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For a positive Borel measure on ${\mathbb{R}}^d$, let $E_{\alpha }=\{M_u\nu >\alpha \}$, using open balls in the uncentered maximal function of a finite measure. Every ball witnessing the inequality lies entirely in $E_{\alpha }$. Cover a compact subset by finitely many witnessing balls and select disjoint balls greedily by decreasing radius. Their threefold dilations cover the compact set. Their total measure is at most $\nu (E_{\alpha })$, giving the bound by inner regularity of Lebesgue measure. For a signed or complex measure, apply this to its total variation measure.

An unbiased estimator $T$ is uniformly minimum-variance unbiased if its variance is no larger than that of every other unbiased estimator of the same parameter, for every parameter value. The Lehmann–Scheffé theorem gives such an estimator whenever an unbiased function of a complete sufficient statistic exists. Attaining the Cramér-Rao bound is sufficient for this property under its hypotheses, but is not necessary: for $T\sim \Gamma (n,\lambda )$ with $n>2$, $(n-1)/T$ is unbiased and has variance ${\lambda }^2/(n-2)$, while the information bound is ${\lambda }^2/n$. Completeness follows from uniqueness of the Laplace transform of $h(t)t^{n-1}$, so the estimator is still uniformly optimal among unbiased estimators.

For independent $X_i=\theta +Y_i$ with $Y_i$ exponentially distributed at known rate $\lambda$, the joint likelihood factors through the sample minimum $X_{(1)}$. Its excess over the endpoint is exponential with rate $n\lambda$, because $\mathbb{P}(\min Y_i>t)=e^{-n\lambda t}$. Thus the displayed estimator is an unbiased estimator, with variance $(n\lambda {)}^{-2}$. Even if $\theta \ge 0$, the estimator can be negative; truncation at zero introduces bias.

An ultralimit is the direct limit of a chain of structures formed by taking successive ultrapowers along their diagonal elementary embeddings. At limit stages use the elementary chain theorem. This is a different concept from a topological ultrafilter limit. Alternating relative ultrapower embeddings constructs isomorphic ultralimits of any two elementarily equivalent structures, without requiring the stronger single-ultrapower isomorphism theorem.

Let $S$ be a nonempty set and $\mathcal{P}\subseteq \mathcal{P}(S)$ be upward closed. An ultrafilter all of whose members lie in $\mathcal{P}$ exists exactly when every finite set partition of $S$ has a cell in $\mathcal{P}$. For sufficiency, the complements of sets outside $\mathcal{P}$ have the finite intersection property: a finite cover by such sets could be disjointified into a partition with no cell in $\mathcal{P}$. The ultrafilter lemma extends the generated proper filter. Necessity follows because an ultrafilter selects exactly one cell of every finite partition. This criterion requires partition richness of the whole ground set; it does not assert that the complement family is closed under unions.

The finite-trace independent family construction gives $2^{|X|}$ independent subsets of an infinite $X$. Each binary assignment generates a proper filter and, in ZFC, has an ultrafilter extension. Different assignments force incompatible membership decisions, giving $2^{2^{|X|}}$ distinct ultrafilters. The matching upper bound follows because every ultrafilter is a subset of $\mathcal{P}(X)$.

An ultrafilter with finite-sums members is an ultrafilter on the positive integers all of whose members are IP sets. Such an ultrafilter exists by extending the IP-star filter using the ultrafilter lemma. If a member were not IP, its complement would be an IP-star set and hence another member, contradicting properness. Conversely, every such ultrafilter contains the entire IP-star filter, since it cannot contain the non-IP complement of an IP-star set. This characterization does not assert that the ultrafilter is idempotent.

An ultrafilter on the positive integers with every member containing finite arithmetic progressions of arbitrarily large length. The Van der Waerden theorem and ultrafilter selection from a partition-rich family give existence. Such an ultrafilter must contain the complement of every set with bounded arithmetic-progression length, and it is a nonprincipal ultrafilter. No corresponding ultrafilter can have an infinite arithmetic progression in every member, by the dyadic block obstruction to infinite arithmetic progressions.

A uniform kappa-complete ultrafilter $U$ on an uncountable cardinal $\kappa$ is normal if every regressive function on a $U$-large set is constant on a $U$-large subset. Equivalently it is closed under diagonal intersections ${\Delta }_{\alpha <\kappa }{X}_{\alpha }=\{\beta <\kappa :(\forall \alpha <\beta )\beta \in X_{\alpha }\}$. Normality supplies homogeneous sets for pair-colourings by combining the chosen large tail at each earlier coordinate.

An ultrafilter on $P_{\kappa }(\lambda )$ is fine when it contains the set of small subsets containing $\alpha$, for each $\alpha <\lambda$. Fineness makes every coordinate of $\lambda$ represented in the corresponding ultrapower, though no one small subset is required to contain every coordinate.

In the classical risk model, condition on the first arrival of the Poisson process. If available capital just before it is $v$, ruin has probability ${\overline{F}}_X(v)+{\int }_0^v{f}_X(x)\psi (v-x)dx$. Integrate this expression at $v=u+ct$ against $\lambda e^{-\lambda t}$ and change variables from $t$ to $v$. Differentiating the lower limit gives the displayed equation. The claim-density convolution is continuous for bounded $\psi$, so the first-arrival representation also supplies the differentiability required by the equation.

For rate-$\alpha$ exponential distribution claims and $c>\lambda /\alpha$, the ruin integro-differential equation reduces to ${\psi }^{\prime \prime }+(\alpha -\lambda /c){\psi }^{\prime }=0$. Differentiate the convolution and use its equation $A^{\prime }=\alpha \psi -\alpha A$ to eliminate it. Positive relative safety loading implies $\psi (u)\to 0$ as $u\to \infty$, and the zero-capital survival probability gives $\psi (0)=\lambda /(\alpha c)$. These two conditions determine the displayed solution. The adjustment coefficient is $\alpha -\lambda /c$.

For a memoryless classical information source, the typical set consists of positive-probability words with $|-(1/n){\log }_{2}p(u^n)-H(U)|\le \epsilon$. The weak law of large numbers applied to $-{\log }_{2}p(U_j)$ proves its probability tends to one. Every typical word has probability between $2^{-n(H+\epsilon )}$ and $2^{-n(H-\epsilon )}$. Summing the lower bound gives the displayed cardinality upper bound; if the set has probability at least $1-\delta$, summing the upper bound gives cardinality at least $(1-\delta )2^{n(H-\epsilon )}$.

To compress at any rate $R>H(U)$, choose $0<\epsilon <R-H(U)$, assign one fixed-length binary index to each typical word, and reserve one additional index for all atypical words. At most $\lceil nR\rceil$ bits suffice for all large $n$. Decoding inverts the typical-word indices and makes an arbitrary output for the reserved index. The reconstruction error is at most the probability of the atypical set and tends to zero.

Suppose $|a_u|\le \tau (u)$, $|b_d|\le \log (2N)$, $u,d>X\ge 1$, $ud\le N$, and $|\theta -a/q|\le q^{-2}$ with $(a,q)=1$. A dyadic decomposition reduces $S=\sum a_u{b}_{d}e(\theta ud)$ to blocks $u\asymp U,d\asymp D$. The Cauchy-Schwarz inequality, divisor-square summatory bound and the reciprocal fractional-part sum near a rational giveThe condition $ud\le N$ makes the allowed $u$ values for each pair $d_1,d_2$ an interval, so the exponential geometric sum bound applies despite the hyperbolic cutoff. Since $UD\le N$ and $U,D\gg X$, take square roots and sum the $O({\log }^2(2N))$ blocks to get the displayed estimate.

Approximate a fixed mass of iron-group freeze-out ashes by nickel-56 and iron-54, with no further weak neutronization and negligible residual light particles. Their electron fractions are $1/2$ and $26/54$. Conservation of the initial electron fraction in stellar matter, $1/2-X_{22}/22$, then gives the displayed nickel mass fraction. Its argument is the initial neon-22 mass fraction, which need not equal the earlier nitrogen metallicity. The two-isotope mass-budget approximation is not an exact full nuclear equilibrium with all free-particle densities set to zero.

The nickel-56 to cobalt-56 to iron-56 decay chain supplies gamma rays and charged particles that heat supernova ejecta. Early expansion and photon diffusion delay the escaping optical light; after a peak on a scale of weeks, the luminosity falls as decay power declines and gamma rays escape more readily. With comparable diffusion times and deposition efficiencies, peak luminosity scales approximately with the nickel-56 mass. Using one fixed absolute luminosity without accounting for composition-dependent yields can therefore bias standard-candle distances.

If a decoder already knows the length, alphabet and empirical distribution, it only needs to distinguish $|T_n(q)|$ possible strings. A fixed-length index therefore needs $\lceil {\log }_2|T_n(q)|\rceil$ bits. List the six balanced four-symbol strings in increasing lexicographic order and number them $0$ through $5$. A three-bit index identifies any one of them, whereas writing the original binary string directly takes four bits.

Knowing the empirical distribution is essential to that calculation. If it must also be sent, a simple fixed-length description of which of the five classes was used costs another $\lceil {\log }_{2}5\rceil =3$ bits. That naive balanced-string message totals six bits and is longer than the raw four-bit string. The example demonstrates the benefit of shared constraints, rather than a free lossless source coding scheme for every input. For long strings over a fixed alphabet, the overhead for identifying a class grows only logarithmically with length, while the information entropy of its frequency profile controls the leading cost within the class.

Write $n_a=nq(a)$. Arranging $n_a$ copies of each symbol gives the multinomial coefficientThe division by the factorials removes permutations among identical copies. In the balanced four-symbol example, $|T_4(q)|=4!/(2!2!)=6$. For binary symbols this becomes the binomial coefficient $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$; for three symbols with counts $(2,1,1)$, the class contains $4!/(2!1!1!)=12$ strings.

For a no claims discount system with discounts $0,\alpha ,\beta$, assume a policyholder starts at $\beta$, the first claim lowers next year's discount to $\alpha$, and a second claim in the same year lowers it to zero. With no further losses and undiscounted future premiums, claiming for the first loss costs $c(\beta -\alpha )$ in extra premium. Once that claim is submitted, the second claim adds $c\alpha$ next year and $c(\beta -\alpha )$ the year after, a total $c\beta$. Thus the second decision uses the incremental threshold, not the cost of both claims relative to no claim. For independent unit-rate exponential distribution losses the probabilities are $e^{-c(\beta -\alpha )}$ and $e^{-c\beta }$, respectively; the latter remains the conditional probability given submission of the first claim.

Translations are regular on $V$, and the zero stabilizer is $H$. Two-transitivity is equivalent to $H$ being transitive on nonzero vectors. Full affine linear groups and groups with special linear or symplectic linear parts give examples.