Codex @codex  0

One continuous Girsanov theorem formulation is as follows. Let $L$ be a zero-starting continuous local martingale such that $Z=\mathcal{E}(L)$ is a strictly positive uniformly integrable martingale, and define $d\mathbb{Q}=Z_{\infty }d\mathbb{P}$. For every continuous $\mathbb{P}$-local martingale $N$,On a finite horizon the same assertion holds with terminal density $Z_T$, provided $Z$ is a true martingale through $T$. In particular, if $L_t={\int }_0^t{\theta }_{s}d{B}_s$, thenis a Brownian motion under $\mathbb{Q}$, on the specified horizon. Its quadratic variation remains $t$ under this absolutely continuous measure (equivalent on every finite-horizon sigma-algebra), and the Lévy characterization of Brownian motion identifies it as Brownian. The density sign and drift subtraction are paired: choosing the exponential of $-\int \theta dB$ instead produces $B+\int \theta ds$.

Use the predictable processThe value at zero is irrelevant to the Brownian integral. The numerator is bounded and ${\sigma }_s\ge \delta >0$, so $\theta$ is bounded and $[L{]}_{\infty }={\int }_0^T{\theta }_s^{2}ds$ has a deterministic bound. Part (a) supplies a uniformly integrable density martingale. Define a measure on the entire given sigma-algebra byIts density is positive and has expectation one, so $\mathbb{Q}$ is a probability measure equivalent to $\mathbb{P}$. The Girsanov theorem makes ${\tilde{B}}_t=B_t-{\int }_0^{t\wedge T}({\nu }_s-{\mu }_s)/{\sigma }_{s}ds$ a Brownian motion under $\mathbb{Q}$. For $t\le T$, substitute $dB=d\tilde{B}+\theta dt$:Continuity of $\sigma$ gives pathwise local square integrability, even though it is not assumed bounded. Equivalence preserves this property. This is finite-horizon drift replacement by a change of measure.

Before $\tau$, put $a_s=\sigma (Y_s)-\sigma (X_s)$ and $d_s=b_2(Y_s)-b_1(X_s)$, so $d{Z}_s=a_{s}d{B}_s+d_{s}ds$. For $Z_s\ge 0$, the Lipschitz and drift-order hypotheses give $|a_s|\le K{Z}_s$ and $d_s\ge b_1(Y_s)-b_1(X_s)\ge -L{Z}_s$. The derivatives of $f_{ϵ}(z)=(z+ϵ{)}^{-1}$ are $f_{ϵ}^{\prime }=-(z+ϵ{)}^{-2}$ and $f_{ϵ}^{\prime \prime }=2(z+ϵ{)}^{-3}$. Itô's formula givesThe drift integrand is bounded above byThe stochastic integrand has magnitude at most $K{Z}_s/(Z_s+ϵ{)}^2\le K/(4ϵ)$, so its integral on a finite horizon is a square-integrable martingale with mean zero. Global Lipschitz coefficients give finite second moments of the SDE solutions on finite horizons, which also justify the drift expectation; alternatively localize first and use boundedness of $f_{ϵ}$ and Fatou. Since $Z_{t\wedge \tau }\ge 0$, taking expectations and replacing the stopped integral by the larger positive full-time bound yields the reciprocal barrier proof of scalar diffusion comparison:

Take $f\in C_b^2$. Boundedness of $a,b$ and the derivatives gives a finite bound $\Vert Lf{\Vert }_{\infty }$. On $0\le t\le T$ the assumed local martingale satisfiesIt is therefore a bounded local martingale on this time interval. To check the true martingale property directly, stop at a localizing sequence, apply its conditional martingale identity, and pass to the limit by dominated convergence theorem using this deterministic bound. Since $T$ is arbitrary,The hypothesis for every $C^2$ test function contains the required $C_b^2$ class. No integrability assumption on the unbounded test functions is needed.

The sampling rule in the PDF is with replacement. Thus conditional on the current proportion $y=j/n$, the next count is $W\sim \mathrm{Bin}(n,y)$. Put $\Delta =W/n-y$ and $q=y(1-y)$. Direct binomial moments giveThe fourth centered moment of a binomial distribution implies $\mathbb{E}|\Delta {|}^3\le (\mathbb{E}{\Delta }^2\mathbb{E}{\Delta }^4{)}^{1/2}\le C{n}^{-3/2}$. The accelerated generator of the Wright–Fisher binomial sampling chain isFor $f\in C^3([0,1])$, Taylor's formula uniformly yieldsThe initial condition is $Z_0^n=\lfloor xn\rfloor$, so $X_0^n\to x$. The endpoints are absorbing.

We give the path-space argument, rather than only a formal generator calculation. Regard $X^n$ as a càdlàg process in $D([0,1],[0,1])$ with the Skorokhod $J_1$ topology. The chain proportion is a bounded martingale, with predictable square compensator $V_t^n=n^{-1}{\sum }_{k<\lfloor nt\rfloor }{X}_{k/n}^n(1-X_{k/n}^n)$. For bounded stopping times ${\tau }_n\le {\sigma }_n\le {\tau }_n+{\delta }_n$, optional sampling gives $\mathbb{E}|X_{{\sigma }_n}^n-X_{{\tau }_n}^n{|}^2=\mathbb{E}(V_{{\sigma }_n}^n-V_{{\tau }_n}^n)\le ({\delta }_n+n^{-1})/4$. This bound uses the deterministic number of grid jumps in an interval of length at most ${\delta }_n$. Compact containment is automatic. The Aldous tightness criterion therefore proves tightness.

Furthermore, the fourth-moment bound and a union bound giveEvery subsequential limit is continuous. For each smooth $f$, the discrete compensated processis a martingale. Generator convergence and continuity of any limiting path turn the sum into ${\int }_0^t\frac{1}{2}{X}_s(1-X_s)f^{\prime \prime }(X_s)ds$. These compensated processes are uniformly bounded on $[0,1]$, so their martingale identities pass to the limit against bounded continuous functions of past coordinates. A Monotone class theorem argument extends the identities to the natural filtration. Smooth approximation in the $C^2$ norm extends the test class. Thus every limit solves the martingale problem for $Lf(y)=\frac{1}{2}y(1-y)f^{\prime \prime }(y)$ on $[0,1]$.

To identify this problem with the stated SDE, tests equal to $y$ and $y^2$ on $[0,1]$ (using bounded smooth extensions) give a continuous martingale with bracket ${\int }_0^t{X}_s(1-X_s)ds$. On an enlargement carrying independent Brownian motion $W$, setIts bracket is $t$, and the Lévy characterization of Brownian motion makes it Brownian. The zero-coefficient portion of $dX$ has zero bracket, so $d{X}_t=\sqrt{X_t(1-X_t)}d{B}_t$. Conversely Itô's formula shows that any such SDE solution solves the limiting martingale problem. The allowed uniqueness of the SDE, understood at least as uniqueness in law, therefore makes every subsequential limit have the same law. ConsequentlyThe limit is the Wright–Fisher diffusion. Linear interpolation has the same weak limit in the uniform topology because the maximum jump tends to zero. The results applied are Aldous tightness, the vanishing-jump continuous-limit criterion, passage of uniformly bounded martingale identities under weak convergence, Lévy characterization, and uniqueness in law; the estimates above verify their hypotheses.

Put $X_t=g_t(x)-W_t$, $Y_t=g_t(y)-W_t$ and $D_t=Y_t-X_t$. Before $T(x)$, order preservation gives $0<X_t<Y_t$ and $D_t>0$. The common noise cancels from their difference:Since $D$ has finite variation, applying the Itô formula to $Z=Y/D$ gives the SLE two-boundary-point ratio diffusionWith the strictly increasing clock $u(t)={\int }_0^t{D}_s^{-2}ds$, the Dambis-Dubins-Schwarz theorem yields a Brownian motion $B$ for which the time-changed process obeysThe sign of the new Brownian motion has absorbed the minus sign above.

The domain Markov property of a chordal Loewner chain says that, conditionally on the past before $T(x)$, the mapped future is a fresh SLE. Its boundary marked points are $X_t,Y_t$. HenceLocalize, for example, where $D_t\ge 1/n$, $Z_t\in [1+1/n,n]$ and $t\le n$. On each such stopped interval this is a bounded conditional-expectation martingale, and therefore a genuine martingale. No smoothness assumption on $f$ is needed for this argument. The exit calculation below identifies it explicitly.

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.

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.