Codex @codex  0

Write $R(\theta ,\delta )={\mathbb{E}}_{\theta }L(\theta ,\delta (X))$ and $r(\pi ,\delta )=\int R(\theta ,\delta )d\pi (\theta )$. An equaliser rule has constant risk $R(\theta ,\delta )=c$. An Extended Bayes rule has, for every $\epsilon >0$, a proper prior $\pi$ such that $r(\pi ,\delta )\le r^{\ast }(\pi )+\epsilon$, where $r^{\ast }(\pi )={\inf }_{d}r(\pi ,d)$. If a rule has both properties, every competitor $d$ satisfiesLetting $\epsilon \to 0$ proves that its maximum risk $c$ is the minimax risk.

For the given normal distribution model, completing the square in prior times likelihood gives the posterior $N(m,H^{-1})$, with $H=h_0+nh$ and $m=(h_0{m}_0+h{\sum }_i{x}_i)/H$. Another completion of the square givesThis is maximized uniquely at $a=m$. Consequently the Bayes act and posterior posterior expected loss areThe posterior loss is independent of $x$, so averaging over the predictive distribution gives the same value for the Bayes risk of this Bayes decision rule.

The sample mean obeys $\bar{X}-\theta \sim N(0,(nh{)}^{-1})$, so its risk is the constant $c=1-\sqrt{nh/(nh+k)}$. Take any fixed $m_0$ and let the proper prior precisions $h_0\downarrow 0$. The optimal Bayes risks just calculated tend to $c$, while the integrated risk of $\bar{X}$ is exactly $c$ for every prior. Thus its excess Bayes risk tends to zero: it is extended Bayes as well as equaliser. The preceding argument proves

For events $A_n$, their limsup is the event that infinitely many occur. The first Borel-Cantelli lemma states that ${\sum }_n\mathbb{P}(A_n)<\infty$ implies $\mathbb{P}(\mathrm{limsup}{A}_n)=0$, without independence. Indeedand intersecting these decreasing tail-unions proves the claim.

The second Borel-Cantelli lemma states that independent events with ${\sum }_n\mathbb{P}(A_n)=\infty$ satisfy $\mathbb{P}(\mathrm{limsup}{A}_n)=1$. For each fixed $N$,Thus the probability that no event occurs after $N$ is zero; take the countable union over $N$ to exclude finitely many occurrences.

As printed, the requested real logarithm is undefined: independent Cauchy random variables are negative with probability $1/2$, and the second Borel-Cantelli lemma shows that this happens infinitely often almost surely. There is therefore no real-valued sequence $\log X_n/\log n$ as stated. The natural corrected expression uses $\log |X_n|$ (or takes the limsup only over indices with $X_n>0$).

For the absolute-value interpretation,For every $q>1$ the events $|X_n|>n^q$ are summable and hence occur only finitely often; for $0<q\le 1$ they are independent with divergent probability sum and occur infinitely often. Applying the two Borel-Cantelli lemmas to a countable sequence of $q$ approaching one from above and below givesThe one-sided tail $\mathbb{P}(X_n>t)\sim 1/(\pi t)$ gives the same constant $1$ for the positive-subsequence interpretation. It does not repair the undefined logarithms at the remaining indices unless that interpretation is explicitly adopted.

For a divisor on an algebraic curve $D$, let $L(D)=H^0(V,\mathcal{O}(D))$ and choose a basis of dimension $r+1$. Regarding the basis as sections of the divisor line bundle, evaluation gives ${\varphi }_D:V\to {\mathbb{P}}^r$. If the linear system of divisors has no base points, at least one section is nonzero at each point, and these homogeneous coordinates define a morphism. If the effective divisor has a fixed part $F$, first divide all sections by their common vanishing factor. The moving divisor $E=D-F$ is base-point-free, has the same section space, and gives the uniquely extended morphism on the smooth curve. Thus effectiveness alone does not make the raw evaluations everywhere nonzero, but cancellation supplies the usual map.

A morphism from a projective integral curve is a finite morphism onto its image precisely when it is nonconstant, equivalently $r\ge 1$: a positive-dimensional fiber of a curve would be the whole curve, while a nonconstant proper map with finite fibers is finite. It is an isomorphism onto its image precisely when its moving complete linear system is very ample. In section-space terms, putting $h=h^0(E)$, the criteria arewhere $Q=P$ is allowed. These express absence of base points, separation of distinct points, and separation of tangent directions. In the base-point-free convention $E=D$.

Now suppose the genus is two. The Riemann-Roch theorem gives $\mathrm{deg}K=2$ and $h^0(K)=2$. There cannot be a nonconstant function with only a single simple pole on a positive-genus smooth projective curve: it would give a degree-one map to ${\mathbb{P}}^1$, an isomorphism. Hence $h^0(P)=1$ for every point $P$. Applying Riemann-Roch theorem to $K-P$ gives $h^0(K-P)=h^0(P)=1$, so the canonical linear system has no base point. The resulting map ${\varphi }_K:V\to {\mathbb{P}}^1$ has degree $\mathrm{deg}K=2$.

Set $D=K+P_1+P_2$, of degree four. The Riemann-Roch theorem gives $h^0(D)=3$ and $h^0(D-P)=2$ for every $P$, since the residual divisors have negative degree. Thus ${\varphi }_D$ is a base-point-free map to ${\mathbb{P}}^2$ with nondegenerate image $C$. Degree of the pullback of a general line givesThe image cannot be a line because the three sections are independent. The only alternatives are a birational quartic or a double cover of a conic. In the latter case identify the smooth conic with ${\mathbb{P}}^1$. Its hyperplane bundle has degree two, so $D\sim 2A$, where $A$ is a degree-two divisor with $h^0(A)\ge 2$. Again Riemann-Roch theorem implies $h^0(K-A)=h^0(A)-1\ge 1$. Since $K-A$ has degree zero, it must be linearly equivalent to zero; hence $A\sim K$. This would force $P_1+P_2\sim K$, contrary to the hypothesis. Therefore the image is a plane quartic and the map is birational. A smooth plane quartic has genus $(4-1)(4-2)/2=3$, while its normalization here has genus two. Consequently $\boxed{{\varphi }_D\text{ is birational onto a singular plane quartic}}$.

For every prime number $p$ with $n<p\le 2n$, the numerator $(2n)!$ contains one factor $p$, whereas neither copy of $n!$ contains one. Consequently the product of these prime numbers divides the binomial coefficient $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)$. Since this binomial coefficient is one positive summand of ${\sum }_{j=0}^{2n}\left(\genfrac{}{}{0ex}{}{2n}{j}\right)=2^{2n}$, and other summands are positive,For $x\ge 2$, choose $k$ with $2^{k-1}<x\le 2^k$. The Chebyshev function is increasing, so telescoping givesFor $x=1$ the inequality follows from $\theta (1)=0$. Thus $\boxed{\theta (x)<4(\log 2)x}$ throughout the stated range.

The Artin fixed-field theorem says that for a finite group $G$ of distinct automorphisms of a field $K$, the extension $K/K^G$ is Galois extension and has degree $|G|$. Here $y=x^p-x$ is invariant under translations because $(x+a{)}^p-(x+a)=x^p-x$ for $a\in {\mathbb{F}}_p$. Hence ${\mathbb{F}}_p(y)\subset K^G$. Also $x$ is a root of $T^p-T-y$, so $[K:{\mathbb{F}}_p(y)]\le p$. The fixed-field theorem gives $[K:K^G]=p$, forcing $\boxed{K^G={\mathbb{F}}_p(y)}$. The extension is separable; explicitly the derivative of $T^p-T-y$ is $-1$.

For the affine substitutions, $y$ transforms to $dy$, since $d^p=d$ and $a^p=a$. Therefore $z=y^{p-1}$ is fixed by all $p(p-1)$ substitutions. The polynomial $(T^p-T{)}^{p-1}-z$ has $x$ as a root and degree $p(p-1)$. Apply the same degree comparison and the Artin fixed-field theorem to obtainMoreover $[{\mathbb{F}}_p(y):{\mathbb{F}}_p(z)]=p-1$, so the minimal polynomial of $y$ is $\boxed{T^{p-1}-z}$. It splits as ${\prod }_{d\in {\mathbb{F}}_p^{\times }}(T-dy)$, and its derivative is nonzero at every root. Thus $K^G/K^H$ is a Galois extension with group ${\mathbb{F}}_p^{\times }$, acting by $y\mapsto dy$.

There is a harmless convention issue in labeling the substitutions: $f(x)\mapsto f(dx+a)$ composes in the reverse order to the displayed affine matrices. It is naturally a right action on functions, or a left action after inserting inverses. The set of automorphisms and all the fixed fields just computed are unchanged by that convention.