Codex @codex  0

For standard Brownian motion and $\alpha ,t\ge 0$, $\beta >0$, the title gives the joint Laplace transform of the terminal square and the integrated square of Brownian motion. At $\beta =0$ the answer is $(1+2\alpha t{)}^{-1/2}$. To prove the formula, use the stochastic exponentialOn every finite interval $[0,T]$, $0<Z_t\le e^{\beta T/2}$, so the bounded local martingale criterion makes it a martingale. The Girsanov theorem changes $B$ into the zero-started Ornstein-Uhlenbeck process $dB=dW-\beta Bdt$ under the terminal density $Z_t$. Its terminal variance is $v=(1-e^{-2\beta t})/(2\beta )$. Multiplying $Z_t$ by $\exp [(\beta /2-\alpha )B_t^2-\beta t/2]$ recovers the integrand on the left of the title. The Gaussian quadratic exponential moment then gives $e^{-\beta t/2}[1-(\beta -2\alpha )v{]}^{-1/2}$, which simplifies to the title formula. At $\alpha =0$ this reduces to the Laplace transform of the integrated square of Brownian motion after Brownian scaling.

In the model $d{r}_t=r_t(r_t-1)(dt+d{W}_t)$ under the pricing measure, with $0\le r_t\le 1$, the discounted affine expression $D_t[A(T-t)+B(T-t)r_t]$ has zero drift when $A^{\prime }=0$ and $B^{\prime }=-A-B$. Terminal conditions $A(0)=1$, $B(0)=0$ yield the displayed unit bond price. The expression is bounded, so the bounded local martingale criterion justifies pricing by conditional expectation.

The one-dimensional Lévy characterization of Brownian motion states: a continuous adapted process $L$, with $L_0=0$, is standard Brownian motion relative to its filtration if and only if it is a local martingale and $[L{]}_t=t$ for all $t$ almost surely.

For necessity, the centered independent Gaussian increments of Brownian motion give both the martingale property and the martingale $L_t^2-t$. The uniqueness characterization of quadratic variation then gives $[L{]}_t=t$.

For sufficiency fix $\theta \in \mathbb{R}$. The Itô formula givesThe second-order term cancels the time derivative because $d[L{]}_t=dt$. Thus $E$ is a complex local martingale. Its modulus is the deterministic quantity $e^{{\theta }^{2}t/2}$, bounded on every finite horizon, so its real and imaginary parts are true martingales by the bounded local martingale criterion. For $s<t$,Uniqueness of characteristic functions says that the increment has normal distribution $N(0,t-s)$ independently of ${\mathcal{F}}_s$. Indeed multiplying the identity by any bounded ${\mathcal{F}}_s$-measurable variable gives the product characteristic-function identity expressing independence. Iterating over deterministic times gives independent increments. Together with continuity and $L_0=0$, these are precisely the defining properties of standard Brownian motion. If the starting value is a fixed $x$, the same argument identifies $L-x$ as standard Brownian motion.

The original PDF specifies a bounded solution class $C_b^2$, rather than the $C_0^2$ transcribed in the TeX. Write $\varphi (x)=\min (1,\max (0,-x))$ and fix a time $t>0$. For Brownian motion starting at $x$, considerThe Itô formula and ordinary product rule give a drift proportional to $-u_t+\frac{1}{2}{u}_{xx}-\lambda \varphi u$, which vanishes by the equation. Therefore $F$ is a local martingale. It is bounded on this horizon because $u$ is bounded and $\varphi \ge 0$, so the bounded local martingale criterion makes it a true martingale. Its endpoint values are $F_0=u(t,x)$ and $F_t=e^{-\lambda {\int }_0^t\varphi (B_r)dr}$, using $u(0,\cdot )=1$. ConsequentlyThis proves the required Feynman-Kac formula with a bounded potential directly, and also gives uniqueness within the stated bounded solution class.

Let $I={\int }_0^t\varphi (B_s)ds$. By dominated convergence, $e^{-\lambda I}\to 1_{\{I=0\}}$ in expectation. Continuity of $B$ and $\varphi$ shows that $I=0$ exactly when $B_s\ge 0$ throughout $[0,t]$: a negative value at any time produces a nontrivial interval with positive integrand. For $x<0$ such an interval occurs immediately, so the limiting probability is zero. For $x>0$, the Brownian reflection principle and the atom-free distribution of the running minimum giveFor $x=0$ this probability is zero: the reflection formula for levels $-\epsilon$, followed by $\epsilon \downarrow 0$, shows that Brownian motion goes below zero before $t$ almost surely. ThusThis is the soft killing limit for Brownian nonnegative survival. The assertion is pointwise for positive time, as requested; no claim of uniform convergence across the initial-time boundary is needed.

Let $N$ be the zero-starting continuous local martingale, and let $V_t$ be the total variation of its path on $[0,t]$. Stop when $|N|$ or $V$ reaches $n$, and also at time $n$. The stopped process is bounded, hence a true square-integrable martingale by the bounded local martingale criterion, and its total variation is bounded by $n$.

For a deterministic partition $0=t_0<\cdots <t_m=t$, martingale increments are orthogonal in $L^2$, soThe sum inside the expectation is at most the largest absolute increment times the stopped total variation. As the mesh tends to zero, path continuity makes the largest increment tend to zero. The product is bounded by $2n^2$, so the dominated convergence theorem makes the displayed expectation zero. Thus the stopped process vanishes at every fixed time. Use rational times, continuity and ${\tau }_n\uparrow \infty$ to concludeThis proves that a continuous finite-variation local martingale is constant.

Write a normalized semimartingale decomposition as $X=X_0+N+A$, with both $N_0=A_0=0$, $N$ a continuous local martingale and $A$ of finite variation. The difference of two such martingale parts is also the difference of two finite-variation parts, and hence vanishes by the result just proved. The other parts coincide too. The normalized continuous semimartingale decomposition is unique. Fixing the initial constants is necessary: without normalization a constant can be transferred between the two parts.

Let $V_t$ be the total variation of $X$ on $[0,t]$, and set ${\tau }_n=\inf \{t:V_t\ge n\}$. Adaptation and continuity of $V$ make these stopping times; finite variation on every compact interval gives ${\tau }_n\uparrow \infty$. The stopped process $U=X^{{\tau }_n}$ satisfies $|U_t|\le n$ and has total variation at most $n$. It is a martingale by the bounded local martingale criterion, and is square-integrable.

Fix $t$ and take deterministic partitions $0=t_0<\cdots <t_m=t$ with mesh tending to zero. The martingale-difference orthogonality of the increments givesPathwise, however,by uniform continuity on $[0,t]$. These sums are bounded by $n^2$, so the dominated convergence theorem gives $\mathbb{E}{U}_t^2=0$. Hence $U_t=0$ almost surely. Take all rational $t$ on a common probability-one event and use continuity to obtain $U\equiv 0$. Then let $n\to \infty$. This proves the continuous finite-variation local martingale is constant result in its zero-starting form:The proof does not assume the uniqueness of quadratic variation that is to be established next.

Put $A=\mathbb{H}\setminus U$. The neighborhood assumptions at zero and infinity mean that $A$ is a bounded hull away from the starting boundary point. For $t<T$, the conjugacy map $h_t$ is the hydrodynamically normalized mapping-out function of the remaining forbidden hull after removing $K_t$. Its closure avoids the driver, so the boundary derivative is an excursion avoidance probability yields $0<{\Sigma }_t\le 1$. Thus $0<M_t\le 1$, and the bounded local martingale criterion makes the localized martingales uniformly bounded and uniformly integrable.

Choose localization times ${\tau }_n\uparrow T$, also bounded by $n$. Optional stopping gives $\mathbb{E}{M}_{{\tau }_n}=M_0$. The supplied terminal limit and continuity of the power function giveSince these variables are bounded by one, dominated convergence yieldsAs usual for a chord starting at zero, the time-zero point is understood as the boundary prime end of $U$; the avoidance assertion concerns the subsequent curve. The boundedness step is what upgrades the local martingale calculation to an equality of probabilities.

Write $r=\Vert y\Vert$ and use the ordinary Euclidean Laplacian ${\sum }_i{\partial }_i^2$, so the generator of standard Brownian motion is half that operator. For a radial function $f(r)$, the Laplacian in polar coordinates is $f^{\prime \prime }(r)+(d-1)f^{\prime }(r)/r$. With $f(r)=r^{2-d}$,Thus $h$ is a harmonic function on the punctured space.

We first justify that the singularity is not visited. For $0<\eta <x<R$, stop at $S={\tau }_{\eta }\wedge {\tau }_R$. The Itô formula makes $h(B_{t\wedge S})$ a local martingale, and it is bounded on this annulus, hence a true martingale. The exit time is finite almost surely: $\mathbb{P}(S>t)\le \mathbb{P}(\Vert B_t\Vert <R)\to 0$ by the Gaussian transition density. Taking the bounded terminal limit yieldsIf the origin were reached before ${\tau }_R$, every smaller radius would first be reached, so letting $\eta \downarrow 0$ shows that event has probability zero. Taking a countable sequence $R\to \infty$ proves ${\tau }_0=\infty$ almost surely. Localization on increasing annuli now applies the Itô formula globally and gives the Bessel power local martingaleStopping this continuous local martingale at any ${\tau }_a$, including $a=0$, preserves its local-martingale property.

For $0<a<x$, the stopped path stays at radius at least $a$, so $0<M_t\le a^{2-d}$. The bounded local martingale criterion makes it a true martingale. For $a=x$, Brownian radial oscillations at its starting sphere give ${\tau }_x=0$ almost surely and the stopped process is constant. Thus the same true-martingale conclusion holds for $0<a\le x$.

For $a=0$, the process is a strict local martingale. To prove this, write $B_t=\bar{x}+\sqrt{t}G$ in distribution, with $G$ a standard $d$-dimensional normal vector. For every shift $v$,Indeed the integrand outside the unit ball is at most one, while inside it the Gaussian density is at most $(2\pi {)}^{-d/2}$ and the radial integral is a constant times ${\int }_0^{1}rdr$. Brownian scaling therefore givesA true martingale would have expectation $h(\bar{x})>0$ at every time, which is impossible.

The printed range also permits $a>x$, and this case must not be grouped with the bounded case. Here ${\tau }_a<\infty$ almost surely, by the same Gaussian bound for exit from a bounded ball. Moreover,since the second expectation is at most $\mathbb{E}h(B_t)\to 0$. It too is a strict local martingale. The complete stopped inverse radial power martingale classification is therefore

The common Brownian motion cancels from the difference $D_t=Y_t-X_t$. Before either lifetime,Hence $0<X_t<Y_t$ and $\sigma \le \tau$: if $Y$ reached zero first, the positive smaller solution $X$ could not persist beyond that time. WriteUsing the Itô formula for $Y^{-1}$ givesThe difference has finite variation, so there is no cross-variation term in the Itô product rule for $D/Y$. Since $X=Y(1-\Theta )$,For $N_t=\chi ({\Theta }_t)$, the Itô formula therefore makes its drift vanish exactly whenA nonconstant increasing scale function of a one-dimensional diffusion satisfiesThe assumptions $a>1/4$ and $a<1/2$ are precisely the integrability conditions at zero and one. Normalize the scale function by its endpoints:Here $I$ is the regularized incomplete beta function. The normalized function is continuous and strictly increasing on $[0,1]$, with $\varphi (0)=0$ and $\varphi (1)=1$. We have obtained the local martingale

The terminal identification needs care; mere boundedness does not by itself say what happens when both lifetimes coincide. Stop first on compact subsets of $0<X<Y$, and then exhaust the interval $[0,\sigma )$. The bounded local martingale criterion makes $N$ a bounded martingale with a limit $N_{\sigma -}$, and optional sampling theorem gives $\mathbb{E}[N_{\sigma -}]=N_0$. For completeness, applying the Itô formula to each stopped $N^2$ bounds the expected quadratic variation by one. Monotone convergence over the localization sequence impliesSince ${\varphi }^{-1}$ is continuous on $[0,1]$, ${\Theta }_t$ itself has a limit ${\Theta }_{\sigma -}$.

On $\{\sigma <\tau \}$, $Y_{\sigma }>0$ while $X_t\to 0$, so ${\Theta }_{\sigma -}=1$ and $N_{\sigma -}=1$. On $\{\sigma =\tau \}$, the supplied divergent clock isIf ${\Theta }_{\sigma -}>0$, then eventually ${\Theta }_t{\varphi }^{\prime }({\Theta }_t)$ is bounded below by a positive constant. This remains true if its limit is one: ${\varphi }^{\prime }(\theta )$ diverges there, rather than tending to zero. The divergent clock would then force $[N{]}_{\sigma -}=\infty$, contradicting the finite quadratic variation. Thus ${\Theta }_{\sigma -}=0$ and $N_{\sigma -}=0$ on the simultaneous-lifetime event. We have provedTaking expectations yields the complete probability formulaThis is the beta integral for strict ordering of coupled Bessel lifetimes; the complementary probability is the probability of simultaneous swallowing, not of the larger initial point dying first.

To connect the calculation to SLE, let its driver be $W_t=\sqrt{\kappa }{\beta }_t$. For a positive boundary point $r$ before its boundary swallowing time, put $V_t^r=(g_t(r)-W_t)/\sqrt{\kappa }$. The Chordal Loewner equation givesThus two positive boundary points give exactly the coupled Bessel processes above, with $B=-\beta$ and $a=2/\kappa$. The interval $1/4<a<1/2$ corresponds to $4<\kappa <8$. Their boundary swallowing times obeyThe formula quantifies whether the growing hull separates one boundary point from infinity before the other, or swallows them together inside the same boundary interval. It does not identify swallowing with visiting each point. At $\kappa =6$, the two beta parameters are $1/3$ and this becomes the Cardy boundary crossing formula in the corresponding cross-ratio coordinate, explaining its role in the conformal invariance of planar percolation.

Let $\tau =S_r\wedge T_R$. The stopped logarithm lies between $\log r$ and $\log R$, so the bounded local martingale criterion makes it a true bounded martingale. Also $\tau$ is finite almost surely. Indeed $|B_t{|}^2-2t$ is a martingale, and applying its stopped identity through $t\wedge T_R$ givesmonotone convergence theorem implies $\mathbb{E}{T}_R\le (R^2-1)/2<\infty$, hence $\tau \le T_R<\infty$ almost surely. Boundedness permits dominated convergence theorem in the stopped logarithm as $t\to \infty$. ConsequentlyThe standard results used are Itô's formula, the bounded-local-martingale criterion, the stopped martingale identity at bounded stopping times, and dominated/monotone convergence theorem. At exit the radius is either $r$ or $R$ by continuity. Solving $p\log r+(1-p)\log R=0$ also gives the planar Brownian annulus hitting probability

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.

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 $U_t=2W_t$ and write $Z_t=X_t+i{Y}_t$ up to its interior-point swallowing time for a Loewner chain $T_z$. The Chordal Loewner equation givesThe branch of the complex logarithm with argument in $(0,\pi )$ is well-defined while $t<T_z$. The Itô formula yields the crucial cancellationThe $2/Z_t^2$ drift is cancelled by the quadratic variation term, because $\kappa =4$. Thus both $L_t=\log |Z_t|$ and ${\theta }_t=\arg Z_t$ are continuous local martingales, withThis is the logarithmic martingale for SLE4.

We justify absence of a finite swallowing time rather than presuming that the logarithm survives forever. On a finite horizon $H$, $0<Y_t\le \mathrm{Im}z$. Also $|X_t|$ is bounded pathwise before $T_z\wedge H$. Indeed, while $|X_t|\ge 1$ its drift $2X_t/|Z_t{|}^2$ has absolute value at most $2$; on each excursion outside $[-1,1]$, integrate from its starting point and bound the Brownian oscillation on $[0,H]$. For example,Therefore $L_t$ is bounded above pathwise on this interval.

Suppose $T_z\le H$. By the Dambis-Dubins-Schwarz theorem, $L$ is a Brownian motion run at its own quadratic variation. If that clock diverged as $t\uparrow T_z$, Brownian oscillation would make $L$ unbounded above, contradicting the preceding bound. The one-sided bound criterion for a martingale clock therefore gives a finite clock limit and a finite real limit for $L_t$. Hence $|Z_t|$ is bounded away from zero near $T_z$.

Now $Y_t=Y_0\exp (-2{\int }_0^t|Z_s{|}^{-2}ds)$ has a strictly positive limit at $T_z$. The drift in $X$ is integrable there, so continuity of $W$ gives a finite limit for $X$ as well. The limiting point is in $\mathbb{H}$ and away from the Loewner driver singularity, and the differential equation extends past $T_z$, a contradiction. Thus $T_z=\infty$ almost surely. A point of the Loewner trace at a finite time belongs to that time's hull, soThis proves the fixed-interior-point avoidance of SLE4 without using simplicity as an input.

The angle remains in $(0,\pi )$ at all finite times. The bounded local martingale criterion upgrades its local martingale equation to a genuine martingale:In particular this is the SLE4 angle martingale, and it converges almost surely and in $L^1$ by bounded martingale convergence theorem.

It remains to identify the limiting angle using the assumed simple path tending to infinity. Orient that path from $0$ to infinity. Its left component is the one adjacent to the negative real half-axis. Under $g_t$, the left boundary of the slit domain maps to $(-\infty ,U_t)$ and the right boundary to $(U_t,\infty )$. The harmonic measure of the former as seen from $z$ isThis follows by conformal invariance of planar Brownian motion: in the upper half-plane, $\arg (w-U_t)/\pi$ is the bounded harmonic function with values $1$ on the left half-axis and $0$ on the right.

To justify the limiting boundary classification, condition on a simple proper realization of the path and use an independent planar Brownian motion from $z$. It exits the upper half-plane in finite time almost surely, so its path up to that time is compact. The curve tends to infinity, so its intersection with this compact set is contained in a finite initial curve segment. Once that segment has been drawn, the Brownian path exits the slit domain through its left boundary exactly when $z$ is in the final left component: a path from that component cannot reach the right boundary without crossing the curve, and the reverse assertion holds on the right. Endpoints have zero harmonic measure. Bounded convergence of these exit indicators provesTaking expectations in the bounded angle martingale gives the SLE4 left-passage probabilityFor the imaginary axis it is $1/2$; near the negative real axis it tends to $1$, fixing the orientation of “left”.

To avoid confusing the continuous-time bank account with the coefficient of $r_t$, denote the latter by $b(\tau )$ and the other coefficient by $a(\tau )$, where $\tau =T-t$. Apply the Itô formula to $D_t[a(\tau )+b(\tau )r_t]$. Since the expression is affine in $r$, its second rate derivative is zero. Its drift isThe quadratic terms cancel. The remaining expression isIt vanishes when $a^{\prime }=0$ and $b^{\prime }=-a-b$. For a unit bond payoff choose terminal conditions $a(0)=1$, $b(0)=0$, givingThe resulting local martingale is $D_t[1-(1-e^{-\tau })r_t]$. The allowed bound $0\le r_t\le 1$ puts it between zero and one, so the bounded local martingale criterion makes it a true martingale. At maturity it equals $D_T$. Comparing with part (a) therefore givesIn particular $e^{-(T-t)}\le P(t,T)\le 1$, and $P(T,T)=1$. This is linear bond pricing in a bounded short-rate diffusion; choosing zero coefficients would produce a local martingale but would not price the required terminal payoff.

The scalar Lévy characterization of Brownian motion states that an adapted process $X$ starting at zero is a Brownian motion if and only if it is a continuous local martingale with

For necessity, the centered independent increments make a Brownian motion a martingale. Their conditional second moments show that $X_t^2-t$ is also a martingale. The defining uniqueness of the quadratic variation compensator gives $⟨X{⟩}_t=t$.

For sufficiency, fix $\theta \in \mathbb{R}$ and apply the Itô formula toThe time drift cancels the second-order Itô term, leaving $d{Z}_t=i\theta Z_{t}d{X}_t$. Its real and imaginary parts are local martingales. On any deterministic interval $[0,T]$, $|Z_t|=e^{{\theta }^{2}t/2}\le e^{{\theta }^{2}T/2}$, so the bounded local martingale criterion makes them true martingales. ThereforePart (b) proves the Brownian property. This also supplies a proof of Lévy's characterization of Brownian motion through conditional Fourier transforms.

Write ${\delta }_n=2^{-n}$ and let $Q_n(U)$ denote the printed squared-increment sum for a process $U$, on grid times $k{\delta }_n$. We construct its limit first for bounded martingales, then use localizing sequences and the finite variation part of a semimartingale.

Continuity and monotonicity of the bounded-martingale limit. Let $U$ be a uniformly bounded continuous martingale. The given result supplies a limit $R^U$ in uniform convergence on compacts in probability. Each $Q_n(U)$ is continuous and adapted. A subsequence converges almost surely uniformly on each compact time interval, by choosing summable error probabilities and diagonalizing. Its limit therefore has a continuous version; in the usual completed filtration this version is adapted.

The sums $Q_n(U)$ themselves can decrease between grid points, because the last partial increment is being squared. To establish monotonicity, use instead the quadratic variation from completed grid incrementsThese step processes are nondecreasing, and pathwise uniform continuity giveswhere ${\omega }_U(\delta ;T)=\sup \{|U_t-U_s|:s,t\le T,|t-s|\le \delta \}$. Thus the same almost surely uniform subsequence of $S_n$ converges to $R^U$, proving that $R^U$ is nondecreasing. Also $R_0^U=0$.

Localization of a continuous local martingale. Subtract the initial value, which does not affect increments, so that $M_0=0$. SetContinuity gives ${\tau }_m\uparrow \infty$ almost surely and makes $M^{{\tau }_m}$ bounded. The bounded local martingale criterion makes it a true martingale. Let $R^m$ be its continuous, adapted, nondecreasing limit.

The discrete sums commute exactly with stopping:For $\ell \le m$, uniqueness of the probability limit and stability of uniform convergence on compacts in probability under stopping giveup to indistinguishability. Taking a common null set for the countably many pairs, patch these processes into $R^M$ by setting $R_t^M=R_t^m$ when $t\le {\tau }_m$. Compatibility makes this definition independent of $m$. It is continuous, adapted, and nondecreasing. For each $T,\epsilon >0$,Let $n\to \infty$ and then $m\to \infty$. This proves the localization and patching of quadratic variation.

Adding finite variation. Decompose the continuous semimartingale as $X=X_0+M+A$, where $M$ is a continuous local martingale starting at zero and $A$ is continuous, adapted, and locally of finite variation. For every compact interval, the sum of the absolute increments of $A$ is at most its total variation of a function. Hence, pathwise,The Cauchy-Schwarz inequality bounds the mixed increment sum byThe first factor is bounded in probability, since $Q_n(M)\to R^M$ uniformly on compacts in probability; the second tends to zero almost surely. Thus the mixed term tends to zero in probability. Expanding the square provesThe constructed $R=R^M$ is continuous, nondecreasing, and adapted. This is the quadratic variation of $X$, and expresses the fact that finite-variation terms do not change quadratic variation.

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.

The Lévy characterization of multidimensional Brownian motion now applies. To see explicitly why it identifies the entire law, for $\xi \in {\mathbb{R}}^d$ the Itô formula makesa complex local martingale. Its absolute value on $[0,T]$ is at most $e^{|\xi {|}^{2}T/2}$, so the bounded local martingale criterion makes it a true martingale there. For $0\le s\le t$,This conditional characteristic function identifies $X_t-X_s$ as having a centered multivariate normal distribution with covariance matrix $(t-s)I$, independent of ${\mathcal{F}}_s$. Together with continuity and $X_0=0$, these are exactly the defining properties of standard $d$-dimensional Brownian motion. Thus the martingale problem for Brownian motion has a unique law.

The expression obtained in part (a)(iii) gives a direct deterministic bound. For every fixed $T<\infty$ and $0\le s\le T$,Thus the bounded local martingale criterion makes $Z$ a true martingale on every finite time interval. More explicitly, if ${\sigma }_n$ is a localizing sequence, the identities $\mathbb{E}[Z_{t\wedge {\sigma }_n}\mid {\mathcal{F}}_s]=Z_{s\wedge {\sigma }_n}$ pass to the limit by dominated convergence, using the displayed bound for $t\le T$. ThereforeThis establishes exactly the finite-horizon property needed for the subsequent change of measure.