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Write $Z_i=(X_i,Y_i{)}^T$ for the independent observations and put $\Omega ={\Sigma }^{-1}$. Apart from constants, the log-likelihood isUsing the stated matrix derivatives, its score function isAt a root of the score, ${\Omega }^{-1}=n^{-1}{\sum }_i{Z}_i{Z}_i^T$. Hence the maximum-likelihood covariance estimator for centered Gaussian data is

Provided $X$ and $Y$ have finite positive variances, their correlation coefficient isIt lies in $[-1,1]$ by the Cauchy--Schwarz inequality.

False. Take the chain from part (b) with $\lambda =1$. It is irreducible and nonexplosive, and it has the invariant distributionHence every state is a positive recurrent state for the continuous-time chain.

Its jump chain is the reflected simple symmetric random walk on $\mathbb{N}$. This chain is recurrent but has only the infinite invariant measureso every state is a null recurrent state. Equivalently, the invariant-measure transfer between a jump chain and a CTMC divides this infinite jump-chain measure by the rapidly growing holding rates and produces the summable continuous-time invariant measure. Thus fast holding rates create continuous-time positive recurrence, and positive recurrence need not pass from a continuous-time chain to its jump chain.

For $i\ge 1$, the total holding rate is $(\lambda +1)2^i$, and the jump chain moves upward and downward with probabilitieswhile it moves from zero to one with probability one. Thus it is a reflected nearest-neighbour random walk. Recurrence is unaffected by the holding times, so it is recurrent exactly when $p\le q$. In the given range,For $\lambda >1$ the jump chain is transient.

The detailed-balance equations arewhere the formula also covers the boundary $i=0$. Hence every candidate invariant measure hasAt $\lambda =1$ this is summable and the chain is nonexplosive, givingFor $\lambda =2$ it is not summable. When $1<\lambda <2$, it is a summable formal solution of $\pi Q=0$, but the chain explodes, so it is not an invariant probability for the minimal process; this is precisely the caveat in invariant distribution of an explosive chain. Consequently

It remains to classify explosion. At $\lambda =1$, the recurrent jump chain visits zero infinitely often. The holding time at zero has rate one, so nonexplosion from recurrent visits to a slow state shows that the accumulated time is infinite almost surely.

For $\lambda >1$, let $V_i$ be the total number of visits of the transient jump chain to $i$. Using the stated uniform visit bound and writing $q_i=(\lambda +1)2^i$ for $i\ge 1$,because ${\sum }_i{2}^{-i}<\infty$. Conditional on the jump path, this is the expected sum of all holding times, so the total lifetime is finite almost surely. This is explosion of an upward-biased walk with geometrically increasing rates. Therefore

A simple birth process with parameter $\lambda$ starting from one individual is the Yule process: when its population is $n$, it waits an exponential time of rate $n\lambda$ and then moves to $n+1$.

For $f(n)=n$, the Markov jump-process generator satisfiesTherefore $\mu (t)=\mathbb{E}{X}_t$ obeysIt follows, as recorded by the mean population of a Yule process, that

PutThis function is bounded near zero and is $O(u^{-2})$ at infinity, so $H\in L^1(\mathbb{R})$ and the defining integral for $g$ converges absolutely.

For the triangular functiondirect integration givesThe Fourier inversion theorem, or equivalently the Fourier transform of a triangular function, therefore yieldsIn particular $g(x)=0$ whenever $|x|>1$. Finally,This also agrees with the Plancherel theorem applied to $h$ and $H$.

Use the conventionThe Fourier inversion theorem says that, when $f,\hat{f}\in L^1({\mathbb{R}}^d)$,for almost every $x$.

Sincethe Fubini's theorem and Fourier inversion giveThus the Plancherel theorem identity in this convention is

The inverse integral is a continuous function of $x$, since $\hat{f}\in L^1$ permits dominated convergence. If $f$ is continuous, it and the inverse integral are continuous and agree almost everywhere. They must agree everywhere: otherwise their continuous difference would be nonzero on an open set of positive measure. This is the continuous version of Fourier inversion.

No. For sufficiently smooth curves,is unchanged by proper Euclidean motions and transforms correctly under translations of the arc-length parameter, so it is a pointwise Euclidean invariant of a curve. It is not determined by the two numbers $\kappa (s)$ and $\tau (s)$.

For a concrete comparison near $s=0$, use the plane curve reconstructed from curvature construction withThe resulting unit-speed planar curves both have $\kappa (0)=1$ and $\tau (0)=0$, but their curvature derivatives at zero are respectively $0$ and $1$. Thus no single function $f$ can satisfy $Q(\gamma ,s)=f(\kappa (s),\tau (s))$ for all curves.

Indeed, the covariance conditions alone even permit nonlocal examples such as $Q(\gamma ,s)={\kappa }_{\gamma }(s+1)$. This is the distinction recorded by pointwise Euclidean invariants need not depend only on current curvature and torsion.

Write a proper Euclidean motion of Euclidean three-space as $E(x)=Ax+b$, where $A\in SO(3)$. ThenBecause $A$ preserves norms, dot products, and cross products, the formulasshow directly that curvature and torsion are unchanged by $E$.

For the shifted curve ${\gamma }_{s_0}(s)=\gamma (s-s_0)$, every derivative at $s$ equals the corresponding derivative of $\gamma$ at $s-s_0$. ConsequentlyBoth are therefore pointwise Euclidean invariants in the stated sense.

If $\kappa \equiv 0$, then $T^{\prime }=0$, so the unit tangent is constant andThus both curves are straight line segments. Equal arc length makes their parameter intervals equally long, and a rotation in $SO(3)$ followed by a translation maps one to the other. Hence they are related by a proper Euclidean motion of Euclidean three-space.

The conclusion fails for $\kappa \equiv 1$, because curvature alone does not determine a space curve. A unit circle has curvature one and torsion zero. The circular helix with $a=b=1/2$,is unit speed and has curvature one but torsion one. Proper Euclidean motions preserve torsion, so equal-length pieces of these two curves cannot be related by one.

Let $\gamma$ be a $C^3$ regular curve parametrized by arc length, so $|{\gamma }^{\prime }(s)|=1$. Define the Frenet frameand define the torsion byThus curvature requires two derivatives, while $N,B,$ and torsion require $\kappa >0$; the displayed classical torsion requires three derivatives. At a point where $\kappa =0$, $N,B,$ and $\tau$ are not defined by this construction.

Since $(T,N,B)$ is an orthonormal frame, differentiating its inner products shows that its derivative matrix is skew-symmetric. The definition gives $T^{\prime }=\kappa N$. Next,so $N^{\prime }=-\kappa T+cB$. As $B^{\prime }=T\times N^{\prime }$ and $T\times B=-N$, we have $B^{\prime }=-cN$; hence $c=-B^{\prime }\cdot N=\tau$. The Frenet-Serret formulas are therefore

For $f\in L^1({\mathbb{R}}^n)$, use the angular-frequency conventionThis is the Fourier transform on $L^1({\mathbb{R}}^n)$.

For a measurable set $E\subseteq U$, definewhere $1_E$ is its indicator function. This is finite because $U$ has finite measure and hence $1_E\in L^p(U)$. If the sets $E_k$ are pairwise disjoint and $E={\bigcup }_{k\ge 1}{E}_k$, thenThe continuity of the positive linear functional $\Lambda$ therefore givesso $\nu$ is a measure by countable additivity. Moreover, a set of zero Lebesgue measure has zero indicator in $L^p$, so $\nu$ is absolutely continuous with respect to Lebesgue measure.

The Radon-Nikodym theorem now supplies a measurable function $\omega \ge 0$ such thatConsequently, first for nonnegative simple functions and then, by approximation by nonnegative simple functions, for every nonnegative $f\in L^p(U)$,Applying the stated characterization of the Lp norm to $\omega$ shows that $\omega \in L^q(U)$ and $\Vert \omega {\Vert }_q\le \Vert \Lambda \Vert$. The Holder inequality makes $f\mapsto {\int }_{U}f\omega$ continuous on $L^p(U)$, while bounded simple functions form a dense subset; henceThis is the positive functional representation on Lp.

Assume $u(0)=0$. Applying midpoint preservation to $2v$ and $0$ givesso $u(2v)=2u(v)$. Consequently,Thus $u$ is additive. It follows successively that

An isometry is continuous. For any $\lambda \in \mathbb{R}$, choose rationals $q_j\to \lambda$. ThenTogether with additivity, this proves that $u$ is real-linear. Hence the origin-fixing case of the Mazur-Ulam theorem givesfor all $a,b\in \mathbb{R}$ and $v,w\in V$.

Here “isometry” is used in the standard Mazur-Ulam theorem sense of a surjective distance-preserving map. Surjectivity is needed; a merely distance-preserving embedding need not preserve midpoints, as shown by a nonsurjective isometry need not preserve midpoints.

Distance preservation and surjectivity giveAssume inductively thatThen the two sets have the same diameter, and the universal distance condition defining the next set transfers through the bijection $u$. Hencefor every $n$.

Because $u$ is injective, it also preserves the intersection of this nested family. Part (a) therefore givesThus

LetThe map $R$ is reflection about $m$ and is an isometry. It interchanges $v,w$, so it preserves $S_1^{vw}$. Inductively, if it preserves $S_{n-1}^{vw}$, then it preserves its diameter and all distances in the defining condition for $S_n^{vw}$; hence it preserves every $S_n^{vw}$.

The midpoint belongs to $S_1^{vw}$ becauseSuppose $m\in S_{n-1}^{vw}$. For every $z\in S_{n-1}^{vw}$, reflection invariance gives $R(z)\in S_{n-1}^{vw}$ andThus $m\in S_n^{vw}$, so all the sets are nonempty and contain $m$.

Put $D_n=\mathrm{diam}(S_n^{vw})$. If $x,y\in S_n^{vw}$ with $n\ge 2$, then $y\in S_{n-1}^{vw}$ and the defining property of $x$ givesThereforeIf $z$ lies in every $S_n^{vw}$, then both $z$ and $m$ lie in $S_n^{vw}$, soHence $z=m$. We have proved the metric extraction of a midpoint by shrinking diameters:

Put $\theta =\sqrt[3]{3}$. The given ring of integers has integral basis $1,\theta ,{\theta }^2$. The discriminant of the power basis, equivalently of $T^3-3$, isThe field has signature $(1,1)$: one real embedding and one conjugate pair of complex embeddings. The Minkowski bound for ideal classes is thereforeEvery ideal class consequently has an integral representative of norm at most four.

It remains to inspect prime ideals above $2$ and $3$. Modulo $2$,The Dedekind factorization theorem giveswhereButso the principal ideal $(\theta -1)$ is contained in ${\mathfrak{p}}_2$ and has the same norm; henceAlsosoBoth primes above $2$ are principal.

Finally,Thus the unique prime above $3$ is the principal ideal $(\theta )$. By unique factorization of ideals in a number field, every integral ideal of norm at most four is built from these principal prime ideals. Every ideal class is therefore trivial, and the Class group of Q of cube root of three is