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The Girsanov theorem construction proves a weak stochastic solution, but does not by itself establish a strong stochastic solution. It constructs $X=x+W$ first, and then constructs a new driving Brownian motion $B=W-\int b(X_s)ds$. Adaptation of $X$ to the original filtration of $W$ does not by itself prove adaptation to the smaller natural filtration of $B$. Inverting this relation requires an additional result; uniqueness in law alone is insufficient in general.

For the particular coefficients in part (b), however, the literal claim that a strong solution may fail is incorrect. The strong existence theorem for additive-noise SDEs with bounded measurable drift gives strong existence and pathwise uniqueness for bounded Borel state-dependent $b$ and unit diffusion. Consequently every weak realization here is determined by its driving noise and initial value. This theorem is deeper than the Girsanov theorem argument and no proof is needed here. Examples with path-dependent drift or with a nonconstant diffusion coefficient do not contradict it.

The valid distinction is therefore: the argument in part (b) has established weak existence and uniqueness in law; strong existence requires an additional theorem, and in this class that theorem is available. A primary reference for the correction is On strong solutions and explicit formulas for solutions of stochastic integral equations.

The Lévy characterization of Brownian motion states that an adapted continuous real process $M$ with $M_0=0$ is standard Brownian motion relative to $({\mathcal{F}}_t)$ if and only if it is a continuous local martingale and $[M{]}_t=t$. The multidimensional version replaces the bracket identity by $[M^i,M^j{]}_t={\delta }_{ij}t$.

For the converse direction, a standard Brownian motion is a continuous martingale, and its independent centered increments give quadratic variation $t$. Distinct independent coordinates have zero quadratic covariation.

For the substantive direction, fix a vector $\theta \in {\mathbb{R}}^d$. The Itô formula applied toshows that its drift is zero: the second-order term is $-\frac{1}{2}|\theta {|}^2{E}_{t}dt$ and cancels the time derivative. Thus $E(\theta )$ is a complex local martingale. On $[0,T]$ its modulus is the deterministic value $e^{|\theta {|}^{2}t/2}\le e^{|\theta {|}^{2}T/2}$; localization and the dominated convergence theorem make it a true martingale. HenceThis conditional characteristic function is the deterministic characteristic function of $N(0,(t-s)I)$. For $A\in {\mathcal{F}}_s$, multiplying by $1_A$ and taking expectations, then using uniqueness of characteristic functions, proves that $M_t-M_s$ has this multivariate normal distribution and is independent of ${\mathcal{F}}_s$. Iterated conditioning gives independent increments with Gaussian distributions. Continuity, the zero initial value and these increment laws are the definition of standard Brownian motion. The case $d=1$ proves the requested statement; the vector form will also be used below.

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.

For a harmonic function $u$ on $\mathbb{C}\cong {\mathbb{R}}^2$, the Itô formula givesStop on leaving increasing discs and at increasing deterministic times. On each such interval the gradient is bounded, so the stochastic integral is a martingale. This proves that $u(B)$ is a continuous local martingale.

If $|u|\le K$, this is a bounded local martingale and hence a true martingale. Start Brownian motion at an arbitrary $z$. Its Gaussian heat kernel givesFor a unit vector $e$, ${\partial }_e{p}_t(v)=-(e\cdot v)p_t(v)/t$, so $\int |{\partial }_e{p}_t(v)|dv=\sqrt{2/\pi }/\sqrt{t}$. Integrating the directional derivative along the segment from $z$ to $w$ yieldsLetting $t\to \infty$ gives the harmonic Liouville theorem:This Gaussian heat-kernel proof of the harmonic Liouville theorem uses the bounded martingale identity established above.

We prove the extension from spatial test functions to time-dependent test functions for a diffusion martingale problem directly, without assuming a driving Brownian motion. Fix $0\le s<t$ and a deterministic partition $s=t_0<\cdots <t_n=t$. For each $j$, $g_j(x)=f(t_j,x)$ lies in $C_b^2$. The defining martingale problem givesCall the sum $S_n$. Telescoping separates the spatial and temporal changes:The sample path $X$ is uniformly continuous on $[s,t]$ and has compact range. Joint continuity of ${\partial }_{t}f$, the spatial first derivatives, and the spatial second derivatives therefore gives, as the mesh tends to zero,almost surely. For bounded $a,b$ and bounded derivatives of $f$, both integrals are uniformly bounded on this finite horizon, as are the two endpoint terms. The dominated convergence theorem permits passage to the limit in the expectation against $1_A$. Thus $M_t^f$ is integrable and $\mathbb{E}[1_A(M_t^f-M_s^f)]=0$ for every $A\in {\mathcal{F}}_s$, provingAdaptation and continuity follow from the formula, so $M^f$ is a continuous martingale. Under the more general integrability condition in part (a), the same proof uses an integrable bound proportional to $1+{\int }_s^t({\sum }_i|b_i(X_u)|+{\sum }_i{a}_{ii}(X_u))du$; positive semidefiniteness bounds every off-diagonal entry by the diagonal ones. With only local assumptions this reasoning proves a local martingale, and an integrability hypothesis is needed to upgrade that conclusion.

The Itô formula for $f(t,x)=e^{x+at}$ gives the stochastic differential equationWrite $N_t={\int }_0^t{Z}_{u}d{B}_u$ for its continuous local martingale part. Its clock is the quadratic variationBecause $Z_t>0$, $A$ is strictly increasing. For $s<A_{\infty }$, its inverse ${\tau }_s$ satisfies $A_{{\tau }_s}=s$ and $d{\tau }_s/ds=Z_{{\tau }_s}^{-2}$. The Dambis-Dubins-Schwarz theorem, or the Lévy characterization of Brownian motion after the optional time-change theorem, makes ${\beta }_s=N_{{\tau }_s}$ a Brownian motion for the time-changed filtration. With $Y_s=Z_{{\tau }_s}$,This is the Bessel process equation with dimension $\boxed{2+2a}$, on the positive stochastic interval. This establishes the Exponential Brownian-to-Bessel time change.

The lifetime convention is important: the inverse is defined only for $s<A_{\infty }$. If $a<0$, the strong law for Brownian motion gives $B_t/t\to 0$, so $Z_t\to 0$ exponentially and $A_{\infty }<\infty$. Then $Y_s\to 0$ as $s\uparrow A_{\infty }$, and this is its first hit of zero after continuous extension. If $a>0$ the clock diverges by the same strong law. If $a=0$ it diverges by infinite occupation time of one-dimensional Brownian motion on a bounded interval, since $e^{2B_t}$ is bounded below there. Thus the lifetime is infinite when $a\ge 0$. Negative dimensions are interpreted by the same positive-state equation killed at zero; no post-zero Bessel process behavior is asserted.

Work before the first zero of $Z$, with $Z_0=z>0$. For the Bessel process equation, the Itô formula applied to $x^r$ givesIts quadratic variation isThis clock is strictly increasing on the positive-state interval. For $s$ before its terminal clock value, let $Y_s=W_{{\sigma }_s}$ and let ${\beta }_s=({\int }_0^{\cdot }r{Z}_u^{r-1}d{B}_u{)}_{{\sigma }_s}$. The Dambis-Dubins-Schwarz theorem makes $\beta$ a Brownian motion there. Since $d{\sigma }_s=ds/(r^2{Z}_{{\sigma }_s}^{2r-2})$, the drift changes toConsequently the Power time change of a Bessel process givesThe initial value is $z^r$, and this equation identifies a Bessel process of dimension ${\delta }^{\prime }$ until its first zero. Localize both equations on compact subintervals of $(0,\infty )$ to justify the inverse and all stochastic integrals for arbitrary $r>0$.

For completeness, the inverse-clock lifetime really is the positive Bessel process lifetime: the map $z\mapsto z^r$ is a homeomorphism of $(0,\infty )$, and exits from nested compact intervals correspond under the clock. There is no finite explosion at infinity for a positive Bessel process, since the Itô formula for $Y^2$ gives $d(Y^2)=2Yd\beta +{\delta }^{\prime }ds$; stopping at zero or level $R$ yields a probability of hitting $R$ before a fixed $T$ bounded by $(Y_0^2+\max ({\delta }^{\prime },0)T)/R^2$. If the transformed lifetime were finite while $Y$ stayed in a compact positive interval, the inverse clock and the original solution could both be continued. Thus a finite lifetime is approach to zero. For $\delta <2$, also ${\delta }^{\prime }<2$ and the positive-state Bessel process hits zero; for $\delta \ge 2$ the transformed lifetime is infinite. No continuation through zero is needed.

If the initial value is already zero, the interval before its first zero is empty; the positive-state argument concerns starts $z>0$.

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:

In the half-plane-capacity parameterization, let $W$ be standard real Brownian motion and put ${\xi }_t=\sqrt{\kappa }{W}_t$. Solve the Chordal Loewner equationThe points for which this flow ceases to exist form the growing compact H-hulls $K_t$, with half-plane capacity $2t$. The process is an SLE trace when its initial segments generate these hulls andThe limit is the continuous trace of a Loewner chain started at zero. If the curve disconnects a region from infinity, that region is included in $K_t$; the hull is not generally just the visited curve. For $\kappa =0$ the driver is identically zero and the trace is the deterministic vertical slit ${\gamma }_t=2i\sqrt{t}$.

For the SLE driver $d{\xi }_t=\sqrt{6}d{W}_t$, apply the Itô formula to ${\tilde{\xi }}_t={\varphi }_t({\xi }_t)$. At a fixed spatial point, ${\varphi }_t$ has finite variation locally in time by the derivative formula in part (c); there is no extra spatial martingale term. ConsequentlyThis exact cancellation is the target-change locality of SLE6. For a general parameter the transformed SLE driving function would have drift $(\kappa /2-3){\varphi }_t^{\prime \prime }({\xi }_t)$.

Define the increasing clock and its inverse byThe derivative $p_t$ is real and strictly positive before $T$, so this is a valid continuous time change of a continuous process. The driver is a continuous local martingale with quadratic variation $[\tilde{\xi }{]}_t=6a(t)$ and initial value zero. The Dambis-Dubins-Schwarz theorem gives a standard Brownian motion $\hat{W}$ such thatIf the terminal clock is finite, the Brownian motion may be extended beyond that stopping time; only its stopped part is used here. Reparameterizing the image maps givesThese are exactly the defining Chordal Loewner equation and half-plane-capacity parameterization for ${\mathrm{SLE}}_6$. The trace is ${\tilde{\gamma }}_{t(s)}$. Hence the image curve is also ordinary chordal ${\mathrm{SLE}}_6$ after this time change, up to the specified stopping time:

The printed assertion is false as a supremum bound uniform over shrinking bandwidths. The supplied hint proves a pointwise stochastic order with constants uniform in $x$. Taking the supremum inside the probability changes the problem: a growing number of spatial windows produces a logarithmic cost. We first establish the valid consequence of the hint, then give a counterexample satisfying the printed assumptions.

For $I=I_{x,h}=[x-h,x+h]$, let $p=\mu (I)$, $A_T={\int }_0^T{1}_I(X_t)dt$ and $N_T={\int }_0^T{1}_I(X_t)\sigma (X_t)d{W}_t$. The Invariant distribution of an Itô diffusion has a probability density function bounded above and bounded away from zero on the required compact interval. Choosing $h_0<M/4$ therefore gives constants $c_0,C_0>0$ such that $c_{0}h\le p\le C_{0}h$, uniformly for $|x|\le M/2$ and $0<h<h_0$.

For the centered function $f=1_I-p$, $\int |f|d\mu =2p(1-p)\le 2p$. Outside $[-M,M]$, $f=-p$. The given second-moment estimate and the Chebyshev inequality yieldOn the event $A_T\ge Tp/2$, the drift coefficient contributes an average whose error is at most $R{h}^{\alpha }$, by Hölder continuity. The Itô isometry and stationarity give $\mathbb{E}{N}_T^2\le \Vert \sigma {\Vert }_{\infty }^{2}Tp$. Another use of the Chebyshev inequality provesThis also accounts for the zero-denominator convention through the first exceptional event. The bound proves pointwise $R{h}^{\alpha }+O_{\mathbb{P}}((Th{)}^{-1/2})$, uniformly in the location's tail probabilities. It supplies no bound for the probability of a supremum over all locations.

For an explicit counterexample, take $\sigma \equiv 1$ and the bounded globally Lipschitz functionIt has Lipschitz constant $R=1$, exponent $\alpha =1$, and satisfies the inward-drift conditions with $M=2$, $\gamma =2$. The stochastic differential equation has a unique strong solution of a stochastic differential equation. Its Invariant distribution of an Itô diffusion has densityThe exponent is zero on $[-1,1]$, equals $-(|x|-1{)}^2$ when $1<|x|<2$, and equals $3-2|x|$ outside $[-2,2]$. This verifies integrability, positivity and a constant density ${\rho }_0=Z^{-1}$ on the central interval. Start the Itô diffusion in this Invariant distribution of an Itô diffusion to obtain the required stationary process.

Set $h=T^{-1/3}$ and choose $N=\lfloor (4h{)}^{-1}\rfloor$ windows with centers $x_j=-1/2+(4j-2)h$, $1\le j\le N$. Their intervals $I_j=[x_j-h,x_j+h]$ are disjoint and lie in $[-1,1]$, where the drift is zero. WriteThen ${\hat{b}}_T(x_j,h)=M_j(T)/A_j(T)$ whenever $A_j(T)>0$. The same occupation bound as above gives $\mathbb{E}(A_j(T)-q{)}^2\le C_{1}T{h}^2$ for $T\ge 1$. With $\delta =T^{-1/6}$, a union bound and the Chebyshev inequality implyThe quadratic variations are $⟨M_j⟩=A_j$, and the cross quadratic covariations vanish since the windows are disjoint. Each clock tends to infinity almost surely: for any fixed window, the occupation estimate at times $2^k$, followed by the Borel-Cantelli lemmas, gives $A_j(2^k)/(2^k\mu (I_j))\to 1$. The Knight theorem for orthogonal martingales therefore represents $M_j(t)=B_j(A_j(t))$ using independent standard Brownian motions $B_1,\dots ,B_N$. This theorem is applied to the finite collection for each $T$; no growing-dimensional central limit theorem is being assumed.

On the preceding clock event, compare each time-changed Brownian motion with its value at deterministic time $q$. The Brownian reflection principle and a union bound show that, for every fixed $\epsilon >0$,This bound does not require the Brownian motions to be independent of their clocks. In particular, ${\max }_j|M_j(T)-B_j(q)|/\sqrt{q}\to 0$ in probability. The variables $B_j(q)/\sqrt{q}$ are independent $N(0,1)$ random variables. Their Gaussian maximum exceeds $\sqrt{\log N}$ with probability tending to one: for $r\ge 1$, integration of the normal density over $[r,r+1/r]$ gives $\mathbb{P}(|Z|>r)\ge c{r}^{-1}{e}^{-r^2/2}$, and henceSince $A_j(T)\le (1+\delta )q$ on the clock event, it follows that, for some $c_3>0$,But $\sqrt{Th}Rh=1$ for this bandwidth. Thus the error after subtraction of $Rh$ and multiplication by $\sqrt{Th}$ is not bounded in probability, contradicting the printed stochastic order. The valid pointwise bound above is provable; the uniform shrinking-bandwidth assertion is not. This counterexample shows why a uniform rate needs at least a $\sqrt{\log (1/h)}$ enlargement of the noise scale. If $h$ is instead held fixed and constants may depend on $h$, this counterexample does not contradict that different interpretation.

The natural filtration of $S$ is the natural Brownian filtration: since $\sigma >0$, the equation $W_t=(\log (S_t/S_0)-\mu t)/\sigma$ recovers the entire Brownian motion history from the stock history. For $s<t$, independent increments and the moment-generating function of a normal distribution giveAll these expectations are finite. Since $S_s>0$, the martingale identity holds for every $s<t$ exactly when the last exponential equals one. The required logarithmic drift isThis is the distinction between the drift of $\log S$ and the drift of $S$: here the latter is zero.

Write $h=T-t$. Conditional on ${\mathcal{F}}_t^S$, independent increments of the Brownian motion givewhere $Y$ has the standard normal distribution and is independent of ${\mathcal{F}}_t^S$. Factoring $S_t$ out of the positive part identifies the normalized Black-Scholes call function:The European call option payoff is integrable, because $0\le (S_T-K{)}^{+}\le S_T$ and $\mathbb{E}{S}_T=S_0$. The tower property of conditional expectation therefore proves that $C$ is a true martingale. At $t=T$, the boundary value $F(0,m)=(1-m{)}^{+}$ gives $C_T=(S_T-K{)}^{+}$. The answer is the conditional payoff martingale

First work with the filtration ${\mathcal{G}}_t={\mathcal{F}}_t^W\vee \sigma (1_{\{u\le \tau \}}:0\le u\le t)$, which records the Brownian motion and whether default has occurred. On $\{s>\tau \}$ all subsequent defaultable stock prices vanish. On $\{s\le \tau \}$, the memoryless property of the exponential distribution gives survival from $s$ to $t$ with conditional probability $e^{-\lambda (t-s)}$. The residual survival event is independent of the future independent increments of the Brownian motion. Thus, for deterministic $s<t$,Also $\mathbb{E}{\hat{S}}_t=e^{\lambda t}{e}^{-\lambda t}{S}_0=S_0$, so this is a true martingale, with no localization needed. Its natural filtration is contained in $\mathcal{G}$; applying the tower property of conditional expectation to the displayed identity yields the requested natural-filtration martingale property:The printed convention keeps the stock alive at $t=\tau$. It is a left-continuous convention at that one random time. Replacing it by $1_{\{t<\tau \}}$ gives the usual càdlàg zero-recovery default model; since $\mathbb{P}(\tau =t)=0$ at each fixed $t$, the fixed-time martingale calculations above are unchanged.

Since $P(T,T)=1$, part (b) shows that the density process of the T-forward measure isIts terminal expectation is one and it is strictly positive, so the stated Radon-Nikodym derivative defines an equivalent probability measure. By the Girsanov theorem,is a Brownian motion under ${\mathbb{Q}}_T$. Substitution of $d{W}_t=d{W}_t^T-b_t^{T}dt$ cancels the entire instantaneous forward rate drift:Because the integrand is deterministic and bounded, its Itô integral is square-integrable on $[0,T]$. With the usual fixed initial instantaneous forward rate curve, the requested true martingale is