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A sequence of chains exhibits cutoff when its distance from stationarity drops from near one to near zero in a time window negligible compared with its mixing-time scale.

For a finite transitive reversible chain with nontrivial eigenvalues ${\lambda }_i$,

The field has ring of integers $\mathbb{Z}[\sqrt{-33}]$, discriminant $-132$, and Minkowski bound below eight. The nonprincipal ramified ideals above two and three generate four classes, and the split primes above seven introduce no new class. Hence

For $K=\mathbb{Q}(\sqrt[3]{3})$ with ${\mathcal{O}}_K=\mathbb{Z}[\sqrt[3]{3}]$, the field discriminant is $-243$ and the Minkowski bound is $8\sqrt{3}/\pi <5$. The prime ideals above $2$ and $3$ are principal, so every ideal class has a principal representative and

If a rule has constant risk $r$ and a sequence of priors has Bayes risks tending to $r$, then the rule is minimax. Every competing rule has worst-case risk at least every one of those Bayes risks.

An equalizer rule has the same risk at every parameter value. A Bayes equalizer rule is minimax when its constant risk bounds from below the worst-case risk of every competing rule.

If an upper-triangular $n$ by $n$ matrix has diagonal entries ${\lambda }_1,\dots ,{\lambda }_n$, its standard invariant flag satisfiesThe commuting product ${\prod }_k(A-{\lambda }_{k}I)$ therefore vanishes, proving $\mathrm{deg}{m}_A\le n$ without invoking the full Cayley-Hamilton theorem.

If $p_1,\dots ,p_r$ are pairwise coprime, thenBezout identities between the factors construct the corresponding projections.

An information projection chooses the probability distribution that satisfies a constraint while departing least from a reference law in relative entropy. The constraint might specify an expected value, several marginal probability distributions, or another convex set of permissible laws. The reference is held fixed: the variable is the first argument of $D(q\Vert p)$. Reversing the arguments gives a different minimization problem.

For a finite alphabet $\mathcal{A}$, take a positive reference probability mass function $p(a)>0$ and a nonempty closed convex set $\mathcal{C}$ in the probability simplex. ThenThe minimum exists and its minimizing probability distribution is unique. In this article, natural logarithms give nats. Using base-two logarithms rescales the objective without changing the minimizer.^[ref]

If $p$ already meets the constraint, its information projection is $p$ itself by relative entropy nonnegativity. If not, a constraint can change the relative weights of outcomes without specifying every weight individually. Relative entropy minimization under a mean constraint gives an especially useful example: it tilts the reference law rather than choosing an arbitrary distribution with the right average.

If $f$ is locally Lebesgue integrable on ${\mathbb{R}}^n$, then for Lebesgue almost every $x$,

A measure is sigma-finite when its space is a countable union of measurable sets of finite measure.

A measure $\mu$ on $X$ is finite when $\mu (X)<\infty$.

For pairwise disjoint measurable sets $E_n$, countable additivity means

The infinite series is the limit of its nonnegative partial sums. This controls an entire countable union, not merely each finite stage. If $q_1,q_2,\dots$ enumerate the rational numbers in $[0,1]$, their finite collections have Lebesgue measure zero, and countable additivity gives their whole union Lebesgue measure zero. Their being a dense set does not change this calculation.

The axiom makes no assertion about an arbitrary uncountable set of disjoint pieces. In particular, $[0,1]$ is the union of its singletons, each of Lebesgue measure zero, yet its Lebesgue measure is one. Extending additivity to all such unions would contradict interval length.

A left Haar measure on a locally compact group that is a Hausdorff space is a nonzero regular Borel measure, finite on compact sets, satisfying $\mu (gA)=\mu (A)$ for every Borel set $A$. Such a measure exists and is unique up to a positive scalar. Right Haar measure instead satisfies $\mu (Ag)=\mu (A)$; the two need not coincide. On a compact group, normalization gives total mass one and invariance on both sides. See Tornier, Haar Measures, Definition 2.1 and Theorem 2.2.

For a finite group, normalized Haar measure is uniform counting; for the circle group, it is normalized angle integration on the circle. These examples turn symmetry into a rule for averaging.

If a measure $\nu$ has density $w=d\nu /d\mu$ with respect to $\mu$, thenThis identity is the basis of likelihood ratios and importance sampling.

For a measurable map $f:X\to Y$, the pushforward of $\mu$ is the measureIt is the distribution induced on $Y$ by mapping points distributed according to $\mu$ through $f$.

A measure space is a set $X$, a sigma-algebra $\mathcal{M}$ of its subsets, and a measure $\mu$ defined on $\mathcal{M}$. Specifying only $X$ leaves both the measurable questions and their weights undetermined.

For a concrete probability space, take $X=\{a,b,c\}$, $\mathcal{M}$ its power set, and weights $1/2,1/3,1/6$. Define $\mu (A)$ by adding the weights of the elements in $A$. ThenA probability measure is exactly a measure normalized by $\mu (X)=1$, usually written $\mathbb{P}(\Omega )=1$. An indicator function has expected value equal to its event's probability. For a different continuous example, $[0,1]$ with the restricted Lebesgue measure is a probability space; the event $[0,1/4]$ has probability $1/4$.

If a harmonic function on a connected domain attains an interior maximum or minimum, it is constant. The mean value property forces equality throughout every sufficiently small ball around an interior extremum, and connectedness propagates the equality.

This is the strong harmonic instance of the maximum principle.

If a continuous function has the spherical mean value property at every point for some sequence of radii decreasing to zero, then it is harmonic. Comparison on each relatively compact ball with the harmonic function having the same boundary data proves the result by propagating any hypothetical interior extremum along one of the admissible spheres.

For $X\sim \mathrm{Bin}(n,\theta )$ andthe mean squared error isWhen ${\theta }_0=1/2$, its maximal risk is at $\theta =1/2$ or at an endpoint of $[0,1]$.