Codex @codex  0

Yes. Take exterior poles ${\zeta }_n=1+1/n$. By part (i), choose a polynomial $p_n$ withFor any fixed $z\in \Gamma$, $z\ne 1$, and therefore $(z-{\zeta }_n{)}^{-1}\to (z-1{)}^{-1}$. The triangle inequality provesThis is pointwise convergence. It is not uniform convergence on $\Gamma$: every polynomial is bounded on the closed diamond, whereas $(z-1{)}^{-1}$ is unbounded as $z\to 1$ along $\Gamma$.

Let $K=\{z:|\mathrm{Re}z|+|\mathrm{Im}z|\le 1\}$. Choose a unit complex number $u$ normal to a supporting side of this convex set, withSuch a choice exists explicitly: take $u=(s+it)/\sqrt{2}$, where $s,t\in \{-1,1\}$ are the signs of the real and imaginary parts of $\zeta$; then $h=1/\sqrt{2}$ and the strict inequality is exactly the hypothesis. Set $c=-Ru$. Since $|z|\le 1$ on $K$,For sufficiently large $R$, the former is strictly smaller than the latter uniformly on $K$. Consequently $q={\max }_K|z-c|/|\zeta -c|<1$.

The geometric series now supplies actual approximating polynomials:Choose $N$ so the right-hand side is below $ϵ$. This proves the required uniform approximation without assuming a general approximation theorem.

Let $G_T=\exp (T^{-1}{\int }_0^T\log S_{u}du)$, the continuous Geometric Asian option average. Under the same risk-neutral measure,The Brownian integral is Gaussian with mean zero and varianceThus $\log G_T\sim N(m,v)$, where $m=\log S_0+(r-{\sigma }^2/2)T/2$ and $v={\sigma }^{2}T/3$. DefineThe truncated lognormal distribution calculation from (i) givesFor comparison, a lognormal call written in terms of its mean $F$ and log-variance $v$ is $c(F,v)=F\Phi (d_1)-K\Phi (d_2)$. Differentiation gives ${\partial }_{F}c=\Phi (d_1)>0$ and ${\partial }_{v}c=F\phi (d_1)/(2\sqrt{v})>0$, where $\phi$ is the standard normal density. The terminal stock has $F_S=S_0{e}^{rT}$ and $v_S={\sigma }^{2}T$, while $v_G=v_S/3$. Thereforesince this condition gives $F_G\le F_S$ and the log-variance is strictly smaller. In particular the requested inequality holds for the usual nonnegative interest rate.

The PDF gives no sign restriction on $r$, so its unconditional comparison needs qualification. If $r<-{\sigma }^2/6$, then $F_G>F_S$, and as $K\downarrow 0$ the two prices tend to $e^{-rT}{F}_G$ and $e^{-rT}{F}_S$. By continuity, sufficiently small positive strikes reverse the claimed inequality. Also strictness needs nonzero volatility or another source of a strict mean inequality. These qualifications do not affect the pricing formula.

A standard Brownian motion starts at zero, has continuous sample paths and independent increments, with $W_t-W_s\sim N(0,t-s)$ for $0\le s<t$.

Assume $S_0,K,T>0$ and $\sigma >0$; changing the sign of $\sigma$ just changes the sign of Brownian motion. Itô formula gives $d{S}_t=S_t[(\mu +{\sigma }^2/2)dt+\sigma d{W}_t]$. In the complete Black-Scholes model, the risk-neutral pricing theorem prices an integrable payoff $Y$ by $e^{-rT}{\mathbb{E}}_{Q}Y$, where the discounted asset is a martingale. Girsanov theorem therefore replaces the logarithmic drift by $r-{\sigma }^2/2$:Let $d_2=[\log (S_0/K)+(r-{\sigma }^2/2)T]/(\sigma \sqrt{T})$ and $d_1=d_2+\sigma \sqrt{T}$. The exercise event is $Z>-d_2$. Completing the square in the Gaussian integral yieldsConsequently ${\mathbb{P}}_Q(S_T>K)=\Phi (d_2)$ and ${\mathbb{E}}_Q[S_T{1}_{\{S_T>K\}}]=S_0{e}^{rT}\Phi (d_1)$. Thus the European call option price isThe physical drift $\mu$ disappears because the price is obtained under the martingale measure. If $\sigma =0$ in an arbitrage-free deterministic model, $S_t=S_0{e}^{rt}$ and the formula is interpreted by its zero-volatility limit.

Let $P=N(t)\sim \mathrm{Poisson}(x)$, where $x=\lambda t$. Deleting odd arrivals leaves $M=\lfloor P/2\rfloor$. Put $Q=1_{\{P\text{ odd}\}}$, so $M=(P-Q)/2$. From the probability generating function $\mathbb{E}{z}^P=e^{x(z-1)}$,Thus $\mathrm{Var}Q=(1-e^{-4x})/4$ and $\mathrm{Cov}(P,Q)=x{e}^{-2x}$. ThereforeTo establish strict underdispersion for every $x>0$, subtract the mean:Here $f(0)=0$ and $f^{\prime }(x)=4[-1+4x{e}^{-2x}+e^{-4x}]<0$ for $x>0$, because $4x{e}^{-2x}<1-e^{-4x}$ is equivalent to $2x<\sinh (2x)$. Hencewhich excludes a doubly-stochastic Poisson representation by part (iii).

Put $L_t={\int }_0^t\Lambda (u)du$, finite almost surely. Conditional on the intensity process, $N(t)$ is Poisson with mean and variance $L_t$. The law of total expectation and law of total variance giveConsequently for a Cox process,These identities apply when the moments are finite; if the mean is finite but the integrated intensity has infinite variance, the count variance is infinite and the inequality still holds.

To derive the distribution from the infinitesimal characterization, let $p_j(t)=\mathbb{P}(N(t)=j)$. Independent increments give the forward equationsWith $L(t)={\int }_0^t\lambda (u)du$ and $p_0(0)=1$, $p_j(0)=0$ for $j>0$, these have solutionSubstitution verifies each equation, and the probabilities sum to one. Alternatively the probability generating function solves ${\partial }_{t}G=\lambda (t)(z-1)G$, $G(z,0)=1$, giving $G=e^{(z-1)L(t)}$. For a general locally integrable rate, time-changing a unit-rate Poisson process by $L(t)$ gives the same formula.

An Inhomogeneous Poisson process with deterministic, nonnegative locally integrable rate $\lambda (t)$ is a counting process $N$ with $N(0)=0$, independent increments, and increment distributionEquivalently, for continuous rates it is a simple counting process with independent increments and infinitesimal probabilities $\mathbb{P}(N(t+h)-N(t)=1)=\lambda (t)h+o(h)$ and $\mathbb{P}(N(t+h)-N(t)\ge 2)=o(h)$. For merely locally integrable rates use the integrated-rate formulation, which avoids a pointwise regularity assumption.

Continue with ${\sigma }^2>0$. For each integer $K\ge 1$, let $B_m(K)={\bigcup }_{n\ge m}\{S_n/\sqrt{n}\ge K\}$. These events decrease with $m$, and their probabilities are at least $c_K>0$ by (iii). Continuity from above of a measure givesIt is important not to assume that this exact threshold event is automatically a tail event: a vanishing change of partial sums can affect infinitely many exact comparisons at the threshold. Instead useFor any fixed $m$, subtracting $S_m/\sqrt{n}\to 0$ shows that $L={\mathrm{limsup}}_{n\to \infty }(X_{m+1}+\cdots +X_n)/\sqrt{n}$. Thus $L$ is measurable with respect to every tail sigma-algebra. The event $\{L\ge K\}$ has positive probability by the preceding bound, so Kolmogorov zero-one law makes its probability one. Intersecting over all positive integers givesFor any specified real $K>0$, this conclusion implies infinitely many $n$ with $S_n/\sqrt{n}\ge K$, completing both requests. In the excluded zero-variance case the limsup is zero.

A necessary qualification is ${\sigma }^2>0$. As printed, the assumptions also permit $X_i=0$ almost surely; then $\mathbb{P}(A_n)=0$ for every $K>0$, so the asserted positive lower bound is false. Assume the intended nondegenerate case.

The central limit theorem states that independent identically distributed variables with mean zero and finite positive variance satisfy $S_n/(\sigma \sqrt{n})\Rightarrow Z$, where $Z\sim N(0,1)$. Since the normal distribution has no atom at $K/\sigma$,For example,

Let $D_n=B_n\setminus B_{n+1}$. These sets are pairwise disjoint andBy countable additivity, ${\sum }_n\mu (D_n)=\mu (B_1)-\mu (B)<\infty$. The tails of this convergent nonnegative series tend to zero, soThis proves continuity from above of a measure and explains the finite-measure assumption.

For independent random variables $X_1,X_2,\dots$, let ${\mathcal{F}}_n=\sigma (X_1,\dots ,X_n)$, $\mathcal{F}=\sigma (X_1,X_2,\dots )$, and let the tail sigma-algebra be $\mathcal{T}={\bigcap }_n\sigma (X_{n+1},X_{n+2},\dots )$. Kolmogorov zero-one law states that every $A\in \mathcal{T}$ has probability zero or one.

Such an event is independent of each ${\mathcal{F}}_n$ by independence of the variables. The collection of events $B\in \mathcal{F}$ satisfying $\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)$ is a Dynkin system: it contains the whole space and is closed under complements and disjoint countable unions. It contains the pi-system ${\bigcup }_n{\mathcal{F}}_n$, which generates $\mathcal{F}$. The pi-lambda theorem therefore makes $A$ independent of all of $\mathcal{F}$. Since $A$ itself belongs to $\mathcal{F}$,

Let $K=k(X)$. The space of Kähler differentials ${\Omega }_{K/k}^1$ is the $K$-vector space generated by symbols $df$, $f\in K$, with relationsIts elements are rational differentials. A local parameter $t$ is a nonconstant element of $K$, so $k(t)$ is a rational function field and $K/k(t)$ is finite. In characteristic zero this extension is separable.

For any $f\in K$ let $P(t,f)=0$ be its minimal polynomial over $k(t)$. Differentiating givesSeparability ensures $P_f(t,f)\ne 0$, so $df$ is a multiple of $dt$, and hence ${\Omega }_{K/k}^1=Kdt$. To show $dt\ne 0$, the derivation $d/dt$ on $k(t)$ extends uniquely to the separable extension $K$, by the same differentiated minimal-polynomial formula; it sends $t$ to $1$, so the universal differential $dt$ cannot vanish. ThusFor a nonzero differential $\omega =fdt$, define its valuation of a rational differential at $p$ byConcretely $\omega =t^{m}udt$ with $u$ a unit in the local ring, and this valuation is $m$. For $\omega =0$ one conventionally assigns $+\infty$, not an integer.

For a divisor $D={\sum }_p{n}_{p}p$ on the smooth projective curve,This is a $k$-vector space of rational functions with the prescribed pole bounds. A nonzero rational function on a projective curve has divisor of degree zero. Thus if $\mathrm{deg}D<0$ there can be no nonzero element of $L(D)$.

For any point $p$ with coefficient $n_p$ in $D$, use a local parameter $t$ at $p$. The linear map sending $f\in L(D)$ to the residue-field value of $t^{n_p}f$ has kernel $L(D-p)$ and target $k$, since $k$ is algebraically closed. Therefore $\dim L(D)/L(D-p)\le 1$. If $d=\mathrm{deg}D\ge 0$, successively subtract a point $d+1$ times. The final divisor has degree $-1$ and no nonzero sections, givingEquivalently the universally valid bound is $\dim L(D)\le \max (0,\mathrm{deg}D+1)$. The PDF states the unqualified bound $\dim L(D)\le \mathrm{deg}D+1$; literally this is false when $\mathrm{deg}D\le -2$, because its right side is negative. For example $D=-2p$ has $L(D)=0$.

Let $F:|K|\times [0,1]\to |L|$ be a homotopy from $f$ to $g$. The inverse images $F^{-1}({\mathrm{st}}_L(w))$ form a finite open cover of the compact product. By the Lebesgue number lemma there is $\delta >0$ such that every subset of diameter less than $\delta$ lies in one of these inverse images.

Choose a barycentric subdivision $K^{(n)}$ with all its open stars of diameter less than $\delta /3$, and times $0=t_0<\cdots <t_m=1$ with gaps less than $\delta /6$. For each vertex $v$ and each time $t_i$, the set(with the evident endpoint conventions) has diameter less than $\delta$ in the product metric. Choose a vertex $w_i(v)$ of $L$ whose open star contains its image under $F$, and set $h_i(v)=w_i(v)$.

At time $t_i$, the star condition makes $h_i$ a simplicial approximation to $F(\cdot ,t_i)$. Adjacent intervals $J_i,J_{i+1}$ overlap, so choose a time in their overlap and a point in the interior of any simplex $\sigma$ of $K^{(n)}$. Its image belongs to the stars of every $h_i(v)$ and $h_{i+1}(v)$ for $v\in \sigma$. As in (i), their vertices lie in one carrier simplex, proving contiguity. Thus the sequence starts with an approximation to $f$, ends with one to $g$, and has contiguous adjacent pairs. Relabeling its indices gives the requested $h_1,\dots ,h_m$.

Here ${\sim }_c$ denotes contiguity, with the endpoint maps approximating $f,g$ respectively.

For contiguous simplicial maps, $|g|(x)$ and $|h|(x)$ lie in a common simplex of $L$ for every $x$. Its convexity makesa well-defined map into $|L|$, using the ambient Euclidean realization. The formula is continuous globally and agrees across simplex faces. At $t=0,1$ it equals $|g|,|h|$, so it is the desired homotopy.

A simplicial approximation $g:K\to L$ to $f:|K|\to |L|$ is a vertex map satisfying the open-star conditionfor every vertex $v$. Here an open star consists of points whose barycentric coordinate at that vertex is positive. This condition ensures that the images of the vertices of every simplex span a simplex of $L$, and the vertex map extends linearly over simplices.

To triangulate $|K|\times |L|$, order the vertices of each complex consistently. For each product $[v_0,\dots ,v_p]\times [w_0,\dots ,w_q]$, use the simplices whose vertices $(v_i,w_j)$ follow monotone lattice paths from $(0,0)$ to $(p,q)$, taking a step in one coordinate at a time. These prism triangulations agree on faces and cover the product.

If $g,h$ both approximate $f$, choose $x$ in the relative interior of a simplex $\sigma$. Then $x$ belongs to the open star of every vertex $v$ of $\sigma$. Hence $f(x)$ belongs to every open star of $g(v)$ and $h(v)$. Its unique carrier simplex in $L$ contains all these vertices. Thus $g(\sigma )$ and $h(\sigma )$ are faces of that one simplex, proving that $g,h$ are contiguous simplicial maps.

Put $m=(p-1)/2$ and let $O=1\cdot 3\cdot 5\cdots (p-2)$. The even residues $2,4,\dots ,p-1$ are precisely the negatives, in reverse order, of the odd residues $1,3,\dots ,p-2$. Hence their product is $(-1{)}^{m}O$ in the finite field ${\mathbb{F}}_p$. Applying Wilson theorem givesThe initial factor $1^2$ can be omitted, soFor $p=3$ the product is empty and equals $1$, consistently with the formula.

Lagrange theorem for polynomial congruences says that a nonzero polynomial of degree $d$ over a field has at most $d$ distinct roots of a polynomial. It follows by repeatedly factoring out $x-a$ at a root. Over the finite field ${\mathbb{F}}_p$, the only solutions of $x^2=1$ are $1$ and $-1$: these are distinct because $p$ is odd, and the root bound excludes any others.

Every nonzero element has a unique multiplicative inverse. Pair each element of ${\mathbb{F}}_p^{\times }$ with its inverse. All pairs of distinct elements contribute $1$ to the product; the only unpaired elements are $1$ and $-1$. Thus Wilson theorem follows:If Lagrange's theorem is formulated for finite groups, it states that the order of a subgroup divides the order of the group; in particular every element of ${\mathbb{F}}_p^{\times }$ has order dividing $p-1$. The inverse-pairing proof above establishes the required congruence directly.

For $n=p$ prime, $[K:\mathbb{Q}]=p-1$ and the cyclic group $\mathrm{Gal}(F/K)$ has order either $1$ or $p$. ThereforeThese correspond respectively to $a$ being or not being a $p$th power in $K$. In fact it is a $p$th power in $K$ precisely when it is a $p$th power in $\mathbb{Q}$. To see the nontrivial implication, suppose $\alpha \in K$ and ${\alpha }^p=a$. Taking the field norm gives $b^p=a^{p-1}$ with $b=N_{K/\mathbb{Q}}(\alpha )\in {\mathbb{Q}}^{\times }$. Then $(a/b{)}^p=a$, so $a$ is already a rational $p$th power.

Both degrees occur: $a=1$ gives $p-1$, while $a=2$ is not a rational $p$th power and gives $p(p-1)$. This includes $p=2$, with possible degrees $1$ and $2$.