For an independent binomial distribution count $N\sim \mathrm{Binomial}(n,p)$, the aggregate can also be represented as ${\sum }_{j=1}^n{B}_j{X}_j$, with independent parameter-$p$ Bernoulli random variables $B_j$ and independent claim sizes. Its cumulant-generating function is $n\log (1-p+p{M}_X(t))$. Its third cumulant is $n(p{m}_3-3p^2{m}_1{m}_2+2p^3{m}_1^3)$ in severity moments. Constant positive claims $d$ give $np(1-p)(1-2p)d^3$, which is negative when $p>1/2$. Thus positive claim sizes do not by themselves force a positive third central moment.

A compound mixed Poisson distribution is a random sum of independent claims whose count has a Poisson mixture law: conditional on a nonnegative random intensity $\Lambda$, $N$ has Poisson distribution with parameter $\Lambda$. The claim sizes are independent and identically distributed random variables, independent of the intensity and count mechanism. If $M_X$ is their moment-generating function, then $M_S(t)=\mathbb{E}\exp (\Lambda (M_X(t)-1))$ wherever finite. A gamma distribution intensity of shape $r$ and rate $\alpha$ gives a negative binomial distribution count with shape $r$ and success probability $\alpha /(\alpha +1)$. Independent intensities with common gamma rate add their shapes by additivity of independent gamma distributions with a common rate.

For independent identically distributed summands, independent also of the count $N$, conditioning on $N=n$ gives $\mathbb{E}[e^{tZ}\mid N=n]=M_Y(t{)}^n$. The law of total expectation gives the composition of the count's probability generating function with the summand's moment-generating function. This identity is valid as an extended nonnegative expectation; finiteness near zero requires additional exponential-integrability assumptions.

For $k=2$, a disconnected graph has a component of size $3\le s\le n/2$. The probability that a fixed $s$-set has no crossing edge is at most $(s/n{)}^{2s}(1-s/n{)}^{2(n-s)}$. The union bound and $\left(\genfrac{}{}{0ex}{}{n}{s}\right)\le (n/s{)}^s[n/(n-s){]}^{n-s}$ bound the failure probability by ${\sum }_{s=3}^{\lfloor n/2\rfloor }(s/n{)}^s=o(1)$. Split this sum at $\sqrt{n}$. The random k-out graph coupling extends the result to every fixed $k\ge 2$.

Every open set $O$ in the ordinary topology on infinite subsets is a Ramsey set of infinite subsets: some infinite $B$ has $[B{]}^{\omega }\subseteq O$ or $[B{]}^{\omega }\cap O=\varnothing$. Use acceptance and rejection of finite stems and fusion to decide every finite stem. If the empty stem is rejected, each rejected stem has only finitely many accepting one-point extensions. Successive avoidance of these finite obstructions gives an infinite set all of whose finite stems are rejected; openness then excludes every infinite subset from $O$. This concerns increasing enumerations of infinite subsets, rather than unrestricted sequences with repetitions.

Suppose ${\alpha }^m=a\in {\mathbb{Q}}^{\times }$ in a number field and $\mathfrak{p}$ lies above $p$. Its normalized integral valuation satisfies $m{v}_{\mathfrak{p}}(\alpha )=e_{\mathfrak{p}/p}{v}_p(a)$. If $m$ and $v_p(a)$ are coprime, then $m\mid e_{\mathfrak{p}/p}$. In a degree-$m$ field the fundamental identity for prime decomposition forces a unique prime, ramification index $m$, and residue degree one. This detects total ramification of a prime even when the displayed defining polynomial is not Eisenstein at $p$.

// Target: algebra.bigb

For an odd prime $p$, this Eisenstein polynomial generates a totally ramified cubic root field. Its discriminant is $-p^2(4p+27)$. For $p>3$ the unit part is a square exactly when $p\equiv 1(\mathrm{mod}3)$. The splitting field then has cyclic Galois group of order three; otherwise it has symmetric group $S_3$, with an unramified extension of degree two and inertia group $C_3$. Both cases are tamely ramified, so all positive ramification groups are trivial.

At $p=3$, adjoining $\beta =\sqrt{-39}$ gives a ramified quadratic subfield, and the splitting field has total ramification index six. A root $\alpha$ has normalized valuation two, while $\beta$ has valuation three, so $\pi =\beta /\alpha$ is a uniformizer. The derivative $3{\alpha }^2+3$ has valuation six, and all distinct root differences have valuation three. A three-cycle fixes $\beta$, giving $v_L(\sigma \pi -\pi )=2$. A transposition fixing $\alpha$ negates $\beta$, giving valuation one; conjugation handles the other transpositions. Hence $G_0=S_3$, $G_1=C_3$, and $G_i=1$ for $i\ge 2$. The different exponent is seven.

For $\sigma \in G_i$, the residue of $(\sigma {\pi }_L-{\pi }_L)/{\pi }_L^{i+1}$ gives a homomorphism to the additive residue field with kernel $G_{i+1}$. On composing two automorphisms, their leading displacement coefficients add because each fixes residues and changes the uniformizer only in degree at least two. These successive quotients are elementary abelian groups in residue characteristic $p$. The filtration terminates, so the wild inertia group is a finite $p$-group.

The Gamma reflection formula gives $(\sin \pi z/\pi )\Gamma (z)=1/\Gamma (1-z)$, analytic at zero with value one. A finite-parameter Watson's lemma representation of Ramanujan's function therefore gives the displayed coefficients. In particular $c_0=1$, $c_1=-\gamma$ and $c_2={\gamma }^2-{\pi }^2/6$, with $\gamma$ Euler's constant. Obtaining this representation requires taking an imaginary part before sending its Mellin exponent to zero; the original complex integral is divergent at that boundary.

A stellar core in which radiative diffusion in a star carries most of the energy, rather than convection. Its formation in a pre-main-sequence star breaks the fully convective assumption behind an idealized Hayashi track.

For a steady point-like astronomical source with isotropy of emission, in transparent space with negligible relativistic or cosmological effects, conservation of energy spreads its luminosity over a sphere of surface area $4\pi d^2$. Its bolometric observed radiative flux isThis inverse-square law reflects geometric dilution: the same emitted power crosses increasingly large spheres. It does not require light to lose energy on the journey. For a fixed luminosity, doubling Euclidean distance quarters the observed radiative flux. ^[ref].

With $L_0=10^{26}\mathrm{W}$ and $d_0=20$ parsecs, the result is $F_0\simeq 2.09\times 10^{-11}\mathrm{W}{\mathrm{m}}^{-2}$. At $40$ parsecs, the same source gives $F_0/4\simeq 5.22\times 10^{-12}\mathrm{W}{\mathrm{m}}^{-2}$. An ideal collecting area of $1{\mathrm{m}}^2$ at $20$ parsecs intercepts approximately $2.09\times 10^{-11}\mathrm{W}$, before detector losses.

Multipole radiation expands radiation from localized time-dependent sources into electric and magnetic multipoles. Electric dipole radiation is the leading electric term, while magnetic dipole radiation and higher electric multipoles contribute when the leading term vanishes or higher precision is needed.

A 2013 experiment by Paul A. Williams and collaborators mounted a 25-mm dielectric mirror on a commercial force-restoration balance. Removing its pan-weight compensation spring allowed horizontal force sensing with a horizontal laser path. They measured mirror optical reflectance and varied incidence from about $7^{\circ }$ to $45^{\circ }$. A 1071-nm laser, capable of 530 W, was switched at 0.1 Hz. Digitized mass-equivalent readings were converted to force using $g$; quadrature demodulation isolated the switched signal from drift. ^[ref]

Measurement calibration used a 20-g mass, with checks down to 1 mg. The optical forces lay below that checked range, limiting the accuracy claim. An angle scan tested the predicted cosine dependence at fixed intercepted power. Replacing the mirror with a carbon nanotube coated absorber increased absorbed power while reducing the predicted optical force; no modulation peak was detected. This bounded gas-heating artifacts under those conditions, rather than proving their universal absence. The fitted slope was about 7% below $2/c$. Inferring laser power requires calibrated force, optical reflectance, radiometric absorptance and angle of incidence, with measurement uncertainty for each. ^[ref]

For radiation in vacuum with isotropy, energy is distributed equally per solid angle: write $u_{\Omega }=u/(4\pi )$ for energy density per unit solid angle. Let $\theta$ be the angle between a ray's path and the wall normal. Its normal velocity $c\cos \theta$ determines the arrival rate, while normal momentum per energy is $\cos \theta /c$. Their product is ${\cos }^2\theta$. For a perfectly reflecting cavity wall,The leading two is the normal momentum reversal; the incoming hemisphere contains half of the field's directional energy density. This is also the normal component of the bulk pressure tensor, $P_{nn}={\int }_{4\pi }{u}_{\Omega }{\cos }^2\theta d\Omega =u/3$. Isotropy makes all three diagonal components equal, while their sum is $u$ because the squared direction components sum to one. The factor one third therefore depends on three-dimensional angular geometry, not on the shape of the blackbody radiation spectrum.

A perfectly absorbing wall exposed to this incoming hemisphere receives only $u/6$ from absorption. If it is in thermal equilibrium with the isotropic radiation, its outgoing emission supplies another $u/6$ of recoil, recovering $u/3$. A cold absorber before equilibrium is a different boundary condition. Similarly, the one-way incident energy flux is $cu/4$, whereas the net energy flux across a plane in the undisturbed isotropic field is zero. Dividing $cu/4$ by $c$ misses the extra normal-momentum projection and does not give the cavity pressure.

For a stipulated isotropic radiation energy density $u=6\mu \mathrm{J}{\mathrm{m}}^{-3}$, the bulk pressure is $2\mu \mathrm{Pa}$ regardless of its spectral shape. In thermal equilibrium, blackbody radiation supplies $u=a_{\mathrm{rad}}{T}^4$ with radiation constant $a_{\mathrm{rad}}=4\sigma /c$, so $p_{\mathrm{rad}}=a_{\mathrm{rad}}{T}^4/3$; Stefan–Boltzmann law defines $\sigma$. The equilibrium photon gas derives this same equation of state from its free energy.^[ref]

For normal incidence in vacuum, a uniform beam delivers energy $IA\Delta t$ to an illuminated surface area $A$ during time $\Delta t$. A perfect absorber receives momentum $IA\Delta t/c$, so force divided by surface area gives $p_{\mathrm{abs}}=I/c$. A perfect stationary mirror sends the photons back with reversed momentum, so momentum conservation gives twice the normal transfer, $p_{\mathrm{refl}}=2I/c$. These are time-averaged pressures; the microscopic arrivals need not be uniform.

For a stipulated model light intensity $I=1000\mathrm{W}{\mathrm{m}}^{-2}$ and the exact SI speed of light $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$,These are computed predictions, not measured data or a universal sunlight value. If the beam fully covers a normally oriented surface area $A=0.010{\mathrm{m}}^2$, the intercepted radiant flux is $\Phi =IA=10\mathrm{W}$ and the absorber's force is $\Phi /c\simeq 33.356\mathrm{nN}$. Using the total source power would be wrong if only part of its beam reaches the receiver. The absorption prediction assumes complete photon absorption, while the reflection prediction assumes complete specular reflection; transmission, scattering of light and directed thermal photon emission require their own outgoing momentum accounting.

Measure the angle of incidence $\theta$ between the incoming path and the surface's normal line, with $0\le \theta <\pi /2$. There are two separate cosine factors. A fixed actual surface area $A$ intercepts only its projected area $A\cos \theta$, so $\Phi =IA\cos \theta$. Each incoming photon supplies normal momentum $(E_{\gamma }/c)\cos \theta$ on absorption, or twice that on specular reflection. Thus, for radiometric absorptance $a$ and reflected power fraction $R$,This model assumes transmitted light continues straight in the same medium and carries away its original momentum. It omits thermal recoil and assumes a stationary receiver. At $\theta =60^{\circ }$, fixed beam light intensity therefore gives one quarter of the normal-incidence pressure. For fixed intercepted radiant flux $\Phi$, however, the normal force has only one cosine factor. Equivalently, if the given quantity is receiving-surface irradiance $E_{\mathrm{inc}}=I\cos \theta$, then $p_{\perp }=(a+2R)E_{\mathrm{inc}}\cos \theta /c$. The distinction matters when interpreting an experiment's angle scan. ^[ref]

Absorption also transfers a tangential momentum component at oblique incidence. It contributes a tangential force, rather than normal pressure. Ideal specular reflection preserves the tangential momentum component, so it transfers only normal momentum.

Simultaneous trilinear baryon number and lepton number violation permits a squark to connect the two interactions and induce a dimension-six proton-decay operator. Its coefficient is of order ${\lambda }^{\prime }{\lambda }^{\prime \prime }/m_{\tilde{q}}^2$. A dimensional decay-width estimate then gives ${\Gamma }_p\sim |{\lambda }^{\prime }{\lambda }^{\prime \prime }{|}^2{m}_p^5/m_{\tilde{q}}^4$, up to phase space and hadronic matrix elements. The flavor antisymmetry of $U^c{D}^c{D}^c$ matters: the two down-type flavor indices must differ. For example, ${\lambda }_{112}^{\prime \prime }$ and ${\lambda }_{112}^{\prime }$ allow strange-squark exchange in $p\to e^{+}{\pi }^0$.

For fixed $u\ne 0$, the parallel affine lines $l_x=x+\mathrm{span}(u)$ are precisely the cosets of the indicated quotient vector space. Changing representatives adds a multiple of $u$, so $l_x+l_y=l_{x+y}$ and $a{l}_x=l_{ax}$ are well-defined. The zero is the whole line $l_0$, not a single point. If $(u,b_1,b_2)$ is a basis, the two line classes $l_{b_1},l_{b_2}$ form a basis of the quotient by spanning and independence modulo $\mathrm{span}(u)$.

For independent Poisson process arrivals and exponential distribution claim sizes of mean $\mu$, minimizing the logarithm of the negative expected exponential utility gives $H^{\prime }(\alpha )=\lambda \beta \mu [(1-\beta \mu \alpha {)}^{-2}-(1+{\theta }_R)]$. Its strictly positive second derivative makes the displayed stationary point unique. On $0\le \alpha \le 1$, the optimum is $\min (1,{\alpha }_{\ast })$. On the strictly open interval $0<\alpha <1$, there is no maximizer when ${\alpha }_{\ast }\ge 1$, only a supremum at full retention. The stationary retention increases with reinsurer loading and is independent of the direct insurer's loading.

If insurer and reinsurer use the same positive relative safety loading $\theta$, retained claim sizes and net premiums both scale by the retention $\alpha$: $X_{\alpha }=\alpha X$ and $c_{\alpha }=\alpha c$. Substitution into the adjustment coefficient equation gives $R_{\alpha }=R_1/\alpha$. Thus there is no maximizing positive retention on $(0,1]$; the supremum is approached as $\alpha \downarrow 0$. At full cession, if allowed, both retained claims and net premiums vanish, and the insurer's nonnegative capital remains constant. The coefficient is then not uniquely defined by the degenerate equation, although ultimate ruin probability is zero in this idealized model.

A flat universe dominated by a canonical homogeneous scalar field with constant equation-of-state parameter $w>-1$ has $\rho ={\rho }_0(a/a_0{)}^{-3(1+w)}$ and ${\dot{\varphi }}^2=(1+w)\rho$. The positive rolling branch gives the displayed relation and $V(\varphi )=(1-w){\rho }_0\exp [-\sqrt{24\pi G(1+w)}(\varphi -{\varphi }_0)]/2$. The opposite rolling branch reverses the exponential slope. For $w=-1$ the field is constant; $w<-1$ cannot be realized by a real scalar with this positive kinetic term. An exponential slope $\lambda \sqrt{8\pi G}$ yields $w={\lambda }^2/3-1$, with accelerating nonphantom expansion when $0<{\lambda }^2<2$.