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For a continuous local martingale $M$, its stochastic exponential isThe Itô formula gives $d{Z}_t=Z_{t}d{M}_t$ and $Z_0=1$, so $Z$ is a strictly positive local martingale. The finite-horizon Girsanov theorem says: if $Z$ is a true martingale on $[0,T]$ and $d\tilde{\mathbb{P}}=Z_{T}d\mathbb{P}$ on ${\mathcal{F}}_T$, then for every continuous local martingale $N$ under $\mathbb{P}$,For an infinite-horizon change of measure, assume $Z$ is a uniformly integrable martingale and use its terminal density $Z_{\infty }$. In particular, if $M=\int \theta dB$, the process $\tilde{B}=B-\int \theta ds$ is Brownian motion under the new probability measure, by the Lévy characterization of Brownian motion.

The quadratic variation cannot change. Adding the continuous finite-variation process $-[N,M]$ leaves $[\tilde{N}]=[N]$. Also the same pathwise partition sums defining $[N]$ have the same limit under an absolutely continuous change of measure; convergence in probability transfers by integrability of the density. Thus the bracket of the unchanged process $N$, now a semimartingale, is also unchanged. The local martingale property itself need not survive. This is quadratic covariation under an absolutely continuous measure change.

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.

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.

The Tanaka equation is an example:It has weak solutions and uniqueness in law, but lacks pathwise uniqueness. The value $s(0)=1$ is deliberate: using the sign function with value zero at zero would admit the identically zero solution and would change the example.

For a quick verification, take a Brownian motion $W$ and put $B=\int s(W)dW$. Its quadratic variation is $t$, so it is a Brownian motion by the Lévy characterization of Brownian motion, and associativity of stochastic integration gives $W=\int s(W)dB$. Every weak solution likewise has quadratic variation $t$ and therefore the law of Brownian motion. On this same enlarged filtration, both $W$ and $-W$ solve the equation driven by $B$: $s(-W)=-s(W)$ away from zero, and the Brownian zero set has zero time measure, so the discrepancy at zero contributes nothing to the Itô integral. These solutions differ with positive probability at any positive time. This proves the claimed distinction without relying on a choice of zero convention left unstated.

Let $W$ be a Brownian motion, set $X=W$, and defineThis is a continuous local martingale with quadratic variation $t$, hence is Brownian by the Lévy characterization of Brownian motion. Since ${\mathrm{sign}}^2=1$,which gives a weak solution.

Suppose a strong solution existed. It has quadratic variation $t$, so $X$ itself is Brownian. The supplied Tanaka formula gives $|X_t|=B_t+L_t$, and $L$ is adapted to the completed filtration of $|X|$. Hence $B$ and $|X|$ generate the same completed filtration. Strongness would make $X$, and therefore $\mathrm{sign}(X_t)$, measurable with respect to the history of $|X|$. But a Brownian excursion has an independent symmetric sign; in particular, conditionally on the reflected Brownian path, the sign at a fixed nonzero time is not measurable. This contradiction proves that no strong solution exists.

Write the semimartingale decomposition as $U=M+A$, where $M$ is a continuous local martingale and $A$ has finite variation. Applying Itô formula to $\log Z_t(z)$ shows that its finite-variation part isIt vanishes for every $z$. Multiplying by $Z_t(z{)}^2$ givesSubtract this identity for two points with distinct $Z_t$ to obtain $d{A}_t=0$; then $d⟨M{⟩}_t=4dt$. Since the curve starts at zero, $U_0=0$. The Lévy characterization of Brownian motion now gives $U_t=2B_t$. Hence the Loewner chain is

The Dambis-Dubins-Schwarz theorem says that there is Brownian motion $W$ such thatWhen $[M]$ is strictly increasing, define its inverseand set $W_s=M_{T_s}$. Optional sampling shows that $W$ is a continuous local martingale, while time change gives $[W{]}_s=s$. The Lévy characterization of Brownian motion makes $W$ Brownian, and inverse time change gives the displayed representation.

The process $M$ is continuous and Gaussian. Since $A$ has finite variation,The Lévy characterization of Brownian motion makes $M$ a Brownian motion in the enlarged filtration.

Moreover,Every finite vector from $M$ is jointly Gaussian with $X_1$, so zero covariance implies independence. Thus the whole process $M$ is independent of $X_1$.

Enlarge the space by an independent Brownian motion $W$ and defineThe two terms have zero cross-variation andThe Lévy characterization of Brownian motion makes $B$ a Brownian motion. The residual $\int 1_{\{A=0\}}dX$ has zero quadratic variation and is therefore constant, so

The integrand $\mathrm{sign}({\beta }_s)$ is a bounded previsible process, so $B$ is a continuous local martingale. The quadratic variation of a stochastic integral isbecause the Brownian zero set has zero Lebesgue measure. Since $B_0=0$, the Lévy characterization of Brownian motion shows that $B$ is a standard Brownian motion.

On $[\tau ,\sigma )$ the Brownian path does not meet zero, so the power function $x\mapsto |x{|}^{\delta }$ is twice continuously differentiable along the path. The Itô formula givesSince $Y_t=|B_t{|}^{\delta }$, this becomeswhere ${\hat{B}}_t={\int }_0^t\mathrm{sign}(B_s)d{B}_s$ is a standard Brownian motion by the Lévy characterization of Brownian motion. Thus the displayed equation in the paper is valid after the customary renaming of $\hat{B}$ as $B$.

As printed, the requested conclusion is false for $\delta \ne 1$. On an interval on which $X$ stays positive, the time change of a continuous process satisfies $d{\tau }_s=ds/({\delta }^2{X}_s^{2(\delta -1)/\delta })$. The time-changed martingale term has quadratic variation $s$, so the Lévy characterization of Brownian motion identifies it with a standard Brownian motion $\tilde{B}$. Dividing the drift in part (a) by the derivative of the clock givesConsequently the construction actually satisfieswhich is the Bessel process equation of dimension $2-1/\delta$. It equals the paper's claimed drift $(\delta -1)/(2X_s)$ only when $\delta =1$. The mismatch between the specified power, clock, and conclusion is therefore a typographical error in the question.

The $d$-dimensional Lévy characterization of Brownian motion says that a continuous local martingale $X$ with $X_0=0$ is a standard $d$-dimensional Brownian motion exactly whenfor all $i,j$ and $t$.

One direction follows directly from independent Gaussian increments. Conversely, fix $\theta \in {\mathbb{R}}^d$. Applying Itô formula and the bracket assumption shows thatis a complex local martingale. After stopping $X$ on leaving large balls it is bounded, so optional sampling and then dominated convergence give, for $s<t$,This is the characteristic function of $N(0,(t-s)I_d)$ and is deterministic. Thus each increment is Gaussian with the required covariance and independent of the past. Together with continuity, these are precisely the defining properties of standard $d$-dimensional Brownian motion.

Apply Itô formula to $Z_t^{(A)}=\exp (A{X}_t-A^{2}t/2)$. Its semimartingale decomposition isThe second term is a continuous finite-variation process. Since $Z^{(A)}$ is assumed to be a local martingale, uniqueness of the semimartingale decomposition makes this term identically zero. Both $A$ and $Z^{(A)}$ are nonzero, so the quadratic variation of $X$ is $[X{]}_t=t$.

The Lévy characterization of Brownian motion now says that $X_t-X_0$ is a Brownian motion. Consequentlyis a constant multiple of an exponential Brownian martingale. It is therefore a true martingale for every $a\in \mathbb{R}$.

Set $K_t=\sqrt{H_t}$ and define the stochastic integralStrict positivity and predictability of $H$ make the integrand locally admissible. The process $W$ is a continuous local martingale starting from zero, and the quadratic variation of a stochastic integral givesBy the Lévy characterization of Brownian motion, $W$ is a Brownian motion. The associativity of stochastic integration then yieldswhich is the required representation.

With $s(x)=1$ for $x\ge 0$ and $s(x)=-1$ for $x<0$, the equation starting at zero has uniqueness in law but not pathwise uniqueness. For existence, take a Brownian motion $W$ and set $B=\int s(W)dW$; the Lévy characterization of Brownian motion and associativity of stochastic integration give $W=\int s(W)dB$. Every solution has quadratic variation $t$ and hence the Brownian motion law. Both $W$ and $-W$ are solutions driven by the same $B$, since the time spent at zero is zero and $s(-W)=-s(W)$ away from zero. Using $s(0)=0$ instead would allow the constant solution and invalidate this uniqueness-in-law example.

The radial part of three-dimensional Brownian motion started away from the origin is a Bessel process of dimension three. The radial Laplacian is $\Delta |x|=2/|x|$, so the Itô formula gives the displayed equation. The radial martingale has unit quadratic variation rate and is Brownian motion by the Lévy characterization of Brownian motion. The reciprocal $1/R$ is a local martingale. Stopping it on $(\epsilon ,M)$ gives ${\mathbb{P}}_x(T_{\epsilon }<T_M)=\epsilon (M-x)/(x(M-\epsilon ))$. Letting $\epsilon \downarrow 0$ shows that zero is not hit.

A dimension-two Bessel process starting at $r>0$ has the displayed marginal law, where $G_1,G_2$ are independent standard normal random variables. Take the norm of a planar Brownian motion starting at $(r,0)$. Planar Brownian motion avoids a fixed point, so the norm stays positive; Itô formula gives drift $1/(2R)$ and a martingale term of bracket $t$. The Lévy characterization of Brownian motion identifies that term as its driving Brownian motion. The squared radius divided by $t$ has a noncentral chi-squared distribution with two degrees of freedom and noncentrality $r^2/t$.