For a uniform distribution on a finite set of $m$ outcomes, direct expansion of relative entropy giveswhere $H_e$ is information entropy in nats. Therefore minimizing relative entropy from this particular reference is equivalent to maximizing information entropy subject to the same constraint. This statement depends on the uniform reference; an arbitrary reference contributes its own weights.

In the three-outcome example, $H_e(q^{\ast })=\ln 3-D(q^{\ast }\Vert p)\approx 0.901235$ nats. The mean fixes one aggregate quantity, and information projection chooses how to distribute the remaining uncertainty. This extends maximum entropy on a finite alphabet, whose unconstrained maximizer is the uniform law itself.

Let the possible numerical outcomes be $-1,0,1$, and give the reference law $p$ uniform distribution on a finite set: $p=(1/3,1/3,1/3)$. We want a law $q$ with expected value $1/2$. Write the three probabilities in outcome order as $(q_{-},q_0,q_{+})$. The mean constraint is $q_{+}-q_{-}=1/2$, while normalization requires $q_{-}+q_0+q_{+}=1$.

These conditions leave a whole line of possible probability distributions:For example, $(0,1/2,1/2)$ and $(1/4,0,3/4)$ both have the required mean. The constraint alone does not choose between them. Information projection selects the point on this line with the smallest relative entropy from the reference.

Exponential tilting gives weights proportional to $e^{-\theta },1,e^{\theta }$. Set $t=e^{\theta }>0$ and multiply these weights by $t$ to obtain $q^{\ast }=(1,t,t^2)/(1+t+t^2)$. Enforcing the mean givesThe positive root is $t=(1+\sqrt{13})/2$, henceThe most favored outcome becomes $1$, but all three remain possible. This is an exact constructed example with a stipulated reference law and constraint, not a fitted distribution from measurements.

For positive slope parameter $A$, the exponential-potential field increases during cosmic inflation. The quantity $aH$ grows by $e^{(1-{ϵ}_H)N}$ with constant ${ϵ}_H=A^2/(16\pi )$ and $N=8\pi ({\varphi }_R-{\varphi }_i)/(A{m}_{\mathrm{pl}})$. Starting from order-one curvature and demanding order-one or smaller present curvature gives the displayed upper bound on the initial field. A specified tighter final curvature tolerance adds one-half its logarithm to the numerator. Thus sufficient cosmic inflation sets an upper, not a lower, initial-field bound for this sign of the potential slope. Instantaneous reheating and subsequent entropy conservation determine $a_R{H}_R/a_0{H}_0$.

For $|q|>1$, define $S_q(z)={\prod }_{k\ge 0}(1-q^{-2k-1}z)(1-q^{-2k-1}{z}^{-1})$ on ${\mathbb{C}}^{\ast }$. The geometric decay gives locally uniform absolute tail convergence there. Its simple zeros are $q^{2k+1}$ and $q^{-2k-1}$ for $k\ge 0$, and reindexing gives $S_q(qz)=-z{S}_q(q^{-1}z)$. The variable zero is outside its domain, and the poles introduced by each reciprocal factor cannot be interpreted as a value of the product at zero.

// Destination: analysis.bigb

For a stress-free, fixed-temperature layer, the infinite-Prandtl number growth rate of the fundamental vertical normal mode is ${\lambda }_{\infty }=R{k}^2/(k^2+{\pi }^2{)}^2-(k^2+{\pi }^2)$. For large Rayleigh number, its maximum has $k^2={\pi }^2-8{\pi }^6/R+O(R^{-2})$ and ${\lambda }_{\max }=R/(4{\pi }^2)-2{\pi }^2+O(R^{-1})$. The infinite-Prandtl number limit must be taken first, or inertia must remain small on the resulting growth-time scale. This optimum differs from the free-slip convection neutral curve minimum, which selects onset rather than fastest growth far above it.

For an axisymmetric normal mode in a rigid rotating sphere, substitute $W=z(A{R}^2+B{z}^2+C)$ into the pressure equation and its impermeability condition. With $\beta =1-4{\Omega }^2/{\sigma }^2$, the interior equation gives $2A+3\beta B=0$. The boundary polynomial gives $3\beta B=(2+\beta )A$ and $\beta C=-(2+\beta )A{r}_0^2$. A nontrivial oscillatory solution has $\beta =-4$, hence the displayed frequency and $B=A/6$, $C=-A{r}_0^2/2$. It satisfies both equations identically.

In the large-diffusivity regime, equate diffusion rate $\alpha /r^2$ with inertial eddy rate ${ϵ}^{1/3}{r}^{-2/3}$. This gives the Corrsin scalar microscale, which exceeds the Kolmogorov length scale when $\alpha \gg \nu$.

The congruence axiom says that an affordable alternative indirectly revealed at least as desirable as a chosen alternative must also be chosen. For a chosen bundle $x$ and an affordable unchosen bundle $y$, the axiom rules out $yRx$; because $xRy$, every strict-part-preserving completion ranks $x$ strictly above $y$.

In the genuine indexed point-set model, spectrification $L$ is the left adjoint to the forgetful functor from indexed spectra to indexed prespectra. An indexed spectrum has adjoint structure maps $E(V)\to {\Omega }^{W\ominus V}E(W)$ that are homeomorphisms. Applying $L$ to a cellular indexed prespectrum supplies the associated genuine topological spectrum.

For compact $X$ over the space of linear isometries $U\to U\prime$, choose a subordinate flag for a family of linear isometries $V_i,W_i$. The vector bundle ${\xi }_i$ has fiber $W_i\ominus f_x(V_i)$. Form $P(W_i)=\mathrm{Th}({\xi }_i)\wedge E(V_i)$. Its structure maps use the orthogonal decompositionand the structure map of $E$. The twisted half-smash product is the spectrification $LP$, after extending from the cofinal flag. Unlike an arbitrary replacement by $X_{+}\wedge E$, this definition retains the specified isometry family at the point-set level.

A rotation through $\pi$ about the third coordinate axis preserves the strain components $e_{11},e_{22},e_{33},e_{12}$ and reverses $e_{13},e_{23}$. A symmetry-invariant elastic stiffness tensor cannot couple the two parity spaces. Its independent symmetric blocks therefore contain $4\cdot 5/2+2\cdot 3/2=13$ coefficients. A twofold axis alone does not impose invariance under every rotation about that axis.

Draw two different labels uniformly from $\{1,2,3,4\}$ using sampling without replacement. Each position has a uniform distribution on a finite set, but $\mathbb{P}(X_1=X_2)=0$. Two independent random variables with those same marginal distributions would match with probability $1/4$. Equal marginal laws therefore do not make this sample an IID sample. Mark some of those four labels as successes. Their count in the two draws has a hypergeometric distribution; independent Bernoulli trials with a common success probability give a binomial distribution.

The IID sample property places no integrability condition on its common probability distribution. An IID sample with the standard Cauchy distribution has no finite expected value or variance. The Cauchy sample mean stability result shows that its averages keep that same Cauchy distribution for every sample size. Thus increasing $n$ need not concentrate an IID sample around a population mean that does not exist. The finite-variance bound below applies only when its hypotheses hold.

In the copied-coin model, $X_i=Z$ for every $i$, so the sample mean is exactly $Z$, however large $n$ is. Its expected value is $1/2$ and its variance is $1/4$, matching each observation's marginal variance. The off-diagonal covariances in the variance of a sum explain the difference:For an IID sample, those off-diagonal terms are zero. Repeating the same random input therefore does not create the reduction in fluctuations obtained from fresh independent draws. The general common-plus-individual structure is described by compound-symmetry covariance.

For $n\ge 1$ independent and identically distributed Bernoulli random variables with success probability $p$, the sum $K={\sum }_i{X}_i$ has a binomial distribution. Every length-$n$ pattern with $k$ successes has probability $p^k(1-p{)}^{n-k}$, and the binomial coefficient counts its $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ possible success positions. ThereforeThe usual endpoint convention includes $p=0$ and $p=1$ as deterministic cases. For $p=1/2$ and $n=4$, the five probability masses are $1,4,6,4,1$ divided by $16$. The sample mean takes values $0,1/4,1/2,3/4,1$, with variance $1/16$ and standard deviation $1/4$. At $n=64$, its variance is $1/256$ and its standard deviation is $1/16$.

A cost-effectiveness acceptability curve displays the probability of positive incremental net monetary benefit as a function of the willingness-to-pay threshold. Under a posterior distribution this is a posterior probability. A frequentist paired bootstrap version displays the fraction of replicates for which the net benefit is positive, a resampling measure of uncertainty rather than literally a probability that a fixed true parameter is positive. Using the sign of net benefit handles all quadrants of the cost-effectiveness plane; counting only replicates whose ratio is below the threshold does not.

The acoustic tensor determines the plane waves of a locally uniform incremental elastic medium. For a unit material direction $\nu$, its eigenvalues are ${\rho }_0{c}^2$ and its eigenvectors are displacement polarizations. Major symmetry of the incremental elastic moduli makes $Q$ a real symmetric matrix. Positivity on every rank-one matrix $m\otimes \nu$ makes all its wave speeds real and nonzero.

Strong ellipticity requires the displayed inequality for every nonzero displacement vector $m$ and material direction $\nu$. Equivalently, every acoustic tensor is positive-definite. It is weaker than positivity of the moduli on all matrices. A quadratic minor null Lagrangian vanishes on rank-one matrices, so adding such a term does not change the acoustic tensor or this condition.

For a static base state with zero incremental body force, homogeneous incremental tractions or homogeneous Dirichlet boundary conditions eliminate the boundary term in integration by parts. The displayed Rayleigh quotient then gives real squared frequencies because of major symmetry. Positive moduli give strictly positive squared frequencies for modes with nonzero displacement gradient. Pure dead loading leaves constant translations at zero; fixed displacement conditions remove those modes.

A compression disk for a two-sided embedded surface $F$ in a three-manifold is an embedded disk meeting $F$ exactly in its boundary, with boundary essential in $F$. Replacing an annular neighborhood of this boundary by two parallel copies of the disk reduces the surface's genus or separates it into simpler pieces.