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Jointly time-change the radial and area martingales by the inverse of $A=\int R^{2}ds$. The bracket matrix becomes $u{I}_2$, so the Lévy characterization of multidimensional Brownian motion gives two independent Brownian coordinates. The clock is strictly increasing; if it were finite at infinity, the radial martingale would converge and $R_t^2=2t+O(1)$ would force its integral to diverge. Thus $R_t^2=2W_{A(t)}+2t$ and $Z_t=B_{A(t)}$ with $W,B$ independent.

If a continuous vector local martingale has bracket matrix $A_{t}I$ with $A_{\infty }=\infty$, inverse-clock time change gives bracket matrix $rI$. The Lévy characterization of multidimensional Brownian motion then gives Brownian motion in the common time-changed filtration. A continuous local martingale is constant on flat intervals of its bracket, so inverses using $A_t\ge r$ and $A_t>r$ give the same martingale values. The filtration at the earlier inverse is smaller; adaptation and independence from the larger past transfer the Brownian property to it. Nonzero initial values are retained by adding the initial vector back.

A continuous adapted process started at a fixed point is Brownian motion exactly when $f(X_t)-f(X_0)-\frac{1}{2}{\int }_0^t\Delta f(X_s)ds$ is a martingale for every $f\in C_c^{\infty }({\mathbb{R}}^d)$. Cutoff coordinate and coordinate-product tests give continuous local martingales with quadratic covariations $[X^i,X^j{]}_t={\delta }_{ij}t$. The Itô formula makes $\exp (iu\cdot (X_t-X_0)+|u{|}^{2}t/2)$ a local martingale, whose deterministic modulus bound on bounded horizons makes it a true martingale. Its conditional expectation identifies every increment as centered Gaussian with covariance $(t-s)I$, independent of the past. This proves the converse via the Lévy characterization of multidimensional Brownian motion. The forward implication is the Itô formula.

Two continuous local martingales are orthogonal when their quadratic covariation is zero. With zero initial values, the martingale product identity says that their product is a local martingale. For two Brownian motions in a common filtration, orthogonality and the Lévy characterization of multidimensional Brownian motion make the pair a two-dimensional Brownian motion and imply independence. Merely having the two marginal Brownian motion laws on the same space does not imply orthogonality.

On a finite horizon $[0,T]$, let $W$ be a standard $d$-dimensional Brownian motion and let $\theta$ be a predictable process with ${\int }_0^T|{\theta }_s{|}^{2}ds<\infty$ almost surely. Suppose the stochastic exponentialis a true martingale with $\mathbb{E}{Z}_T=1$. The Novikov condition $\mathbb{E}\exp (\frac{1}{2}{\int }_0^T|{\theta }_s{|}^{2}ds)<\infty$ is a sufficient condition. The Girsanov theorem says that under $dQ=Z_{T}dP$,Because $Z_T>0$, the two measures are equivalent on the terminal sigma-algebra.

For the proof sketch, Itô formula gives $d{Z}_t=-Z_t{\theta }_t\cdot d{W}_t$. Apply the product rule to $Z_t{W}_t^{Q,i}$. Its finite-variation term $Z_t{\theta }_t^{i}dt$ is cancelled by $d[Z,W^i{]}_t=-Z_t{\theta }_t^{i}dt$. After localization the product is a martingale under $P$, and the conditional change-of-measure formula makes $W^{Q,i}$ a local martingale under $Q$. Adding finite variation leaves its covariations unchanged: $[W^{Q,i},W^{Q,j}{]}_t={\delta }_{ij}t$. The Lévy characterization of multidimensional Brownian motion completes the proof. Localization is essential if the integrands are not bounded; the true-martingale hypothesis on $Z$ permits the final measure change.

For the deal, write $T=t_0$ and use constant interest rate $\rho$, volatility $\sigma >0$ and no stock dividends. Under the risk-neutral measure, the Black-Scholes model isFor example this follows from the preceding theorem with $\theta =(\mu -\rho )/\sigma$ when the physical stock drift is $\mu$. The delivered shares have terminal cash value $S_T\max (r,c/A)$, so its initial value isThe factor $S_T$ must be included: the deal is a stock delivery, not a cash payment of the random number of shares.

Use the stock-numeraire measure in the Black-Scholes model, whose density isThe Girsanov theorem gives $W_t^S=W_t^Q-\sigma t$ as Brownian motion under $Q^S$. ThereforeThe discrete geometric average under the stock-numeraire measure has normally distributed logarithm withThe last identity uses the printed descending order of sampling times. Each $t_i$ is the minimum in exactly $2i-1$ ordered pairs. In particular $v>0$.

Assume the ordinary positive deal parameters $r,c>0$ and put $k=\log (c/r)$, $d=(m-k)/\sqrt{v}$. The change of numeraire gives $V_0=S_0{\mathbb{E}}_{Q^S}\max (r,c{e}^{-Y})$, where $Y=\log A$. Split the integral at $k$. Completing the square in $e^{-y}$ times the normal probability density givesThus the stock-delivery option with a geometric average has valueHere $\Phi$ is the standard normal distribution function. Delivery time $T$ does not enter separately once the sampling dates are fixed: the stock numeraire cancels the final stock factor and discounting. As a check, for one sample at $T$, the payment is $\max (r{S}_T,c)$ and the formula becomes the value of $r$ shares plus a put on their terminal value. If $c=0$ the value is $r{S}_0$; if $r=0$ the value is $c{S}_0{e}^{-m+v/2}$.

The Lévy characterization of multidimensional Brownian motion says that a continuous adapted ${\mathbb{R}}^d$-valued process $Z$, with $Z_0=0$, is standard Brownian motion relative to the given filtration if and only if its coordinates are local martingales andFor Brownian motion, independent centered Gaussian increments give the martingale property of each coordinate and of $Z_t^i{Z}_t^j-{\delta }_{ij}t$. The defining uniqueness of quadratic covariation consequently gives the bracket condition.

Conversely, suppose the local martingale and bracket conditions hold. For $\theta \in {\mathbb{R}}^d$, the Itô formula givessince the bracket term cancels the time derivative. On $[0,T]$, $|E_t|\le e^{|\theta {|}^{2}T/2}$, so its real and imaginary parts are bounded local martingales and hence true martingales. Thus, for $s\le t$,This is the characteristic function of $N(0,(t-s)I_d)$ and is independent of ${\mathcal{F}}_s$. Uniqueness of characteristic functions shows that $Z_t-Z_s$ has that Gaussian distribution and is independent of the past. Iterating over ordered times gives independent Gaussian increments; together with continuous paths and $Z_0=0$, these are precisely the defining properties of standard Brownian motion.

Fix $t>0$. For $0\le s\le t$, the residual observed path isThis Brownian bridge is independent of $(\lambda ,Y_t)$. Indeed it is independent of $\lambda$, while its covariance with $W_t$ is $sI-(s/t)tI=0$; zero covariance implies independence in the joint Gaussian law. The same argument for every finite collection of bridge times, followed by continuity and generation from rational times, gives independence of the entire bridge. Since the full observed path is determined by its endpoint and this independent bridge, Brownian endpoint sufficiency for a constant drift gives

First suppose the prior covariance matrix $V_0$ is positive definite. The Bayes' theorem multiplies the prior Gaussian density by the conditional observation density, whose dependence on $\lambda$ is $\exp ({\lambda }^T{Y}_t-t|\lambda {|}^2/2)$. Thus the posterior exponent, up to terms not depending on $\lambda$, isCompleting the square gives the Gaussian Brownian drift filter:For invertible $V_0$ these are equivalently $V_t=(V_0^{-1}+tI{)}^{-1}$ and ${\lambda }_t=V_t(V_0^{-1}{\lambda }_0+Y_t)$. At $t=0$ they reduce to the prior parameters.

The boxed formulas remain valid for a singular positive semidefinite matrix $V_0$. To see this directly without inverse prior matrices, the joint Gaussian random vector $(\lambda ,Y_t)$ hasThe latter matrix is invertible for $t>0$. The conditional multivariate normal distribution therefore has meanand covariance matrixAlternatively, subtract this mean from $\lambda$; the resulting Gaussian random vector has zero covariance with $Y_t$ and is therefore independent of it, which proves these conditional parameters.

Write the proposed innovation process in the observable formIt is adapted to the observation filtration and continuous. To prove the martingale property, take $0\le u<v$. Future increments of $W$ are independent of the joint information generated by $\lambda$ and the past of $W$, soAlso the tower property of conditional expectation gives $\mathbb{E}[{\lambda }_s\mid {\mathcal{G}}_u]={\lambda }_u$ for $s\ge u$. Conditional integration is legitimate because $\mathbb{E}|{\lambda }_s{|}^2\le \mathbb{E}|\lambda {|}^2<\infty$. Subtracting the integral proves $\mathbb{E}[{\hat{W}}_v-{\hat{W}}_u\mid {\mathcal{G}}_u]=0$. Removing a continuous finite variation process does not change quadratic covariation, henceThe Lévy characterization of multidimensional Brownian motion now applies: a continuous zero-start martingale with this bracket matrix is a standard $d$-dimensional Brownian motion relative to its given filtration. The observable innovation is a Wiener process in the observation filtration.

Differentiating the posterior formulas yields $V_t^{\prime }=-V_t^2$ andFor the final expression, the printed inverse matrices require the extra assumption that $V_0$ is positive definite. Under that assumption put $A_t=V_t^{-1}=V_0^{-1}+tI$. The Itô formula givessince $A_t{V}_t=I$ and $\mathrm{tr}(A_t{V}_t^2)=\mathrm{tr}{V}_t$. On the other hand,The trace terms cancel in the innovation exponential for Gaussian drift filtering, leavingThus $Z_0=1$ and $Z$ is a positive stochastic exponential, hence a nonnegative local martingale. To obtain the requested supermartingale conclusion explicitly, choose localizing stopping times ${\tau }_n\uparrow \infty$ for which $Z^{{\tau }_n}$ is a true martingale. The Conditional Fatou lemma givesThe same argument at $u=0$ gives $\mathbb{E}{Z}_v\le 1$, supplying integrability. The process is a supermartingale in the observation filtration. No unproved exponential-integrability condition on the random drift is required.

For a singular prior the literal inverse-matrix expression in the PDF is undefined, but there is a natural extension with the same conclusion. Define, for fixed $y$,For positive definite $V_0$, a Gaussian integral shows this reciprocal is exactly the printed expression. For any positive semidefinite $V_0$, the integral is positive and finite, $L_t=-{\Delta }_{y}L/2$, and ${\nabla }_y\log L(t,Y_t)={\lambda }_t$, by the same posterior weighting. The Itô formula consequently gives $d\log L(t,Y_t)={\lambda }_t^{T}d{Y}_t-|{\lambda }_t{|}^{2}dt/2={\lambda }_t^{T}d{\hat{W}}_t+|{\lambda }_t{|}^{2}dt/2$. Its reciprocal satisfies the boxed stochastic differential equation and the same supermartingale proof. This completes the singular-covariance extension while explicitly separating it from the undefined printed formula.

The Dambis-Dubins-Schwarz theorem states that if $N$ is a continuous local martingale with $N_0=0$ and $[N{]}_{\infty }=\infty$ almost surely, then for ${\sigma }_r=\inf \{t:[N{]}_t>r\}$, $W_r=N_{{\sigma }_r}$ is Brownian motion in ${\mathcal{F}}_{{\sigma }_r}$ and $N_t=W_{[N{]}_t}$. If the terminal bracket is finite, Brownian motion can be continued beyond it on an extension of the probability space. For a nonzero initial value, apply the theorem to $N-N_0$.

Apply time change to the vector $(M^1-M_0^1,M^2-M_0^2)$ with the common bracket $A$. The time-change theorem for continuous local martingales says that at inverse bracket times the processes remain continuous local martingales in the time-changed filtration and their quadratic covariations are composed with those times. Since $A$ is continuous and unbounded, $A_{{\sigma }_r}=r$, givingThe Lévy characterization of multidimensional Brownian motion states that a zero-starting continuous vector local martingale with bracket matrix $rI$ is standard multidimensional Brownian motion. Thus $\tilde{B}$ is standard two-dimensional Brownian motion; this step establishes the joint property, not merely the one-dimensional property of each coordinate.

The printed inverse is ${\tau }_r=\inf \{t:A_t\ge r\}$. For $r>0$ it is a stopping time, since $\{{\tau }_r\le t\}=\{A_t\ge r\}$, and ${\tau }_0=0$. The martingale is constant on each flat interval of $A$, as follows from its Dambis-Dubins-Schwarz representation. Hence $M_{{\tau }_r}^i=M_{{\sigma }_r}^i$, including any initial flat interval. Moreover ${\mathcal{F}}_{{\tau }_r}\subseteq {\mathcal{F}}_{{\sigma }_r}$. The common Brownian process is adapted to the smaller filtration, and its increments are independent of the larger past, so they are independent of the smaller past as well. Therefore the common-clock time change of orthogonal local martingales givesIf Brownian motion is understood to start at zero, the centered vector $B_r-B_0$ is the standard Brownian motion. Retaining the initial vector is the convention required in part (c).

A martingale describes an observable after its predictable drift has been removed. For a Markov chain the drift is the change produced by its transition operator; for Brownian motion it is one half of the Laplacian. Requiring the resulting processes to be martingales for a sufficiently rich class of observables determines the transition law itself. The same identities, combined with optional stopping, determine hitting probabilities, expected exit times and recurrence or transience. The duplicated Q5 paragraph in the PDF is a repeated printing of this one essay request.

For a finite or countable-state Markov chain with transition matrix $P$, define $(Pf)(x)={\sum }_{y}P(x,y)f(y)$ for a bounded test function $f$. The Markov property says $\mathbb{E}[f(X_{n+1})\mid {\mathcal{F}}_n]=(Pf)(X_n)$. Consequentlyis a martingale, since each increment has conditional expected value zero. Conversely, if these processes are martingales for every bounded $f$, taking the conditional expectation of an increment recovers that identity. With $f=1_{\{y\}}$ it gives $\mathbb{P}(X_{n+1}=y\mid {\mathcal{F}}_n)=P(X_n,y)$ for every state $y$. Thus, given the initial law, these identities characterize the Markov chain, not merely a few of its moments. This is the discrete-time Markov chain martingale characterization.

The drift operator is $P-I$. If $(P-I)h=0$ away from a target set, $h(X_{n\wedge T})$ is a stopped martingale, where $T$ is the first hit of that set. Under boundedness or suitable uniform integrability, the optional stopping theorem expresses $h$ in terms of boundary hitting probabilities. Similarly, if $(P-I)v=-1$ away from the target and $v=0$ on it, then $v(X_{n\wedge T})+(n\wedge T)$ is a martingale, giving $v(x)={\mathbb{E}}_{x}T$ when the terminal limit is justified.

For a concrete calculation, take a simple symmetric random walk on $\{0,1,\dots ,N\}$ stopped at $T$, its first hit of $0$ or $N$. In the interior, $h(i)=i$ has zero drift, while $v(i)=i(N-i)$ satisfiesThe stopped identity gives ${\mathbb{E}}_i(T\wedge n)=v(i)-{\mathbb{E}}_{i}v(X_{T\wedge n})\le v(i)$, proving $T$ finite with finite mean. Bounded convergence in both stopped identities now yields the gambler's ruin answersThis illustrates why the integrability justification is part of using a martingale, rather than an automatic consequence of writing an equation.

For a finite-state continuous-time Markov chain with generator matrix $Q=(q_{ij})$, the corresponding martingale problem for a continuous-time Markov chain usesThese are martingales. Conversely, suppose an adapted càdlàg finite-state process satisfies these identities. Let $P_t=e^{tQ}$, fix $T$, and set $g(s,i)=(P_{T-s}f)(i)$. Then ${\partial }_{s}g+Qg=0$. Apply deterministic integration by parts to the finite collection of indicator martingales to obtain the time-dependent Dynkin formula: $g(s,X_s)-g(0,X_0)-{\int }_0^s({\partial }_{r}g+Qg)(r,X_r)dr$ is a martingale. ThereforeThis recovers the Markov property and the entire semigroup, proving the characterization. On countable spaces one uses nonexplosion and localization to justify the analogous unbounded-rate formulas.

For $d$-dimensional Brownian motion, the Itô formula shows that, for $f\in C_c^{\infty }({\mathbb{R}}^d)$,is a true martingale, since its bounded derivatives make the integral square-integrable on each bounded horizon. This is the martingale problem for Brownian motion with generator $L=\frac{1}{2}\Delta$.

Here is why it also characterizes Brownian motion. Suppose a continuous adapted process $X$, with $X_0=x$, satisfies that martingale problem. Use smooth cutoff test functions equal to $x_i$ and $x_i{x}_j$ inside successively larger balls, and stop before leaving those balls. The coordinate tests show that $X^i-x_i$ are continuous local martingales; the product tests show that $X^i{X}^j-x_i{x}_j-{\delta }_{ij}t$ are local martingales. The Itô product rule consequently givesthe difference between the two proposed compensators is a continuous finite-variation local martingale, hence constant and zero at time zero. For each $u\in {\mathbb{R}}^d$, another application of the Itô formula gives the exponential local martingaleIts modulus is the deterministic quantity $e^{|u{|}^{2}t/2}$, so it is a true martingale on every bounded time interval. Taking its conditional expectation and dividing by its known value at time $s$ givesThe conditional characteristic function identifies a centered multivariate normal distribution of covariance $(t-s)I$, independent of the past. Together with continuity, this is the defining law of Brownian motion. This proves the Lévy characterization of multidimensional Brownian motion in the form needed for the generator characterization.

The analytic use of these martingales parallels the chain case. A harmonic function $h$ satisfies $\Delta h=0$, so it yields a stopped local martingale until the process leaves its domain of harmonicity. Bounded stopped observables permit optional stopping and solve boundary hitting problems. A solution of $\frac{1}{2}\Delta v=-1$ with zero boundary values gives an expected exit time. For example, in the ball of radius $R$, $v(x)=(R^2-|x{|}^2)/d$ gives ${\mathbb{E}}_x{T}_R=(R^2-|x{|}^2)/d$; stopping $|B_t{|}^2-dt$ first at $T_R\wedge n$ proves finiteness and justifies the limiting calculation. The Brownian local time construction in Q4 is another example: the predictable compensators of approximations to $|B_t|$ recover its increasing occupation term.

We now prove the required three-dimensional transience of Brownian motion in dimension at least three, in the strong sense $|B_t|\to \infty$ almost surely. Start first at a point $x$ with $r<|x|<R$, and let $T$ be the first exit from this annulus. The radial Laplacian in dimension three givesThus $|B_{t\wedge T}{|}^{-1}$ is a bounded martingale. The exit time is finite: stopping $|B_t{|}^2-3t$ gives $3\mathbb{E}(t\wedge T)\le R^2-|x{|}^2$. At exit, continuity puts $|B_T|$ at either $r$ or $R$. The optional stopping theorem therefore yieldsand hence the Brownian sphere-hitting probability in dimension threeAs $R\to \infty$, the events $\{{\tau }_r<{\tau }_R\}$ increase to $\{{\tau }_r<\infty \}$: a continuous path is bounded on the finite interval preceding any finite inner hit. This argument assumes no transience. ConsequentlyTo pass from a positive escape probability to almost-sure eventual escape, take Brownian motion started at zero. Let ${\sigma }_R$ be its first hit of the sphere of radius $R$. Stopping the squared-norm martingale proves $\mathbb{E}{\sigma }_R=R^2/3<\infty$. At ${\sigma }_R$, the Strong Markov property and the preceding hitting formula giveThe Strong Markov property here follows from independent Brownian increments by approximating a stopping time from above with dyadic-valued stopping times and using path continuity. Choose $R_m=2^{m}r$. The exit times ${\sigma }_{R_m}$ increase to infinity, since a continuous path is bounded on every compact time interval. The events of a return to the closed ball of radius $r$ after ${\sigma }_{R_m}$ decrease and have probabilities $2^{-m}$. Their intersection has probability zero. Thus almost surely there is some finite exit time after which the path never again enters that ball. Apply this simultaneously to all positive integer $r$. This proves the last visit to a bounded ball for three-dimensional Brownian motion conclusion and therefore

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.

Let $V_t=⟨M{⟩}_t$, and first suppose $V_{\infty }=\infty$ almost surely in addition to strict increase. The Dambis-Dubins-Schwarz theorem says thatdefine a Brownian motion $B$ in the time-changed filtration and giveEach $T_u$ is a stopping time and is finite. Continuity and strict increase of $V$ make $u\mapsto T_u$ continuous, $V_{T_u}=u$ and $T_{V_t}=t$.

Here are the martingale details behind this inverse-clock proof of the Dambis-Dubins-Schwarz theorem. Stopping a continuous local martingale when its bracket reaches $n$ makes it an L2-bounded continuous martingale. This follows from the stopped Itô isometry or from the $p=2$ estimate proved in Question 1(a), applied after localization. In particular it is uniformly integrable, and optional sampling is valid even at an unbounded stopping time by taking limits. Applying this to $M$ stopped at $T_n$ shows that $(M_{T_{u\wedge n}}{)}_{u\ge 0}$ is a martingale. Thus $B$ is a continuous local martingale. Time-changing $M^2-V$ in the same way shows $B_u^2-u$ is a local martingale, so $⟨B{⟩}_u=u$.

For completeness, the Lévy characterization of Brownian motion follows directly from the Itô formula. If a continuous local martingale $N$, starting at zero, has bracket $u$, thenis a complex local martingale. Its modulus is bounded on each deterministic finite horizon, so it is a true martingale there. ConsequentlyConditional characteristic functions give Gaussian increments independent of the past. Iterating this identity gives independent increments, and continuity completes the Brownian characterization. The identical vector argument proves the Lévy characterization of multidimensional Brownian motion when the bracket matrix is $uI$.

The printed strict-increase hypothesis does not imply $V_{\infty }=\infty$. For example, $M_t={\int }_0^t{e}^{-s}d{W}_s$ has strictly increasing bracket $(1-e^{-2t})/2$. To state the theorem under exactly the printed hypothesis, allow an independent enlargement of the probability space if the terminal clock $L=V_{\infty }$ can be finite.

On $\{L<\infty \}$ the martingale $M$ has a finite terminal limit. Indeed, stopping at each bracket level $n$ gives an L2-bounded continuous martingale which converges; on $\{V_{\infty }<n\}$ the stopped process is the original one. This proves the finite-bracket convergence lemma. Set $T_u=\infty$ when $u\ge L$ and continue $N_u=M_{T_u}$ by that terminal limit. The optional-sampling argument just given makes $N$ a continuous local martingale with bracket $u\wedge L$. Moreover $L$ is a stopping time in ${\mathcal{G}}_u={\mathcal{F}}_{T_u}$.

On a product extension add an independent Brownian motion $\beta$ in clock time, and putThe two summands have zero quadratic covariation, and their brackets are $u\wedge L$ and $(u-L{)}^{+}$. Thus $⟨\tilde{B}{⟩}_u=u$; the proved characterization makes $\tilde{B}$ Brownian. Since $V_t<L$ at every finite $t$ when $L$ is finite, $M_t={\tilde{B}}_{V_t}$ still holds. This is the finite-lifetime extension of the Dambis-Dubins-Schwarz theorem. An infinite clock gives Brownian motion on the original space; a finite clock may require the independent extension.

Interpret $f$ as a conformal bijection onto $D^{\prime }$. If it is merely an injective map into a larger target, the correct exit domain is $f(D)$ instead. Let $\zeta$ be the exit time from $D$, write $f=u+iv$, and identify planar Brownian motion with $B^1+i{B}^2$. The Cauchy-Riemann equations give $u_x=v_y$, $u_y=-v_x$, and both $u,v$ are harmonic functions. The Itô formula, localized on compact subsets of $D$, yieldsThus both coordinates are continuous local martingales, withSince $f^{\prime }$ never vanishes, $A$ is strictly increasing before $\zeta$. Let ${\eta }_s=\inf \{t:A_t>s\}$, defined for $s<A_{\zeta -}$. The optional time-change theorem gives continuous local martingale coordinates for ${\beta }_s=f(B_{{\eta }_s})$, with bracket matrix $sI$. The Lévy characterization of multidimensional Brownian motion makes $\beta$ a standard planar Brownian motion started at $f(z)$, up to its lifetime. Therefore conformal invariance of planar Brownian motion takes the formTo identify the lifetime, exhaust $D$ by relatively compact open sets $D_n$ with nested closures. The images $f(D_n)$ exhaust $D^{\prime }$ because $f$ is a homeomorphism onto $D^{\prime }$. Each stopped transformed path exits $f(D_n)$ at precisely the clock value of its exit from $D_n$. Increasing these exit times gives the exit lifetime of Brownian motion from $D^{\prime }$, including a possibly infinite lifetime. This identifies $A_{\zeta -}$ with that lifetime and needs no global extension of $f$ to the boundary.

Take the convention that the coordinates of planar Brownian motion are independent standard real Brownian motions, so its infinitesimal generator is $\frac{1}{2}\Delta$. Define the forward conformal Brownian clockSince a conformal bijection has nonzero derivative, $A$ is a strictly increasing continuous map from $[0,T)$ onto $[0,\tilde{T})$. The time used inside the original path is its inverse:Thus the interval direction for $\tau$ is the inverse-clock direction.

Write $\varphi =u+iv$. The Cauchy-Riemann equations and the Itô formula make $u(B_s),v(B_s)$ continuous local martingales, withIndeed both components are harmonic functions, and their gradients are orthogonal with the same squared norm. After the time change of a continuous process by $\tau$, their quadratic variations are $t$ and their quadratic covariation is zero. The Lévy characterization of multidimensional Brownian motion therefore identifies $\tilde{B}$ as planar Brownian motion started at $z^{\prime }$, up to its lifetime.

It remains to identify that lifetime as the exit time, rather than merely produce a local Brownian path. Boundedness of $D$ gives $T<\infty$ almost surely. For every compact subset $C\subset D^{\prime }$, its inverse image under $\varphi$ is compactly contained in $D$. As $s\uparrow T$, continuity gives $B_s\to B_T\notin D$, so $\varphi (B_s)$ eventually leaves $C$. Consequently the transformed path leaves every compact subset of $D^{\prime }$ at its lifetime. If $\tilde{T}<\infty$, its Brownian extension has a finite limit, and that limit is outside $D^{\prime }$; if $\tilde{T}=\infty$, the path never exits. In both cases its maximal lifetime is precisely the exit time from $D^{\prime }$.

The killed paths, together with their lifetimes, have the same law:

TakePart (i) proves that Brownian motion solves this martingale problem. For the converse, let $X$ be any continuous adapted process satisfying this martingale problem and $X_0=0$.

Set ${\tau }_n=\inf \{t\ge 0:|X_t|\ge n\}$. Choose a smooth cutoff function ${\chi }_n$ equal to one on a neighbourhood of the closed ball of radius $n$. Apply the martingale problem to ${\chi }_n(x)x_i$ and ${\chi }_n(x)x_i{x}_j$, and stop at ${\tau }_n$. Inside that ball, their Laplace operators are respectively $0$ and $2{\delta }_{ij}$. The optional sampling theorem therefore shows thatare martingales. Continuity makes ${\tau }_n\uparrow \infty$ almost surely. Hence every $X^i$ is a continuous local martingale. The Itô product rule and the fact that a continuous finite-variation local martingale is constant identify its quadratic covariations:Here ${\delta }_{ij}$ is the Kronecker delta.