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Let the active block have size $m$. Using a stored skew triangle, the product $Dv$ costs $m(m-1)$ scalar multiplications: two for each unordered off-diagonal pair. Updating its stored triangle with $v{w}^T-w{v}^T$ costs another $m(m-1)$ products. Reflector construction, scalings and the active column take only $O(m)$ additional products. Summing over $m=n-1,\dots ,2$ gives

For a symmetric block, an optimized matrix-vector calculation costs $m^2$ products, and the symmetric rank-two update of one triangle costs $m(m+1)$ products. The additional scalar $v^{T}Dv$ and its correction to the update vector cost only $O(m)$ per step. ThusThe leading terms are the same under the requested product-count convention. Skew symmetry removes the diagonal work and the scalar correction, giving lower-order savings, not a factor-of-two leading advantage over an already symmetry-aware implementation. Counting additions as well would change the reported flop constants; it is not the convention requested here.

At step $k=1,\dots ,n-2$, isolate the active column tail $a=A_{k+1:n,k}$. If its subtail is already zero, no reflection is needed. Otherwise choose a Householder reflection $H=I-2v{v}^T/(v^{T}v)$ on the active coordinates such that $Ha$ has only its first component nonzero. A cancellation-resistant choice is $v=a+\mathrm{sgn}(a_1)\Vert a\Vert e_1$, with an arbitrary positive sign if $a_1=0$. Embed $H$ as $P=\mathrm{diag}(I_k,H)$ and replace $A$ by $P^{T}AP$. Previous reduced columns are untouched, and column $k$ now has no entries below the first subdiagonal.

Similarity by an orthogonal matrix preserves skew symmetry. Consequently an upper-Hessenberg result also has zeros above the first superdiagonal, so it is tridiagonal with zero diagonal and opposite-sign corresponding off-diagonal entries.

For the active skew-symmetric block $D$, compute $w=2Dv/(v^{T}v)$. Because $v^{T}Dv=0$, expanding the two-sided reflection simplifies toStore and update just one triangle, obtain the other by negation, and keep the diagonal exactly zero. Matrix-vector multiplication can likewise use each stored entry once to contribute to two output coordinates. This is skew-symmetric Householder tridiagonalization; there is no need for two dense matrix-matrix multiplications.

An elementary geometric-series construction suffices, without assuming a general polynomial-approximation theorem. Rotate the square by a multiple of $\pi /2$ so that $\Re w>1$. Put $c=-R$ with $R>0$. Every point of the square then satisfies $|z-c{|}^2\le (R+1{)}^2+1$, whereasfor sufficiently large $R$. Thus the whole square lies inside a disk centred at $c$ whose radius is less than $|w-c|$. The other three ways $w$ can lie outside the square are handled by the indicated rotations, after which rotating back preserves polynomial form.

Now expandLet $\rho ={\max }_{z\in K}|z-c|/|w-c|<1$. The tail after degree $m$ is uniformly bounded by ${\rho }^{m+1}/[|w-c|(1-\rho )]$, tending to zero. Hence the displayed partial sums are the required uniform polynomial approximations. This is the Taylor series about a carefully chosen distant centre, rather than a potentially inadequate series about $0$.

The complex Taylor theorem says that an analytic function has the expansionthroughout every disk centred at $z_0$ contained in $D$. In particular it converges for $|z-z_0|<\mathrm{dist}(z_0,\mathbb{C}\setminus D)$, and uniformly on every smaller closed disk. The coefficients are uniquely its derivatives divided by factorials. The convergence radius can be larger if the function extends beyond the boundary of $D$; the theorem does not say that the given domain determines its exact radius.

Assume a positive barrier $B\le K<S_0$. Under the zero-rate risk-neutral Black-Scholes model, $X_t=\log (S_t/B)$ starts at $x=\log (S_0/B)>0$ and is Brownian motion with variance rate ${\sigma }^2$ and drift $\mu =-{\sigma }^2/2$. The drifted transition density killed at zero isTo see the factor, multiply the zero-drift reflected Gaussian difference by $\exp [\mu (y-x)/{\sigma }^2-{\mu }^{2}t/(2{\sigma }^2)]$; its second term is the ordinary drifted Gaussian from starting point $-x$ times $e^{-2\mu x/{\sigma }^2}$.

Integrate the payoff $(B{e}^y-K{)}_{+}$ against this density. The first term is $C(S_0,K)$. Since $K\ge B$, the payoff is zero for $y\le 0$, so the second integral may be extended to all real $y$ and equals the ordinary call price from reflected spot $B{e}^{-x}=B^2/S_0$. The reflection weight is $e^x=S_0/B$. Thus the down-and-out European call value isThis assumes continuous monitoring and knock-out on first hitting, with no rebate.

At zero interest, interchanging spot and strike gives $d_1(K,S_0)=-d_2(S_0,K)$ and $d_2(K,S_0)=-d_1(S_0,K)$. Substitution into the put formula therefore givesScaling both positive arguments by $\alpha$ leaves $\log (S_0/K)$ and both $d$ values unchanged, while multiplying every price prefactor by $\alpha$. ThusThese are Put-call symmetry in the Black-Scholes model and degree-one homogeneity.

For positive spot, strike, volatility and maturity, the Black-Scholes formula iswhere $d_1=[\log (S_0/K)+(r+{\sigma }^2/2)T]/(\sigma \sqrt{T})$ and $d_2=d_1-\sigma \sqrt{T}$. Degenerate zero-volatility or zero-maturity cases are obtained by limits.

Assume $\alpha ,\beta ,\kappa >0$, with the given gamma using shape and rate. PutThe joint posterior density before normalization isComplete each square and integrate over all ${\theta }_i$. The Gaussian integral contributes ${\tau }^{-n/2}$, givingThis hierarchical form completely specifies the normal-gamma distribution of the posterior. For comparison, $\tau \mid \theta ,X$ has shape $\alpha +n$ and rate $\beta +\frac{1}{2}{\sum }_i[{\theta }_i^2+\kappa (X_i-{\theta }_i{)}^2]$; confusing this conditional shape with the marginal shape would lose the Gaussian-integration factor.

Marginally $\theta$ has a multivariate Student distribution with $2A=2\alpha +n$ degrees of freedom and scale matrix $B/[A(1+\kappa )]I_n$. The coordinates are conditionally independent given $\tau$, but are generally dependent after integrating their shared precision. This is conjugate shrinkage with uncertainty in the overall noise scale.

Complete the square in the independent normal prior and likelihood:Thus, conditional on the full observations, the hospital parameters remain independent andIf $j$ is the observed minimum index and $X_j=x$, its conditional distribution is exactly this formula with $X_j=x$. No extra truncation is introduced because the selection index is already a function of the conditioned data. Its posterior mean is shrunk toward zero, so the worst measured performance is not simply identified with the worst true performance.

For level $k$, let $H_k$ be the horizontal generator: neighboring horizontal sites exchange at rate $1/6$, with reflecting endpoints. It is symmetric and has zero row sums, so the uniform row vector $u_k=(1/(k+1),\dots ,1/(k+1))$ satisfies $u_k{H}_k=0$. Removing paths as soon as they jump vertically gives the killed horizontal generator $H_k-(2/3)I$. ThereforeConditional on any elapsed waiting time without a vertical jump, the horizontal position remains uniform. This proves the exit-position uniformity as well as independence of that position from the vertical holding time.

Now consider the upward rate block $U_k$ from level $k$ to $k+1$. For $k>0$ each of its $k+2$ destination columns has sum $1/3$: an endpoint receives its single doubled edge jump, and each other site receives two jumps of rate $1/6$. HenceLikewise every destination column of the downward block $D_k$ has sum $1/3$, so $(u_k{D}_k{)}_j={\mu }_k/k$. Thus, conditional on either next vertical level, the arrival position is uniform on that new level. At the corner the two upward rates are equal, giving uniform arrival on level one directly.

Starting with the single site at level zero, induction now proves the uniform conditional law for a triangular-wedge walk: conditional on every attainable past-level sequence, arrival is uniform; the killed semigroup makes departure uniform; and the rate-block column sums make arrival at the following level uniform again. This proves both equalities of the requested property and justifies the averaging in part (b). The statement does not condition departure on the future level, which would in general bias the exit positions.

For $k>0$, an interior horizontal position has total upward and downward rates $1/3$ each. At either edge the upward rate is $1/3+1/6=1/2$ and the downward rate is $1/6$. Thus every position has total vertical-jump rate $2/3$. Averaging over the uniform conditional horizontal law givesAt zero there are two upward jumps of rate $1/3$, so ${\lambda }_0=2/3$ and ${\mu }_0=0$, agreeing with these formulas. Dividing by $2/3$ yields the two embedded transition probabilities in part (a).

The constant vertical hazard makes every vertical holding time exponential with rate $2/3$. Conditional uniformity at a new level, together with preservation of uniformity during horizontal motion, makes the next direction depend only on that level. Consequently the vertical process has the Markov property in its own history and the generator in part (a). Part (c) proves the preservation assertion and the required conditioning on past levels, rather than assuming strong lumpability for every possible horizontal starting point.

The embedded birth-death chain has upward probability $p_k=(k+2)/(2(k+1))$ and downward probability $q_k=k/(2(k+1))$. To calculate a return probability, solve the harmonic difference equation on $\{0,\dots ,m\}$ with boundary values one at zero and zero at $m$. Successive differences have ratio $q_k/p_k=k/(k+2)$, so their weights areTheir total infinite sum is $2$. Letting $m\to \infty$ gives the probability of ever hitting zero from $k$:In particular the chain leaves zero to one and returns with probability $1/2<1$, so it is transient. This finite-interval calculation selects the minimal nonnegative hitting-probability solution, rather than merely observing that $h_k$ is harmonic.

The continuous-time chain has the same embedded transitions and constant total exit rate $2/3$, so it is nonexplosive and has exactly the same return events. It too is transient, neither positive recurrent nor null recurrent.

Let $\zeta =u+iv$. Direct differentiation givesThese are holomorphic polynomials and their squared sum isMoreover direct calculation of the metric gives $E=G=(1+u^2+v^2{)}^2>0$ and $F=0$, so the map is genuinely an immersion everywhere. The preceding characterizations prove that it is an isothermal parametrization of a minimal surface, the Enneper surface.

The null condition follows by expansion:It vanishes exactly when $E=G$ and $F=0$; the immersion hypothesis ensures their common value is positive. Thus it characterizes isothermal coordinates for an immersion.

With complex coordinate $\zeta =u+iv$, mixed derivatives commute andand similarly for $y,z$. By the Cauchy-Riemann equations, each ${\phi }_j$ is holomorphic exactly when the corresponding coordinate is harmonic. Combining with the preceding part gives an isothermal immersion is minimal exactly when all three ${\phi }_j$ are holomorphic. Without immersion the null condition could also describe a degenerate map, so the nondegeneracy hypothesis matters.