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Let $J$ be rotation by ninety degrees in the oriented tangent plane, $Jv=N\times v$. In local coordinates seek a curve withThis smooth second-order ordinary differential equation, written as a first-order system for $(x,\dot{x})$, has a unique local solution with the specified initial position and velocity. Its speed is constant, sinceThe initial speed is one, so the parameter is arc-length and the signed geodesic curvature is $⟨{\nabla }_s\dot{\gamma },J\dot{\gamma }⟩=\lambda$. Conversely every requested curve satisfies this system, which proves the claimed uniqueness as well as existence on some $(-ϵ,ϵ)$.

The contraction mapping theorem says that if $(X,d)$ is a nonempty complete metric space and $T:X\to X$ satisfies $d(Tx,Ty)\le qd(x,y)$ for a constant $0\le q<1$, then $T$ has exactly one fixed point.

For the proof, start at $x_0\in X$ and set $x_n=T^n{x}_0$. Iterating the contraction bound gives $d(x_{n+1},x_n)\le q^{n}d(x_1,x_0)$. The triangle inequality then gives, for $m>n$,Thus $(x_n)$ is a Cauchy sequence and has a limit $x_{\ast }$ by completeness. Since the contraction is continuous, $T{x}_{\ast }=\lim T{x}_n=\lim x_{n+1}=x_{\ast }$. If $y_{\ast }$ is another fixed point, then $d(x_{\ast },y_{\ast })\le qd(x_{\ast },y_{\ast })$, forcing distance zero. This proves existence and uniqueness.

For the integral map, the fundamental theorem of calculus shows that $Tf$ is continuous whenever $f$ is continuous. In the supremum norm,Hence $\Vert Tf-Tg{\Vert }_{\infty }\le \frac{3}{2}\max (a^2,b^2)\Vert f-g{\Vert }_{\infty }$. We can chooseThe space $C[a,b]$ with this supremum norm is complete: a uniformly Cauchy sequence has pointwise real limits, the Cauchy bound passes to those limits to give uniform convergence, and a uniform limit of continuous functions is continuous. The contraction mapping theorem therefore gives a unique continuous $y$ with $y(x)=1+{\int }_0^{x}3ty(t)dt$. The fundamental theorem of calculus makes this $y$ differentiable, with $y^{\prime }=3xy$ and $y(0)=1$. Conversely every solution of that initial-value problem obeys the integral equation, so uniqueness transfers to the ordinary differential equation. Explicitly $y(x)=e^{3x^2/2}$ on $[-1/2,1/2]$, as direct differentiation confirms.

Write $F=\mathcal{L}f$. Using the Laplace transform of a derivative and the initial data givesThe transformed equation is $(s^4+2s^3+s^2)F=1$, so $F=1/[s^2(s+1{)}^2]$. The preceding inverse Laplace transform therefore givesFor a direct check, $f^{\prime }=1-(t+1)e^{-t}$, $f^{\prime \prime }=t{e}^{-t}$, $f^{(3)}=(1-t)e^{-t}$, and $f^{(4)}=(t-2)e^{-t}$. These satisfy both the ordinary differential equation and all four initial conditions.

Work on a finite interval $[0,T]$ with a fixed initial point $y_0\in {\mathbb{R}}^m$. To give a clean global statement, assume $V_0,V_1,\dots ,V_d$ are bounded vector fields with bounded continuous derivatives through order three. Consider the Stratonovich integral form of the stochastic differential equationFor a control in the Cameron-Martin space of Wiener measure,let $y^h$ solve the controlled ordinary differential equationThe Stroock-Varadhan support theorem says, in the uniform metric on continuous paths,Thus a path is in the support of a measure of the diffusion precisely when it can be approximated by these deterministic controlled trajectories. There is no ellipticity assumption: the allowed controls and vector fields already encode any restrictions on reachability.

The central point of the rough-path proof is that ordinary uniform approximation of the driver does not make the solution map continuous. Small planar loops can have small amplitude but accumulated area of fixed size, and vector fields with a nonzero Lie bracket detect that area. The correct driver is enhanced Brownian motion $\mathbf{B}$, including Lévy area. Fix $2<p<3$. The universal limit theorem gives a continuous mapfrom the geometric $p$-variation rough-path space to solution paths. It agrees with the controlled ordinary differential equation when $\mathbf{X}=S_2(h)$. Adding the time coordinate, with its mixed integrals defined by integration against $dt$ and integration by parts, incorporates $V_0$ without changing the continuity argument. This is the role of the universal limit theorem: it turns a support problem for all these nonlinear diffusions into a support problem for one enhanced driving process.

For $\mathbf{X}=\mathbf{B}$, the rough differential equation is the stated Stratonovich equation. Its second-order increment includesReplacing the Stratonovich integral in the second level by its Itô integral expression contributes $\frac{1}{2}{\sum }_{i}D{V}_i{V}_{i}dt$, which is exactly the Stratonovich-to-Itô correction. Polygonal convergence from Question 3 and the universal limit theorem therefore give the Wong-Zakai approximation: the ordinary equations driven by $B^n$ converge to this Stratonovich diffusion.

It remains to establish the support of enhanced Brownian motion:One inclusion follows immediately from Question 3. Every polygonal $B^n$ belongs to $H$, and $S_2(B^n)\to \mathbf{B}$ almost surely in the rough path metric. Since $K$ is closed, $\mathbf{B}\in K$ almost surely, whence its support is contained in $K$. This alone does not prove the reverse inclusion: convergence of random approximations does not automatically give positive mass near every control.

For that inclusion, take dyadic mesh ${\delta }_n=T{2}^{-n}$ and writeOn each grid interval $R^n$ is a pinned Brownian bridge. The bridges are mutually independent and independent of the grid values defining $B^n$, by the Gaussian projection formula. Their canonical Stratonovich enhancement ${\mathbf{R}}^n$ is therefore independent of $B^n$. It is important to enhance the bridge residual itself, including its area, rather than infer area smallness from its uniform size.

The bridge enhancement tends to the zero rough path in probability. Here are the estimates behind this assertion. A bridge on an interval of length ${\delta }_n$ scales as $\sqrt{{\delta }_n}$ times a unit bridge; its full antisymmetric area has mean zero, moments of order ${\delta }_n$, and second moment of order ${\delta }_n^2$. Since its endpoint displacement is zero, at grid points the cumulative second level is just the sum of the independent bridge areas. The Doob L2 maximal inequality givesPartial intervals at the two ends contribute bridge increments and at most two local areas. Bridge scaling, Gaussian moment bounds and the corresponding iterated-integral estimates yield, uniformly in $n$ and for arbitrarily large $q$,For an interval crossing several grid cells, the full-cell second levels are independent centered areas, whose sum has moments bounded by $C_q((t-s){\delta }_n{)}^{q/2}$; the two boundary pieces and their cross product satisfy the displayed bound as well. For an interval inside one cell these are the usual scaled bridge increment and area bounds. At each fixed pair $(s,t)$ both levels converge to zero in probability: the first-level variance is $O({\delta }_n)$, and the second-level second moment is $O((t-s){\delta }_n+{\delta }_n^2)$ for sufficiently fine grids. The Kolmogorov continuity theorem gives uniformly bounded moments of a Hölder rough-path norm at any exponent $\beta <1/2$. Tightness, convergence on a dense set of pairs and interpolation to $\alpha <\beta$ now give ${\rho }_{\alpha }({\mathbf{R}}^n,e)\to 0$ in probability. Taking $1/p<\alpha <1/2$ gives $d_p({\mathbf{R}}^n,e)\to 0$ in probability.

Fix $h\in H$. Its polygonal interpolations $h^n$ converge to $h$ in the total-variation norm: their derivatives are the cell averages of $\dot{h}$, which converge in $L^2$, hence in $L^1$. Moreover $|h^n{|}_{1\text{-var}}\le |h{|}_{1\text{-var}}$. Thus $S_2(h^n)\to S_2(h)$ in the rough path metric.

The rough path translation operation satisfiesFor a bounded variation path $b$, translation adds $b$ at level one and, at level two, the mixed termsThe second mixed term is defined by integration by parts. Their variation bounds show that translation is continuous, uniformly when $|b{|}_{1\text{-var}}$ is bounded, near the zero rough path. More explicitly the mixed second-level norms are bounded by a constant times $\Vert R{\Vert }_{p\text{-var}}|b{|}_{1\text{-var}}$; the other new second-level term is the path signature of $b$.

Given a ball of radius $\epsilon$ around $S_2(h)$, choose a sufficiently large $n$ so that $S_2(h^n)$ is close andfor the small $\eta$ supplied by translation continuity. The relevant variation bounds can be chosen uniformly in $n$, since the variations of the $h^n$ are uniformly bounded. For this fixed $n$, the event $|B^n-h^n{|}_{1\text{-var}}<\eta$ has positive probability: the finite vector of Brownian increments has an everywhere positive Gaussian density, and sufficiently small deviations of its vertices imply this variation bound. These two events are independent. On their intersection, translation continuity gives $d_p(\mathbf{B},S_2(h))<\epsilon$. Every ball around $S_2(h)$ therefore has positive probability. This proves $S_2(h)$ lies in the support, and taking closures gives $K$.

Finally, a continuous map between the separable path spaces in use satisfiesFor one direction the measure is concentrated on its support. For the other, an open neighborhood of $\Phi (x)$ pulls back to an open neighborhood of a support point $x$, which has positive measure. Using continuity of $\Phi$ and the identity $\Phi (S_2(h))=y^h$ proves the boxed Stroock-Varadhan support theorem. This separates the two ingredients: the probabilistic support calculation for Brownian motion and area, and the deterministic continuity supplied by the universal limit theorem.

If the diffusion is instead presented as an Itô integral equation with drift $b$, the controlled equation must useUsing $b$ unchanged in the controlled equation would generally give the wrong support description. The result above states explicit sufficient smoothness and nonexplosion assumptions; extensions to more general coefficients require the usual localization, not an unqualified assertion for arbitrary vector fields.

A compact H-hull is a bounded relatively closed subset $K$ of $\mathbb{H}$ with simply connected complement, considered together with its compact closure in $\overline{\mathbb{H}}$. Its unique mapping-out function has hydrodynamic normalization at infinityThe nonnegative coefficient $a_K$ is its half-plane capacity. Thus the half-plane-capacity parameterization here gives $g_t(z)=z+2t/z+O(z^{-2})$. Strict increase means $K_s⊊K_t$ whenever $s<t$; the strictly increasing capacity already excludes equality.

The Loewner local growth property means that the new growth becomes small after removing the old compact H-hull. More precisely set $K_{t,t+h}=\overline{g_t(K_{t+h}\setminus K_t)}$, taking the closure from the complex upper half-plane. On each bounded time interval, the diameters of these image compact H-hulls tend uniformly to zero as $h\downarrow 0$. Equivalently the new growth can be separated from infinity by crosscuts of vanishing diameter in the mapped domain. It is this single-point local growth, together with capacity continuity, that gives a continuous real Loewner driver.

The Loewner transform ${\xi }_t$ is the limiting point at which this mapped growth occurs:The limit is independent of the chosen new-growth points. The Loewner correspondence theorem says that the resulting mapping-out functions satisfyConversely, given a continuous real $\xi$, solve this ordinary differential equation until the maximal lifetime $T_z$ of each $z\in \mathbb{H}$. The surviving set $H_t=\{z:T_z>t\}$ is mapped conformally onto $\mathbb{H}$ by $g_t$; its complement, with the appropriate boundary closure, is $K_t$. The Laurent coefficient is $2t$, and the composition law reconstructs the local growth. A continuous Loewner driver always determines the compact H-hulls; generation by a continuous Loewner trace is an additional property, not a conclusion for every continuous Loewner driver.

A chordal SLE Loewner trace in the complex upper half-plane with parameter $\kappa \ge 0$ is the continuous curve associated with the Loewner driver ${\xi }_t=\sqrt{\kappa }{B}_t$, where $B$ is standard real Brownian motion started at zero. It starts at zero, its compact H-hull at time $t$ is the curve's past with all pockets cut off from infinity filled, andThe capacity clock has $\mathrm{hcap}(K_t)=2t$. For $\kappa =0$ this gives the vertical slit $\gamma (t)=2i\sqrt{t}$.

To prove Scaling invariance of SLE, fix $r>0$ and putThe Laurent series of ${\hat{g}}_t$ has coefficient $2t$, and direct differentiation givesBrownian scaling says $(r^{-1}{B}_{r^{2}t}{)}_{t\ge 0}$ has the same law as $B$. The Loewner drivers therefore agree in law, and uniqueness of the Chordal Loewner equation gives equality in law of the maps, compact H-hulls, and continuous Loewner traces:This identifies the time change $t\mapsto r^{2}t$ as well as the spatial scaling.

For a simply connected Jordan domain $D$ with distinct marked boundary points, choose a conformal bijection $f:\mathbb{H}\to D$ taking zero to the initial point and infinity to the target, and use $f(\gamma )$ as an unparameterized curve. The Caratheodory boundary extension theorem supplies the boundary values of $f$. Any second such map is $f\circ (z\mapsto rz)$ for $r>0$, because the automorphisms of $\mathbb{H}$ fixing zero and infinity are positive dilations. The Scaling invariance of SLE just proved makes the two image laws identical, up to the corresponding deterministic change of capacity clock. Thus this defines SLE consistently in marked Jordan domains. The assumed $|\gamma (t)|\to \infty$ and the continuous boundary extension give $f(\gamma (t))\to f(\infty )$, the specified target. The law is conformally natural and does not impose a distinguished half-plane-capacity parameterization on an arbitrary domain.

Write $A=\sigma {\sigma }^T$ and let $V(x,t)$ denote the optimal remaining expected reward from positive portfolio wealth $x$ at time $t$. The Hamilton-Jacobi-Bellman equation and terminal condition areThe controls are progressively measurable, the investment integrals exist, consumption is nonnegative and locally integrable, and the resulting portfolio wealth stays in the utility domain. At zero consumption use the right limit of the utility function if needed. This is the usual admissible trading strategy formulation; without restrictions preserving the domain the reward need not be defined.

For the verification bound, assume $V$ is a nonnegative classical solution, continuous through the terminal time, with one time derivative and two wealth derivatives in the interior. Fix any admissible controls and putThe Itô formula givesThe drift is nonpositive because the actual controls cannot exceed the supremum in the Hamilton-Jacobi-Bellman equation. Localize by stopping times on which the stochastic integral is a true martingale, the derivatives and wealth stay in their interior domain, and the accumulated reward is finite. For each such stop ${\tau }_n$,Both utilities and $V$ are nonnegative. Thus the Fatou lemma, continuity, and the terminal condition giveFor admissible paths strictly positive on the closed horizon, the domain-localizing stops eventually reach $T$. If zero wealth is allowed, the same proof uses the continuous boundary extension of $V$ and the utilities there. Nonnegativity is what permits removing the stops without assuming the uncontrolled stochastic term already has zero expectation. This is verification by a nonnegative control supermartingale.

The square-root case needs an additional no-arbitrage qualification because the printed volatility matrix need not be invertible. If there is $v\in \ker {\sigma }^T$ with $v^T\mu >0$, investment $\theta =Kv$ has zero volatility and arbitrarily large positive drift. With any fixed positive consumption, sufficiently large $K$ preserves positive wealth and makes terminal square-root utility tend to infinity. For example, in one dimension $\sigma =0$, $\mu =1$, $\theta =K$, $C=1$ gives $X_s=x+(K-1)(s-t)$, so the value is infinite when $T>t$. A finite separated solution does not exist for all matrices allowed by the printed hypotheses.

The needed condition is $\mu \in \mathrm{Range}A$, equivalent to $\mu$ being orthogonal to $\ker {\sigma }^T$, because $A$ is a symmetric matrix and $\ker A=\ker {\sigma }^T$. Assume this condition and let $B=A^{+}$ be its Moore-Penrose inverse; if $A$ is invertible, simply use $B=A^{-1}$. Put $q={\mu }^{T}B\mu \ge 0$. Seek the square-root investment-consumption value in the formThen $V_x=g/\sqrt{x}$ and $V_{xx}=-g/(2x^{3/2})$. Completing the quadratic investment expression, using $AB\mu =\mu$, shows that its maximum is attained at ${\theta }^{\ast }=2xB\mu$ and has value $qg\sqrt{x}$:Any investment component in $\ker A$ has no effect. For consumption, set $z=\sqrt{c}$. Its contribution is the concave quadratic $2z-(g/\sqrt{x})z^2$, whose maximum is $\sqrt{x}/g$ at $c^{\ast }=x/g^2$. Substituting both maxima into the Hamilton-Jacobi-Bellman equation givesWith $h=g^2$, this becomes $h^{\prime }+qh+1=0$, $h(T)=1$. Solving this linear ordinary differential equation givesIn particular $h\ge 1$, so $g=\sqrt{h}$ stays positive, and the two controls areThese controls are admissible and actually attain the value. Indeed their portfolio wealth satisfiesSince $\Vert 2{\mu }^{T}B\sigma {\Vert }^2=4q$, the explicit positive solution starting at $(x,t)$ isIts finite-horizon moments of every positive order are finite. More directly,Differentiating $g(s)\mathbb{E}\sqrt{X_s}$ and using the equation for $g$ gives $\frac{d}{ds}(g(s)\mathbb{E}\sqrt{X_s})=-\mathbb{E}\sqrt{X_s}/g(s)$. Integrating from $t$ to $T$ therefore yieldsThis proves attainment without relying on an unjustified expectation of a local martingale. When $\mu$ is outside $\mathrm{Range}A$, the singular-volatility drift arbitrage counterexample applies instead.

For a regular two-parameter model, write $\psi$ for the parameter of interest and $\lambda$ for the nuisance parameter. If $u_{\psi },u_{\lambda }$ are its score functions, the parameters are orthogonal statistical parameters when the cross entry of the expected Fisher information matrix vanishes:This condition may hold at a particular parameter value or throughout the parameter domain. It concerns the information geometry, not exact independence of finite-sample estimators. Under regular asymptotic efficiency, a diagonal information matrix gives zero first-order asymptotic covariance.

An interest-respecting reparametrization leaves $\psi$ unchanged, or replaces it by a one-to-one function of $\psi$ alone, while changing the nuisance coordinate by a smooth invertible transformation which may depend on both old parameters. Thus it preserves the level sets of the inferential target. Without loss of generality, keep $\psi$ itself and write the old nuisance as $\lambda =g(\psi ,\eta )$, where $\eta$ is the new nuisance coordinate and $g_{\eta }\ne 0$.

The chain rule for the log-likelihood gives new score functionsTheir expected cross product isSetting this to zero, with $I_{\lambda \lambda }>0$, gives the orthogonalization equation for two statistical parametersSolve this ordinary differential equation along curves of constant $\eta$, using $\eta$ to label distinct initial conditions. Smooth local solutions with nonzero transverse derivative produce an invertible interest-respecting reparametrization. This informal construction identifies precisely which derivative is held at fixed new nuisance coordinate.

For the given positive-shape generalized Pareto distribution, its one-observation log-likelihood isDefine $L=\log (1+\psi Y/\sigma )$ and $D=e^{-L}$. The density transformation $y=\sigma (e^L-1)/\psi$, $dy/dL=\sigma e^L/\psi$, givesThus $L$ has an exponential distribution of mean $\psi$. Direct differentiation of the original log-likelihood, holding the other original parameter fixed, givesThe following exponential integrals determine their expected products:For example, $\mathbb{E}{e}^{-kL}=(1+k\psi {)}^{-1}$ follows by integrating the exponential density, and differentiating this integral in $k$ gives $\mathbb{E}(L{e}^{-kL})=\psi /(1+k\psi {)}^2$. These formulas also give $\mathbb{E}{u}_{\psi }=\mathbb{E}{u}_{\sigma }=0$.

For clarity, the nuisance information and cross information can be calculated explicitly. Put $a=1+\psi$. ThenUsing $\mathbb{E}[L-a+aD]=0$, the cross product isThe same moments give $I_{\psi \psi }={\psi }^{-4}\mathbb{E}[L-a+aD{]}^2=2/[(1+\psi )(1+2\psi )]$. Therefore the Fisher information matrix for one observation isIn particular, $I_{\psi \sigma }>0$ for every admissible pair, proving the original shape and scale parameters are not orthogonal. No finite-mean assumption on $Y$ was needed: the transformed logarithm $L$ and the bounded variable $D$ have the required integrable score products even when the Pareto mean is infinite.

The orthogonalization equation for two statistical parameters now becomesIntegration along a constant-$\eta$ curve gives $\sigma (1+\psi )=\eta$. Hence a global interest-respecting reparametrization isIt is smoothly invertible on the entire positive parameter domain, with inverse $\sigma =\eta /(1+\psi )$. The Fisher information matrix transforms as $J^{T}IJ$, whereSubstitution givesThis verifies the orthogonal shape and scale for a generalized Pareto distribution directly, as well as by the differential equation. For an iid sample, every information entry is multiplied by the sample size, so the same parameter orthogonality remains valid.

The surplus in the classical risk model is $U(t)=u+ct-S(t)$. With ruin defined as a strictly negative surplus, its ultimate ruin probability isLet ${\overline{F}}_X(v)={\int }_v^{\infty }{f}_X(x)dx$. The first arrival time has exponential distribution of rate $\lambda$, independent of the first claim. Before that claim no ruin occurs, because the premium rate is positive. Given its arrival at $t$ and size $x$, available capital just before it is $u+ct$. If $x>u+ct$ ruin is immediate; otherwise the independent increments and fresh claim sizes make subsequent ruin probability $\psi (u+ct-x)$. Thus first-claim conditioning givesThis is the ruin counterpart of the first-claim decomposition for survival probability. DefineThe convolution term is continuous: extend the bounded function $\psi$ by zero to negative arguments and use continuity of translations in $L^1$ for the integrable density $f_X$. The tail function is also continuous. Changing variables to $v=u+ct$ yieldsDifferentiating this exponentially weighted integral establishes differentiability rather than presuming it. The fundamental theorem of calculus givesReversing the integration variable in the convolution therefore proves the ruin integro-differential equation:At zero the derivative is the right derivative; the same representation supplies it.

For the exponential claim law, introduceDifferentiating the explicit convolution gives $A^{\prime }=\alpha \psi -\alpha A$. Henceusing the proved integral equation. Differentiating that equation once more givesso the required ordinary differential equation isPositive relative safety loading means $c>\lambda /\alpha$, so $\kappa =\alpha -\lambda /c>0$. The general solution is $a+b{e}^{-\kappa u}$.

The condition at large capital must also be used. The strong law of large numbers for the Poisson process and the independent claims gives $S(t)/t\to \lambda /\alpha$ almost surely. Thus $S(t)-ct\to -\infty$. Its path has only finitely many jumps on each bounded time interval, so its supremum over all time is finite almost surely. Therefore $\psi (u)\to 0$ as $u\to \infty$, forcing $a=0$. The supplied initial value $\psi (0)=\lambda /(\alpha c)$ then gives ultimate ruin with exponential claims:The decay rate is also the exponential-claim adjustment coefficient.

For the maximum aggregate loss in a classical risk model, the ruin event is exactly $\{L>u\}$. Since $L(0)=0$, $L\ge 0$, and its full cumulative distribution function isIn particular it has an atom of mass $1-\lambda /(\alpha c)$ at zero. Conditional on $L>0$, its distribution is exponential with rate $\kappa$; omitting that atom would give an incorrectly normalized law.

A collocation Runge-Kutta method approximates an ordinary differential equation by a polynomial whose derivative agrees with the vector field at selected times. Let $s$ distinct nodes be $c_i\in [0,1]$, and consider the possibly nonautonomous equation $y^{\prime }=f(t,y)$. On a step from $(t_n,y_n)$, choose a polynomial $U(\tau )$ of degree at most $s$, with $U(0)=y_n$, such that $U^{\prime }(c_i)=hf(t_n+c_{i}h,U(c_i))$. Here $\tau =(t-t_n)/h$ is scaled time; the numerical endpoint is $y_{n+1}=U(1)$.

Let ${\ell }_j$ be the Lagrange interpolation polynomial equal to one at $c_j$ and zero at the other nodes, and set $k_j=f(t_n+c_{j}h,U(c_j))$. Interpolation of the polynomial derivative givesEvaluating at the nodes and endpoint derives the Runge-Kutta method coefficients:Thus collocation is a specific family of generally implicit RK methods, not a separate update mechanism. The identity ${\sum }_j{\ell }_j=1$ gives $A1=c$ and ${\sum }_j{b}_j=1$, the collocation tableau row-sum identity. For sufficiently small $h$, a vector field with Lipschitz constant $L$ gives a contraction of the stage map when $hL{\max }_i{\sum }_j|a_{ij}|<1$, on a suitable bounded neighborhood. The Banach fixed-point theorem then gives a unique nearby stage solution. Equivalently smooth stage equations have identity derivative in the stage variables at $h=0$, so the inverse function theorem gives a smooth nearby branch. Arbitrarily large-step solvability is a separate issue for an implicit Runge-Kutta method.

Order analysis has two levels: polynomial accuracy at stages and extra accuracy at the endpoint. Integration of interpolated powers gives the stage moment identitiesAn induction in powers of $h$, using these identities in the stage equations, shows that the collocation polynomial agrees coefficientwise with the exact Taylor polynomial through degree $s$:For example, the coefficient of each new stage term is integrated exactly because its node dependence is a polynomial of degree at most $s-1$. This proves stage order $s$ and bounded actual-time derivatives of the collocation polynomial up to degree $s$. Assuming the vector field is sufficiently smooth on a fixed compact solution neighborhood, its composition with this polynomial has uniformly bounded derivatives of all finite orders needed below.

The endpoint weights define the interpolatory quadrature rule $Q(v)={\sum }_i{b}_{i}v(c_i)$. Suppose it is exact through degree $q-1$, but not through degree $q$. Necessarily $s\le q\le 2s$: interpolation integrates degrees below $s$ exactly, while the nonzero polynomial $\omega (\tau {)}^2$, with $\omega (\tau )={\prod }_i(\tau -c_i)$, has positive integral but quadrature value zero and degree $2s$. Exactness also implies the nodal-polynomial criterion for quadrature exactnessbecause the integrand has degree at most $q-1$ and vanishes at every node.

Here is the endpoint-error argument underlying the collocation order theorem. Let $v(t)=U((t-t_n)/h)$ and $\delta (t)=v^{\prime }(t)-f(t,v(t))$ be its residual. It vanishes at the collocation times, and if $F_h(\tau )=f(t_n+h\tau ,U(\tau ))$, then $\delta ={\mathcal{I}}_s{F}_h-F_h$, where ${\mathcal{I}}_s$ is interpolation at the nodes. Taylor expansion in actual time givesThe vector coefficients $d_r$ are bounded. Each bracket has the factor $\omega (\tau )$, with quotient degree $r-s$. In particular the residual is $O(h^s)$ throughout the step.

Write $\varphi (t_{\ast },t,z)$ for the exact flow from time $t$ and state $z$ to $t_{\ast }=t_n+h$. Differentiating $\varphi (t_{\ast },t,v(t))$ gives $D_z\varphi (t_{\ast },t,v(t))\delta (t)$: the derivative with respect to the starting time cancels the vector field by the flow composition identity. Integration therefore gives the exact local endpoint errorFor $q=s$, the $O(h^s)$ residual already gives $O(h^{s+1})$. For $q>s$, Taylor-expand the smooth matrix weight as $\Psi (t_n+h\tau )={\sum }_{\ell =0}^{q-s-1}{h}^{\ell }{\Psi }_{\ell }{\tau }^{\ell }+O(h^{q-s})$. Every product term with $r+\ell <q$ integrates to zero: its polynomial factor is $\omega$ times a polynomial of degree $r-s+\ell \le q-s-1$. All remaining terms are $O(h^q)$ before the time integration, including both remainders. HenceThis is collocation endpoint superconvergence by quadrature orthogonality. For the global assertion, the stage contraction estimate gives a one-step Lipschitz bound $1+Ch$ on a fixed solution neighborhood. Thus the propagated error satisfies $e_{n+1}\le (1+Ch)e_n+C{h}^{q+1}$, and the discrete Gronwall inequality gives $e_n=O(h^q)$ on a fixed time interval. The order cannot generically exceed $q$: for $y^{\prime }=t^q$, each collocation step reduces to the same quadrature rule, whose first nonexact moment yields a nonzero $O(h^{q+1})$ local error and $O(h^q)$ composite error.