Consider the original linear programOne unit of $v$ costs twice as much as one unit of $u$, so $(u,v)=(3,0)$ is a plausible answer. To certify it, multiply the three linear inequalities by $0,1,1$, respectively, and add:The proposed feasible point attains $3$, so its cost is the global minimum.

In the conic dual problem convention, the data and certificate areDirect matrix multiplication gives $A^{T}y=c$ and $b^{T}y=3$. At $x=(3,0{)}^T$, the slack variable is $s=(3,0,0{)}^T$, so $s^{T}y=0$. The first linear inequality has positive slack and zero multiplier; the two positively weighted linear inequalities are tight. This is complementary slackness.

The entire dual feasible set can also be read off: write $y=(1-r,2-r,r{)}^T$ with $0\le r\le 1$. Then $A^{T}y=c$ and $b^{T}y=3r$. Maximizing this lower bound gives $r=1$. A point such as $r=1/2$ is still a valid certificate of the weaker bound $u+2v\ge 3/2$.

The difference of the two Levi-Civita connections is ${\tilde{\Gamma }}^a{}_{bc}-{\Gamma }^a{}_{bc}={\delta }_b^a{\partial }_c\log \Omega +{\delta }_c^a{\partial }_b\log \Omega -g_{bc}{g}^{ad}{\partial }_d\log \Omega$. Along a null geodesic with tangent $k$, the last term vanishes and the remaining acceleration is $2k(\log \Omega )k$. It is parallel to the tangent, so changing the affine parameter removes it. Thus a positive conformal factor preserves unparametrized null geodesics, although it generally does not preserve timelike or spacelike geodesics.

For the complete observer $\chi =0$ in two-dimensional de Sitter spacetime, let $d(\chi ,0)\in [0,\pi ]$ be shortest angular distance. An event can signal to the observer exactly when $d(\chi ,0)<\pi /2-T$, and can receive its signals exactly when $d(\chi ,0)<T+\pi /2$. Equality describes its future and past observer event horizons. Their intersection is the static patch of de Sitter spacetime containing the observer. These are observer-dependent cosmological horizons; constant-$T$ circles are Cauchy hypersurfaces, so there are no Cauchy horizons in the full spacetime.

The Projective Minkowski compactification has infinite cyclic fundamental group, while the conformal compactification of Euclidean space is $S^4$ and simply connected. Thus the affine substitution $t\mapsto it$ cannot extend to an invertible smooth identification of these real compactifications. Both are real slices of the same complex Klein quadric, which permits local analytic continuation without equating their global real topologies.

In affine cone coordinates $[U,V,t,\mathbf{x}]$ with $2UV+t^2-|\mathbf{x}{|}^2=0$ and $U=1$, the substitution is represented on the complex quadric by the rational map $[U,V,t,\mathbf{x}]\mapsto [U^2,UV+t^2,iUt,U\mathbf{x}]$. It preserves the quadric equation but is indeterminate at some infinity points, including $[0,1,0,0]$. This is a direct algebraic manifestation of the obstruction to a global regular Wick map.

Inverse stereographic projection identifies Euclidean four-space with a sphere minus one point. Multiplying its flat metric by $4/(1+|x{|}^2{)}^2$ gives the round metric, which extends smoothly over the added point. In the Klein quadric description, the compact space is the projectivized real null cone of signature $(1,5)$. It differs topologically from Minkowski conformal compactification.

A boundary obtained by extending the rescaled metric $\tilde{g}={\Omega }^{2}g$ smoothly and nondegenerately to a hypersurface on which $\Omega$ vanishes with nonzero derivative. Its causal character is measured using the extended Lorentzian metric, not the divergent physical metric. In signature $(+---)$, a timelike normal makes the boundary spacelike, a spacelike normal makes it timelike, and a null normal makes it null. These cases lead to different causal and initial-value properties. The conformal cylinder of two-dimensional de Sitter spacetime has spacelike past/future boundaries; the conformal strip of two-dimensional anti-de Sitter spacetime has two timelike boundaries.

For the pressure-neglecting confined plume with compensating return flow, define specific plume momentum flux $M=b^2{w}^2$ and area fraction $s=Q^2/(R^{2}M)$. Then $J=M/(1-s)$ andThe branch issuing from a pure plume reaches a fold at $s=1/2$, where plume and ambient have equal areas, opposite velocities and equal specific momentum fluxes. The numerator is nonzero on this branch, so $M^{\prime }$ diverges although $M,Q,w$ remain finite. This signals failure of the steady uniform-profile description, not infinite physical velocity.

A half-edge records an incidence of an edge at a vertex. Two half-edges form each edge, including two incidences at the same vertex for a self-loop. In the configuration model these incidences are individually labelled before random pairing.

For bounded vertex degrees, total degree $M\to \infty$, and a forbidden graph $G_0$ of bounded maximum degree, put $\lambda =M^{-1}{\sum }_i{d}_i(d_i-1)$ and $\mu =M^{-1}{\sum }_{ij\in E(G_0)}{d}_i{d}_j$. Count prescribed loop pairs, double-edge pairs, and forbidden-edge pairs in the configuration model. Their means are respectively $\lambda /2+o(1)$, ${\lambda }^2/4+o(1)$, and $\mu +o(1)$. A fixed collection of $t$ disjoint prescribed pairs occurs with probability $[(M-1)(M-3)\cdots (M-2t+1){]}^{-1}$. Bounded degrees make intersecting bad configurations contribute $O(M^{-1})$ to each fixed factorial moment. Thus the total bad-pattern count has factorial moments $(\lambda /2+{\lambda }^2/4+\mu {)}^j+o(1)$. The Bonferroni inequalities and uniformly bounded means give the displayed zero-pattern probability, without requiring the parameters to converge.

For independent observations from a continuous distribution function, the probability integral transform makes the Kolmogorov-Smirnov statistic distribution-free. Choose $c_{n,\alpha }$ from its uniform-sample distribution to have probability $1-\alpha$ of not exceeding that critical value. Clipping the empirical distribution plus/minus this value to the unit interval gives simultaneous coverage at every argument. This is a band for the entire distribution, rather than separate pointwise intervals.

// Target: statistical-modelling.bigb

A cone point on a metric surface has total angular measure $\theta$ about its tip. The flat model is $d{r}^2+(\theta /2\pi {)}^2{r}^{2}d{\phi }^2$, with $0\le \phi <2\pi$; the curvature-minus-one model replaces $r$ by $\sinh r$. In these models the tip is smooth when $\theta =2\pi$. For integer $m>1$, angle $2\pi /m$ gives a cyclic orbifold point of order $m$. Under a degree-$q$ local branched lift, the total angle is multiplied by $q$, and the residual orbifold order, when defined, is $m/q$.

For a finite abelian extension of number fields, its relative discriminant ideal is the product of the finite conductor ideals of the characters of its Galois group, with the trivial character contributing one. In a multiquadratic extension of $\mathbb{Q}$, the nontrivial characters correspond to its quadratic subfields, and their conductors are the absolute values of the corresponding fundamental discriminants. This gives an efficient discriminant computation and can prove that a compositum is unramified over a quadratic subfield.

The periodic zero-mean corrector $w_j$ adjusts a unit imposed gradient so microscopic flux is divergence-free. For bounded uniformly elliptic coefficients, the weak problem has a unique solution modulo constants. The averaged flux is $a^{\ast }{e}_j=⟨a(e_j+\nabla w_j)⟩$. For symmetric coefficients this also gives a symmetric positive-definite effective tensor.

Relative to a uniform reference conductivity $a_0$, this polarization is the excess positive conductivity flux. The equation becomes $a_0\Delta u+\nabla \cdot P=0$. For a prescribed mean gradient $E$, averaging gives $a^{\ast }E=a_{0}E+⟨P⟩$. This sign convention differs from the physical heat-flux sign.

With appropriate probability density functions and finite differential entropies, $h(X)-h(X\mid Y)=I(X;Y)\ge 0$, with equality iff independence. Differential entropy does not reverse the inequality; infinite or undefined quantities cannot justify subtraction.^[ref]

However, differential entropy can be negative and depends on the choice of coordinates. For $X$ with a uniform distribution on an interval of length $a>0$, $h(X)={\log }_{2}a$; in particular $a=1/2$ gives $-1$. Rescaling $X$ by a constant $c>0$ adds ${\log }_{2}c$ to both $h(X)$ and $h(X\mid Y)$, whenever they are finite, leaving the mutual information unchanged.^[ref]

If a continuously distributed $X$ is revealed exactly, its conditional distribution is a Dirac measure with no ordinary probability density function. The discrete identity $H(X\mid X)=0$ has no finite differential entropy counterpart. Mutual information, defined through relative entropy, is infinite here.^[ref]

For finite discrete random variables $X,Y,Z$, retaining $Y$ and also observing $Z$ givesEquality means conditional independence of $X$ and $Z$ given $Y$. Both sides average over reports; the additional observation is retained alongside $Y$.^[ref]

A short proof of the original inequality uses Kullback-Leibler divergence:Gibbs inequality makes each term nonnegative, vanishing exactly when $p_{X\mid Y=y}=p_X$: independence for every possible report. Applying this within each $Y=y$ and averaging proves the additional-conditioning inequality.^[ref]

Let $X$ be a Bernoulli random variable with $\mathrm{Pr}(X=1)=p$. Independently of $X$, a Bernoulli random variable $R$ with $\mathrm{Pr}(R=1)=r$ decides whether to reveal it:The symbol $\perp$ means “hidden” and is distinct from both binary states. This is a binary erasure channel with erasure probability $1-r$. The hidden report leaves the posterior distribution equal to the prior distribution, because hiding is independent of the state. Either visible report determines $X$. Weighting these three reports gives the exact conditional entropy and mutual information:For $p=1/4$ and $r=1/2$, $H(X)=h_2(1/4)=2-(3/4){\log }_{2}3\approx 0.811278$ bits, while $H(X\mid Y)=I(X;Y)\approx 0.405639$ bits. With $r=0$ the report tells us nothing; with $r=1$ it reveals the state completely. If $0<p<1$, every positive reveal probability makes the conditioning inequality strict. At $p=0$ or $p=1$, the state was already known and every value of $r$ gives equality.

For a Bayesian inference problem, this model distinguishes an uninformative missing report from a decisive observed state. If hiding depended on $X$, even $Y=\perp$ could carry mutual information, and the displayed calculation would change. In lossless source coding, suppose both encoder and decoder already know the reveal pattern for repeated independent and identically distributed pairs. Only the hidden states require descriptions. Applying Shannon's source coding theorem to those states gives an asymptotic average rate $(1-r)h_2(p)$ bits per original state. A single hidden state need not have a codeword of that fractional length.

The reveal pattern itself has information entropy $H(R)=h_2(r)$. If that pattern must also be sent, the chain rule for information entropy gives $H(R,X)=h_2(r)+h_2(p)$, since $R$ and $X$ are independent random variables. Equivalently, $H(Y)=h_2(r)+r{h}_2(p)$ and $H(X,Y)=H(Y)+H(X\mid Y)=h_2(r)+h_2(p)$. Thus a saving in the description of $X$ given $Y$ assumes that $Y$ is already available; it is not a free saving in the description of the whole pair.

Consider a binary state $X$ and a report $Y$ with the following joint probability mass function:Before the report, $X=0$ is strongly favored: its information entropy is $H(X)=h_2(1/10)\approx 0.468996$ bits, where $h_2$ is binary entropy. A report $Y=1$ makes the two states equally likely, so $H(X\mid Y=1)=1$ bit, strictly larger than before. This report occurs with probability $1/5$. The more common report $Y=0$ determines $X=0$, giving $H(X\mid Y=0)=0$. Therefore the averaged conditional entropy isThe resulting mutual information is $I(X;Y)=h_2(1/10)-1/5\approx 0.268996$ bits. The rare report increases uncertainty about the state; that increase does not make the report useless. Together the reports remove uncertainty on average. These are stipulated probabilities in an exact model, not measured frequencies.

If a finite group $\Gamma$ acts by measure-preserving maps on a probability space, let $\mathcal{I}$ be its invariant sigma-algebra. For an integrable random variable $Z$, the displayed group average is $\mathcal{I}$-measurable. For $A\in \mathcal{I}$, changing variables under each group element gives ${\int }_{A}Z\circ T_{g}d\mathbb{P}={\int }_{A}Zd\mathbb{P}$. The average therefore satisfies the defining integral identity for conditional expectation. For permutations of $N$ exchangeable coordinates, averaging a product attached to $d$ distinct indices yields the sum over all ordered distinct $d$-tuples divided by the falling factorial $(N{)}_d=N!/(N-d)!$.

For an integrable function on $[0,1]$, average $f$ on each dyadic interval of length $2^{-n}$. The resulting step function $f_n$ converges to $f$ almost everywhere and in the L1 norm. If $U$ has the uniform distribution on $[0,1]$, then $f_n(U)=\mathbb{E}[f(U)\mid \sigma (b_n(U))]$, where $b_n$ rounds down to a dyadic endpoint. Refinement makes these conditional expectations a uniformly integrable martingale. The uniformly integrable martingale convergence theorem gives almost-sure and convergence in L1 to a limit $Y$.

Since $b_n(U)\to U$, the generated limiting sigma-algebra contains $U$ up to null sets. For every event $A$ in any finite-stage sigma-algebra, convergence in the L1 norm gives $\mathbb{E}[Y{1}_A]=\mathbb{E}[f(U)1_A]$. The pi-lambda theorem extends this equality to the limiting sigma-algebra, identifying $Y=f(U)$. The uniform distribution of $U$ translates the two convergence conclusions into the assertions for $f_n$. Choosing a Borel representative of $f$ and assigning arbitrary values at the endpoint one makes all random variables well defined without changing either conclusion.