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Fit a quadratic mean at three temperatures $x_0,x_0\pm 1$, with equal replication $n$ at the off-center temperatures and any positive replication $m$ at the center. The three parameters fit the three group means exactly, giving the displayed estimates. Independent Gaussian errors make these estimates jointly normal. Their covariance is $\frac{1}{4}[\mathrm{Var}({\bar{Y}}_{+})-\mathrm{Var}({\bar{Y}}_{-})]=0$, so they are independent. Equal off-center replication is essential to this cancellation.

An ideal linear polarizer transmits the fraction ${\cos }^2\theta$ of incident linearly polarized light when its axis makes angle $\theta$ to the polarization. For a single photon, the corresponding conditional transmission probability is $|⟨\theta |0⟩{|}^2={\cos }^2\theta$, by the Born rule; rejection has probability ${\sin }^2\theta$. The classical field calculation projects its electric-field amplitude onto the transmission direction and squares that amplitude to obtain intensity. This is a conditional probability given the incident polarization and is not automatically the joint probability of passing one analyzer and failing another.

An operator on a complex Hilbert space is accretive when the displayed inequality holds throughout its domain. Its negative is a dissipative operator. For a bounded everywhere-defined operator, expansion gives $\Vert (I+T)x{\Vert }^2\ge \Vert x{\Vert }^2$. Its adjoint operator satisfies the same accretivity condition. Consequently $I+T$ has closed range and trivial kernel, and its range is dense because its adjoint has trivial kernel. Thus $I+T$ is onto and $\Vert (I+T{)}^{-1}\Vert \le 1$. Boundedness and the full domain are material here; for unbounded operators accretivity alone need not imply surjectivity.

A linear interpolant is the unique polynomial of polynomial degree at most one taking prescribed values $y_a,y_b$ at distinct interpolation nodes $a,b$. It is an affine function, including the constant function case. For $a\le x\le b$, its two displayed coefficients are nonnegative and sum to one, so changing each sampled value by at most $\delta$ changes every interpolated value by at most $\delta$. Differentiation instead uses the difference of the two samples divided by $b-a$, causing noise amplification by differentiation of a linear interpolant.

A sequence with errors $e_k=x_k-x^{\ast }$ converges Q-linearly in a specified norm if there is a constant $q<1$ with $\Vert e_{k+1}\Vert \le q\Vert e_k\Vert$ eventually. The limsup of consecutive nonzero error ratios is its asymptotic factor. This is stronger than merely having a bound $\Vert e_k\Vert \le C{q}^k$, which is called R-linear convergence. The norm and whether an iteration is one coordinate step or an entire sweep must be specified when quoting a factor.

A progressively measurable real diffusion coefficient of absolute value one gives a continuous local martingale with quadratic variation ${\int }_0^t{\sigma }_s^{2}ds=t$. With zero initial value, Lévy characterization of Brownian motion makes the solution standard Brownian motion. This applies to every weak solution of $d{X}_t=\mathrm{sgn}(X_t-a)d{B}_t$. The Brownian law does not imply that the solution equals the given driver or establish strong existence for a discontinuous coefficient.

If the sum of Lebesgue outer measures on the right is infinite, the inequality is automatic. Otherwise, fix $\epsilon >0$ and cover each $A_n$ by a countable family of intervals whose total length is less than $m^{\ast }(A_n)+\epsilon 2^{-n}$. These covers exist by the definition of infimum. The combined countable family covers ${\bigcup }_n{A}_n$ and has total length at most ${\sum }_n{m}^{\ast }(A_n)+\epsilon$. Take the infimum over covers of the union, then let $\epsilon$ tend to zero. No measurability assumption on the $A_n$ is needed.

If $A\subseteq B$, every countable interval cover of $B$ is also a cover of $A$. Taking the infimum of the lengths over the larger family of covers can only decrease the result. Thus Lebesgue outer measure is monotone on all subsets, with no measurability assumption.

Translating every interval of a countable cover of $A\subseteq \mathbb{R}$ by $a$ produces a cover of $A+a$ with the same lengths. Taking the infimum of cover lengths gives $m^{\ast }(A+a)\le m^{\ast }(A)$. Translating back by $-a$ gives the reverse inequality. This proves translation invariance for Lebesgue outer measure on every subset, without first assuming the set is Lebesgue measurable.

Translating a Lebesgue measurable set by a real number preserves measurability and its Lebesgue measure. Translation invariance of Lebesgue outer measure and Carathéodory's criterion give the measurability assertion and equality. Applying the result to indicator functions, then simple functions, and finally increasing simple approximations gives ${\int }_{\mathbb{R}}u(x-t)dx={\int }_{\mathbb{R}}u(x)dx$ for every nonnegative Lebesgue measurable function. For a Lebesgue integrable function, apply the identity to its positive part and negative part.

The Lebesgue inner measure approximates a subset of the real line from within by compact sets, while Lebesgue outer measure approximates it from outside by countable interval covers. For a finite closed interval $I$ containing $E$,Indeed, every compact $K\subseteq E$ gives $m^{\ast }(I\setminus E)\le m(I)-m(K)$. Conversely, an open cover of $I\setminus E$ with total length arbitrarily close to its outer measure has compact complement in $I$ lying inside $E$, giving the reverse inequality in the limit.

For $m^{\ast }(E)<\infty$, equality of inner and outer measures holds precisely when $E$ is Lebesgue measurable. For an unbounded set, measurability is equivalently checked on all bounded intersections $E\cap [-n,n]$. Equality $m_{\ast }(E)=m^{\ast }(E)=\infty$ alone is insufficient: adjoining a bounded nonmeasurable subset to a disjoint half-line makes both quantities infinite without making the union measurable.

A bound $\mathbb{E}[B_{j+1}\mid B_j]\le f(B_j)$ implies $\mathbb{E}{B}_{j+1}\le \mathbb{E}f(B_j)$. It does not by itself imply $\mathbb{E}{B}_{j+1}\le f(\mathbb{E}{B}_j)$. For $f(b)=b^2/n$, Jensen's inequality goes in the opposite direction. The missing contribution is the variance: $\mathbb{E}[B_j^2]=(\mathbb{E}{B}_j{)}^2+\mathrm{Var}(B_j)$.

A null weight contributes zero to a Laplace integral on its domain of convergence. It can be added to a weight without changing the represented function, so an integration-by-parts differential equation for the weight is not an unconditional uniqueness theorem.

For example, let $\Gamma$ enter along angle $\pi /3$ from infinity to zero and leave along angle $2\pi /3$ to infinity. The entire weight $h(\zeta )=e^{{\zeta }^2}$ is null for every complex $z$. Throughout the wedge between those rays, $\mathrm{Re}({\zeta }^2)\le -r^2/2$; the factor $e^{z\zeta }$ is dominated by this quadratic decay. Closing the contour by a large circular arc and applying the Cauchy integral theorem therefore proves the zero integral. All derivatives with respect to $z$ converge locally uniformly as well.

On the same contour, the quartic weight $e^{{\zeta }^4/4}$ gives a nontrivial entire solution. Its rays can be deformed within the quartic decay sectors for a contour integral to angles $\pi /4$ and $3\pi /4$, where the value at $z=0$ is $-\sqrt{2}{4}^{-3/4}\Gamma (1/4)\ne 0$. Adding the null weight changes neither this integral nor its vanishing endpoint terms, but the sum does not satisfy $g^{\prime }={\zeta }^{3}g$. Thus the usual Laplace method selects a convenient weight by cancelling the interior integrand separately; a necessity claim needs additional injectivity assumptions.

Retaining integer assignment equations while dualizing extra capacity inequalities gives an easily evaluated upper bound for a maximization problem. With multipliers $\lambda \ge 0$ and capacities $b$, it is $\theta (\lambda )={\max }_{x\text{ assignment}}[c^{T}x+{\lambda }^T(b-Ax)]$. It is a convex maximum of affine functions. Minimizing it exploits the convex hull of assignments, so it can have a finite optimum when only a fractional mixture satisfies the side constraints and no integer assignment does. This separates a useful relaxation bound from a certificate of integer feasibility.

For the real-coefficient differential operator $Mv=(p{v}^{\prime }{)}^{\prime }+qv$, integration by parts givesIf $p$ tends to zero at both endpoints and $u,v,u^{\prime },v^{\prime }$ remain bounded there, the boundary term vanishes without prescribing endpoint values. Thus regular endpoint behaviour can supply the domain condition that makes the expression symmetric. Endpoint degeneracy alone is insufficient for arbitrary singular functions: for $p=x(1-x)$ and $v=\log x$, one has $p{v}^{\prime }=1-x\to 1$ at zero. A self-adjoint differential operator requires a specified admissible domain, rather than only a formal differential expression.

Continuous square-integrable martingales on a fixed finite horizon form a complete space under the displayed terminal norm, modulo indistinguishability of stochastic processes. A terminal-norm Cauchy sequence admits a subsequence whose successive differences have norms below $2^{-n}$. The Doob L2 maximal inequality and Markov's inequality make the probabilities that these differences have suprema above $2^{-n/2}$ summable. The Borel-Cantelli first lemma then gives an almost surely uniformly convergent telescoping series, so the limit has continuous paths.

Conditional-expectation contractivity gives convergence in $L^2$ at every fixed time, and passage to the limit in the martingale identities proves that the limit is a martingale. Adaptedness can be retained by taking the pointwise finite subsequence limit when it exists, and zero otherwise. The original Cauchy sequence converges in terminal norm to this limit. This completeness allows the Itô isometry to extend stochastic integration from simple predictable integrands to the full square-integrable integrand space while retaining a continuous martingale version.

Suppose measurable $a_n$ on a Lebesgue measurable set $E$ satisfy ${\sum }_n{\int }_E|a_n|<\infty$. The nonnegative partial sums $S_N={\sum }_{n\le N}|a_n|$ increase to a measurable extended-valued $S$. The monotone convergence theorem gives ${\int }_{E}S={\sum }_n{\int }_E|a_n|<\infty$, forcing $S<\infty$ almost everywhere. Each $a_n$ is finite almost everywhere too; discard the union of these countably many null sets when defining signed partial sums. The series ${\sum }_n{a}_n$ then converges absolutely outside a null set.