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A test channel is a proposed conditional distribution of a reconstruction given the source symbol. It describes a candidate joint probability distribution for evaluating mutual information and expected distortion. It need not be a physical transmission channel, or an implementable one-symbol encoder; the Shannon rate-distortion theorem relates this optimization to long block codes.

A distortion measure assigns a nonnegative cost to replacing a source symbol $x$ by a reconstruction $\hat{x}$. It need not be a metric: the two alphabets may differ, and neither symmetry nor the triangle inequality is required. Choosing the cost is part of the model. For numerical symbols, $d(x,\hat{x})=(x-\hat{x}{)}^2$ gives mean squared error after averaging; for exact symbol agreement, use Hamming distortion.

In free molecular flow, gas–gas collisions are negligible over the trajectories relevant to the observable. Molecules still encounter walls, whose scattering of momentum and energy must be modeled. In a long narrow tube, ${\ell }_{\mathrm{mfp}}$ exceeding the diameter does not by itself prove that every complete transmission path is collision-free: repeated wall reflections can create much longer paths. The approximation must be checked against those paths as well as the width. ^[ref]

A Knudsen layer is the gas region adjoining a boundary where the molecular distribution departs from the bulk near-equilibrium description over distances of order a few mean free paths. Continuum slip and jump conditions describe an effective boundary for the outer solution; they do not resolve this layer. ^[ref]

For endomorphisms $S,T$ of a finite-dimensional space of dimension $d$, $r(S+T)\le r(S)+r(T)$ and $n(ST)\le n(S)+n(T)$. If both equalities hold, the first gives $r(S)+r(T)\le d$, while the second gives $2d-r(S)-r(T)\le d$. Thus the rank sum is $d$, $S+T$ is an isomorphism, and $ST=0$. Conversely, $ST=0$ implies $\mathrm{im}T\subseteq \ker S$ and hence rank sum at most $d$, whereas invertibility of $S+T$ gives rank sum at least $d$. This proves both equalities.

Divide participants into strata defined by prespecified baseline prognostic variables and randomize independently within each stratum. Permuted blocks can keep treatment counts close throughout recruitment. This supports balance of the stratification variable between arms by design, while allocation concealment keeps upcoming assignments hidden. Analyses should account for important design strata where appropriate.

For each new eligible participant, compare the imbalance that would result from each treatment assignment across prespecified baseline covariate margins. Assign with a probability favoring the less imbalanced choice, while retaining a random component and allocation concealment. This targets marginal balance when a large number of joint strata makes stratified randomization with blocks inefficient. It is a sequential allocation rule, and inference should respect its design.

For discrete random variables $X,Y$, their joint probability mass function assigns a probability to each possible ordered pair. Its entries are nonnegative and sum to one. Marginalization gives the probability mass functions $p_X(x)={\sum }_y{p}_{X,Y}(x,y)$ and $p_Y(y)={\sum }_x{p}_{X,Y}(x,y)$. In a table with rows indexed by $x$ and columns by $y$, these are the row and column totals. The joint probability distribution records how the coordinates occur together, while the marginal distributions describe each separately.

For independent and identically distributed random variables $X_j$ independent of a nonnegative integer count $N$, set $S={\sum }_{j=1}^N{X}_j$, including $S=0$ when $N=0$. Conditioning on the count gives $\mathbb{E}[e^{tS}\mid N=k]=M_X(t{)}^k$, hence the displayed composition of the moment-generating function with the probability generating function. For positive summands the identity holds for all $t\le 0$; positive $t$ requires convergence of the composed series. This permits Laplace transforms of nonnegative random variables to identify the law even when positive exponential moments do not exist.

Let $n$ be a positive integer, let $0<p<1$, and put $q=1-p$ and let the original independent count have negative binomial distribution with probability mass function $\left(\genfrac{}{}{0ex}{}{n+k-1}{k}\right)p^n{q}^k$ on $k\ge 0$. With unit-rate exponential distribution severities, its moment-generating function is $[p+qp/(p-t){]}^n$ for $t<p$. Thus it also has a random sum of independent claims representation with independent binomial distribution count $B$ and rate-$p$ exponential distribution steps. For $n=2$ the law is $p^2{\delta }_0+2pq\mathrm{Exp}(p)+q^2\mathrm{Erlang}(2,p)$. The hurdle decomposition of a positive random sum therefore has zero mass $p^2$ and positive density $p^2{e}^{-px}(2q+q^{2}x)/(1-p^2)$.

The positive-integer restriction on $n$ is essential for the finite binomial distribution representation. A negative binomial distribution also permits noninteger positive shape through its generalized probability mass function, but that broader model cannot be represented by a binomial distribution with a noninteger number of trials.

Take an independent geometric distribution count with $\mathbb{P}(N=k)=p(1-p{)}^k$, $k\ge 0$, and sum its independent positive claims. Its moment-generating function is the displayed composition wherever finite. With $q=1-p$ and severity moments $m_j$, the third cumulant is $(q/p)m_3+3(q/p{)}^2{m}_1{m}_2+2(q/p{)}^3{m}_1^3$, which is strictly positive for nonzero positive claims and $0<p<1$. The law has an atom of mass $p$ at zero. Its positive-count convention would instead add a factor $M_X(t)$ to the numerator, as in the geometric-sum moment-generating function.

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.