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If the column player has an optimal strategy at value $v$ and a row probability vector $p$ satisfies $p^{T}A\ge v{e}^T$, then $p$ is optimal: it guarantees $v$, while the optimal column strategy prevents every row strategy from exceeding $v$.

The row player solves

On its half-plane of convergence,Indeed, applying the transform to $y^{\prime }=Ay$, $y(0)=y_0$, gives $(sI-A)Y=y_0$ for every $y_0$.

Balancing outward radiative acceleration against gravity givesIt is an approximate upper luminosity for a hydrostatic star of opacity $\kappa$.

Applying Markov's inequality to the positive random variable $e^{tX}$ givesHere $M_X(t)=\mathbb{E}[e^{tX}]$ is the moment-generating function, and the useful bound requires it to be finite at the chosen $t$. Optimizing over $t>0$ gives the Chernoff bound method: an exponential function weights large values more strongly before applying the same average-budget argument.

For example, let $N$ have the Poisson distribution with parameter one. Its expected value and variance are both one, and its moment-generating function is $M_N(t)=\exp (e^t-1)$. At threshold four, choosing $t=\log 4$ givesThis $t$ minimizes the exponent $e^t-1-4t$ because its derivative is $e^t-4$, with positive second derivative. The direct Markov's inequality bound is $1/4$, and the two-sided Chebyshev's inequality bound, applied to a deviation of at least three from the mean one, is $1/9$. The exact Poisson distribution tail probability is $1-e^{-1}(1+1+1/2+1/6)\approx 0.0189882$. These compare three valid bounds obtained from progressively different information; none is an exact tail probability formula for general random variables.

For a joint density $f_{XY}$, a two-coordinate Gibbs update draws

The effective sample size discounts a correlated chain's raw length by its integrated autocorrelation time.

Random-walk Metropolis proposes a symmetric increment from the current state and accepts a proposal $y$ from $x$ with probability $\min \{1,\pi (y)/\pi (x)\}$.

Suppose $U_f$ agrees with the reversible identity-function oracle except at $x_0$, where its answer differs by a fixed one-bit string $a$. Composing $U_f$ with the known identity oracle and preparing the affected answer qubit in $|-⟩$ kicks back the phase $-1$ exactly on $|x_0⟩$, using one query to $U_f$.

For a homogeneous Poisson point process of intensity $\lambda$ in ${\mathbb{R}}^d$, mapping $x$ to $v_d|x{|}^d$, where $v_d$ is the unit-ball volume, gives a rate-$\lambda$ Poisson process on $[0,\infty )$. Mapping to $|x{|}^d$ instead gives rate $\lambda v_d$.

If $H$ is the rank-$p$ orthogonal projection onto a normal linear model's column space, thenAdding $2p{\sigma }^2$ gives $\Vert (I-H)\mu {\Vert }^2+(n+p){\sigma }^2$, so Mallows' $C_p$ is unbiased for independent-copy prediction error even when the projection model is misspecified.

If $Y,Y^{\ast }$ are independent $N(\mu ,{\sigma }^2{I}_n)$ vectors and $H$ is a rank-$p$ orthogonal projection, thenThe first term is squared approximation bias, while $p{\sigma }^2$ is fitted-model variance and $n{\sigma }^2$ is irreducible new-response noise.

If a rigid body's inclined magnetic dipole rotates at angular speed $\Omega$, its radiated power is $C{\Omega }^4$. Conservation of rotational energy $I{\Omega }^2/2$ then gives

This is the finite special-linear-group instance of the general Bruhat decomposition.

For the upper-triangular subgroup $B$ of $S{L}_2({\mathbb{F}}_p)$ andthe two double cosets are $B$ and $BwB$, and $B\cap wB{w}^{-1}=T$, the diagonal subgroup.

For an irreducible $H$-representation $V$, the induced representation ${\mathrm{Ind}}_H^{G}V$ is irreducible exactly when, for every nonidentity double coset representative $x\in H\backslash G/H$, the restrictions of $V$ and its $x$-conjugate to $H\cap xH{x}^{-1}$ have inner product zero.

On $(0,1)$, $x^{-a}$ is integrable exactly when $a<1$, providing sharp counterexamples between finite-measure $L^p$ spaces.

The essential supremum is the upper bound for a measurable function after disregarding null sets for a specified measure $\mu$. Its definition iswith values allowed in the extended real number line. A supremum controls every value; an essential supremum controls values almost everywhere. It therefore depends on the chosen measure, as well as on the function. ^[ref].

For a worked example, use Lebesgue measure on $[0,1]$ and putIts supremum is ten, attained at $1/2$. But its essential supremum is one: $f\le 1$ outside the singleton $\{1/2\}$, which is a null set. Every candidate $M<1$ fails, because the interval $(\max (0,M),1)$ has positive Lebesgue measure and includes values larger than $M$; changing one point cannot remove that interval. In particular, $\mu (\{f>0.8\})=0.2$, while $\mu (\{f>1\})=0$.

Why is this useful? Lebesgue integrals ignore changes on null sets, and Lp spaces identify functions that agree almost everywhere. The L-infinity norm uses $\mathrm{esssup}|f|$ so it respects that identification. For this nonnegative example the supremum norm is ten and the L-infinity norm is one; replacing $f(1/2)$ by $1/2$ changes the former while leaving the latter unchanged. ^[ref].

The measure qualification is essential. For $\mu ={\delta }_{1/2}$, the Dirac measure concentrated at the exceptional input, that singleton has measure one and this function has essential supremum ten. A set ignored by one measure need not be ignored by another. The lower counterpart is the essential infimum, with $\mathrm{essinf}f=-\mathrm{esssup}(-f)$.

IfthenInitial segments of colexicographic order attain equality.

If $A$ is an infinite $L$-structure and $|L|+{\aleph }_0\le \kappa \le |A|$, then $A$ has an elementary substructure of cardinality $\kappa$. One proof adds Skolem functions for existential formulas and closes a chosen $\kappa$-element subset under them; the closure still has size $\kappa$, and the Tarski--Vaught test makes it elementary.

If an $L$-theory has an infinite model, then it has models of arbitrarily large cardinality. For a cardinal $\kappa$, add constants $c_{\alpha }$ for $\alpha <\kappa$ and all inequalities $c_{\alpha }\ne c_{\beta }$. Every finite fragment is satisfiable in the given infinite model, so compactness gives a model with at least $\kappa$ elements. Applying the downward theorem produces a model of exactly $\kappa$ when $\kappa \ge |L|+{\aleph }_0$.