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The finite-horizon discrete-time fundamental theorem of asset pricing states that the market has no arbitrage if and only if there exists a probability measure $\mathbb{Q}$ equivalent to $\mathbb{P}$ under which every integrable discounted asset price $X^i$ is a martingale. Such a $\mathbb{Q}$ is an equivalent martingale measure.

Let $V_t$ denote wealth measured in units of the strictly positive numéraire. A predictable self-financing portfolio is an arbitrage ifEquivalently, one may permit $V_0\le 0$ provided one of the two gains is strict.

The multivariate Wald statistic for a proposed value $\theta$ isBecause ${\hat{i}}_n\to I_p$ andSlutsky theorem gives

If $q_{p,1-\alpha }$ is the $(1-\alpha )$-quantile of the chi-squared distribution with $p$ degrees of freedom, an asymptotic $(1-\alpha )$ confidence region isIn this known-identity-covariance model one may replace ${\hat{i}}_n$ by $I_p$; then the confidence ball is exact. For an individual coordinate, the corresponding exact confidence interval is

The maximum-likelihood estimator isUnder the score-outer-product convention stated in the question,This is the sample covariance matrix with divisor $n$. The strong law of large numbers, applied entrywise, givesThus the empirical score outer-product information consistently estimates the per-observation Fisher information.

For one observation with density $f(x,\theta )$, the Fisher information matrix at ${\theta }_0$ isHereSince ${\mathrm{Cov}}_{{\theta }_0}(X)=I_p$, the Fisher information of a multivariate normal location model isThe information in the full independent sample is $n{I}_p$ by Tensorization of Fisher information.

First intersect $E$ with a sufficiently large bounded interval to obtain a finite positive-measure set. By the Lebesgue density theorem, this set has a density-one point $x$. Hence, for every $\epsilon >0$, some sufficiently small interval $I=(a,b)$ centred at $x$ satisfieswhich is the high-density interval in a positive-measure subset of the real line.

To obtain the final claim, choose $\epsilon <1/2$, put $L=b-a$, and translate $A=E\cap (a,b)$ into $[0,L]$. WritePart (f), after rescaling the interval length, givesThus $E-E$ contains an open interval around zero, which is the Steinhaus theorem.

Let $|t|<2\alpha$. Both $E$ and its translate $E+t$ lie in an interval of length $1+|t|$. By finite additivity and translation invariance of Lebesgue measure,Choose $x\in E\cap (E+t)$. Then $x\in E$ and $x-t\in E$, so $t=x-(x-t)\in E-E$. Therefore the difference-set overlap bound on an interval gives

Yes. In fact, measurability of $f$ is unnecessary. Define the oscillationFor every $\epsilon >0$, the set $\{x:{\omega }_f(x)<\epsilon \}$ is open, and $f$ is continuous at $x$ exactly when ${\omega }_f(x)=0$. Hence the continuity set of a function isa $G_{\delta }$ set and therefore a Borel set.

A function $f:\mathbb{R}\to \mathbb{R}$ is Borel measurable whenis a Borel set for every Borel set $B\subseteq \mathbb{R}$. It is enough to require this for every open set, or for every interval of the form $(-\infty ,a)$.

The rational numbers are countable, sois a countable union of closed sets and hence a Borel set; its Lebesgue measure is zero. Its complement is also Borel and has infinite measure:

The intervals are nested in the relevant directions, and the strongest lower and upper restrictions both occur at $n=2$:Thus this set is closed, hence Borel, and

A set $E\subseteq \mathbb{R}$ is Lebesgue measurable if it satisfies the Caratheodory criterionfor every $A\subseteq \mathbb{R}$, where $m^{\ast }$ is Lebesgue outer measure. Equivalently, $E$ belongs to the completion of the Borel sigma-algebra with respect to Lebesgue measure.

A subset of $\mathbb{R}$ is a Borel set when it belongs to the Borel sigma-algebra, the smallest sigma-algebra containing every open set of $\mathbb{R}$.

No. The missing point is not assumed to complete $S$ as a smooth surface, so the hypotheses of the global Gauss-Bonnet theorem do not apply to $S\cup \{p\}$.

For a concrete counterexample, rotate a smooth simple arc $(r(s),z(s))$, parametrized by arc length for $0\le s<L$, about the $z$-axis. Arrange that it starts smoothly on the axis,stays in $r>0$ for $0<s<L$, and converges to another point of the axis as $s\uparrow L$, withThe endpoint is omitted. The resulting surface of revolution is diffeomorphic to ${\mathbb{S}}^2$ minus one point, and adjoining its limiting endpoint makes the set compact.

For an arc-length surface of revolution,Therefore the total Gaussian curvature of a punctured surface with a singular compactification point iswhich can take values different from $4\pi$. Compactness of the one-point closure alone therefore does not determine the integral.

The global Gauss-Bonnet theorem for a compact oriented surface without boundary isFor a connected closed orientable surface of genus $g$, this is

For a local parametrization $\varphi (u,v)$, write the first fundamental form and second fundamental form asThe Gaussian curvature is

To prove the Theorema Egregium, set $g_{ij}=⟨{\varphi }_i,{\varphi }_j⟩$ and decomposeDifferentiating $g_{ij}$ and solving for the tangential coefficients givesso every Christoffel symbol depends only on $E,F,G$ and their first derivatives. Equality of mixed third derivatives yields the Gauss equationThe Riemann curvature tensor is formed from first derivatives and quadratic products of the Christoffel symbols, and hence from the first and second derivatives of $E,F,G$. Consequentlydepends only on the first fundamental form.

DefineIf $D$ is a divisor class of degree zero, Riemann--Roch gives $\ell (D+Q)=1$, because $K_E\sim 0$ and the divisor $-D-Q$ has negative degree. A nonzero section therefore yieldsfor some point $P$, proving that $\alpha$ is surjective. If $\alpha (P)=\alpha (P^{\prime })$, then $P-P^{\prime }$ is a principal divisor. For $P\ne P^{\prime }$ its defining function would have one simple pole and give a degree-one map $E\to {\mathbb{P}}^1$, impossible for a genus-one curve. Thus $P=P^{\prime }$. The map and its inverse are regular, so this proves the Abel-Jacobi map of a genus-one curve is an isomorphism.

Transport the addition of the divisor class group through $\alpha$: define $P⊞R$ byThen $Q$ is the identity, and this is the chord-and-tangent group law on the plane cubic.

For the displayed equation, inversion is $(x,y)\mapsto (x,-y)$. Thus $P⊞P=Q$ exactly when $P=-P$. Besides the point at infinity $Q$, this requires $y=0$. Therefore the complete set of two-torsion points is

Choose a point $Q\in E$. Since $g(E)=1$, a nonzero regular differential has an effective divisor of degree $2g-2=0$, so $K_E\sim 0$. The Riemann-Roch theorem therefore givesChooseThe rational functions $x$ and $y$ have their only poles at $Q$, of orders two and three respectively.

The seven functionsbelong to the six-dimensional Riemann-Roch space $L(6Q)$, so they satisfy a nontrivial linear relation. Comparing pole orders shows that the coefficients of both $x^3$ and $y^2$ are nonzero. After rescaling, the relation has the Weierstrass formThe divisor $3Q$ is very ample, so its complete linear system embeds $E$ in ${\mathbb{P}}^2$ and the displayed relation cuts out its image; this is the Weierstrass construction from Riemann-Roch in genus one.

Because $\mathrm{char}k\ne 2$, completing the square removes the terms linear in $y$, and because $\mathrm{char}k\ne 3$, translating $x$ removes the quadratic term. Thus the equation becomesSince $k$ is algebraically closed, the cubic factors asSmoothness of $E$ says that the elliptic-curve discriminant is nonzero, equivalently the three roots are distinct. Hence

For a divisor on an algebraic curve $D$ on a smooth projective curve $C$ of geometric genus $g$, and a canonical divisor $K_C$, the Riemann-Roch theorem states