Conjugation preserves order of a group element, because $(gx{g}^{-1}{)}^k=g{x}^k{g}^{-1}$, and preserves the sign of a permutation, because sign is a group homomorphism into $\{1,-1\}$. These invariants do not determine cycle type. In $S_6$, $(123)$ and $(123)(456)$ both have order three and positive sign, but have three and zero fixed points respectively. Conjugation relabels cycles, so these permutations are not conjugate.

The Lorentz force per unit area on an ideal current sheet is computed using the mean of its limiting magnetic fields. The sheet's own local magnetic field is equal and opposite on its two sides and cancels in this mean. Equivalently, integrate the Lorentz force density through a thin regularized layer: its self-field contribution averages to zero, leaving the external field acting on the surface current density.

For a twice continuously differentiable function, the curl of its gradient vanishes. In components, ${ϵ}_{ijk}{\partial }_j{\partial }_{k}f=0$ because the second derivatives are symmetric in $j,k$ while the Levi-Civita symbol is antisymmetric. Thus adding a gradient to a vector potential preserves the represented curl. No topological assumption is needed for this identity, unlike the converse assertion that every curl-free vector field has a global potential.

For distinct ordered observations $x_i$, let $s_u$ be their natural cubic spline interpolant with values $u_i$. Its roughness is a quadratic form, $\int (s_u^{\prime \prime }{)}^2=u^{T}Ku$, where $K$ is positive semidefinite and its nullspace consists of vectors sampled from affine functions. Minimizing penalized least squares $\Vert y-u{\Vert }^2+\lambda u^{T}Ku$ gives $\hat{u}=(I+\lambda K{)}^{-1}y$. Thus the smoothing matrix is $A_{\lambda }=(I+\lambda K{)}^{-1}$ and its effective degrees of freedom are $\mathrm{tr}(A_{\lambda })$. As $\lambda \downarrow 0$ the fit approaches interpolation and the trace approaches $n$; as $\lambda \to \infty$ it approaches the least-squares affine line and the trace approaches $2$.

Integrating the Crooks fluctuation theorem over work and using normalization of the reverse probability distribution givesConsequently Jensen's inequality gives $⟨W{⟩}_F\ge \Delta F$. Some realizations may nevertheless have $W<\Delta F$. This is a work relation with equilibrium endpoint preparations, while the integral fluctuation theorem for total entropy production uses the actual final forward distribution. ^[ref]

A wave trough is a local minimum of the wave's displacement.

For independent and identically distributed exponential random variables of rate $\lambda >0$, the Sample mean has large-deviation speed $n$ and the displayed good rate function, infinite for $x\le 0$. The cumulant-generating function is $\log [\lambda /(\lambda -\theta )]$ for $\theta <\lambda$. Its Legendre-Fenchel transform has maximizing parameter $\theta =\lambda -1/x$. For $x\le 0$, sending $\theta \to -\infty$ gives infinite cost. The cost has its unique zero at $1/\lambda$ and diverges at both ends of its positive domain.

For sample means of Poisson random variables of mean $\lambda >0$, the cumulant-generating function is $\lambda (e^{\theta }-1)$. Its Legendre-Fenchel transform is the displayed good rate function, with $0\log 0=0$, and is infinite for $x<0$. Near its zero, $I_P(\lambda +h)=h^2/(2\lambda )+O(h^3)$. This local quadratic explains the Poisson moderate deviation principle.

If the Cramér–Lundberg ruin asymptotic has a finite positive constant, compare it with the leading nonzero term of a finite exponential representation. An adjustment coefficient smaller than that term's decay rate would give limit zero; a larger coefficient would give infinity. Thus the two rates agree and the prefactor is the leading coefficient. Terms with zero coefficient are absent and do not determine a decay rate.

Subtracting the uniform-flow pressure $-\mu Ux/k$ leaves $(\mu Ua/k)\alpha e^{-\xi }\cos \eta /(1+\alpha )$. The coordinate map has $r\sim a{e}^{\xi }/2$, so $e^{-\xi }\sim a/(2r)$, giving the displayed dipole coefficient. The one-half is essential. In the perfect-conductor limit the independent expression $p=-(\mu U/k)\mathrm{Re}\sqrt{z^2-a^2}$ has the same far-field expansion.

For $d{C}_t=a(b-C_t)dt+\sigma \sqrt{C_t}d{W}_t$, the Itô formula applied to $Z=\sqrt{C}$ gives the displayed dynamics before hitting zero. In particular $2ab={\sigma }^2$ does not eliminate the reciprocal drift. The cancellation condition is $4ab={\sigma }^2$. Under that condition $C$ can be represented as the square of a signed Ornstein-Uhlenbeck process with drift $-aY/2$ and volatility $\sigma /2$; its principal square root is the reflected process $|Y|$, not an unrestricted signed OU process. The mean of $C$ is $b+(C_0-b)e^{-at}$ for all admissible positive parameters.

The Pearson correlation coefficient is the covariance divided by the product of the standard deviations. For real random variables with finite, strictly positive variances,It is the expected value of the product of the two centered variables after dividing each by its standard deviation. The Cauchy-Schwarz inequality gives $-1\le {\rho }_{X,Y}\le 1$. This removes measurement units, while retaining the direction and strength of the linear relationship. Constant random variables have zero standard deviation, so their Pearson correlation coefficient is undefined.

For the covariance model $Y=2X+Z$, with $X,Z$ independently uniform on $\{-1,0,1\}$, the calculation givesThe positive correlation is less than one because the independent $Z$ term introduces extra variation. For $Y=-2X+Z$ the correlation is $-2/\sqrt{5}$. Both examples appear in the covariance diagram.

For nonzero real $a,c$, the correlation of $aX+b$ and $cY+d$ is $\mathrm{sgn}(ac){\rho }_{X,Y}$. Positive rescaling and added offsets therefore leave the correlation unchanged; reversing one scale reverses its sign. Equality $|{\rho }_{X,Y}|=1$ means that one random variable is an affine function of the other almost surely, with a nonzero slope of the corresponding sign. This is the equality condition in the Cauchy-Schwarz inequality, and includes perfect positive correlation.

Zero correlation says that the centered variables are orthogonal in their square-integrable inner product. It permits nonlinear statistical dependence, as the example $Y=X^2$ in uncorrelated random variables demonstrates. More generally, a correlation is a statement about a joint probability distribution; identifying effects of interventions is the separate task of causal inference. This distinction keeps an observed association from being mistaken for a demonstrated mechanism.

An early source for standardized association is Francis Galton's 1888 paper “Co-relations and their Measurement, chiefly from Anthropometric Data”. He compared stature and forearm measurements and changed each measurement to units of its own variability so the two regression directions could be compared. His scale used probable error, so it should not be identified word for word with the modern standard deviation formula above. ^[ref].

For a positive-definite covariance matrix, this is Euclidean distance after whitening. It accounts for both scales and correlations and is invariant under consistent nonsingular linear changes of coordinates. Estimating the covariance can be unstable with small samples, and a singular covariance requires restricting to a nonsingular subspace rather than claiming an ordinary metric on the full space.

For $p\ge 2$ variables with unit variances and common pairwise correlation $r$, the equicorrelation covariance matrix isOn the constant-vector direction, $11^T$ has eigenvalue $p$; on its orthogonal complement, it has eigenvalue zero. Hence the eigenvalues of $\Sigma$ are $1+(p-1)r$ and $1-r$, the latter with multiplicity $p-1$. Positive semidefiniteness is therefore equivalent toAt $r=1$, the matrix has matrix rank one. At $r=-1/(p-1)$, the constant direction has zero variance and the matrix has matrix rank $p-1$. In the interior it is positive-definite. For $r>0$, the first population principal component is the normalized sum of the centered variables; for $r<0$, the largest eigenvalue instead occurs on the orthogonal complement.

The dimension restriction matters. For $p=1$, the formula gives $\Sigma =[1]$ for every $r$, so the lower bound involving $p-1$ does not apply: there is no distinct variable pair whose correlation could determine $r$.

For a random vector with covariance $\Sigma$ and positive coordinate variances, let $D=\mathrm{diag}(\Sigma )$. Its correlation matrix is the covariance of $D^{-1/2}(X-\mathbb{E}X)$, so it is positive semidefinite with unit diagonal. Each entry is the correlation of the corresponding pair of variables. In particular $\mathrm{tr}(C)=p$, explaining why an eigenvalue in correlation-based principal component analysis represents that eigenvalue divided by $p$ of the total variance. The empirical version is the sample correlation matrix.

For a two-dimensional positive-definite covariance matrix $\Sigma$, the centered covariance ellipse is the image of the unit circle under its positive definite square root of a matrix. Equivalently, it satisfies $x^T{\Sigma }^{-1}x=1$. Its semi-major and semi-minor axes have lengths given by the square roots of the larger and smaller eigenvalues, along the corresponding eigenvectors. The ellipse encodes a second-moment geometry; a probability-coverage interpretation needs additional information about the probability distribution and the chosen scale.

In independent trials with two outcomes of probabilities $p$ and $q=1-p$, let $N$ include the first trial and the trial that first completes the pair. Conditional on the first outcome, $N-1$ has a geometric distribution of parameter $q$ or $p$, with mixture weights $p,q$. The law of total expectation and law of total variance giveThe final term in the unsimplified variance is the variance of the two conditional means; omitting it is incorrect unless $p=q$. Both outcomes must have positive probability for these finite formulas.

For independent coupon labels arriving at rate $\lambda$, Poisson thinning makes the type-specific first arrival times independent exponentials of rates $\lambda p_j$. Completion time is their maximum. Its distribution is $\mathbb{P}(T<t)={\prod }_j(1-e^{-\lambda p_{j}t})$. The completion count $L$ depends only on the labels and is independent of the exponential interarrival sequence. Therefore $T={\sum }_{k\ge 1}{E}_k{1}_{\{L\ge k\}}$ and nonnegative integration gives the displayed identity. Completion count and its expectation do not depend on arrival rate.

For a Poisson random measure on ${\mathbb{R}}^d$ with Lebesgue measure intensity and a nonnegative compactly supported continuous function $f$, put $m_r=\mu (B(0,r))$ and $a_r={\int }_{B(0,r)}{e}^{-f}d\mu$. Continuity gives $a_r/m_r\to e^{-f(0)}$. With $L=\mathbb{E}{e}^{-M(f)}$, the count-weighted Laplace functional of a Poisson random measure and independent inside/outside counts give respectivelyBoth converge to the displayed value. The conditioning is meaningful for every $r>0$; it is not conditioning on a positive-probability point at the origin.

If a Banach space $Y$ has cotype 2 constant $C$, every bounded linear operator $T:{\ell }_{\infty }^N\to Y$ satisfies ${\pi }_2(T)\le 3^{1/4}C{\pi }_4(T)$ and ${\pi }_2(T)\le 4\sqrt{3}{C}^2\Vert T\Vert$, uniformly in $N$. The first bound follows from Pietsch factorization theorem and the fourth moment bound for Rademacher sums. For the second, 2-summing domination gives a weighted coordinate $L^2$ norm; clipping coordinates at a variable threshold yields ${\pi }_4(T)\le 2\sqrt{\Vert T\Vert {\pi }_2(T)}$, which absorbs the first bound.