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If $M$ is a continuous local martingale and $f\in C^2(\mathbb{R})$, then finite variation of $f(M)$ forces $f(M_t)=f(M_0)$. The Itô formula and the quadratic variation of a stochastic integral give ${\int }_0^t{f}^{\prime }(M_s{)}^{2}d[M{]}_s=0$. The occupation-times formula implies that $L_t^a(M)=0$ almost everywhere where $f^{\prime }(a)\ne 0$. On the zero set of $f^{\prime }$, its derivative $f^{\prime \prime }$ vanishes almost everywhere: zeros at which $f^{\prime \prime }\ne 0$ are isolated and hence countable. Thus ${\int }_0^t|f^{\prime \prime }(M_s)|d[M{]}_s=\int |f^{\prime \prime }(a)|L_t^a(M)da=0$. Both terms in the Itô formula vanish.

Work in the usual joint natural filtration of $B$ and $W$, in which both are Brownian motions. The integrand is $\mathrm{sgn}(B_s)$, as printed in the PDF. It is predictable and bounded, so $X$ is a zero-starting continuous local martingale. The quadratic variation of a stochastic integral givesIf the sign is defined to be zero at zero, the final equality still holds: Tonelli's theorem gives $\mathbb{E}{\int }_0^t{1}_{\{B_s=0\}}ds={\int }_0^t\mathbb{P}(B_s=0)ds=0$. Thus the Brownian zero set has zero Lebesgue time on every compact interval. The Lévy characterization of Brownian motion proves that $X$ is standard Brownian motion. This is the Brownian sign transform.

The correct quadratic variation of a stochastic integral identity isThe brackets around the final integrator are missing in the PDF. Without them the assertion is false: for $H=1$ and $M$ Brownian motion, the left side is $t$ while the printed right side is $B_t$.

For elementary predictable $H$, the integral is obtained by scaling each martingale increment by its known coefficient. Quadratic variation on each interval is therefore scaled by the square of that coefficient, giving the displayed formula. To pass to a general $L^2(M)$ integrand on $[0,T]$, take elementary $H_n\to H$ in the integrand norm and put $N_n=H_n\cdot M$, $N=H\cdot M$. The Itô isometry gives $\mathbb{E}(N_n-N{)}_T^2\to 0$, hence $\mathbb{E}[N_n-N{]}_T\to 0$. The bracket Cauchy-Schwarz boundand polarization show that $[N_n]\to [N]$ uniformly in probability. On the other side, Cauchy-Schwarz for the quadratic-variation measure givesTaking limits proves the formula in the square-integrable case. Stopping and pasting as in part b proves it for locally bounded predictable integrands and continuous local martingales.

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

Let $t(u)$ be the inverse image-capacity clock. Part (a) and the quadratic variation of a stochastic integral giveThe Dambis-Dubins-Schwarz theorem, or the Lévy characterization of Brownian motion, now shows that ${\tilde{\xi }}_u/\sqrt{6}$ is standard Brownian motion up to the terminal image time. If that terminal clock is finite, it can be extended by a fresh Brownian continuation; this does not assert anything about the subsequent image trace. The initial value is zero because $\Phi (0)=0$.

Differentiating the image Loewner chain with respect to its own half-plane capacity time givesThus, until its first hit of $[1,\infty )$, the reparametrized image $\Phi (\gamma )$ has the ordinary chordal SLE law from $0$ to infinity. This follows from the uniqueness of the Chordal Loewner equation for a given continuous driver; the continuous Loewner trace is recovered from the inverse-map limit. Localization can be removed by exhausting times strictly before this boundary contact. The terminal event is defined from the trace itself, so equality of the stopped trace laws follows as well.

On the other hand, the definition of Conformal invariance of SLE says that the full unparameterized image $\Phi (\gamma )$ has the law of chordal ${\mathrm{SLE}}_6$ from $\Phi (0)=0$ to $\Phi (\infty )=1$, namely the law of ${\gamma }^{\prime }$. Its stopping event at $[1,\infty )$ is exactly the image of the original stopping event at $(-\infty ,-1]$. Therefore the same stopped image has both laws: ordinary ${\mathrm{SLE}}_6$ from $0$ to infinity stopped at $T$, and ${\mathrm{SLE}}_6$ from $0$ to $1$ stopped at $T^{\prime }$. We concludeThe statement does not identify $T$ with $T^{\prime }$ in their original clocks. It is target-change locality of SLE6, derived from the zero drift and capacity time change rather than assumed as a theorem.

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.

Because $K$ is continuous, adapted, and strictly positive, $K^{-1/2}$ is predictable and locally integrable against $M$. DefineThe quadratic variation of a stochastic integral isThus $W$ starts at zero and is a continuous local martingale with bracket $t$. The Lévy characterization of Brownian motion makes it a Brownian motion in the original filtration.

The process $H$ is continuous and adapted, and ${\int }_0^t{H}_s^{2}ds={\int }_0^t{K}_{s}ds<\infty$ almost surely. The associativity of stochastic integration givesThis is the stochastic-integral representation from absolutely continuous quadratic variation.

Put $S=H\cdot M$ and $A_t={\int }_0^t{H}_s^{2}d[M{]}_s$. The candidate $A$ is continuous, adapted, nondecreasing, starts at zero and has integrable terminal value. We prove $S^2-A$ is a martingale, rather than assuming the desired quadratic variation of a stochastic integral.

The assumed boundedness of $M$ also makes $M^2-[M]$ a true martingale on every finite horizon. Indeed, apply its local martingale identity at a localizing sequence; bounded stopped squares and the Fatou lemma show that $[M{]}_t$ is integrable. The stopped expressions are then dominated by ${\sup }_s|M_s{|}^2+[M{]}_t$, so the dominated convergence theorem gives the unrestricted conditional identity on that horizon.

First let $H$ be a bounded simple predictable process. On each interval with coefficient $h$, an increment of $S$ is $h(M_t-M_a)$. The defining quadratic variation identity for $M$, together with the martingale property of $M$, makesa martingale for $t\ge a$. Expanding $S_t^2-S_a^2$ gives a term $2S_{a}h(M_t-M_a)$ and the square term above; their conditional expectations yield the desired identity on that interval. Integrability follows because the variables are square-integrable; concatenating the finitely many intervals gives $S^2-A$ a martingale.

For general bounded predictable $H$, choose bounded simple predictable processes $H^n\to H$ in $L^2(M)$, which is possible since they generate the predictable sigma-algebra. Set $S^n=H^n\cdot M$ and $A_t^n={\int }_0^t(H_s^n{)}^{2}d[M{]}_s$. The Itô isometry and Doob L2 maximal inequality imply $S_t^n\to S_t$ in $L^2$; the Cauchy-Schwarz inequality gives $(S_t^n{)}^2\to S_t^2$ in $L^1$ andPass the conditional expectation identity for $(S^n{)}^2-A^n$ to the $L^1$ limit. Thus $S^2-A$ is a martingale. Uniqueness in the defining quadratic variation property now gives

By the Itô formula,To compute the Stratonovich integral correction, apply the Itô formula also to $f^{\prime }$, which is twice continuously differentiable because $f$ is $C^3$. The continuous local martingale part of $f^{\prime }(X)$ is $\int f^{\prime \prime }(X_s)d{L}_s$, where $L$ is the continuous local martingale part of $X$. A finite-variation process has zero quadratic covariation with $X$, and the quadratic variation of a stochastic integral together with polarization identity givesAll these statements can be localized to compact ranges of $X$, so unbounded derivatives create no global integrability requirement. Substitute this identity into the Stratonovich integral to obtain the Stratonovich chain rule:This is the integrated meaning of the requested differential identity; there is no extra second-order term after the Stratonovich integral correction has been included.

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.

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.