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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

Set $N_t={\int }_0^t{Z}_s^{-1}d{Z}_s$, the stochastic logarithm of $Z$. On each finite horizon, a strictly positive continuous path has its reciprocal bounded; stopping at levels of $Z$, $Z^{-1}$ and $[Z]$ rigorously localizes this stochastic integral. Thus $N$ is a zero-start continuous local martingale, withThe Itô formula for the ordinary logarithm givesAccordingly, with the exponential convention printed in this question,The initial constant in $M$ contributes no quadratic variation. Under the standard normalized Doléans-Dade exponential convention, which subtracts the driver's initial value in the exponent, the same formula is written $Z=Z_0\mathcal{E}(N)$.

For uniqueness, suppose another continuous driver $L$ gives the printed exponential. Evaluating at zero forces $L_0=\log Z_0$. Applying the Itô formula to its exponential gives $d{Z}_t=Z_{t}d{L}_t$. Since $Z_t>0$, associativity of stochastic integration yields $d{L}_t=Z_t^{-1}d{Z}_t$. Thus $L_t=L_0+N_t=M_t$, up to indistinguishability of stochastic processes.

There is an initial-integrability qualification if “local martingale” is used with the usual integrable-initial-value definition. The displayed $M$ is a continuous local martingale when $\log Z_0\in L^1$, in particular when $Z_0$ is a positive deterministic number. An arbitrary positive integrable $Z_0$ does not ensure this. For example, take an ${\mathcal{F}}_0$-measurable positive integer $W$ with $\mathbb{P}(W=n)=6/({\pi }^2{n}^2)$ and put $Z_t=e^{-W}$ for all $t$. This is a bounded strictly positive martingale, but any representing driver must have $M_0=-W$, which is not integrable. Thus the unqualified claim is false under that definition. For general $Z_0$, the always-valid local-martingale statement is the zero-start representation $Z=Z_0\mathcal{E}(N)$; the printed representation additionally requires the stated integrability or the convention that allows arbitrary finite initial constants.

Work on a common stochastic basis on which $B,B^{\prime }$ are independent Brownian motions and the solutions are adapted; independent weak solutions can be put on the product stochastic basis. Independence gives $[B,B^{\prime }]=0$. Put $S=Z+Z^{\prime }\ge 0$ and define the predictable weightsThey are bounded and satisfy $u_t^2+v_t^2=1$ everywhere. DefineThis is a zero-start continuous local martingale, and its quadratic variation is $[\beta {]}_t={\int }_0^t(u_s^2+v_s^2)ds=t$. By the Lévy characterization of Brownian motion, $\beta$ is a Brownian motion in the common filtration.

On $\{S_t>0\}$, $\sqrt{S_t}{u}_t=\sqrt{Z_t}$ and $\sqrt{S_t}{v}_t=\sqrt{Z_t^{\prime }}$. On $\{S_t=0\}$, nonnegativity forces both $Z_t,Z_t^{\prime }$ to be zero, so the same identities hold. Associativity of stochastic integration therefore givesThis proves the addition law for square-root branching diffusions: $Z+Z^{\prime }$ is a weak solution with initial value $z+z^{\prime }$. Specifying the weights at zero is necessary to keep $[\beta {]}_t=t$ after extinction; simply setting both weights to zero would fail to produce a Brownian driver.

The previsible sigma-algebra $\mathcal{P}$ on $\Omega \times [0,\infty )$ is generated by $E\times \{0\}$ with $E\in {\mathcal{F}}_0$ and by $E\times (s,t]$ with $E\in {\mathcal{F}}_s$, $0\le s<t$. Equivalently it is the smallest sigma-algebra making every left-continuous adapted process measurable. A previsible process is a process measurable for $\mathcal{P}$.

A bounded simple predictable process has finitely many time intervals and bounded coefficients known at their left endpoints. One may use deterministic endpoints, or the customary more general elementary predictable process with stopping-time intervals:A bounded ${\mathcal{F}}_0$-measurable value at time zero may be added; it does not affect integration against a continuous martingale. A process with finite support is included by adding a zero coefficient on its terminal interval. Both versions of simple processes satisfy the arguments below. The elementary stochastic integral isWhen a stopping time is infinite, use the terminal value $M_{\infty }$, which exists in $L^2$ and almost surely for an L2-bounded continuous martingale. Refining intervals leaves this definition unchanged.

For a continuous local martingale $M$, its quadratic variation $[M]$ is the unique continuous adapted nondecreasing process starting at zero such that $(M_t-M_0{)}^2-[M{]}_t$ is a local martingale. It is predictable because it is continuous and adapted. To recover it from the path, use refining deterministic grids, for example $t_k^{(n)}=k{2}^{-n}$, and formOnly finitely many increments are nonzero on a compact time interval. ThenIn particular each fixed-time sum converges in probability, and a suitable subsequence converges uniformly on compact intervals almost surely. The partial last grid interval is included in this path approximation.

For $x\le 0$, $f_n(x)=|x|$; for $x>0$, only the interval $(0,1/n)$ contributes an error to the integral of the derivative. HenceBy the Itô formula, the continuous processesare nondecreasing, since $f_n^{\prime \prime }\ge 0$. Parts (a) and (b), together with the uniform approximation of absolute value, show that $C^{(n)}$ converges uniformly on compacts in probability toThe limit is continuous: the martingale integral is continuous, and the bounded-integrand finite-variation integral is continuous because $A$ is continuous. For any rational $s<t$, convergence in probability and $C_t^{(n)}-C_s^{(n)}\ge 0$ imply $Z_t-Z_s\ge 0$ almost surely. Intersect these countably many events and use continuity to extend to all real $s<t$. ThusSince $|X|={\mathrm{sgn}}_{-}(X)\cdot M+({\mathrm{sgn}}_{-}(X)\cdot A+Z)$ is a continuous local martingale plus continuous finite variation, $|X|$ is a continuous semimartingale. This is the smooth convex approximation proof of the Tanaka formula. With the specified sign value $-1$ at zero, $Z$ is the right local time at zero, rather than an unspecified symmetric normalization.

For an arbitrary continuous semimartingale starting at zero, use its continuous decomposition $X=M+A$, with both parts starting at zero, and letThe continuity and local finiteness of variation imply ${\rho }_k\uparrow \infty$ almost surely. Up to ${\rho }_k$ the required deterministic bound holds, and the stopped local martingale is bounded, hence a true $L^2$-bounded martingale. Apply the previous argument to $X^{{\rho }_k}$ for each integer $k$. Stopping the defining integrals gives precisely $Z_{t\wedge {\rho }_k}$. On the intersection of their full-probability events, each stopped $Z$ is nondecreasing; every finite interval is eventually contained before ${\rho }_k$, so $Z$ is nondecreasing globally. Therefore

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).

For $X=e^{\beta }\cos \theta$ and $Y=e^{\beta }\sin \theta$, apply the two-variable Itô formula. Independence gives $[\beta ,\theta ]=0$, while each individual bracket is $t$. The second derivatives in the two variables cancel, yieldingTheir integrands are continuous adapted and locally bounded, so both coordinates are continuous local martingales. The quadratic variation of a stochastic integral gives

Here is a full proof that the clock diverges. Let $S_0=0$, and successively defineBy recurrence of one-dimensional Brownian motion, all these times are finite almost surely. By the Strong Markov property at the returns $S_j$, the variables $D_j=(U_j-S_j)\wedge 1$ are independent identically distributed, lie in $(0,1]$ almost surely, and have a strictly positive mean. The strong law of large numbers gives ${\sum }_j{D}_j=\infty$ almost surely. On $[S_j,U_j]$, ${\beta }_s\ge -1$, soThe segments are disjoint, and their divergent duration also ensures they cannot accumulate at a finite time. This proves Divergence of a driftless Brownian exponential clock and

For $0\le s\le t$ and $E\in {\mathcal{F}}_s$, the definition of the Radon-Nikodym derivative givesThe density is nonnegative and integrable; with the probability normalization used here its expectation is one. Henceso $Z$ is a martingale. When $Z_t=h(X_t,t)$, the Itô formula under Wiener measure givesThe first term is a continuous local martingale, by localization on compact space-time sets. Since $Z$ is a martingale, uniqueness of the continuous semimartingale decomposition makes ${\int }_0^{t}g(X_s,s)ds=0$ identically. The integrand is continuous in time along every continuous path, so $g(X_t,t)=0$ for every $t$, outside one common null event.

For each fixed $t>0$, $X_t$ has a Gaussian distribution with strictly positive density everywhere on $\mathbb{R}$. Since $g(\cdot ,t)$ is continuous, $g(X_t,t)=0$ almost surely implies $g(x,t)=0$ for every real $x$. ThereforeIf the stated smoothness includes time zero, continuity extends the equation there. This is the backward heat equation. The full-support argument is needed to pass from an identity along Brownian paths to an equation at every spatial point.

Define $k(x,t)=h_x(x,t)/h(x,t)={\partial }_x\log h(x,t)$. Positivity and smoothness of $h$ make $k$ continuous and bounded on compact space-time sets. Brownian paths remain in compact spatial sets on each finite horizon, so ${\int }_0^{T}k(X_s,s{)}^{2}ds<\infty$ almost surely for every $T$. Thusis a well-defined continuous local martingale with $M_0=0$. The backward heat equation from part (b) removes the drift in Itô's formula and yieldsBy the allowed fact $Z_0=1$ and the uniqueness in part (a),

The finite-horizon form of the Girsanov theorem used here says: if $M_0=0$ is a continuous local martingale and its stochastic exponential $Z$ is a strictly positive true martingale of mean one on $[0,T]$, then under $d\mathbb{Q}{|}_{{\mathcal{F}}_T}=Z_{T}d\mathbb{P}{|}_{{\mathcal{F}}_T}$ every continuous $\mathbb{P}$-local martingale $N$ becomes the $\mathbb{Q}$-local martingale $N-[N,M]$. Quadratic variation is unchanged by this locally equivalent change of measure.

All the assumptions hold. The integrand defining $M$ is locally square integrable as checked above; $Z$ is strictly positive; and it is already a true martingale because it is the density process. On each finite horizon $Z_t=\mathbb{E}[Z_T\mid {\mathcal{F}}_t]$, so it is uniformly integrable on that horizon. There is no need to establish a separate Novikov condition, nor to assume equivalence on the infinite-time sigma-algebra.

Take $N=X$. Its covariation with $M$ is $[X,M{]}_t={\int }_0^{t}k(X_s,s)ds$. Hence $B_t=X_t-{\int }_0^{t}k(X_s,s)ds$ is a continuous $\mathbb{Q}$-local martingale starting at zero with $[B{]}_t=t$. The Lévy characterization of Brownian motion makes it Brownian under $\mathbb{Q}$. Since this argument holds on every finite horizon for the same process $B$,This is the Brownian change of measure by a positive space-time harmonic function, a Doob h-transform with drift ${\partial }_x\log h$.

Apply the Itô formula to $Y_t=e^{B_t+t/2}$. The time derivative contributes $Y_{t}dt/2$, and the quadratic-variation correction contributes another $Y_{t}dt/2$. ThusBoth the drift $b(y)=y$ and diffusion coefficient $\sigma (y)=y$ are globally Lipschitz. The global existence theorem for stochastic differential equations with Lipschitz coefficients gives a unique strong stochastic solution for a fixed driving Brownian motion and initial value; in particular pathwise uniqueness holds. The explicit solution is strictly positive.

Write $N_t={\int }_0^t{Y}_{s}d{B}_s$. This is a continuous local martingale withThe strong law for Brownian motion gives $B_t/t\to 0$ almost surely, so $\log Y_t/t\to 1/2$, $Y_t\to \infty$, and $H_t\to \infty$. The Dambis-Dubins-Schwarz theorem therefore supplies a Brownian motion $\beta$ in the inverse-clock filtration with $N_t={\beta }_{H_t}$ for all $t$. Finally $d{H}_s=Y_s^{2}ds$, hence $Y_{s}ds=Y_s^{-1}d{H}_s$. Integrating the stochastic equation givesThe infinite clock lifetime verifies that this Brownian motion is defined for every nonnegative clock time without an additional extension.

Put $R_t=|W_t|$. First verify that the origin is not reached, so that differentiating the radius is legitimate. On ${\mathbb{R}}^3\setminus \{0\}$ the function $F(w)=|w{|}^{-1}$ is harmonic. For $T_{\epsilon }=\inf \{t:R_t=\epsilon \}$ with $0<\epsilon <1$, the Itô formula makes $F(W_{t\wedge T_{\epsilon }})$ a martingale on each finite horizon: its stochastic integrand has norm at most ${\epsilon }^{-2}$. Since $F(W_0)=1$,Let $t\to \infty$, then $\epsilon \downarrow 0$. Any path hitting zero must first hit every positive $\epsilon$, so $\mathbb{P}(\exists t:R_t=0)=0$.

For $r(w)=|w|$, ${\partial }_{i}r=w_i/r$ and $\Delta r=2/r$ in three dimensions. Itô's formula, initially localized away from zero, yieldsThe continuous local martingale ${\gamma }_t={\sum }_i{\int }_0^t(W_s^i/R_s)d{W}_s^i$ has bracket $[\gamma {]}_t={\int }_0^t{\sum }_i(W_s^i/R_s{)}^{2}ds=t$ and starts at zero. By the Lévy characterization of Brownian motion, $\gamma$ is Brownian motion. ThusThis proves that the radial process is a three-dimensional Bessel process solving the same equation as $X$ in part (b). The permitted uniqueness in distribution identifies their laws as processes.

The strong law in part (a) gives $Y_s\to \infty$, and the inverse clock satisfies $J_t\to \infty$. Hence $X_t=Y_{J_t}\to \infty$. Equality of process laws transfers this almost sure path property to $R$.

For the further logarithmic rate, fix $0<\eta <1/2$. Almost surely, for all sufficiently large $s$, $|B_s|\le \eta s$. With a finite random lower cutoff $S$, for $u>S$,Taking logarithms and dividing by $u$ gives lower and upper asymptotic bounds $1-2\eta$ and $1+2\eta$. Letting $\eta \downarrow 0$ along a countable sequence proves the logarithmic growth of an exponential Brownian clock with positive drift, $\log H_u/u\to 1$. Also $\log Y_u/u=B_u/u+1/2\to 1/2$. Since $t=H_{J_t}$,Again transfer the full path event by uniqueness in distribution. We obtain the logarithmic escape rate of three-dimensional Brownian motion

For an independent way to identify the value once existence of the almost sure limit is granted, use Brownian scaling. At each fixed $t$, $W_t/\sqrt{t}$ has the law of $G+(1/\sqrt{t},0,0)$, where $G$ is standard three-dimensional Gaussian. Its norm converges in distribution to the strictly positive finite variable $|G|$, so $\log (|W_t|/\sqrt{t})$ is tight. ThereforeAn existing almost sure limit must agree with this probability limit. This guesses and identifies the value through scaling, while the inverse-clock argument proves that the almost sure limit actually exists.

Fix $T<\infty$. By uniform continuity of each path of the continuous local martingale on $[0,T+1]$, its modulustends to zero almost surely. For every $s\le T$, comparison of powers of each increment givesThe squared-increment sum converges in probability to $[X{]}_T$, by the dyadic definition of quadratic variation. Including the last full grid interval instead of the last partial interval changes that sum by a quantity bounded by $2{\eta }_n^2$, so the ceiling convention has the same limit. In particular these squared-increment sums are bounded in probability.

Since $p-2>0$, a quantity tending to zero almost surely multiplied by a family bounded in probability tends to zero in probability. HenceThis is uniform convergence on compacts in probability, and proves the superquadratic case of dyadic power variation of a continuous local martingale.

Fix the specified time $t$. With the path modulus ${\eta }_n$ on $[0,t+1]$ as above, comparison of increment powers now givesThe hypothesis makes the second factor eventually bounded almost surely, while the first tends to zero almost surely because $2-p>0$. Thus the squared-increment sums tend to zero almost surely. They also converge in probability to the quadratic variation $[X{]}_t$, so uniqueness of the limit in probability implies $[X{]}_t=0$ almost surely.

The quadratic variation is nondecreasing, hence it vanishes throughout $[0,t]$. To see explicitly that the continuous local martingale must then be constant, localize it to square-integrable stopped martingales. The Itô isometry givesRemove the localization, apply the conclusion at every rational $s\le t$, and use continuity. Since $X_0=0$, this provesThus $X$ is indistinguishable from the zero process on that interval. This is the subquadratic case of dyadic power variation of a continuous local martingale.

Let $g_t=P_{1-t}(\nabla f)(X_t)$. The triangle inequality for the conditional average givesThe independent coordinate Brownian motions have zero cross-variation, so the quadratic variation of a stochastic integral givesThe Dambis-Dubins-Schwarz theorem says that a continuous local martingale $N$ starting at zero whose quadratic variation diverges admits a Brownian motion $B$, with the inverse-clock filtration, such that $N_t=B_{[N{]}_t}$. If the total clock is finite, one may enlarge the probability space and append independent Brownian motion after the clock terminates. This gives the same identity up to the original lifetime.

Apply this theorem to $N=M-M_0$. The clock bound implies the event inclusionThe latter event is contained in the union of the upper and lower crossing events. Symmetry of Brownian motion givesThis uses only an event inclusion; independence between the Brownian motion and its random clock is neither assumed nor needed.

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.