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Under the risk-neutral measure, writeLet $M_T={\max }_{t\le T}{X}_t$ and $m_T={\min }_{t\le T}{X}_t$. ThenThe time-reversed increment processhas the same finite-dimensional distributions as $X$, because $X$ has stationary independent increments. Moreover,Choosingtherefore makes $Y_2=S_0{e}^{X_T-m_T}$ equal in distribution to $Y_1$. Their discounted expectations, and hence their time-zero Black--Scholes prices, are equal. This is Brownian time reversal for fixed-strike lookback extrema.

Risk-neutral valuation givesSince $W_T^Q\sim N(0,T)$,Hence the Power payoff in the Black-Scholes model has price

Setand define $Q$ on ${\mathcal{F}}_T$ byThe Cameron-Martin theorem for a linear drift says thatis a Brownian motion under $Q$. ThereforeThe discounted stock is an exponential Brownian martingale, so $e^{-rt}{S}_t$ is a $Q$-martingale. Thus $Q$ is a Risk-neutral measure for the Black-Scholes model.

For $0\le s\le t$, the increment $W_t-W_s$ is independent of ${\mathcal{F}}_s$ and is $N(0,t-s)$. HenceAlso $\mathbb{E}{M}_t=1$, so $M$ is the exponential Brownian martingale.

The given identities implyThus, writingwe have the exact identityThe sample variance satisfies $s_n^2\stackrel{P}{\to }1$, soThe Slutsky theorem and part (c) now give

The central limit theorem givesApplying the delta method to $g(x)=x^2$ yieldsFor $\mu =0$ this denotes the degenerate distribution at zero, which is the correct limit at the $\sqrt{n}$ scale.

Since ${\overline{X}}_n\sim N(\mu ,1/n)$,soEvery leave-one-out mean has variance $1/(n-1)$, and henceTherefore

Each leave-one-out statistic has the same distribution as $T_{n-1}$, soConsequently the bias of the jackknife bias correction isThe assumed expansion givesandTheir difference is therefore $O(n^{-2})$, as required.

Let the Poisson process $X$ have rate $\lambda$, let $\overline{F}(t)=\mathbb{P}({\xi }_1>t)$ be the survival function of an inter-renewal time of $N$, and let $\overline{H}$ be the survival function of the first event time of $Y$. Independence givesPut

At a large time, the excess of $Y$ is the minimum of the excesses of $X$ and $N$. The Poisson excess is an independent rate-$\lambda$ exponential random variable. Applying the renewal excess limit theorem to $N$ and to the assumed renewal process $Y$ givesSubstituting $\overline{F}(y)=e^{\lambda y}\overline{H}(y)$ yields the integral equation

Set $A(x)={\int }_x^{\infty }\overline{H}(y)dy$ and $c={\mu }_H/{\mu }_F<1$. Differentiating the equation almost everywhere givesSince $A^{\prime }=-\overline{H}$, it follows thatUsing $\overline{H}(0)=1$ shows that $\overline{H}(x)=e^{-x/{\mu }_H}$. Thus the first event time of $Y$ is exponential. This is the exponential first interarrival in a renewal superposition containing a Poisson process theorem.

The residual lifetime process, or excess process, isIf the inter-renewal distribution is non-arithmetic and $0<\mu =\mathbb{E}{\xi }_1<\infty$, then the renewal excess limit theorem saysEquivalently, $E_{\infty }$ has the law of $U\hat{\xi }$, where $U$ is uniform on $(0,1)$ and independent of the size-biased inter-renewal time $\hat{\xi }$.

If $0<\mu =\mathbb{E}{\xi }_1<\infty$, the size-biased distribution of ${\xi }_1$ is the law of $\hat{\xi }$ defined byfor bounded measurable $h$. Equivalently,

For independent identically distributed nonnegative random variables ${\xi }_1,{\xi }_2,\dots$, define the renewal epochsThe associated renewal process isthe number of renewals by time $t$.

The Gaussian curvature $K$ is the determinant of the shape operator, equivalently the product of the two principal curvatures. Yes,This is the Theorema Egregium: Gaussian curvature depends only on the first fundamental form, which a local isometry preserves.

The first fundamental form of an embedded surface $S\subseteq {\mathbb{R}}^3$ is the inner product induced on each tangent plane:For a tangent vector field $V$ along a curve $\alpha$, its surface covariant derivative isthe tangential projection of its ordinary derivative. A curve is a geodesic when

A local isometry preserves the first fundamental form, so its Christoffel symbols and Levi--Civita covariant derivatives satisfyTaking $V={\alpha }^{\prime }$ shows that if $\alpha$ is geodesic, then $\varphi \circ \alpha$ is geodesic.

The converse is false. The dilationmaps every affinely parametrized straight-line geodesic to another such geodesic, butso it does not preserve the first fundamental form. This is the geodesic-preserving homothety that is not a local isometry.

Let $\zeta =e^{2\pi i/n}$ and define Möbius transformationsThey satisfyand the $2n$ maps $r^k$ and $s{r}^k$ are distinct. They therefore give a faithful action of the dihedral group $D_{2n}$ on the Riemann sphere.

The rational functionis invariant under both $r$ and $s$, so it is constant on every orbit. Conversely, for nonzero finite $z_1,z_2$, put $u=z_1^n$ and $v=z_2^n$. Equality of the function values givesIf $u=v$, then $z_2={\zeta }^k{z}_1=r^k{z}_1$ for some $k$. If $uv=1$, then $z_2={\zeta }^k/z_1=r^{k}s{z}_1$ for some $k$. Finally, zero and infinity both map to infinity and are interchanged by $s$. HenceThis is the orbit-separating invariant for the standard dihedral action on the Riemann sphere.

Let $q\in {\mathbb{C}}_{\infty }$ and let $G_q$ be its stabilizer subgroup. Choose a sufficiently small disc $U$ about $q$ such thatPart (a) says that $G_q$ is cyclic and supplies a coordinate in which $U\to G_q\backslash U$ is $z\mapsto z^{|G_q|}$. Since the global orbit map identifies no additional points within $U$, this is also a chart on a neighbourhood of the orbit $Gq$ in

Use these quotient charts at every orbit. On overlaps, a local inverse branch of one power map followed by a group translate and the other power map gives the transition function; it is holomorphic. The charts therefore define a conformal structure on $R$, and the local expressions $z\mapsto z^{|G_q|}$ show directly that the quotient map ${\mathbb{C}}_{\infty }\to R$ is holomorphic. This is the finite conformal quotient of a Riemann surface construction.

Choose a local coordinate $w$ with $w(p)=0$. Differentiation at $p$ gives a homomorphismIt is injective. Indeed, a nonidentity Möbius transformation fixing $p$ and having derivative one there is parabolic, hence conjugate to a nonzero translation and of infinite order; no such element can belong to the finite group $H$. Every finite subgroup of ${\mathbb{C}}^{\times }$ is a cyclic group of roots of unity, so $H$ is cyclic. Write $n=|H|$.

The averaged coordinatehas derivative one at $p$, so it is a coordinate on a smaller neighbourhood. For $g\in H$, reindexing by $k=hg$ givesAfter shrinking to an $H$-invariant disc $U$, the group therefore acts as all rotations $z\mapsto {\zeta }^{j}z$, where $\zeta$ is a primitive $n$th root of unity. The invariant coordinateidentifies $H\backslash U$ with a disc and gives it a Riemann-surface chart. In these coordinates the quotient map is exactlyThis is the local cyclic quotient of a Riemann surface.

If $X$ is compact and $f:X\to \mathbb{R}$ is continuous, then $f(X)$ is compact by the continuous image of a compact space theorem. Every compact subset of $\mathbb{R}$ is bounded, so $f$ is bounded.

Conversely, suppose the metric space $X$ is not compact. For metric spaces, compactness is equivalent to sequential compactness, so there is a sequence of distinct points $(x_n)$ having no convergent subsequence. The setis closed: an accumulation point would supply a convergent subsequence. It is also discrete for the same reason. Consequently the functionis continuous. By part (a), $X$ is normal, and the Tietze extension theorem extends $g$ to a continuous $G:X\to \mathbb{R}$. Since $G(x_n)=n$, this extension is unbounded. Therefore, if every continuous real-valued function on $X$ is bounded, $X$ must be compact. This is the bounded-continuous-function characterization of compact metric spaces.