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The conformal invariance of planar Brownian motion is a time-change statement. If $\varphi :D\to D^{\prime }$ is a conformal bijection of planar domains and $B$ is planar Brownian motion started in $D$, let ${\tau }_D$ be its exit time and defineBefore exit this clock is strictly increasing. Its inverse $C_u$ makes $\varphi (B_{C_u})$ a planar Brownian motion started at $\varphi (B_0)$, up to its transformed lifetime $A_{{\tau }_D}$. The image of the stopped trajectory is therefore invariant in law after this change of time. A holomorphic map with nonzero derivative gives the corresponding local statement even without global injectivity.

Here is the proof. Write $\varphi =u+iv$ and $B=(B^1,B^2)$. The real and imaginary parts are harmonic. The Itô formula therefore gives the two continuous local martingalesstopped on compact subsets of $D$. The Cauchy-Riemann equations implyThus both coordinate quadratic variations equal $A_t$ and their quadratic covariation is zero. After inverse time change, the covariance matrix of the two continuous local martingales is $u{I}_2$. The multidimensional Lévy characterization of multidimensional Brownian motion states that a continuous vector local martingale starting at zero with this covariance matrix is standard vector Brownian motion. Apply it after subtracting $\varphi (B_0)$, and exhaust $D$ by compact subsets. This proves the theorem, with killing at the transformed exit lifetime. The clock is the conformal Brownian clock; omitting it would generally give an incorrect speed.

Use the local convention for the martingale problem, which is appropriate for arbitrary measurable coefficients. A continuous adapted process $X$ solves $\mathbf{M}(a,b)$ if the coefficient integrals are locally finite and, for every $f\in C_c^{\infty }(\mathbb{R})$,is a local martingale. Here locally finite means ${\int }_0^T(|a(X_s)|+|b(X_s)|)ds<\infty$ almost surely for each $T<\infty$. Suitable localization upgrades the identities to true martingale identities. With bounded coefficients and bounded test derivatives, this distinction disappears on finite horizons.

A Markov diffusion generated by $L$ is a time-homogeneous Markov process with continuous paths whose infinitesimal generator on the stated test domain is $L$. Its transition operators satisfy the generator limit, and Dynkin formula gives precisely the preceding martingale problem identities. Some terminology uses L-diffusion for the test-function identities alone; the Markov property is an additional assertion unless uniqueness supplies it.

Suppose now $a={\sigma }^2>0$. Use smooth compactly supported functions agreeing with $x$ and $x^2$ on $[-R,R]$, and stop before exiting this interval. Removing these stops shows thatis a continuous local martingale. Consequently $X$ is a continuous semimartingale. The martingale problem applied to the second cutoff function says thatis a local martingale. The Itô formula for $X^2$ also says thatis a local martingale. Subtracting, the continuous finite-variation process $[N{]}_t-{\int }_0^{t}a(X_s)ds$ is a local martingale and starts at zero. The theorem that a continuous finite-variation local martingale is constant givesDefineThe integrand is predictable and is stochastically integrable, since its integral against the bracket is exactly $t$. Thus $B$ is a continuous local martingale with $B_0=0$ and $[B{]}_t=t$. The Lévy characterization of Brownian motion states that these conditions make $B$ a Brownian motion relative to the given filtration. Associativity of stochastic integrals recovers $N=\int \sigma (X)dB$, and thereforeThe original process is a weak solution of a stochastic differential equation, with its driving Brownian motion constructed on the original space.

Conversely, the Itô formula applied to any weak solution with locally integrable coefficients gives the martingale problem identities. A Markov diffusion with generator $L$ gives them by Dynkin formula. A well-posed martingale problem, with the usual measurable path-space setup, identifies a unique family of diffusion laws and yields the Strong Markov property. Without uniqueness, the test-function identities alone do not assert that every chosen solution is Markov. This specifies both the equivalence with weak equations and the role of uniqueness in identifying a diffusion.

On a fixed horizon $T$, an arbitrage is an admissible trading strategy that is self-financing, has $X_0=0$, and satisfies $X_T\ge 0$ almost surely with $\mathbb{P}(X_T>0)>0$. Here admissibility means that wealth in units of the strictly positive riskless numéraire has a deterministic lower bound throughout the horizon. An equivalent martingale measure $Q$ is equivalent to the physical measure on ${\mathcal{F}}_T$ and makes each discounted risky price ${\tilde{S}}^i=S^i/B$ a martingale.

For a predictable, stochastically integrable risky holding vector $\pi$, the discounted self-financing identity isIndeed, since the riskless account is a finite-variation process, its quadratic covariations vanish, and the Itô formula for $X/B$, combined with $X=\varphi B+\pi \cdot S$ and $dX=\varphi dB+\pi \cdot dS$, gives $d(X/B)=\pi \cdot d(S/B)$.

Use two standard facts from stochastic calculus: a stochastic integral against a continuous local martingale is a local martingale, and a local martingale bounded below by a deterministic constant is a supermartingale. The latter follows by adding the constant to make the process nonnegative, stopping at a localizing sequence, and applying the conditional Fatou lemma. Thus admissible discounted wealth is a $Q$-supermartingale, and an arbitrage would satisfyIt follows that $X_T=0$ $Q$-almost surely, hence also physically almost surely by equivalence, a contradiction. Therefore an equivalent martingale measure rules out arbitrage among admissible self-financing strategies. The lower-bound restriction is essential to this argument; stochastic integrals need not be true martingales for arbitrary unbounded holdings.

Use the discounted wealth $\tilde{X}=X/B$ and the discounted dividend gainsThe integral uses $ds$: the printed $dt$ inside an integral indexed by $s$ is a differential typo. Since $B$ is a positive finite-variation process, the Itô product rule givesThe bank-account terms cancel using $X_t={\varphi }_t{B}_t+{\pi }_t\cdot S_t$.

If $Y$ is a $Q$-martingale, discounted self-financing wealth is a $Q$-local martingale. The admissible trading strategy lower bound makes it a supermartingale, exactly as in the previous no-arbitrage proof. A zero-cost portfolio with nonnegative terminal wealth must therefore have terminal wealth zero $Q$-almost surely and, by equivalence, physically almost surely. Hence an equivalent martingale measure for discounted dividend gains excludes arbitrage. It is the gains process, not just the ex-dividend price ratio, that must have zero martingale drift.

Use the Black-Scholes equation with the claim's terminal condition $V(T,s)=g(s)$:For $t<T$, the Itô formula under the physical measure givesChoose the delta hedge and its bank-account holding byTheir portfolio value is $V(t,S_t)$, andThus the portfolio is self-financing and, with initial capital $V(0,S_0)$, has terminal wealth $g(S_T)$.

Localizing to compact stock-price intervals and times below $T$ justifies these stochastic integrals from the classical smoothness of $V$. The extension to maturity also follows from boundedness. Under the Risk-neutral measure for the Black-Scholes model, discounted $V(t,S_t)$ is a bounded local martingale, hence a true square-integrable martingale. Its quadratic variation has finite expectation, so the stock-integral coefficient is square integrable through $T$. Equivalence of measures preserves its almost-sure finiteness, and the physical drift integral is finite by Cauchy-Schwarz. The terminal limit is $g(S_T)$.

Finally, $B_t=B_0{e}^{rt}$ is bounded away from zero on this finite horizon, and bounded $V$ makes $V(t,S_t)/B_t$ bounded below by a deterministic constant. The hedge is therefore an admissible trading strategy. The displayed holdings replicate the payoff admissibly, even if the bounded payoff can be negative.

Put $\tau =T_r\wedge T_R$. First establish almost sure finiteness. The Itô formula gives the martingale $|B_t{|}^2-2t$ for planar Brownian motion. Stop it at $t\wedge T_R$. The stopped path has norm at most $R$, soBy monotone convergence, $\mathbb{E}{T}_R<\infty$, and hence $\tau <\infty$ almost surely.

The function $h(y)=\log |y|$ is a harmonic function away from $0$: in polar coordinates its Laplacian is $h^{\prime \prime }(\rho )+h^{\prime }(\rho )/\rho =-{\rho }^{-2}+{\rho }^{-2}=0$. The Itô formula shows $h(B_{t\wedge \tau })$ is a local martingale. It is bounded between $\log r$ and $\log R$, so it is a true martingale and passage to $\tau$ is justified by dominated convergence. Writing $p={\mathbb{P}}_x(T_r<T_R)$ and using continuity at exit givesSolving yields the planar Brownian annulus hitting probabilityThe stopping argument stays away from the singularity of the logarithm until the calculation is complete.

For $h(x,y)=\frac{1}{2}\log (x^2+y^2)$ away from the origin,The Itô formula for planar Brownian motion therefore has no drift term before the radius reaches $r$. Since the initial logarithm is zero,The integrand's squared norm is $|B_s{|}^{-2}\le r^{-2}$ on this stopped interval, so this is a square-integrable martingale on every finite horizon, in particular a local martingale. One may first stop in an outer annulus to apply the formula on a compact smooth domain and then remove that localization.

Before $T(x)$, the centered image satisfies the stochastic differential equationChanging the sign of the Brownian motion shows that $X/\sqrt{\kappa }$ is a Bessel process of dimension $1+4/\kappa$. Here is a direct proof of the relevant hitting property, including finite lifetime.

Put $p=1-4/\kappa >0$. By the Itô formula, the drift of $X^p$ isThus $X^p$ is a local martingale and a scale function of a one-dimensional diffusion. Fix $R>x$. The nonnegative functionvanishes at both endpoints and satisfies $\mathcal{L}{u}_R=-1$, for $\mathcal{L}=(\kappa /2){\partial }_{vv}+(2/v){\partial }_v$. Stop first on $[ϵ,R]$, and also at a deterministic time. The Itô formula gives $\mathbb{E}(t\wedge {\tau }_{ϵ,R})\le u_R(x)$. Let $t\to \infty$ and then $ϵ\downarrow 0$. It follows that the first exit ${\tau }_{0,R}$ is finite almost surely, not merely that the path approaches a boundary at infinite time.

Optional stopping applied to $X^p$ on this bounded interval givesOn $\{{\tau }_0=\infty \}$ the process must therefore hit every $R>x$ before zero. Letting $R\to \infty$ makes that probability zero. ConsequentlyAt $\kappa =4$ the dimension is two; zero is not hit from a positive starting point. This endpoint must not be included in the finite-swallowing assertion.

For any $C^2$ function $h$, the Itô formula and the ratio equation giveThe bracket is zero for the functions just found. On compact localized intervals the stochastic integrand is bounded, so the stochastic integral is a martingale. Removing localization proves that $h(Z_t)$, on $0\le t<T(x)$, is a local martingale. It is nonnegative because $Z_t>1$ and $C>0$.

In particular it is a nonnegative local martingale, hence a supermartingale after the usual stopping and extension by its endpoint limit. This is the justification for using its maximal inequality; it is not automatically an unstopped uniformly integrable martingale.

The time-changed diffusion has generator $\mathcal{A}=(\kappa /2){\partial }_{zz}+2(1/z+1/(z-1)){\partial }_z$. For $1<z<R$, its boundary hitting probability from a diffusion scale function isTo see this directly, stop the scale local martingale on $[1+ϵ,R]$ and pass to $ϵ\downarrow 0$. Exit from $(1,R)$ occurs in finite clock time: the scale density behaves as $(z-1{)}^{-4/\kappa }$ and the speed density as $(z-1{)}^{4/\kappa }$, so the usual scale-times-speed exit integral is finite at $1$. More explicitly, with $m(v)=2/(\kappa s_{\kappa }^{\prime }(v))$ the finite-interval diffusion exit Green kernel givesThe equation $\mathcal{A}u=-1$ follows by differentiation and the derivative jump at $v=z$, so stopping the corresponding Itô identity gives this exit bound. The integrand near $1$ is $O(v-1)$; elsewhere it is bounded. At large $z$ the drift is bounded, excluding explosion at infinity in finite clock time.

Here is why these exits describe the original swallowing event. In clock time,If $\hat{Z}$ hits $1$ at finite $u$, its integrated stochastic differential equation shows ${\int }_0^{{\tau }_1}({\hat{Z}}_v-1{)}^{-1}dv<\infty$: the Brownian term has a finite limit and both positive drift integrals must then have finite limits. Consequently $D({\tau }_1)>0$, $X$ hits zero and $Y$ stays positive. This gives $T(x)<T(y)$.

Conversely $T(x)$ is finite by part (b). If its clock ends at a finite value, the diffusion must hit $1$; an interior endpoint with positive $D$ would leave $X$ positive. If the clock runs forever without hitting $1$, the standard one-dimensional scale classification gives escape to infinity whenever $S_{\kappa }<\infty$. This cannot represent a strict swallowing event, which has $D\to Y_{T(x)}>0$ and $Z\to 1$. Hence the surviving event represents $T(x)=T(y)$. This reasoning also explains why a simultaneous collision must be treated separately rather than assigned the boundary value at $1$.

For $\kappa \ge 8$, the scale is unbounded. The maximal inequality for a nonnegative supermartingale gives $\mathbb{P}(\sup s_{\kappa }(\hat{Z})\ge s_{\kappa }(R))\le s_{\kappa }(z)/s_{\kappa }(R)$; letting $R\to \infty$ rules out escape. Finite-interval exit then forces a hit at $1$. For $4<\kappa <8$, taking $R\to \infty$ givesThe strict-event probability is also positive, since $0<s_{\kappa }(z)<S_{\kappa }$ for finite $z$.

For $\kappa \ge 8$, first intersect the probability-one strict-order events over rational $0<a<b$. The real boundary flow gives nondecreasing swallowing times on $(0,\infty )$. For any $0<x<y$, choose rational $x<a<b<y$; then $T(x)\le T(a)<T(b)\le T(y)$. This proves the simultaneous assertionThe range is positive points, as in the definition of $F$; reflection reverses the ordering on the negative axis. At the printed endpoint $\kappa =4$, each fixed positive point has $T=\infty$. Thus $\mathbb{P}(T(x)=T(y))=1$ if equality of extended times is allowed, but there is no finite simultaneous-swallowing assertion there. The preceding formulas assume $\kappa >4$.

One precise relation is to radial SLE with parameter $2$. Its orientation is from the boundary point $1$ to the interior target $0$. With conformal-radius time it is defined byThere is a coupling of the conditioned Brownian path $B$ with this simple radial curve $\eta$ such that, for each initial curve segment, the first point at which $B$ encounters it is its tip. More precisely, if $s_t=\inf \{s:B_s\in \eta [0,t]\}$, then $B_{s_t}=\eta (t)$. Reversing the Brownian path turns these first-hit times into the last-visit times that characterize loop-erasure of planar Brownian motion. ThusThis is an existence statement for a coupling, not a pointwise equality of the two processes or a claim that a naive finite-loop deletion algorithm applies to Brownian motion.

Here is the martingale mechanism behind the coupling. Put $a_t=e^{i{\theta }_t}$ andThis is the positive slit-domain Poisson kernel at the tip, normalized by $Q_t(0)=1$. Writing $q_t=g_t(z)/a_t$, the radial equation givesThe drift of $(1+q_t)/(1-q_t)$ is $(2-\kappa )q_t(1+q_t)/(1-q_t{)}^3$, so $Q_t(z)$ is a local martingale at $\kappa =2$.

Start independent stopped $B$ and radial SLE2, and weight their joint laws, before meeting, byFor fixed $s$ this is an SLE local martingale. For fixed $t$ it is a Brownian-transform local martingale, because ${\mathcal{L}}^h(Q_t/h)=(2h{)}^{-1}\Delta Q_t=0$. Since $L(0,t)=L(s,0)=1$, localized two-parameter weighting preserves both original marginals. For a fixed slit, the changed Brownian law is the transform by $Q_t$, so it hits that slit at the pole, its tip. Compatible stopped couplings, followed by exhaustion, give the first-hit-at-tip property. This completion is the substantive continuum coupling step. Discrete loop-erased random walks and their radial-SLE2 limit give a parallel approximation picture.

For example, in this coupling avoidance by $B$ implies avoidance by $\eta$, since the radial curve is contained in the Brownian trace. The probability from part (b) therefore supplies a lower bound $|{\varphi }^{\prime }(1)|\le \mathbb{P}(\eta \subset U)$. The two avoidance laws are not identical: unweighted simple radial ${\mathrm{SLE}}_2$ does not have the Brownian radial-restriction property. This is a way to transfer path-containment information while retaining the distinct geometry and orientation of the two random curves.

The Gaussian free field is a random distribution, so its purported level curve cannot be defined by evaluating its height at every point. A coupling must instead be stated using conditional means and covariances. The SLE4 coupling with a Gaussian free field does exactly this.

Choose the Green-function normalizationA Zero-boundary Gaussian free field $H_0$ is characterized on real compactly supported smooth test functions byEquivalently its energy normalization is the Dirichlet inner product $(2\pi {)}^{-1}\int \nabla f\cdot \nabla gdA$. This fixes the otherwise convention-dependent height constant.

Set $\lambda =\pi /2$ andThe field $H$ has prescribed Dirichlet values $-\lambda$ on the negative half-line and $+\lambda$ on the positive half-line. Subtracting $m_0$ recovers the zero-Dirichlet field $H_0$. It is the shifted field $H$, rather than an unshifted zero-boundary field, whose zero-height interface has ordinary chordal ${\mathrm{SLE}}_4$ law.

Let $\eta$ be chordal ${\mathrm{SLE}}_4$ from $0$ to $\infty$, with Chordal Loewner equation ${\partial }_t{g}_t=2/(g_t-W_t)$ and $W_t=2{\beta }_t$. Write $D_t=\mathbb{H}\setminus \eta [0,t]$ and $f_t=g_t-W_t$. The coupling statement is:where, conditionally on the curve, $H_0^{D_t}$ is an independent zero-boundary field in the slit domain. The equality is for restrictions to test functions in that domain, and extends to suitable stopping times. Its covariance is $G_{D_t}(z,w)=G_{\mathbb{H}}(f_t(z),f_t(w))$ by conformal invariance. The two sides of the revealed slit have heights $-\lambda$ and $+\lambda$, in the order specified by their real images under $f_t$. This is the continuum meaning of a Gaussian free field level line.

The fundamental calculations explain why the parameter is four. For $Z_t=f_t(z)$, the Itô formula givesAt $\kappa =4$ the drift vanishes. ThusEach harmonic mean is a bounded martingale before approaching its point. Differentiating the explicit conformally transformed Green function gives the Loewner variation of the Dirichlet Green functionThereforeIn words, the variance learned from the evolving mean exactly equals the covariance lost when the slit is removed. For a Green kernel $cG$ the same calculation gives $\lambda =(\pi /2)\sqrt{c}$; using a different Green normalization changes the height constant, not the SLE parameter.

To construct the coupling, first sample the SLE curve. In the two components to its left and right, sample independent zero-boundary fields and add the corresponding constant heights. Extend these as distributions to obtain the candidate full field. The following martingale identity proves its marginal law rather than merely matching its first two moments.

For a real test function $\phi$, put $M_t=(m_t,\phi )$ andLocalization permits integration of the pointwise identities. They yield $d⟨M{⟩}_t=-d{V}_t$. Henceis a complex local martingale: the $-\frac{1}{2}d{V}_t$ drift cancels the $-\frac{1}{2}d⟨M{⟩}_t$ Itô correction. Since $V_t\ge 0$, $|{\mathcal{Z}}_t|\le 1$, making it a true martingale. At the complete-curve limit, $m_t$ becomes the constant height in each component and the remaining Green kernel becomes that of those components. Thus ${\mathcal{Z}}_{\infty }$ is the conditional characteristic function of the sampled candidate field. Taking expectations givesApplying this to every linear combination of test functions proves the entire Gaussian law, not only its covariance. The conditional version $\mathbb{E}({\mathcal{Z}}_{\infty }\mid {\mathcal{F}}_t)={\mathcal{Z}}_t$ proves the displayed conditional-field statement. Exhaustion by compact test supports justifies passage across the slit and the limiting distributional extensions.

This revealed curve is a local set of a Gaussian free field: conditionally on it, the remaining field is a zero-boundary field plus a specified harmonic function. The spatial Markov property is thereby preserved at random domains, a property not available for arbitrary field-dependent sets. One can strengthen the coupling to a curve measurable from the field. The proof explores compatible interfaces in small subdomains and uses the conditional boundary heights and monotonicity to show that two such interfaces for the same field cannot separate; a countable exhaustion gives uniqueness. This supplies a rigorous replacement for the informal phrase “draw the zero contour.” It does not assert that the distribution has pointwise values.

The coupling is useful in both directions: SLE martingales give exact conditional Gaussian free field data, while the Gaussian Markov structure explains the SLE domain Markov property and its distinguished parameter. With the above normalization, the height jump is $2\lambda =\pi$, and the corresponding interface is chordal SLE4.

Use the usual normalization $Z_0=1$, finite deterministic initial wealth $x=X_0\ge 0$, positive state-price density, and nonnegative consumption. The deflated asset prices $ZP$ are local martingales, so predictable stochastic integrability makes $M=\int H\cdot d(ZP)$ a zero-starting local martingale. Integrating part (b) givesA nonnegative local martingale is a supermartingale: localize to martingales, apply Conditional Fatou lemma, and obtain the conditional expectation inequality and integrability. Therefore $N$, and hence $M=N-x$, are supermartingales. In particular,This is supermartingale control of deflated consumption gains. Nonnegative consumption is essential to the lower bound; it is the ordinary meaning of a consumption rate here. Wealth nonnegativity alone would not supply that bound if arbitrary signed cash flows were allowed.

Write $f(z)=1/\cosh z$. Differentiating givesTherefore the diffusion operator satisfies $\frac{1}{2}{f}^{\prime \prime }+\tanh z{f}^{\prime }=-\frac{1}{2}f$. Applying the Itô formula to $Y_t=e^{t/2}f(X_t)$ cancels its drift:So $Y$ is a positive local martingale. On every finite interval $[0,R]$, $0<Y_t\le e^{R/2}$. The bounded local martingale criterion makes it a true martingale on that interval. Since $R$ is arbitrary,

Put $F_t=\exp (-i\theta X_t-\frac{1}{2}{\theta }^2{A}_t)$. The Itô formula givesThe explicit decreasing exponential contributes half the finite-variation term; the other half is the quadratic correction from $e^{-i\theta X_t}$. Now the Itô product rule and part (b) yieldThe finite-variation terms cancel because $-i\theta (i\theta )={\theta }^2$. This proves the product is a local martingale. Moreover $|F_t|=e^{-{\theta }^2{A}_t/2}\le 1$ and $|M_t|\le 1$. The bounded local martingale criterion therefore proves the product is a true martingale.

Let $a(x)=\sigma (x)\sigma (x{)}^{\mathsf{T}}$. The multidimensional Itô formula gives the second-order diffusion generatorIndeed,The integrand is locally bounded along the continuous path after stopping on compact sets, because the coefficients and derivatives are continuous. Thus the final stochastic integral is a local martingale. No global growth or uniqueness assumption on this already-given weak solution of a stochastic differential equation is needed.

Continuity and adaptedness make the first boundary hit a stopping time. A continuous path starting in the open domain cannot leave it before meeting its boundary. Thus $X_{t\wedge T}\in \mathcal{D}\cup \partial \mathcal{D}$, with the usual interpretation when $T=\infty$. If $|u|\le K$ on this set, thenThe stopped process is a bounded local martingale; the bounded local martingale criterion makes it a martingale and, in fact, a uniformly integrable martingale. The martingale convergence theorem gives almost sure and $L^1$ convergence as $t\to \infty$.

Its limit can also be identified pathwise. On $\{T<\infty \}$ the stopped process is eventually constant at $e^{-\lambda T}u(X_T)$. On $\{T=\infty \}$ its absolute value is at most $K{e}^{-\lambda t}$ and hence tends to zero. Thus the limit is $e^{-\lambda T}u(X_T)1_{\{T<\infty \}}$.

Use the Chordal Loewner equation with Loewner driving function ${\xi }_t=\sqrt{\kappa }{W}_t$, and set $Z_t=g_t(z)-{\xi }_t=X_t+i{Y}_t$. Before the Loewner swallowing time,The upper-half-plane branch of the logarithm has imaginary part $h_t$. The Itô formula givesand therefore the SLE angle process satisfiesAt $\kappa =4$ the drift vanishes. Since $0<h_t<\pi$, the stopped local martingale is a true bounded martingale, and the SLE4 angle martingale is global for each fixed point almost surely, using fixed-interior-point avoidance of SLE4 for the simple parameter-four Loewner trace.

Conversely, take $\kappa >0$. If this semimartingale were a local martingale, uniqueness of its finite-variation decomposition would force $(\kappa -4)X_t{Y}_t=0$ throughout every compact interval before swallowing. Because $Y_t>0$, for $\kappa \ne 4$ this would force $X_t$ to be identically zero on such an interval. Its Brownian component has quadratic variation $\kappa t$, which makes that impossible. Thus for positive $\kappa$, the martingale parameter is exactly four.

If the degenerate value $\kappa =0$ is admitted, there is one exception to a fixed-point reading of the assertion: for $z$ on the positive imaginary axis the deterministic flow stays on that axis until swallowing, and $h_t=\pi /2$ is constant. For all starting points simultaneously, the unique parameter giving the martingale property is still four.

The multidimensional Itô formula includes a mixed second derivative multiplied by the quadratic covariation, with no extra factor $1/2$. HereConsequently the Itô formula for $M_t=U(t,{\sigma }_t,X_t)$ has drift equal to the left side of the stated backward partial differential equation. That drift vanishes, leavingA stochastic integral against Brownian motion with locally square-integrable predictable integrand is a continuous local martingale. The smoothness of $U$ and localization of the diffusion and its coefficients give this integrability on the model's lifetime. Thus $M$ is a local martingale, as required. The partial differential equation cancellation alone does not establish a true martingale or justify replacing $U$ by a terminal-payoff expectation without an additional integrability argument.

Assume the usual positive initial asset values. The coefficients determine a unique normalized local martingale deflator. The state-price density and local deflator distinction matters here: calling it a state-price density uses the local convention, while true expectation pricing needs an additional qualification, addressed below.

Let $ZB$ and $ZS$ be local martingales, with $Z_0=1$. The Brownian martingale representation theorem says that every local martingale in the usual natural Brownian filtration is continuous and is a stochastic integral against $W$. Applying it to $ZB$, and dividing by the positive finite-variation account, forcesfor a locally square-integrable predictable $\eta$. The Itô product rule for $ZS$ has driftIt must vanish. Since $S,Z,\sigma$ are positive, this yieldsConversely this drift choice makes both $ZB$ and $ZS$ local martingales. Continuity and strict positivity of $\sigma$ make $\lambda$ bounded along each path on every finite time interval, so its pathwise square integral is finite. The unique linear SDE solution isUniqueness follows from the forced drift and diffusion coefficients and uniqueness of this linear stochastic differential equation.

Continuity alone does not make the density a true martingale. For a true equivalent martingale measure, the stochastic exponential $D_t=Z_t{B}_t/B_0$ must have expectation one, for example under the Novikov condition on each horizon. True martingale pricing of all desired deflated payoffs also requires the relevant integrability. These stronger conclusions do not follow just from pathwise continuity.