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For each $d$-element coordinate set $J\subseteq \{1,\dots ,p\}$, let ${\mathcal{G}}_J$ contain the lower orthants whose finite thresholds occur precisely in the coordinates in $J$. Ignoring the remaining coordinates identifies ${\mathcal{G}}_J$ with the lower-orthant class in ${\mathbb{R}}^d$, so part (a) gives $\mathrm{VC}({\mathcal{G}}_J)=d$. Moreover,is a union of $r=\left(\genfrac{}{}{0ex}{}{p}{d}\right)$ such classes. Part (c) and $\left(\genfrac{}{}{0ex}{}{p}{d}\right)\le p^d$ give

If $x_{1:n}$ is shattered by ${\bigcup }_{k=1}^r{\mathcal{F}}_k$, then all $2^n$ label vectors belong to the union of the label-vector sets generated by the individual classes. The shattering coefficient of a union and the Sauer-Shelah lemma therefore giveTaking base-two logarithms yieldsFor $r>1$, apply the stated elementary bound with $\alpha ={\log }_{2}r$ and $\beta =v\ge 3$:If $r=1$, the assertion follows directly from $n\le v$. Taking the largest shattered sample size proves

The Sauer-Shelah lemma states that if $\mathrm{VC}(\mathcal{F})=v<\infty$, thenIn particular, the Sauer-Shelah growth bound gives $s(\mathcal{F},n)\le (n+1{)}^v$.

A sample $x_{1:n}=(x_1,\dots ,x_n)$ is shattered by $\mathcal{F}$ when every binary labeling is realized:The shattering coefficient and VC dimension are

For the lower orthants in ${\mathbb{R}}^d$, the $d$ points $e_1,\dots ,e_d$ are shattered: to realize a subset $I$, choose $a_j=1$ for $j\in I$ and $a_j=1/2$ otherwise. Conversely, among any $d+1$ points choose, for each coordinate, one point attaining the maximum in that coordinate. At most $d$ points have been chosen, so there is an unchosen point $x$. Any lower orthant containing every chosen coordinate maximizer must contain every sample point, including $x$. It cannot realize the labeling that includes all chosen points and excludes $x$. This proves

Suppose the claim $H=(S_1{)}^2$, payable at time $1$, were replicable. Holdings chosen at time zero are constants, so the time-$1$ value of any self-financing portfolio has the affine formBut $S_1=S_0{Z}_1$ has a continuous positive density on $(0,\infty )$. No affine function of $S_1$ can equal the quadratic function $S_1^2$ almost surely. Therefore $H$ is not attainable and the market is not complete.

The same argument applies if $H$ is viewed as a time-$T$ claim: once $H$ is known at time $1$, its required discounted value at time $1$ is fixed, and the first-period portfolio value would still have to reproduce a nonaffine function of $S_1$.

There is no arbitrage. Indeed, part (iii) supplies an equivalent martingale measure, so the fundamental theorem of asset pricing implies absence of arbitrage. Directly, the discounted terminal value of a zero-cost admissible self-financing strategy is a ${\mathbb{P}}^{\ast }$-martingale with mean zero; it cannot be nonnegative almost surely and positive with positive probability.

Set $Y_n=\log Z_n$ andDefine a probability measure ${\mathbb{P}}^{\ast }$ by the positive densityThe density has expectation one and is strictly positive, so ${\mathbb{P}}^{\ast }$ is equivalent to $\mathbb{P}$. By exponential tilting, the variables $Y_k$ remain independent under ${\mathbb{P}}^{\ast }$ and have distribution $N({\mu }_{\ast },{\sigma }^2)$. Thus $Z_k$ has a log-normal distribution andFor the discounted stock ${\tilde{S}}_n=S_n/S_n^0$,Hence ${\mathbb{P}}^{\ast }$ is an equivalent martingale measure.

The model is a complete market when every contingent claim in the specified class, usually every ${\mathcal{F}}_T$-measurable payoff satisfying the required integrability condition, is the terminal value of some admissible self-financing portfolio.

Let $\varphi =({\varphi }_n^0,{\varphi }_n^1,\dots ,{\varphi }_n^d)$ be a predictable self-financing portfolio, and let $V_n^{\varphi }={\varphi }_n\cdot S_n$ be its value. An arbitrage is such a strategy with

Here the total holding rate isand therefore the jump chain has state-independent transitionsIt is an upward biased random walk on $\mathbb{Z}$. By the strong law of large numbers, $Y_n/n⟶1/3$ almost surely, so $Y_n\to +\infty$. The jump chain, and hence $X$, is transient.

Counting measure is invariant for this jump chain: for every $j$, its incoming mass is $1/3+2/3=1$. The invariant-measure transfer between a jump chain and a CTMC therefore gives an invariant measure proportional to $q_i^{-1}=3^{-|i|-1}$. Sincethe normalized invariant distribution of a continuous-time Markov chain isThis does not contradict transience because an invariant distribution of an explosive chain need not imply positive recurrence.

It remains to verify explosion. Starting from zero, the probability that the biased walk ever reaches $-k$ is $2^{-k}$. Once it reaches any state, its probability of returning after departure isso its expected total number of visits to that state, conditional on reaching it, is $1/(1-2/3)=3$. If $V_i$ is the number of visits to $i$, thenConsequently the expected sum of all holding times isThe nonnegative random variable $T_{\infty }$ is therefore finite almost surely. By explosion by summable holding times, $X$ makes infinitely many jumps by time $T_{\infty }$, so it is explosive.

Write $q_i=-g_{ii}$ for the total rate of leaving $i$. The jump chain of a continuous-time Markov chain $X$ is the discrete-time chain $Y=(Y_n)$ recording the states occupied after successive jumps. Its transition probabilities arewhenever $q_i>0$.

The chain $X$ is irreducible if every state can be reached from every other state with positive probability. A state $i$ is recurrent if, after leaving $i$, the process returns to $i$ with probability one; an irreducible chain is recurrent when one, and hence every, state is recurrent.

The continuous-time Markov chain and its jump chain have exactly the same successive states. If $Y$ is transient, it visits each state only finitely often, so $X$ is transient. If $Y$ is recurrent, it visits a fixed state $i$ infinitely often. At those visits, $X$ has independent exponential holding times of rate $q_i<\infty$. Their sum diverges almost surely, so explosion cannot occur before all those returns. Hence $X$ returns to $i$ infinitely often and is recurrent. Thus, for an irreducible chain,

The Caratheodory extension theorem states that a premeasure ${\mu }_0$ on an algebra of subsets of $X$ extends to a measure on the generated sigma-algebra. If ${\mu }_0$ is sigma-finite, this extension is unique. In particular, a finite premeasure has a unique extension.

A sigma-algebra $\mathcal{F}$ on $X$ contains $\varnothing$, is closed under complements in $X$, and is closed under countable unions. A measure on the measurable space $(X,\mathcal{F})$ is a functionsuch that $\mu (\varnothing )=0$ and, for every pairwise disjoint sequence $(A_n)$,A probability measure is a measure normalized by

LetThis is another unit-speed embedded curve. The assumed identities say that $\alpha$ and $\beta$ have the same positive curvature and the same torsion. By the uniqueness part of the fundamental theorem of regular space curves, there is a proper Euclidean motion $\varphi$ such thatSince $s\mapsto \pm s+{\sigma }_0$ is a bijection of $\mathbb{R}$, this motion preserves $\alpha (\mathbb{R})$.

It is nontrivial. For the plus sign, ${\sigma }_0\ne 0$ and $\varphi =\mathrm{id}$ would imply $\alpha (s)=\alpha (s+{\sigma }_0)$, contradicting embeddedness. For the minus sign, the identity would imply $\alpha (s)=\alpha (-s+{\sigma }_0)$ for every $s$, and injectivity would force $s=-s+{\sigma }_0$ for every $s$, which is impossible. Thus the required nontrivial proper Euclidean motion exists.

For $R>r>0$, the closed curveis embedded on a torus and satisfiesAfter arc-length parametrization, it meets all the requirements of part (b). Its bending and twisting vary around the two waves, so neither curvature nor torsion is constant.

The map $\alpha :\mathbb{R}\to \alpha (\mathbb{R})$ is a local arc-length coordinate and, when the image is closed, a covering parametrization. The restriction of $\varphi$ to the one-dimensional submanifold $\alpha (\mathbb{R})$ therefore lifts to a map $h:\mathbb{R}\to \mathbb{R}$ satisfyingBoth curves have unit speed and $\varphi$ is an isometry, so $|h^{\prime }(s)|=1$. Continuity makes the sign constant on the connected domain, hence $h^{\prime }(s)=\pm 1$ throughout andfor some constant ${\sigma }_0$.

A proper Euclidean motion of Euclidean three-space preserves both curvature and torsion. Reversal of an arc-length parameter also preserves them: in the determinant formula for torsion, both the first and third derivatives change sign. Comparing the two expressions for the transformed curve givesand, wherever $\kappa (s)>0$,

If the sign is plus and ${\sigma }_0=0$, then $\varphi$ fixes every point of the curve. A nonidentity proper Euclidean motion with any fixed points is a rotation, whose fixed set is its rotation axis. Thus $\alpha (\mathbb{R})$ lies on that axis and is a straight line.

The fundamental theorem of regular space curves states that smooth functions $\kappa (s)>0$ and $\tau (s)$ determine a unit-speed regular curve in ${\mathbb{R}}^3$ with that curvature and torsion, uniquely up to a proper Euclidean motion of Euclidean three-space. Existence follows by solving the Frenet-Serret formulas for an oriented orthonormal frame and then integrating its tangent vector.

Suppose an affine line lies in $W$. Parametrize it asThe defining equation would give the polynomial identityThe only units of $k[t]$ are nonzero constants. Since a product is a unit only when every factor is a unit, all three linear polynomials must be constant, so $b_1=b_2=b_3=0$. The alleged line is a point, a contradiction. Therefore