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Let $X_i=\theta +E_i$ for $n\ge 2$ independent and identically distributed observations, with $E_i$ having an exponential distribution of unknown scale $\beta >0$. Test the null hypothesis $\theta =0$ within $\theta \ge 0$. For nondegenerate data, the likelihood function isMaximizing over the scale gives $\hat{\beta }(\theta )=\overline{X}-\theta$, and the resulting profile likelihood increases with $\theta$. Thus the unrestricted maximum-likelihood estimates are $\hat{\theta }=X_{(1)}$ and $\hat{\beta }=\overline{X}-X_{(1)}$, while the null estimate is $\tilde{\beta }=\overline{X}$. The likelihood ratio is consequently

Under the null, put $A=n{X}_{(1)}$ and $D={\sum }_i(X_i-X_{(1)})$. The memoryless property gives independent gamma distributions of shapes $1$ and $n-1$ and common scale $\beta$. For a direct verification, fix which observation is the minimum and write its value as $z$ and the other observations as $z+e_j$. Their joint density is proportional to $e^{-nz/\beta }{\prod }_j{e}^{-e_j/\beta }$, which factors into an exponential density for the minimum and independent exponential densities for the excesses. The minimum's index does not affect the resulting laws of $A,D$.

By beta sampling by gamma ratios, $D/(A+D)\sim \mathrm{Beta}(n-1,1)$. ThereforeAn exact test size $\alpha$ is obtained by rejecting for $R\le {\alpha }^{n/(n-1)}$. Equivalently,Although the alternative adds one endpoint parameter, a regular one-parameter Wilks theorem calibration is incorrect: the support depends on the endpoint, and the null lies on its boundary. In particular, $-2\log R$ tends to a chi-squared distribution with two statistical degrees of freedom, not one.

Put $\kappa ={\aleph }_{\omega }$. ZFC gives ${\kappa }^{\kappa }=2^{\kappa }$. König's theorem and the shift of product coordinates give $\kappa <{\prod }_n{\aleph }_n\le {\kappa }^{{\aleph }_0}\le 2^{\kappa }$. The single instance $2^{\kappa }={\kappa }^{+}$ of the generalized continuum hypothesis therefore fixes all three displayed values. The strict lower bound explains why countable exponentiation cannot leave this singular cardinal unchanged.

Let $G=⟨x,y\mid y^2=1⟩$. For each $n\ge 2$, a free basis of a subgroup of rank $n$ isA covering-space proof uses $r=n-1$ spheres double-covering the real projective plane, with antipodal points $a_j,b_j$ over the wedge point. Add lifted $x$ edges in a cyclic order through the $a_j$, and an $x$ loop at each $b_j$. This is a connected $2r$-sheeted cover of a circle wedged with a projective plane. After collapsing one connecting arc in each simply connected sphere, its fundamental group is that of a graph with $r$ vertices and $2r$ edges, hence free of rank $r+1$.

A spanning tree along the first $r-1$ cyclic edges gives the displayed words: the closing cycle is $x^r$, and the loop reached through $a_j,b_j$ is $x^{j}yxy{x}^{-j}$. Injectivity of the induced fundamental-group map of a covering map proves they form a free basis in $G$.

For $I(u)={\int }_a^{b}f(x,u,u^{\prime })dx$ on $W^{1,\infty }$, continuity of $f,f_u,f_v$ on $[a,b]\times {\mathbb{R}}^2$ gives a bounded Fréchet derivative in the displayed form. Fixing $u$ and restricting $\Vert w{\Vert }_{W^{1,\infty }}$ to a bounded neighborhood confines all intermediate pairs $(u+tw,u^{\prime }+t{w}^{\prime })$ to a compact range up to null sets. Uniform continuity of the partial derivatives bounds the integrated first-order remainder by $(b-a)\omega (\Vert w\Vert )(\Vert w{\Vert }_{\infty }+\Vert w^{\prime }{\Vert }_{\infty })$, where $\omega (s)\to 0$. This is little-oh of the Sobolev norm. The result needs no continuous derivative representative for $u$; essential boundedness suffices.

For a positive integer $m$, the negative matrix power is differentiable on the open general linear group. The Neumann series gives $(A+H{)}^{-1}=A^{-1}-A^{-1}H{A}^{-1}+O(\Vert H{\Vert }^2)$ locally around any invertible matrix. Applying the product rule to the $m$ factors of this inverse gives the displayed Fréchet derivative. Each summand chooses one inverse factor to differentiate while keeping the others in order.

The factor order matters. Only when $A$ and $H$ commute does the sum simplify to $-m{A}^{-(m+1)}H$. This is a useful example of why scalar differentiation rules for powers cannot be transferred to matrices without accounting for noncommutativity.

Let $c(u)$ be the Black-Scholes formula value divided by stock price for an at-the-money call of duration $u$. On a discrete set of possible entry dates, homogeneous function and the discounted stock martingale reduce the Snell envelope to $b_i=\max (c(T-T_i),b_{i+1})$, with zero terminal value. Thus the displayed maximum is exact. When $r>0$ and volatility is positive, $c$ increases with maturity, so immediate entry is optimal and the initial value equals the ordinary at-the-money European call option. The maximum recursion remains useful if a different parameter regime loses that monotonicity.

For adequate arithmetic provability predicates, the arithmetized deduction theorem identifies inconsistency of $S+\neg {\mathrm{Con}}_S$ with $S$-provability of ${\mathrm{Con}}_S$. Formalized second incompleteness gives ${\square }_S{\mathrm{Con}}_S\to {\square }_S\perp$. Contraposition yields the displayed conditional. This is an internal theorem and does not license removal of its consistency premise.

Take Lebesgue decompositions relative to a fixed finite positive measure $\kappa$. The finite union $S$ of null supports of their singular components is still $\kappa$-null. On $S$ each original measure equals its singular component; on $S^c$ it equals its absolutely continuous component. Restriction commutes with the partition formula defining the finite supremum of positive measures. Thus the supremum restricts to ${\bigvee }_i{\mu }_i^s$ on $S$ and to ${\bigvee }_i{\mu }_i^a$ off $S$. The latter is absolutely continuous with respect to $\kappa$ because it is bounded by the sum of the absolutely continuous components. Uniqueness of the Lebesgue decomposition proves the displayed identities, and the Radon-Nikodym derivative of the absolutely continuous component is the pointwise maximum of the individual derivatives.

Attempt scalar multiplication on an elliptic curve modulo a composite integer. An addition denominator that is not invertible yields a greatest common divisor with the modulus; a value strictly between one and the modulus is a factor. The method exploits different orders of the reduced point at the unknown prime factors. For $y^2=x^3+5x-5$ modulo $221$, doubling $(1,1)$ gives $(14,-53)$; adding the original point requires division by thirteen, revealing $221=13\cdot 17$.

For the displayed integral equation, reduce its polynomial modulo a prime number $p$. Singular points satisfy $y^2-x^3-c=-3x^2=2y=0$. For $p>3$, they exist exactly when $p\mid c$, and the sole point is $(0,0)$. At $p=2$, the sole point over ${\mathbb{F}}_2$ is $(0,\overline{c})$; at $p=3$, it is $(-\overline{c},0)$ because cubing is the identity on ${\mathbb{F}}_3$. For $c=17$, this gives primes $2,3,17$ and points $(0,1),(1,0),(0,0)$ respectively. This concerns the specified plane equation; intrinsic good or bad reduction of an elliptic curve can require choosing a different integral model, especially at primes two and three.

Every invertible square matrix is a product of elementary matrices. Gaussian elimination reduces $A$ to the identity matrix by elementary row operations, giving $F_r\cdots F_{1}A=I$. Reverse their order and invert them to write $A=F_1^{-1}\cdots F_r^{-1}$. Each inverse is again elementary. Conversely every elementary matrix is invertible, so every such product is invertible. The identity may be represented by the empty product.

Stop the four-ball Ehrenfest urn model upon reaching count zero or four. From count $r$, the transitions are to $r-1$ with probability $r/4$ and to $r+1$ with probability $(4-r)/4$. First-step analysis with absorbing boundary values gives the displayed vector $h$ of probabilities of absorption at zero and vector $t$ of expected hitting times of either endpoint. For instance, $t_1=1+3t_2/4$ and $t_2=1+(t_1+t_3)/2$, with $t_1=t_3$, give $t_1=7$. From any interior state, absorption has probability at least $1/4$ within two steps; the resulting geometric tail bound proves both almost-sure absorption and finiteness of the expectations.

Let $A$ be a doubly stochastic matrix, let $q$ be a probability vector, and put $w=Aq$. In row $i$ with $w_i>0$, define $r_i(j)=a_{ij}{q}_j/w_i$ on the support where $a_{ij}>0$. The positive row entries ${\alpha }_i=(a_{ij})$ form a probability vector. Expanding relative entropy and using the column sums givesZero-output rows contribute zero because all their $a_{ij}{q}_j$ vanish. The row sums supply the normalized comparison laws, while the column sums preserve the input entropy contribution. Equality holds exactly when $q_j=w_i$ for every positive matrix entry $a_{ij}$, including the automatic zero values in zero-output rows.

For a Lebesgue measurable set $E\subseteq \mathbb{R}$ of finite measure, the displayed function satisfies $|H_E(t)-H_E(s)|\le |t-s|$. Indeed, for $s<t$, finite additivity and monotonicity of a measure give $H_E(t)-H_E(s)=m(E\cap (s,t])\le t-s$. Thus $H_E$ is a nondecreasing Lipschitz function and is uniformly continuous even if $E$ is unbounded.

Continuity from above of a measure gives $H_E(t)\to 0$ as $t\to -\infty$, while continuity from below of a measure gives $H_E(t)\to m(E)$ as $t\to \infty$. Every strictly intermediate value is attained by the intermediate value theorem. The corresponding half-line cut of $E$ realizes any prescribed measure between zero and $m(E)$; the endpoints are realized by the empty set and $E$ itself.

A smooth embedded invariant two-torus in three-dimensional Euclidean phase space bounds a solid torus of positive volume. The vector field is tangent to its boundary, so its normal flux is zero. If the divergence is strictly negative throughout the enclosed volume, the divergence theorem gives a strictly negative flux, a contradiction. Thus a globally negative-divergence three-dimensional vector field has no such quasiperiodic invariant torus. This argument is stronger than merely noting contraction of transported phase volume; an attractor's own zero volume is not a contradiction.

For a normalized two-dimensional discrete Laplacian satisfying the discrete maximum principle, suppose quadratic functions are differentiated exactly and $L_{h}1=0$. In a domain of diameter $D$, choose $x_0$ in the domain and $\varphi (x)=|x-x_0{|}^2$. Then $0\le \varphi \le D^2$ and $L_h\varphi =4$. Applying the discrete elliptic barrier argument to zero-boundary error $L_{h}e=\tau$ gives the displayed bound. A mere sign pattern of the stencil does not determine this constant: rescaling the whole operator changes the error-to-residual ratio without changing the domain diameter.

A functional on a Banach space is Gâteaux differentiable at $v$ when the directional limits exist and form a bounded linear functional of $h$. Fréchet differentiability is stronger, requiring a remainder uniformly small compared with the norm of the increment. A quadratic norm bound on the Taylor remainder gives the stronger property directly. Mere existence of unrelated directional limits would not establish linear differentiability.