ray is a measure-zero boundary of the capture set. The finite capture cross-section is not the horizon area, which vanishes in this limit. Nor is it a claim about a low-frequency quantum absorption cross-section or an unspecified reflection law at the singularity.

The Brans-Dicke theory couples a scalar field $\Phi$ to the Ricci scalar. Assume $\Phi \ne 0$, constant $\omega$, matter independent of $\Phi$, and variations of compact support so boundary terms may be discarded. Use the paper's curvature sign convention and signature $(+---)$. Set $X=g^{ab}{\partial }_a\Phi {\partial }_b\Phi$ and $\square \Phi ={\nabla }_a{\nabla }^a\Phi$. Define the stress-energy tensor consistently byThis convention fixes the sign of the matter term.

Vary with respect to $g^{ab}$. The variation of metric volume density contributes $-g_{ab}/2$ times each scalar Lagrangian, while $\delta R=R_{ab}\delta g^{ab}+g^{ab}\delta R_{ab}$. Applying the Palatini identity followed by two rounds of integration by parts gives the metric variation of a scalar-curvature coupling,The differentiated $\Phi$ terms remain because the curvature multiplier is not constant. Varying the kinetic term at fixed $\Phi$ givesCombining these terms with the matter variation, the Euler-Lagrange equation isRaising both free indices reproduces the contravariant form, because ${\nabla }_a{g}_{bc}=0$.

The independent scalar variation givesor $R+\omega {\Phi }^{-2}X-2\omega {\Phi }^{-1}\square \Phi =0$. The trace of the metric tensor equation in four dimensions isMultiplying the scalar equation by $\Phi$ eliminates the same combination $\Phi R+\omega {\Phi }^{-1}X$. Thus the scalar equation of Brans-Dicke theory isAt the degenerate coupling of Brans-Dicke theory, $\omega =-3/2$, the combined equations impose $T=0$ instead; division by $3+2\omega$ is unavailable. If matter depended directly on $\Phi$, its scalar variation would add a source and the displayed scalar equation would change.

The following lettered sections justify the supplied variation identities; they are supporting identities in the PDF, rather than three independent field-equation problems.

The weak topology $\sigma (E,E^{\ast })$ is the coarsest topology making every continuous linear functional in $E^{\ast }$ continuous. A basic neighborhood of $x$ imposes finitely many inequalities $|f_j(z-x)|<\epsilon$, and a sequence converges weakly precisely when each $f(x_n)$ converges to $f(x)$. The Hahn-Banach theorem makes the dual separate points, so the weak topology is Hausdorff.

Here is a proof of the requested weak subsequence property, the compact-to-sequential direction of the Eberlein-Šmulian theorem. Let $Y$ be the closed linear span of the countable set $\{x_n:n\ge 1\}$. It is a separable Banach space. The Hahn-Banach theorem separates points outside a closed vector subspace from that subspace, so $Y$ is weakly closed in $E$. Therefore $C=K\cap Y$ is weakly compact.

Choose a countable dense set $(u_j)$ in the unit sphere of $Y$, and choose norming functionals $f_j\in Y^{\ast }$ with $\Vert f_j\Vert =1$ and $f_j(u_j)=1$. They separate points: for $y\ne 0$, a sufficiently close $u_j$ to $y/\Vert y\Vert$ gives $|f_j(y)|>0$. This is a countable norming family. Extend the functionals to $E$ by the Hahn-Banach theorem. The evaluation mapis weakly continuous and injective. Its domain is compact and its target Hausdorff, so it is a homeomorphism onto its image: images of closed subsets are compact, hence closed. A countable product of real or complex lines is metrizable, for example by ${\sum }_j{2}^{-j}\min (1,|a_j-b_j|)$. Thus $C$ has a compact metrizable weak topology. Compact metric spaces are sequentially compact, giving a subsequence with $\boxed{x_{n_j}\rightharpoonup x\in C\subseteq K}$. If $Y=\{0\}$ the result is immediate. We did not assert that a separable Banach space has a separable dual or that its weak topology is generally metrizable.

Work in three-dimensional projective space, with four homogeneous coordinates indexed $0,1,2,3$. Let $A$ and $B$ be independent representatives of two points. The Plücker coordinates of their line areThis antisymmetric matrix has rank two and is defined only up to a nonzero scalar. Replacing the pair by another basis of the same two-dimensional subspace multiplies $L$ by the determinant of that basis change. Its six independent entries satisfy $L^{01}{L}^{23}-L^{02}{L}^{13}+L^{03}{L}^{12}=0$, since the exterior product $(A\wedge B)\wedge (A\wedge B)$ vanishes.

To describe the same line as an intersection of planes, choose a volume orientation with ${\epsilon }_{0123}=1$ and define the dual Plücker coordinates byThe independent components pair asConversely, $L^{ij}=\frac{1}{2}{\epsilon }^{ijkl}{L}_{kl}$ with the matching volume convention. The lower-index form annihilates both defining points: $L_{ij}{A}^j=L_{ij}{B}^j=0$. Its columns span the pencil of planes containing their line. This dualization uses the volume form, not a Euclidean metric for raising and lowering indices. Two-index point and plane representations are dual in projective dimension three; in higher dimensions the corresponding dual exterior form has a different degree.

For a general plane covector $F_i$, contract the upper-index form:This is a linear combination of $A,B$, so it is on the line, and $F_i{P}^i=0$ by antisymmetry, so it is on the plane. Provided the resulting vector is nonzero, its homogeneous coordinates give the unique intersection, including intersections at infinity. If $F(A)=F(B)=0$, the whole line lies in the plane and the zero result correctly indicates that there is no unique intersection. These are the line-plane incidence in Plücker coordinates operations used in the subparts.

The spatial reflection $x\mapsto -x$ acts on the two complex amplitudes as $(A,B)\mapsto (A,-B)$, because the first spatial eigenfunction is even and the second is odd. The first equation must therefore be even in $B,\overline{B}$, and the second odd. For a nonresonant double Hopf bifurcation, the temporal phases give independent actions $A\mapsto e^{i{\phi }_1}A$, $B\mapsto e^{i{\phi }_2}B$ in the cubic normal form of a dynamical system. A cubic monomial in the first equation must have net $A$ phase weight one and net $B$ phase weight zero. Only $A|A{|}^2,A|B{|}^2$ satisfy this; the second equation similarly has $B|B{|}^2,B|A{|}^2$. Parametrizing the independent cross-saturation coefficients givesAll coefficients can be complex: their real parts determine amplitude growth and their imaginary parts frequency shifts. These are cubic truncations of a family of reduced vector fields on the four-real-dimensional extended centre manifold, with the two unfolding parameters entering ${\nu }_1,{\nu }_2$.

The nonresonance assumption must be made explicit. Merely having ${\omega }_1\ne {\omega }_2$ does not imply this form: for ${\omega }_1=2{\omega }_2$, the quadratic terms $B^2$ in $\dot{A}$ and $A\overline{B}$ in $\dot{B}$ satisfy both temporal resonance and spatial parity. Thus the initial form is the generic nonresonant double-Hopf form, rather than a statement for every possible unequal frequency pair.

At the one-to-one reflection-symmetric one-to-one Hopf resonance, only the common temporal phase action remains. Cubic terms must then have total phase weight one. Spatial reflection retains, in the first equation, the three terms $A|A{|}^2,A|B{|}^2,B^2\overline{A}$; the second retains $B|B{|}^2,B|A{|}^2,A^2\overline{B}$. Common phase covariance excludes all quadratic terms. The additional resonant terms are consequently $-{\gamma }_1{B}^2\overline{A}$ and $-{\gamma }_2{A}^2\overline{B}$, with independent complex coefficients. This is the reflection-symmetric one-to-one Hopf resonance.

In the specified equal-coefficient specialization, put $r=R^2$. A nonzero pure-$A$ periodic branch exists for $\mathrm{Re}{\nu }_1>0$ and hasIts radial eigenvalue is $-2{\alpha }_{R}r<0$, and its phase is neutral. For a transverse $B$ perturbation the linear equation isIts time dependence can be removed exactly by $B=C{e}^{i\Omega t}$. Using ${\nu }_1-\alpha r=i\Omega$ and ${\nu }_2={\nu }_1-d$ givesWriting $C=u+iv$, $d=d_R+i{d}_I$, $\beta ={\beta }_R+i{\beta }_I$ yields the constant real stability matrixLet $Q=\mathrm{Re}(\beta \overline{d})={\beta }_R{d}_R+{\beta }_I{d}_I$. The transverse characteristic polynomial isBoth transverse eigenvalues have negative real part exactly whenThese hold at $r=0$ because $d_R>0$. Thus a sufficiently small pure branch is transversely stable; it is stable modulo its neutral temporal phase until one of the two inequalities fails. The secondary steady and Hopf thresholds of a resonant pure mode are derived in the two parts below. They refer to accessible points on the pure branch; in a multi-parameter problem $d$ may also vary, so a proposed parameter path must actually cross the corresponding surface.

Let $X,Y$ be prey and predator abundances. With constant positive demographic and predation coefficients, the Lotka-Volterra equations areTheir positive equilibrium is $X_{\ast }=\gamma /\delta$, $Y_{\ast }=\alpha /\beta$. Its Jacobian matrix and eigenvalues areThe linearized populations oscillate with period $2\pi /\sqrt{\alpha \gamma }$, with the predator peak following the prey peak. This is a center equilibrium, not an attracting equilibrium. The conclusion is supported beyond linearization by the Logarithmic first integral of the Lotka-Volterra equations: in these units one may useDirect differentiation gives $\dot{V}=0$. Its Hessian matrix is a positive-definite matrix in the positive quadrant, and its minimum is at $(X_{\ast },Y_{\ast })$. Nearby level curves are closed periodic orbits, so there is no damping toward a preferred oscillation amplitude.

For a stationary solid skeleton, the Darcy velocity $\mathbf{u}$ is liquid volume transport per unit total cross-sectional area, while the pore velocity $\mathbf{v}$ is the intrinsic mean fluid velocity within the connected pore space. ThusFor an incompressible liquid in a saturated region, conservation of mass is ${\varphi }_t+\nabla \cdot \mathbf{u}=0$. With time-independent but spatially varying porosity, it becomesnot generally $\nabla \cdot \mathbf{v}=0$. Neglecting dispersion, a fluid interface is advected by the pore velocity; its mean normal speed is $V_n=\mathbf{v}\cdot \mathbf{n}=\mathbf{u}\cdot \mathbf{n}/\varphi$.

For the thin layer on the inclined plane, the gravitational acceleration components are $g\sin \theta$ downslope and $-g\cos \theta$ normal to the plane. The hydrostatic approximation and constant pressure $p_a$ above the current giveWith permeability of a porous medium $k(z)$, Darcy's law therefore givesHere $u_x,u_y$ denote the components of the Darcy velocity. Integrate over depth, using the prescribed power laws:Both integrals are finite for the positive exponents in the model. Conservation of the stored fluid gives ${\mathcal{V}}_t+{\nabla }_{x,y}\cdot \mathbf{Q}=0$, orFor $0<\theta <\pi /2$, set $X=x\tan \theta$, $Y=y\tan \theta$ andSince $\cos \theta {\tan }^2\theta =\sin \theta \tan \theta$, the depth-dependent inclined porous-current equation isThe longitudinal flux carries fluid downslope, while the thickness-gradient terms spread it. The degenerate horizontal or vertical slope limits need their own coordinate scaling.

The Rosseland mean opacity is the transport opacity appropriate to an optically thick region close to local thermodynamic equilibrium. If ${\kappa }_{\nu }$ is monochromatic transport opacity and $B_{\nu }(T)$ the Planck function, thenIt is a weighted harmonic mean: low-opacity frequency windows carry much of the diffusive flux. One must combine the monochromatic processes before averaging, rather than adding their separately averaged opacities.

In hot fully ionized material, electron-scattering opacity supplies an almost density- and temperature-independent contribution at fixed composition. In the nonrelativistic Thomson scattering limit, ${\kappa }_{\mathrm{es}}={\sigma }_T{n}_e/\rho \simeq 0.2(1+X)\mathrm{c}{\mathrm{m}}^2{\mathrm{g}}^{-1}$. Free-free opacity, the absorption counterpart of thermal bremsstrahlung, comes from photon absorption during encounters between free electrons and ions. Its approximate mean obeys Kramers' opacity law, ${\kappa }_{\mathrm{ff}}\propto \rho T^{-7/2}$, with composition and quantum correction factors.

When ions retain bound electrons, bound-free absorption is photoionization; edges, occupations and ionization fractions strongly affect its frequency dependence. Bound-bound absorption produces absorption lines. Numerous metal lines can close low-opacity windows and produce important opacity features in partially ionized layers. Line populations depend on excitation as well as ionization, while Doppler broadening, pressure broadening and overlapping lines change the transport windows. Stimulated inverse processes must be included consistently with detailed balance and the local thermal radiation field.

In cooler hydrogen-rich atmospheres, negative hydrogen ion opacity from both bound-free and free-free processes is often important. Its strong temperature dependence helps determine convective surface relations. In still cooler layers, molecular bands, and at sufficiently low temperatures dust absorption/scattering, may be important. These processes cannot be represented by one Kramers power law over the whole density-temperature plane.

An opacity calculation therefore needs the chemical abundances, the equation of state, electron density and bound-state populations. Saha ionization equation and thermal excitation populations give useful dilute LTE estimates; high-density pressure ionization, nonideal effects and altered level populations require corrections. Composition matters even when one writes only $\kappa (\rho ,T)$: hydrogen, helium and the metal abundance pattern are additional fixed inputs.

In dense degenerate cores, thermal conduction by electrons can dominate energy transport. An equivalent conductive opacity may be introduced so that $1/{\kappa }_{\mathrm{eff}}=1/{\kappa }_R+1/{\kappa }_{\mathrm{cond}}$, because the two diffusive fluxes add; conduction is not itself a photon absorption process. Very hot regimes also require departures from the simple nonrelativistic scattering approximation. Near the optically thin photosphere, a Rosseland diffusion description alone is inadequate, and frequency-dependent radiative transfer with appropriate boundary conditions is required. These regime changes are why realistic stellar opacity tables, rather than one global power law, enter stellar structure.

A reduction must encode primal solutions and infeasibility or unboundedness information; merely declaring a known optimum or subtracting an unknown optimum from the objective function is not a construction. The following homogeneous self-dual embedding of a linear program works without assuming the input problem has an optimum.

First remove linearly dependent equality rows by Gaussian elimination. If an eliminated row has an inconsistent right-hand side, the original problem is already certified infeasible. Return the one-variable special-form program $\min 0$ with $A^{\ast }=0$, $y=1$, and retain that certificate; no original mathematical optimization remains. Otherwise the remaining matrix has full row rank; write it as $A$ again. Put $e=1\in {\mathbb{R}}^n$ andUse nonnegative variables $x,s\in {\mathbb{R}}^n$ and $\tau ,\kappa ,\theta ,\rho \in \mathbb{R}$, with a temporary unrestricted multiplier $\lambda \in {\mathbb{R}}^m$. Impose the homogeneous linear systemAll coefficients are computable from the input. The temporary free multiplier can be eliminated without introducing cancelling positive/negative variables. DefineReplace the second equation by $(I-A^{T}H)q=0$, and substitute $Hq$ in the other equations. Because $A$ has full row rank, $A^T\lambda =q$ is equivalent to these conditions. For no remaining equality rows, simply omit $\lambda$ and its empty equations.

Let $y=(x,s,\tau ,\kappa ,\theta ,\rho )$ and $n^{\ast }=2n+4$. Collect these substituted equations into $A^{\ast }y=0$, impose $1^{T}y=1$, and let $c^{\ast }$ select the coordinate $\theta$. The resulting Karmarkar standard form isThe next two parts prove its known positive starting point and zero optimal value. Notice that normalization is legitimate for any nonzero nonnegative embedding vector because all the equations are homogeneous; it does not require a prior bound on the original optimizer.

Code the two levels of each factor as $\pm 1$. For a factor subset $A$, its factorial contrast column is the character ${\chi }_A(x)={\prod }_{j\in A}{x}_j$. Products of columns correspond to symmetric differences of subsets, because $x_j^2=1$.

Let $G_1,\dots ,G_k$ be defining words whose incidence vectors are linearly independent over ${\mathbb{F}}_2$. Write $x_j=(-1{)}^{u_j}$, with $u_j\in {\mathbb{F}}_2$. Specifying signs ${\chi }_{G_i}(x)=s_i$ is then a rank-$k$ system of binary linear equations. Every one of its $2^k$ right-hand sides has $2^{n-k}$ solutions. Thus $k$ independent defining contrasts split the full $2^n$ design into $2^k$ equal fractions. Independence is necessary: dependent defining words do not yield $2^k$ distinct sets.

The defining contrast subgroup has $2^k$ words. On a selected fraction its word $G$ has a constant sign $s_G$, soHence the coefficients of $A$ and $AG$ cannot be estimated separately without assumptions or extra runs: this is aliasing in a fractional factorial design. Conversely, averaging a character over the affine binary solution space vanishes unless its word is in the defining subgroup, so these signed subgroup cosets give all aliases. Here a lowercase treatment label records the factors at their high levels, and $(1)$ puts all five at their low levels; signs must be retained when listing aliases.

A balanced incomplete block design has $t$ treatments and $b$ blocks, each containing $k$ distinct treatments with $1<k<t$. Every treatment occurs in $r$ blocks and every distinct treatment pair occurs together in $\lambda >0$ blocks. A symmetric balanced incomplete block design has $b=t$, which implies $r=k$ by the first counting identity below.

For the separate even-order condition, let $N$ be the square treatment-by-block incidence matrix of a set system of a symmetric design. Its diagonal overlap counts are $r$ and its off-diagonal counts are $\lambda$, givingThe counting identities proved in the next slots show that $r=k$ and $r+(t-1)\lambda =r^2$. Consequently the eigenvalues of $N{N}^T$ are $r^2$ on the constant vector and $r-\lambda$ on its $(t-1)$-dimensional orthogonal complement. Taking determinants givesIf $b=t$ is even, the exponent $t-1$ is odd. The exponent of every prime in $\det (N{)}^2$ and in $r^2$ is even; therefore its exponent in the positive integer $r-\lambda$ must be even. This proves the even-order symmetric design square obstruction:It is a necessary condition, not a sufficiency claim.

A credibility factor is the weight assigned to a particular risk's observed experience when blending it with a population or manual premium. In the Bühlmann model, the blend is the best affine mean-square predictor, rather than automatically the full posterior mean.

Let $v=\mathbb{E}[{\sigma }^2(\theta )]$, the expected process variance, and assume the displayed second moments are finite. The law of total expectation gives $\mathbb{E}{X}_i=m$. The law of total variance and law of total covariance giveFor the second formula, the conditional covariance is zero by conditional independence, while the covariance of the conditional means is $\mathrm{Var}\mu (\theta )=a$, the variance of hypothetical means. Likewisebecause $\mathbb{E}[X_i-\mu (\theta )\mid \theta ]=0$.

To derive the Bühlmann credibility premium, minimize $\mathbb{E}[(\mu (\theta )-b_0-{\sum }_i{b}_i{X}_i{)}^2]$. Differentiating in $b_0$ sets the predictor's mean to $m$. The remaining normal equations areFor $v>0$, subtracting any two rows forces all $b_i$ to be equal, then gives $b_i=a/(v+na)$. Thus the premium and factor areThe same affine premium predicts $X_{n+1}$: conditional independence makes $\mathbb{E}[(X_{n+1}-\hat{\mu }{)}^2]=v+\mathbb{E}[(\mu (\theta )-\hat{\mu }{)}^2]$, so minimizers agree. If $v=0$, every year's amount equals $\mu (\theta )$ almost surely, and $Z=1$ gives the exact risk mean even though the coefficient vector need not be unique.

For fixed $a>0,n$, $\partial Z/\partial v=-na/(na+v{)}^2<0$. Thus more within-risk variation decreases credibility: individual experience becomes less informative, $Z\to 0$ as $v\to \infty$, and the premium tends to $m$. Increasing the number of observed years has the opposite effect.

For numerical estimation from a balanced panel, assume the $k$ risks are independent draws from the same risk population, with comparable exposures, and $n,k\ge 2$. SetConditionally, each within-risk sample variance has mean ${\sigma }^2({\theta }_s)$; hence $\mathbb{E}\hat{v}=v$. The risk sample means have variance $a+v/n$, so $\mathbb{E}B=a+v/n$. Therefore $\tilde{a}=B-\hat{v}/n$ is an unbiased method of moments estimate of $a$. For a usable nonnegative variance estimate take $\hat{a}=\max (0,\tilde{a})$, noting that this truncation sacrifices exact unbiasedness.

The balanced-panel method-of-moments credibility estimate for risk $s$ is consequentlyIf $\hat{a}=0$ and $\hat{v}>0$, use zero credibility. If both estimates are zero, the within and between sums of squares are zero and all the observed amounts coincide; any blend gives that same premium. With only one year per risk or only one risk, these within/between estimators are unavailable, so external structural estimates or additional assumptions are necessary.

Write the predictable sigma-algebra on positive times asIt is equivalently the smallest sigma-algebra making all left-continuous adapted processes measurable. Indeed, the indicator of each generating rectangle is left-continuous and adapted, and approximating any left-continuous adapted process by its values at the preceding points of deterministic partitions gives the reverse inclusion. Previsibility records information available immediately before observation. If time zero is included, add $B\times \{0\}$ with $B\in {\mathcal{F}}_0$. The displayed description in part (b) uses positive times, as indicated by its product space.

For the exit law, the two identified processes have the same argument. Starting at zero, neither can skip $-a$ or $b$: Brownian motion is continuous, and the symmetric Poisson difference process has unit jumps with integer boundary levels. Therefore $|X_{t\wedge T}|\le K=\max (a,b)$, and if $T$ is finite, $X_T\in \{-a,b\}$.

To justify finiteness rather than presume it, apply the optional stopping theorem to the martingale $X_t^2-t$ at the bounded stopping time $t\wedge T$:The first boundary hitting time is a stopping time in both cases: in the continuous case use the continuous-path hitting criterion, and in the jump case inspect the successive jump times. Monotone convergence now gives $\mathbb{E}T\le K^2$, so $T<\infty$ almost surely. The stopped position tends to $X_T$ as $t\to \infty$ and is bounded, making dominated convergence applicable. Stopping $X$ at $t\wedge T$ and taking this limit yields $\mathbb{E}{X}_T=0$.

If $\pi =\mathbb{P}(X_T=-a)$, then $0=-a\pi +b(1-\pi )$. Thus in both cases the entire distribution isThis is the same balanced gambler's ruin law despite the different one-time distributions. As a further consistency check, passing to the limit in the stopped square-martingale identity givesAll uses of an unbounded exit time were obtained by limits of bounded-time sampling with an explicit bound on the stopped position.