When $b=0$, put $D=e^{-2px+8p^{3}t-2x_0}>0$. Since $c=2ip{e}^{-2x_0}$, evaluation at $ip$ gives $N(ip)=1-DN(ip)$, henceDifferentiate the coefficient of $1/k$ with respect to $x$, using $D_x=-2pD$, and apply the reconstruction formula. The result isThe positive sign follows from the source's $-2i$ reconstruction convention. The soliton has amplitude $2p^2$, speed $4p^2$, and centre $x=4p^{2}t-x_0/p$.

Suppress $(x,t)$, put $R(k)=b(k)/a(k)$ and $C=c{e}^{-2px+8p^{3}t}N(ip)$. Removing the pole gives the additive jumpThe upper analytic function is $\mathcal{M}-1$ and the lower analytic function is $N(-k)-1$. The Cauchy integral formula and the Sokhotski–Plemelj formula solve this jump as the upper and lower boundary values of $(2\pi i{)}^{-1}{\int }_{\mathbb{R}}G(s)/(s-z)ds$. At the lower point $z=-k$, with $\Im k>0$, this gives $N(k)-1$.

The rational part can be evaluated by closing the contour upward: ${\int }_{\mathbb{R}}[(s-ip)(s+k){]}^{-1}ds=2\pi i/(k+ip)$. Thus the required linear integral equation isIt is supplemented by its value at $k=ip$, and follows under the usual decay or principal-value hypotheses of the stated Riemann-Hilbert problem.

A power option need not have bounded payoff, but all real powers of a positive lognormal stock have finite moments, so risk-neutral valuation still applies. The Gaussian moment-generating identity gives $\mathbb{E}[e^{n\sigma B_T}]=e^{n^2{\sigma }^{2}T/2}$. ThereforeFor $n=1$ this reduces to $S_0$, and for $n=0$ to the time-zero value $e^{-rT}$ of one unit paid at expiry.

With continuous dividend yield $D$, it is the discounted total gain from the stock, including dividends, that must be a martingale. The ex-dividend stock drift under the risk-neutral measure is therefore $r-D$. Replace $r-{\sigma }^2/2$ in the exponential representation by $r-D-{\sigma }^2/2$ and repeat the chain rule. The discount rate remains $r$, givingThe terminal payoff condition is unchanged.

Under the given risk-neutral measure, the option price is $V(t,S)=e^{-r(T-t)}\mathbb{E}[\varphi (S_T)\mid S_t=S]$. Thus $e^{-rt}V(t,S_t)$ is a martingale. This valuation also follows from delta hedging: cancel the Brownian exposure with $V_S$ units of stock, after which absence of arbitrage makes the locally riskless remainder earn $r$.

Set $f(t,b)=e^{-rt}V(t,S_0{e}^{\sigma b+(r-{\sigma }^2/2)t})$. The stated Brownian martingale identity and the chain rule giveThe drift vanishes because discounted option value is a martingale. Consequently the Black-Scholes equation isFor bounded measurable payoffs, the conditional-expectation solution is smooth for $t<T$; terminal values are interpreted in the usual limiting sense at continuity points, or in the bounded measurable pricing sense.

The proposal must connect the whole positive posterior support by moves with nonzero reverse probability. Use the full Metropolis–Hastings acceptance probability for asymmetric proposals; dropping the reverse proposal factor generally targets the wrong distribution. Compute likelihoods and acceptance ratios on a logarithmic scale to avoid overflow and underflow, and handle zero weights explicitly. A lazy step prevents periodicity when distributional convergence is needed.

Finite-chain convergence does not guarantee a useful run length: separated modes and narrow bottlenecks can make mixing extremely slow. Tune proposals to explore efficiently, inspect several starting points and chains, and assess autocorrelation and Monte Carlo uncertainty rather than treating correlated draws as independent observations. Discarding an initial segment can reduce initialization effects, but diagnostics do not prove that every posterior mode has been visited. Thinning is not required for correctness and often throws away useful information. The model and prior must themselves be specified correctly; an exact sampler from an incorrect posterior cannot repair the model.

Restrict to the positive-weight support. From a state $\theta$, propose ${\theta }^{\prime }$ according to $q(\theta ,{\theta }^{\prime })$ and accept with probabilityOtherwise stay at $\theta$. For distinct states, the accepted transition kernel satisfiesThus detailed balance holds, and summing over $\theta$ proves stationarity. Irreducibility of the accepted kernel makes this stationary distribution unique. Proposal irreducibility by itself is insufficient if proposed edges have zero reverse probability and are always rejected.

For convergence of the distribution from an arbitrary start, also require aperiodicity, which can always be supplied by making the chain lazy. In a finite irreducible aperiodic chain, some power $P^m$ has all entries positive. Its rows then share a fixed positive probability component, so repeated application contracts total variation distance geometrically. This proves that the chain's distribution approaches $\pi$.

The stronger fact needed for averages isOne can justify this directly by returns to a chosen state. Successive excursions are independent identically distributed cycles by the Strong Markov property. In a finite irreducible chain, return times have finite mean: from every state there is a path of uniformly bounded length and uniformly positive probability to the chosen state, giving a geometric tail bound. Apply the strong law of large numbers to cycle lengths and accumulated rewards. Their ratio converges; the mean visits per cycle, normalized by the mean cycle length, form a stationary distribution and hence equal $\pi$. The unfinished final cycle is negligible. This proves the displayed average law, even when the chain is periodic. Under these conditions, Metropolis–Hastings delivers the required posterior expectations.

In Bayesian inference, a finite parameter space has posterior probabilities $\pi (\theta )=w(\theta )/Z$, with $w(\theta )$ the product of the prior probability and likelihood and $Z={\sum }_{\theta }w(\theta )$ the evidence. Although each weight may be easy to evaluate, summing a large parameter space can be impractical. The Metropolis–Hastings algorithm constructs a Markov chain with posterior stationary distribution, allowing posterior expectations, probabilities and credible regions to be estimated from its trajectory. Its acceptance ratios cancel $Z$, so direct enumeration of the normalizing constant is unnecessary. Posterior sampling alone does not automatically provide the evidence for comparing models.

Condition on $X$. The normal characteristic function gives $\mathbb{E}(e^{iuXY}\mid X)=e^{-u^2{X}^2/2}$, whose Gaussian integral is $(1+u^2{)}^{-1/2}$. A second Gaussian integral gives $\mathbb{E}(e^{-iu{Z}^2})=(1+2iu{)}^{-1/2}$, taking the square-root branch continuous from one at $u=0$. Independence of $(X,Y)$ and $Z$ therefore gives

Independence gives ${\varphi }_{X-Y}(u)={\varphi }_X(u){\varphi }_X(-u)=|{\varphi }_X(u){|}^2=1$ for $|u|<\epsilon$. Taking real parts shows $\mathbb{E}[1-\cos (u(X-Y))]=0$. The integrand is nonnegative, so it vanishes almost surely for each such $u$.

Choose a countable sequence of positive $u_j\to 0$ in that interval, and intersect the corresponding probability-one events. On their intersection, a nonzero difference $D=X-Y$ would give $0<|u_{j}D|<2\pi$ for large $j$, contradicting $\cos (u_{j}D)=1$. Thus $X=Y$ almost surely, and part (a) proves $X$ is almost surely constant.

Choose a bounded strictly increasing function, for example $h(x)=\arctan x$. Independence and identical distribution imply $\mathrm{Var}(h(X)-h(Y))=2\mathrm{Var}(h(X))$. If $X=Y$ almost surely, the left side is zero. Therefore $h(X)$ is almost surely constant; injectivity of $h$ makes $X$ almost surely constant. Using a bounded function avoids any unassumed second moment of $X$.

If the Gauss map is a diffeomorphism onto $S^2$, the surface is compact and $dN$ is invertible everywhere. Its Gaussian curvature $K=\det (dN)$ is continuous and nowhere zero, so has constant sign on the connected surface. To determine that sign, maximize a height function $r\mapsto ⟨r,a⟩$. At a maximum, $a$ is parallel to the normal and the Hessian of the height function is semidefinite. Thus the second fundamental form is semidefinite, up to a sign. Its determinant is nonnegative and cannot be zero because $dN$ is invertible. Hence $K>0$ at that point and therefore everywhere.

The converse is false under the stated hypotheses. An open hemisphere of the unit sphere is a connected oriented surface with $K=1$, but its outward Gauss map has only that hemisphere as its image, and is not a diffeomorphism onto all of $S^2$. The missing compactness hypothesis matters.

The Gauss map sends each point of an oriented surface to its chosen unit normal $N(p)$. Both $T_{p}S$ and $T_{N(p)}{S}^2$ are the plane orthogonal to $N(p)$, so the derivative is an endomorphism of that plane. In local coordinates $r(u,v)$, differentiating $⟨N,r_j⟩=0$ gives $⟨N_i,r_j⟩=-⟨N,r_{ij}⟩$. Since $r_{ij}=r_{ji}$, this equals $⟨N_j,r_i⟩$. Bilinearity then gives $⟨dN(v),w⟩=⟨v,dN(w)⟩$ for all tangent vectors. The derivative of the Gauss map is self-adjoint.

The dual representation on $V^{\ast }$ is $(g\ell )(v)=\ell (g^{-1}v)$. The bilinear form $\epsilon (z,w)=\det (z,w)$ on ${\mathbb{C}}^2$ is nondegenerate and invariant under SU(2), because every group matrix has determinant one. Thus it gives an equivariant isomorphism ${\mathbb{C}}^2\to ({\mathbb{C}}^2{)}^{\ast }$. Taking the $n$th symmetric power gives an equivariant isomorphism between its $n$th symmetric power and its dual. Since homogeneous polynomials realize $V_n={\mathrm{Sym}}^n(({\mathbb{C}}^2{)}^{\ast })$, this proves

Use the Clebsch-Gordan decomposition for SU2, which states that $V_m\otimes V_n\cong {⨁}_{j=0}^{\min (m,n)}{V}_{m+n-2j}$. Here this givesTo determine symmetry, realize the tensor product as polynomials in two pairs of variables. A highest-weight vector in the $j$th summand can be chosen proportional to $(x_1{y}_2-y_1{x}_2{)}^j{x}_1^{3-j}{x}_2^{3-j}$. Exchanging the two factors multiplies it by $(-1{)}^j$, and hence multiplies the entire irreducible summand by that sign. Even $j$ belong to the symmetric square, odd $j$ to the exterior square. ThusTheir dimensions are respectively $7+3=10$ and $5+1=6$, as required for a four-dimensional space.

For $z=(x,y{)}^T$, define $(\rho (g)f)(z)=f(g^{-1}z)$. This preserves the space of degree-$n$ homogeneous polynomials and satisfies $\rho (g_1)\rho (g_2)=\rho (g_1{g}_2)$. The diagonal torus of SU(2) acts on the monomials $x^{n-j}{y}^j$ with distinct weights $2j-n$.

If a nonzero subspace $W$ is invariant, averaging against each torus character projects $W$ onto the individual weight spaces. Therefore $W$ contains at least one monomial. Invariance also implies invariance under differentiated Lie-algebra operators and their complex linear combinations. The complexified algebra contains the operators $x{\partial }_y$ and $y{\partial }_x$, up to harmless signs. They carry each monomial to its adjacent monomial with a nonzero coefficient unless already at the appropriate endpoint. Repeated application produces every monomial, so $W=V_n$. The representation is irreducible.

For each positive integer $n$, the finite ordered interval $[-n,n]\cap \mathbb{Z}$ has a greatest and a least element, and these intervals form a chain of substructures. Their union is $\mathbb{Z}$, which has neither. An inductive first-order theory must be preserved by such unions, as proved above. The theory with both endpoints cannot be axiomatized inductively, even though it has an ordinary first-order axiomatization with existential–universal endpoint sentences.

The reflexivity, antisymmetry, transitivity and totality axioms for a total order are universal with quantifier-free formula matrices, and hence inductive with an empty existential block. Add $\forall x\exists y(y<x)$ and $\forall x\exists z(x<z)$, with strict inequalities expressed using $\le$ and inequality. These exclude least and greatest elements. This theory is inductive.

For fixed $\theta$, the Euler-Lagrange equation for the meridional coordinate requires ${\dot{\varphi }}^2\cos \theta =g/\ell$. Combining this with $v=\ell \sin \theta \dot{\varphi }$ and $\cos \theta =D/\ell$ givesThe resulting motion is a conical pendulum. The positive-depth condition is necessary for a nondegenerate constant-latitude orbit in gravity; the bottom point is the limiting zero-speed equilibrium.

Grazing the equator makes it a turning point, so $\dot{\theta }=0$ at $\theta =\pi /2$. Conservation of energy gives $\frac{1}{2}m{v}^2-mgD=h^2/(2m{\ell }^2)=\frac{1}{2}m{v}^2(1-D^2/{\ell }^2)$. ThereforeTo check that no earlier turning point prevents reaching the equator, put $c=\cos \theta$. On $0\le c\le D/\ell$, $U(c)=h^2/[2m{\ell }^2(1-c^2)]-mg\ell c$ is strictly convex. Its two endpoint values are equal for this speed, so its interior values are smaller. The trajectory can therefore move between the launch point and the equator. The equator is genuinely a turning point since $dU/d\theta =mg\ell$ there.