For a series with constant mean $\mu$, write $X_t=x_t-\mu$. An autoregressive moving-average process has the formwhere $w_t$ is zero-mean white noise with variance ${\sigma }_w^2>0$. The backshift operator is $B{X}_t=X_{t-1}$. Define the autoregressive polynomial and the moving-average polynomial byThe autoregressive operator and moving-average operator give the concise equationOrders are normally specified minimally, with nonzero last coefficients and no common polynomial factor. A displayed unreduced equation can have larger orders than the process actually needs.

Causality means that the stationary process can be recovered from present and past noise:Invertibility means that the noise can be recovered from present and past observations:The causality and invertibility root criteria for an ARMA model, for the reduced representation, areIndeed the transfer series are $\sum {\psi }_j{z}^j=\Theta (z)/\Phi (z)$ and $\sum {\pi }_j{z}^j=\Phi (z)/\Theta (z)$. When all denominator roots lie outside the closed unit disc, each rational function is analytic on a larger disc, so its Taylor coefficients decay geometrically and are absolutely summable. Conversely, a genuine denominator zero in the closed unit disc gives a pole incompatible with such a convergent filter. The reduced-representation proviso prevents a cancelled root from being mistaken for an obstruction.

For order identification by autocorrelation cutoffs, first estimate the sample autocorrelation function and sample partial autocorrelation function from a suitably stationary, centred series. The autocorrelation function is $\rho (h)=\mathrm{Cov}(X_t,X_{t-h})/\mathrm{Var}(X_t)$. The lag-$h$ partial autocorrelation function is the last coefficient in the best linear regression of $X_t$ on $X_{t-1},\dots ,X_{t-h}$, equivalently the correlation left after removing the intervening lags.

For a pure moving-average model of order $q$, observations more than $q$ lags apart share no noise, so the population ACF is zero beyond $q$, while the PACF usually tails off. For a pure autoregressive model of order $p$, the PACF is zero beyond $p$, while the ACF usually decays, possibly with damped oscillation. Thus a clear sample ACF cutoff suggests an MA order and a clear PACF cutoff suggests an AR order. For a mixed ARMA process both usually tail off, so these plots guide a small set of candidate orders rather than determine $p,q$ uniquely. Sampling variation blurs cutoffs; compare fitted candidates and check that their residual ACF resembles white noise. Common-factor cancellation can also disguise the orders in a displayed equation.

One sufficient maximum likelihood estimation statistical consistency theorem is the following. Let the observations be independent random variables with a common density $f_{\theta }$ relative to a fixed measure, let $\Theta$ be a compact metric parameter set containing the true value ${\theta }_0$, and suppose that the expected log-likelihoodis finite, continuous, and uniquely maximized at ${\theta }_0$. Assume also the uniform law of large numbersand assume that a measurable global maximum-likelihood estimator exists. Then every such maximum-likelihood estimator converges in probability to the true parameter. A useful sufficient condition for this uniform law of large numbers is almost-sure continuity of $\theta \mapsto \log f_{\theta }(Y_1)$ on the compact parameter set and an integrable envelope ${\sup }_{\theta }|\log f_{\theta }(Y_1)|\le M(Y_1)$, ${\mathbb{E}}_{{\theta }_0}M(Y_1)<\infty$. Under these conditions, identifiability supplies uniqueness through the Kullback-Leibler divergence identitywith equality only at $\theta ={\theta }_0$.

To see the statistical consistency conclusion, fix $\epsilon >0$. Compactness and the unique maximum give a strictly positive separation gapwhenever the set outside the neighbourhood is nonempty. If ${\Delta }_n$ denotes the supremum discrepancy in the uniform law of large numbers, maximization of the empirical log-likelihood yieldsHence the probability of $d({\hat{\theta }}_n,{\theta }_0)\ge \epsilon$ is at most $\mathbb{P}(2{\Delta }_n\ge {\delta }_{\epsilon })$, which tends to zero. This is the argmin consistency under uniform convergence in probability argument applied to the negative log-likelihood. These are sufficient conditions, not necessary ones; parameter-dependent support can require a direct argument, and independent observations need not have a common distribution in other models.

Put $G=X^{T}X$ and $H=X{G}^{-1}{X}^T$. The rank of a matrix assumption makes $G$ invertible, and $H$ is the orthogonal projection matrix onto the column space of the design matrix. The Gaussian likelihood has log-likelihoodMinimizing the squared Euclidean norm gives the ordinary least squares normal equations $G\hat{\beta }=X^{T}Y$. Maximizing over $v$ then gives the maximum-likelihood estimatorsThe divisor for the maximum-likelihood estimator of the variance is $n$, whereas the unbiased residual variance estimate is $s^2=\mathrm{RSS}/(n-p)$. The exceptional event $\mathrm{RSS}=0$ has probability zero under the Gaussian linear model with ${\sigma }^2>0$; on that event the likelihood is unbounded as $v\downarrow 0$.

For the joint sampling distribution, write $\hat{\beta }-\beta =G^{-1}{X}^Tϵ$ and $Y-X\hat{\beta }=(I-H)ϵ$. Their cross-covariance matrix vanishes because $X^T(I-H)=0$. These two vectors are jointly multivariate normal, so they are independent random variables. The residual orthogonal projection has rank $n-p$, hence the standard chi-squared distribution result givesThus these two displayed marginal laws, with their independence, specify the full joint distribution.

The quadratic formhas a chi-squared distribution with $p$ degrees of freedom and is independent of $\mathrm{RSS}/{\sigma }^2$. Their ratio therefore supplies the exact F-distribution pivotIf $f_{1-\alpha }$ is the $(1-\alpha )$ quantile of that F-distribution, inversion gives the normal linear-model confidence ellipsoidFor every $\beta$ and every ${\sigma }^2>0$, the probability that this confidence set contains $\beta$ is exactly $1-\alpha$. To test the null hypothesis, reject precisely when ${\beta }_0$ lies outside this confidence ellipsoid, equivalently when $T({\beta }_0)>f_{1-\alpha }$. Under the null hypothesis the rejection probability is exactly $\alpha$, independent of the unknown variance.

For the numerical calculation, the second column of the design matrix sums to zero and its sum of squares is $60$. Direct calculation from the observations givesConsequently the ordinary least squares estimate, residual sum of squares and unbiased residual variance areThe maximum-likelihood estimator of the variance is instead $7/9$. At the proposed null value the F-distribution statistic isTherefore reject the null hypothesis at the 5% level. The relevant critical value has $2$ numerator and $7$ residual degrees of freedom.

Put $D=\mathbb{Q}/\mathbb{Z}$. The character module $M^{\ast }={\mathrm{Hom}}_{\mathbb{Z}}(M,D)$ has the natural actionAdditivity is immediate, and $a(b\varphi )(m)=\varphi (bam)=\varphi (abm)$, using commutativity of $A$. The identity of $A$ acts as the identity. We establish the extension and duality tools used in the four implications, so no flatness or injectivity criterion is assumed without proof.

An injective module $E$ is one for which every homomorphism from a submodule $U\subseteq V$ extends to $V$. The Baer criterion says it suffices to test the inclusions of ideals $I\subseteq A$. Necessity follows directly from the definition. For sufficiency, suppose every $I\to E$ extends to $A$. Given $h:U\to E$, use Zorn's lemma to choose a maximal extension $(W,h_W)$ inside $V$; extensions along a chain have a well-defined union. If $v\in V\setminus W$, letThis is an ideal and an $A$-linear map. Its extension to $A$ has the form $a\mapsto ae$ for some $e\in E$. Thenis well-defined: if two expressions represent the same element, their coefficient difference lies in $I$, where the two prescribed values agree. It extends $h_W$ to the strictly larger module $W+Av$, a contradiction. Thus $W=V$, proving the Baer criterion for arbitrary modules and ideals.

The group $D$ is divisible. A homomorphism $n\mathbb{Z}\to D$, for $n>0$, extends to $\mathbb{Z}$ by choosing $d\in D$ with $nd$ equal to its value at $n$; the zero ideal causes no restriction. The proved Baer criterion over $\mathbb{Z}$ therefore makes $D$ injective. It is also a cogenerator: a nonzero element $x$ of an abelian group generates either a finite cyclic group of order $n>1$, which maps $x$ to $1/n+\mathbb{Z}$, or an infinite cyclic group, which maps $x$ to $1/2+\mathbb{Z}$. Injectivity extends this character to the whole group. Thus every nonzero element is detected by an additive character.

It follows that a group homomorphism $u:T\to U$ is injective exactly whenis surjective. For an injection, extension into the injective group $D$ proves surjectivity. Conversely, a character nonzero on an element of $\ker u$ cannot be the pullback of a character on $U$. This is the detection property of the injective cogenerator of abelian groups.

Finally, for any $A$-module $N$ there is a natural character form of the Tensor-hom adjunction:An $A$-linear map $h$ gives the pairing $(n,m)\mapsto h(n)(m)$. It is balanced because $h(an)(m)=(ah(n))(m)=h(n)(am)$. Conversely a balanced additive pairing determines $h$ by this formula. The universal property of the tensor product of modules makes these constructions mutually inverse, and compatible with restriction maps.

The common Brownian motion cancels from the difference $D_t=Y_t-X_t$. Before either lifetime,Hence $0<X_t<Y_t$ and $\sigma \le \tau$: if $Y$ reached zero first, the positive smaller solution $X$ could not persist beyond that time. WriteUsing the Itô formula for $Y^{-1}$ givesThe difference has finite variation, so there is no cross-variation term in the Itô product rule for $D/Y$. Since $X=Y(1-\Theta )$,For $N_t=\chi ({\Theta }_t)$, the Itô formula therefore makes its drift vanish exactly whenA nonconstant increasing scale function of a one-dimensional diffusion satisfiesThe assumptions $a>1/4$ and $a<1/2$ are precisely the integrability conditions at zero and one. Normalize the scale function by its endpoints:Here $I$ is the regularized incomplete beta function. The normalized function is continuous and strictly increasing on $[0,1]$, with $\varphi (0)=0$ and $\varphi (1)=1$. We have obtained the local martingale

The terminal identification needs care; mere boundedness does not by itself say what happens when both lifetimes coincide. Stop first on compact subsets of $0<X<Y$, and then exhaust the interval $[0,\sigma )$. The bounded local martingale criterion makes $N$ a bounded martingale with a limit $N_{\sigma -}$, and optional sampling theorem gives $\mathbb{E}[N_{\sigma -}]=N_0$. For completeness, applying the Itô formula to each stopped $N^2$ bounds the expected quadratic variation by one. Monotone convergence over the localization sequence impliesSince ${\varphi }^{-1}$ is continuous on $[0,1]$, ${\Theta }_t$ itself has a limit ${\Theta }_{\sigma -}$.

On $\{\sigma <\tau \}$, $Y_{\sigma }>0$ while $X_t\to 0$, so ${\Theta }_{\sigma -}=1$ and $N_{\sigma -}=1$. On $\{\sigma =\tau \}$, the supplied divergent clock isIf ${\Theta }_{\sigma -}>0$, then eventually ${\Theta }_t{\varphi }^{\prime }({\Theta }_t)$ is bounded below by a positive constant. This remains true if its limit is one: ${\varphi }^{\prime }(\theta )$ diverges there, rather than tending to zero. The divergent clock would then force $[N{]}_{\sigma -}=\infty$, contradicting the finite quadratic variation. Thus ${\Theta }_{\sigma -}=0$ and $N_{\sigma -}=0$ on the simultaneous-lifetime event. We have provedTaking expectations yields the complete probability formulaThis is the beta integral for strict ordering of coupled Bessel lifetimes; the complementary probability is the probability of simultaneous swallowing, not of the larger initial point dying first.

To connect the calculation to SLE, let its driver be $W_t=\sqrt{\kappa }{\beta }_t$. For a positive boundary point $r$ before its boundary swallowing time, put $V_t^r=(g_t(r)-W_t)/\sqrt{\kappa }$. The Chordal Loewner equation givesThus two positive boundary points give exactly the coupled Bessel processes above, with $B=-\beta$ and $a=2/\kappa$. The interval $1/4<a<1/2$ corresponds to $4<\kappa <8$. Their boundary swallowing times obeyThe formula quantifies whether the growing hull separates one boundary point from infinity before the other, or swallows them together inside the same boundary interval. It does not identify swallowing with visiting each point. At $\kappa =6$, the two beta parameters are $1/3$ and this becomes the Cardy boundary crossing formula in the corresponding cross-ratio coordinate, explaining its role in the conformal invariance of planar percolation.

Write $\eta :1_{\mathcal{C}}\to GF$ and $\epsilon :FG\to 1_{\mathcal{D}}$ for the unit and counit of an adjunction. The induced monad hasThe Eilenberg-Moore comparison functor sends $D\in \mathcal{D}$ to the algebra for a monadand sends $v:D\to D^{\prime }$ to $Gv$. The triangle identities for an adjunction give the algebra unit law, and naturality of the counit gives the algebra multiplication law and the algebra-morphism equation for $Gv$.

If $\mathcal{D}$ has coequalizers, define the Left adjoint to the Eilenberg-Moore comparison functor at an algebra $(A,a:TA\to A)$ byThe parallel pair is reflexive through $F{\eta }_A$. A morphism $L(A,a)\to D$ is equivalently an arrow $b:FA\to D$ with $bFa=b{\epsilon }_{FA}$. Under $F⊣G$, its transpose is $\bar{b}=Gb{\eta }_A:A\to GD$. The transpose of $bFa$ is $\bar{b}a$, while the transpose of $b{\epsilon }_{FA}$ is $G{\epsilon }_{D}T\bar{b}$. Hence the coequalizer condition is exactlythe equation for a morphism of algebras for a monad $(A,a)\to K(D)$. This supplies the natural bijection proving $L⊣K$, and also defines $L$ on algebra morphisms by the coequalizer universal property.

The finite representation type of a quiver condition means there are only finitely many isomorphism classes of indecomposable finite-dimensional $kQ$-modules. Equivalently, every finite-dimensional module is a direct sum of members of a finite list of indecomposables.

For the $E_6$ case, we give the reflection argument underlying the Gabriel theorem, rather than using its conclusion as the proof. Since the underlying graph is a tree, every orientation is acyclic. At a sink $i$, writefor the sum of the incoming arrow maps. If an indecomposable $V$ is not the simple $S_i$, then $h$ is surjective: a nonzero complementary subspace to its image at the sink would be a direct summand supported only at $i$. The Bernstein–Gelfand–Ponomarev reflection functor reverses the arrows and replaces $V_i$ by $\ker h$, with reversed maps the component projections. The source reflection replaces that kernel by the cokernel of its inclusion into $⨁V_j$, recovering $V_i$. The inclusion is injective, so the reflected representation has no direct summand $S_i$ at the new source. If it decomposed, applying the inverse would decompose $V$. Thus reflection preserves indecomposability except that it annihilates $S_i$.

On a dimension vector of a quiver representation, it acts by the integral simple reflectionLet $B=2I-\mathrm{Adj}(E_6)$. These reflections preserve the positive-definite quadratic form $d^{\mathsf{T}}Bd$. Positivity can be checked without a classification theorem: label the graph by the chain $1-2-3-4-5$ and attach vertex $6$ to $3$. The leading principal minors are $2,3,4,5,6,3$, all positive.

Choose a topological ordering so arrows point from smaller to larger indices. Reflect successively at sinks $6,5,\dots ,1$. This restores the orientation after one full cycle. The transformation of dimension vectors is the Coxeter element $C=s_1{s}_2\cdots s_6$. If $U$ has diagonal entries one and entry $-1$ above the diagonal at an arrow, direct multiplication of the reflection matrices givesThus $C$ has no fixed vector. It is an integral isometry of a positive definite form, so it has finite order $h$: the image of each lattice basis vector lies in a finite set of vectors of the same norm, giving only finitely many possible isometries. Since $C^h=I$ and $I-C$ is invertible,If an indecomposable survived all $h$ reflection cycles, its dimension vectors $d,Cd,\dots ,C^{h-1}d$ would all be nonzero and componentwise nonnegative, contradicting their zero sum. Therefore it reaches a simple at one of the finitely many reflection steps and is then annihilated. Applying the inverse reflections uniquely reconstructs it from that simple. There are at most $6h$ such stopping positions, provingThis supplies the requested finite-type proof sketch.

Sunada's theorem states that if a finite group $T$ acts isometrically on a closed Riemannian manifold $X$, and two Gassmann equivalent subgroups $H_1,H_2$ act freely, then $H_1\backslash X$ and $H_2\backslash X$ are isospectral for the function Laplace-Beltrami operator. Gassmann equivalence means $|H_1\cap C|=|H_2\cap C|$ for every conjugacy class $C$ of $T$. For each Laplace-Beltrami operator eigenspace $E_{\lambda }$, its invariant subspace has dimensionThe character of a representation is a class function, so the two multiplicities agree. A necessary condition for nonisometry is that the subgroups are not conjugate in $T$: conjugation by an ambient deck transformation would induce an isometry. This is not sufficient when the common covering has additional geometric symmetries.

Use the oriented hyperbolic triangle groupand the epimorphism $\rho :\Gamma \to T$ given by $a\mapsto s$, $b\mapsto d$. Its base orbifold is the doubled hyperbolic triangle with angles $\pi /3,\pi /3,\pi /6$, having signature $(0;3,3,6)$. Let $K=\ker \rho$ and $K_i={\rho }^{-1}(H_i)$.

The groups $K_i$ are torsion free. Indeed, torsion elements in an oriented triangle group are conjugate to nontrivial powers of its three vertex rotations. Such powers map either to elements of order three, which cannot lie in a subgroup of order $96/12=8$, or to the order-two cube $z=(sd{)}^3$. The latter is central and absent from both $H_i$, so it is absent from every conjugate as well. Thus the degree-twelve orbifold coversare smooth, closed, connected orientable surfaces. The common cover $X=K\backslash {\mathbb{H}}^2$ is smooth too; $H_i$ act freely on it. Orbifold Euler characteristic givesFor example, this formula follows by triangulating the base and weighting a vertex of order $r$ by $1/r$ before multiplying by the cover degree. Equivalently, each doubled triangle has hyperbolic area $\pi /3$, so $N_i$ has area $4\pi$ and Gauss-Bonnet theorem gives the same genus.

There is one extra issue in obtaining the same hyperbolic limit on a single marked surface: the two limiting hyperbolic covers must themselves be isometric. Nonconjugacy and the displayed element orders alone do not prove this. An explicit realization of the supplied example resolves the issue. The group seed is recorded in the original genus-two construction; the following coset actions and reflection identities give a direct verification. Label the twelve sheets so that the two right-coset actions of $s,d$ areThese are concrete permutation data for an order-$96$ realization with the required two subgroups. The permutationsatisfies, by direct substitution,Reflection of the doubled triangle reverses both vertex rotations. Apply this reflection on every fundamental domain and relabel sheets by $P$. The two identities ensure that identified edges go to identified edges, so this descends to a hyperbolic isometry $F_0:N_1\to N_2$. Fix $N=N_1$, let $g$ be its hyperbolic metric, and use $F_0$ to mark $N_2$ by the same smooth surface. Thus both pulled-back limiting Riemannian metrics are exactly $g$. The explicit reversed-action check is what makes the common limit justified, rather than treating it as a consequence of Sunada's theorem.

Choose smooth orbifold Riemannian metrics $h_k$ on the base converging to its hyperbolic metric $h_0$, with no nonidentity local isometry on the regular part. The density assertion in Sunada's local isometry lemma applies in orbifold charts as well: use equivariant perturbations at the finitely many cone points and ordinary local perturbations elsewhere. Smoothness here is orbifold smoothness, so the lifted Riemannian metrics on $X$ and the two smooth covers are ordinary smooth Riemannian metrics. Such choices can be made in successively smaller smooth neighbourhoods of $h_0$; cone orders and the topological coverings stay fixed.

For area separation of convergent Sunada families, adjust each $h_k$ by a constant positive factor so thatExplicitly, multiply the initially chosen Riemannian metric by the ratio of the desired area to its actual area. In dimension two this ratio is exactly the area multiplier. The ratios tend to one, so smooth convergence persists, and constant scaling preserves the absence of local isometries. Lift these adjusted Riemannian metrics to $N_i$ and pull them back by the fixed markings to obtain $g_{ik}$, for $i=1,2$. The three required properties follow below.

A proper morphism of schemes is a separated morphism of finite type which is a universally closed morphism: after every base change, the resulting map of underlying topological spaces takes closed subsets to closed subsets.

The valuative criterion for properness, in the Noetherian setting used here, says that a finite type morphism $f:X\to Y$ is a proper morphism if and only if the following existence and uniqueness condition holds. For every valuation ring $V$, its fraction field $K$, and compatible morphisms of schemes $a:\mathrm{Spec}K\to X$ and $b:\mathrm{Spec}V\to Y$ with $f\circ a=b{|}_{\mathrm{Spec}K}$, there is exactly one morphism of schemes $h:\mathrm{Spec}V\to X$ such that $h{|}_{\mathrm{Spec}K}=a$ and $f\circ h=b$. All valuation rings are allowed; no discreteness assumption is needed in this formulation. The following arguments also use that each of the three defining conditions for a proper morphism is preserved by base change.

A smooth rank-$r$ vector bundle over a smooth manifold $M$ consists of a smooth total space $E$, a smooth projection $\pi :E\to M$, and an $r$-dimensional real vector space structure on each fibre $E_p={\pi }^{-1}(p)$. Each point has a neighborhood $U$ with a vector bundle trivializationthat is a diffeomorphism, commutes with projection to $U$, and is a linear isomorphism on every fibre. The overlap maps have the form $(p,v)\mapsto (p,h(p)v)$ with $h:U\cap V\to G{L}_r(\mathbb{R})$ smooth. For a complex vector bundle, replace $\mathbb{R}$ by $\mathbb{C}$ and require complex-linear fibre maps.

A connection on a vector bundle is an $\mathbb{R}$-linear mapwhere $\Gamma (E)$ is the module of smooth sections, ${\Omega }^1(M;E)$ consists of vector-bundle-valued differential forms of degree one, and $f$ is a smooth function. Equivalently, evaluating on a vector field $X$ gives operators ${\nabla }_X$ with ${\nabla }_{fX}s=f{\nabla }_{X}s$ and ${\nabla }_X(fs)=X(f)s+f{\nabla }_{X}s$. A connection on a vector bundle is local: its value at a point depends on the germ of the section of a vector bundle, so the same definition applies to local sections.

For axisymmetric incompressible flow without swirl, the Stokes streamfunction satisfiesSubstitution into the divergence in spherical coordinates makes incompressibility an identity. Its only vorticity component isTaking the curl of the Stokes equation eliminates pressure and gives ${\nabla }^2\mathbf{\omega }=0$. For an axisymmetric azimuthal vector, its scalar component obeys $(\Delta -1/(R^2{\sin }^2\theta )){\omega }_{\varphi }=0$. Direct differentiation verifiesApply this to $h={\mathcal{D}}^2\Psi$ to obtain

Take a uniformly random ordering of the vertices and select each vertex that precedes every neighbor. The selected set is independent: if two adjacent vertices were both selected, each would have to precede the other. Among the $d(v)+1$ vertices consisting of $v$ and its neighbors, each is equally likely to come first, so the selection probability is $1/(d(v)+1)$. Linearity of expectation gives expected size ${\sum }_{v}1/(d(v)+1)$. At least one ordering attains at least this average; its integer size is therefore at least the ceiling. This proves the Caro-Wei bound

The Miller-Rabin primality test first handles $2,3$ as prime and $n<2$ or even $n>2$ as composite. For odd $n>3$, write $n-1=2^{s}d$ with $d$ odd and choose a base $2\le a\le n-2$. A nontrivial $\gcd (a,n)$ proves compositeness. Otherwise compute $x=a^d\mathrm{mod}n$: the round passes if $x=1$ or $x=-1$. If neither holds, square successively, at most $s-1$ times, and pass if a value $-1$ occurs; otherwise declare composite. An occurrence of $1$ before $-1$ exhibits a nontrivial square root of one and is also a failure. For prime $n$, Fermat's theorem and the fact that a field has only the square roots $\pm 1$ force a pass. The standard strong-liar bound says that an odd composite passes for at most one quarter of the bases; independent random rounds therefore have error probability at most $4^{-r}$ after $r$ rounds. Passing is a probable-prime verdict, not a proof of primality. A Fermat pseudoprime is composite and passes $a^{n-1}\equiv 1(\mathrm{mod}n)$; a strong pseudoprime is composite and passes the stronger round above.