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For solutions with independent Brownian drivers in a common filtration, combine the drivers with weights $\sqrt{Z/(Z+Z^{\prime })}$ and $\sqrt{Z^{\prime }/(Z+Z^{\prime })}$ where the sum is positive. At zero use one original driver. The resulting continuous local martingale has bracket $t$ and is Brownian motion by the Lévy characterization of Brownian motion. Multiplying its differential by $\sqrt{Z+Z^{\prime }}$ gives exactly the sum of the original stochastic differentials.

For independently driven nonnegative squared Bessel processes $X,Y$, put $Z=X+Y$. Where $Z>0$, use Brownian weights $\sqrt{X/Z}$ and $\sqrt{Y/Z}$. Where $Z=0$, fill the weights with $(1,0)$. Their squared norm is always one and their cross-variation is zero, so their stochastic integral is Brownian motion by the Lévy characterization of Brownian motion. Multiplying by $2\sqrt{Z}$ recovers the sum's martingale part even at zero, and the drifts add to $\alpha +\beta$.

For independent Brownian motions $W,B$, the Itô formula cancels the drift after the arctangent transform. Normalizing the two noise coefficients gives a continuous local martingale $Z$ with quadratic variation $t$, hence another Brownian motion by the Lévy characterization of Brownian motion.

For $X_t=W_t+\alpha t$ with an independent equiprobable drift $\alpha \in \{-a,a\}$, the likelihood ratio between the two drifts is $e^{2a{X}_t}$. Multiplying the normal distribution densities of successive increments proves this ratio for every observation partition; continuity then gives the full observed-path posterior. The Bayes' theorem gives $q_t=\mathbb{P}(\alpha =a\mid {\mathcal{F}}_t^X)=(1+e^{-2a{X}_t}{)}^{-1}$. Thus the observed drift is $a(2q_t-1)=a\tanh (a{X}_t)$. The process ${\hat{W}}_t=X_t-{\int }_0^{t}a\tanh (a{X}_s)ds$ is an observed-filtration martingale with quadratic variation $t$, so the Lévy characterization of Brownian motion makes it a Brownian motion. The observed state is a scaled diffusion with hyperbolic tangent drift.

The process $M_t=B_{t^2}$ is a martingale in the filtration ${\mathcal{G}}_t={\mathcal{F}}_{t^2}$ and has quadratic variation $t^2$. The integral $W_t={\int }_0^t(2s{)}^{-1/2}d{M}_s$ is well defined because its bracket is ${\int }_0^t(2s{)}^{-1}d(s^2)=t$. The Lévy characterization of Brownian motion makes $W$ Brownian in this filtration, and stochastic-integral associativity gives the displayed representation. The resulting integrand is continuous at zero despite the singular inverse integrand.

The sign transform of standard Brownian motion is itself standard Brownian motion. The integrand is predictable because it is a Borel function of a continuous adapted process. Brownian motion spends zero Lebesgue time at a specified level: integrate $\mathbb{P}(B_s=0)=0$ over positive times. Consequently the transform's quadratic variation is $t$, regardless of the chosen sign value at zero. The Lévy characterization of Brownian motion applies.

Suppose $X$ is a continuous local martingale starting at zero, with conditionally symmetric increments, and terminal conditional expectations have continuous martingale versions. ThenTo prove this, set $M_t=\mathbb{E}[e^{i\theta X_T}\mid {\mathcal{F}}_t]$. Symmetry makes $e^{-2i\theta X_t}{M}_t$ a martingale, and the Itô product rule gives $d⟨M,X⟩=i\theta Md⟨X⟩$. Consequently $e^{-i\theta X_t-{\theta }^2⟨X{⟩}_t/2}{M}_t$ is a bounded local martingale, hence a martingale. Evaluating at the endpoints proves the formula. If also $X_T\sim N(0,T)$ for all $T$, the values of the bracket Laplace transform at $1$ and $2$ force $⟨X{⟩}_T=T$ almost surely. Continuity and the Lévy characterization of Brownian motion then identify $X$ as Brownian motion.

The polynomial $X^4-6X^2⟨X⟩+3⟨X{⟩}^2$ has zero Itô drift. Finite fourth maximal and second bracket moments make it a true martingale. If $\mathbb{E}{X}_t^2=t$ and $\mathrm{Cov}(X_t^2,⟨X{⟩}_t)=0$, then $\mathbb{E}{X}_t^4=3t^2-3\mathrm{Var}(⟨X{⟩}_t)$. Equality at every time forces a deterministic clock and hence Brownian motion by the Lévy characterization of Brownian motion.

For a continuous strictly increasing unbounded bracket $V$, the inverse times $T_u=\inf \{t:V_t>u\}$ satisfy $V_{T_u}=u$ and $T_{V_t}=t$. Optional sampling of the original martingale stopped at bounded bracket levels makes $M_{T_u}$ a continuous local martingale in ${\mathcal{F}}_{T_u}$. Time-changing the square-minus-bracket martingale gives its bracket $u$; the Lévy characterization of Brownian motion identifies it as Brownian.

A scale-invariant law of unparameterized chords from zero to infinity in the complex upper half-plane is local when a boundary-preserving conformal map between neighborhoods of zero, fixing zero, transports its curve stopped on leaving the first neighborhood to the same-law curve stopped on leaving the second. Equality concerns the stopped curves up to increasing reparameterization. For SLE with parameter six, the transformed SLE driving function is a continuous local martingale with bracket six times the image capacity clock. The Lévy characterization of Brownian motion makes that driver Brownian in the new clock, proving the local stopped-curve invariance.

By the usual definition of a Lévy process, $X_0=0$ almost surely. The two martingale assumptions giveFor a continuous Lévy process, the jump measure in the Lévy–Khintchine formula is zero, and the process has the form of a Brownian Gaussian component plus a deterministic drift. Thus its characteristic function isThe first moment makes $m=0$, and the second moment makes ${\sigma }^2=1$. ConsequentlyEquivalently, continuity and the martingale identity $X_t^2-t$ identify the quadratic variation as $t$, so the Lévy characterization of Brownian motion gives the same conclusion. A continuous nonzero drift is excluded by the mean-zero martingale assumption.

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.

Put $C_u={\int }_0^u{e}^{2W_s}ds$. Its paths are continuous and strictly increasing. We also need $C_{\infty }=\infty$ to ensure that the printed inverse is finite for every $t$.

Here is a duration argument for Divergence of a driftless Brownian exponential clock. Start at zero and choose successive return times $T_j$ to zero, each at least one time unit after the preceding return. Recurrence of one-dimensional Brownian motion makes all these times finite. By the Strong Markov property, each unit interval after $T_j$ has the same fixed positive probability that the Brownian path stays above $-1$. The intervals are disjoint, and repeated strong Markov conditioning makes these trials independent. Infinitely many succeed almost surely, by the Borel-Cantelli lemma. Each successful interval contributes at least $e^{-2}$ to $C_{\infty }$, proving divergence.

Therefore ${\tau }_t$ is the unique finite inverse of $C$, with $C_{{\tau }_t}=t$. It is an increasing family of stopping times, since $\{{\tau }_t\le u\}=\{C_u\ge t\}$. The continuous-local-martingale time-change theorem states that inverse-bracket time change preserves the local-martingale property in the filtration ${\mathcal{G}}_t={\mathcal{F}}_{{\tau }_t}$ and transforms the bracket by composition. Applied to $Y$, this givesThe Lévy characterization of Brownian motion now proves that $Z$ is Brownian motion starting at one; $Z-1$ is standard Brownian motion, in the time-changed filtration. If the term Brownian motion is reserved for a process starting at zero, $Z$ itself fails only that normalization. The substantive Brownian property, including independent Gaussian increments in the new time, holds.

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.

For the first equation use an integrating factor. The Itô product rule gives $d(e^{\lambda t}{X}_t)=\sigma e^{\lambda t}d{B}_t$, soThe Gaussianity of deterministic Brownian stochastic integrals follows by approximating its deterministic integrand by step functions and using the Itô isometry. HenceThe variance is positive also when $\lambda <0$, since numerator and denominator then both have negative sign. This is the explicit Ornstein-Uhlenbeck solution with initial value one.

For the second equation, take independent standard Brownian motions $U,V$ and define $R_t=\sqrt{(1+U_t{)}^2+V_t^2}$. The associated planar Brownian motion does not hit the origin. To check this rather than silently differentiate at zero, stop its radius upon hitting $\epsilon$ or $L$, with $0<\epsilon <1<L$. The function $\log r$ is harmonic on that annulus, so the optional stopping theorem givesLet $\epsilon \downarrow 0$: hitting zero before $T_L$ has probability zero. Taking integer $L\to \infty$ covers every possible finite hitting time, since continuous paths are bounded on compact intervals. Thus planar Brownian motion avoids a fixed point and $R$ stays positive.

The radial Itô formula is now legitimate and givesThe sum of the two martingale terms has bracket $t$ and starts at zero, so it is a standard Brownian motion $\beta$ by Lévy characterization of Brownian motion. Thus $dR=d\beta +(2R{)}^{-1}dt$ with $R_0=1$. The law-determination assumption permitted in the question identifies the law of $Y$ with that of this dimension-two Bessel process. For independent standard normal random variables $G_1,G_2$,This is the two-dimensional Bessel transition law and answers the requested distribution without needing a special-function density.

If $X$ is a P-Brownian motion, its starting value is deterministic, so Q5(a) applies. The correction $[X,M]$ has finite variation and does not alter quadratic variation; Q1(b) gives $[\tilde{X}{]}_t=t$ under Q. The Lévy characterization of Brownian motion therefore identifies $\tilde{X}$ as a Q-Brownian motion with the same starting value.

For the likelihood, use Wiener measure ${\mathbb{Q}}^x$ as the reference measure and write its coordinate as ${\omega }_t=x+W_t$. Setand let $L=\mathcal{E}(M)$ be its stochastic exponential. We must prove it is a true martingale; unverified large-horizon Novikov estimates would not suffice.

Fix $T$ and stop at ${\tau }_n=\inf \{s:|{\omega }_s|\ge n\}$, with $n>|x|$. The bounded integrand satisfies the Novikov condition, so $L_{T\wedge {\tau }_n}$ defines a probability $P_n$ on ${\mathcal{F}}_T$. By the result just established,is Brownian under $P_n$ on this horizon. Therefore $d{\omega }_s=d{W}_s^{(n)}-\mu 1_{\{s\le {\tau }_n\}}{\omega }_{s}ds$. The Gronwall inequality givesThe Doob L2 maximal inequality then bounds its squared expectation by $C_T=2e^{2|\mu |T}(x^2+4T)$, uniformly in $n$. In particular $P_n({\tau }_n\le T)\le C_T/n^2$. On $\{{\tau }_n>T\}$ the stopped and unstopped densities agree, soContinuous Brownian paths are bounded on $[0,T]$, and the indicators increase to one. The monotone convergence theorem proves $\mathbb{E}{L}_T=1$. Since a nonnegative local martingale is a supermartingale, constant expectation at every time makes $L$ a true martingale.

Under its tilted measure, ${\hat{W}}_t=W_t+\mu {\int }_0^t{\omega }_{s}ds$ is Brownian and $d{\omega }_t=d{\hat{W}}_t-\mu {\omega }_{t}dt$. The explicit linear solution gives uniqueness of this law, identifying it with ${\mathbb{P}}^x$. Consequently the Ornstein-Uhlenbeck likelihood relative to Wiener measure isThis is a finite-horizon density statement; it does not assert absolute continuity of the full infinite-horizon path laws.

Put $A_t=[M{]}_t$. Its continuous, strictly increasing paths start at zero and tend to infinity. Thus the inverse $T_s$ is finite and continuous, andso each $T_s$ is a stopping time. Set ${\mathcal{G}}_s={\mathcal{F}}_{T_s}$ and ${\beta }_s=M_{T_s}$. This is the inverse-clock construction in the Dambis-Dubins-Schwarz theorem.

To verify its martingale property, stop $M$ at times ${\tau }_n$ bounding its path, its bracket and time itself. The stopped $M$ and $M^2-A$ are bounded martingales. Optional sampling at $T_s\wedge {\tau }_n$ and $T_t\wedge {\tau }_n$ shows that the time-changed process, stopped at $A_{{\tau }_n}$, is a martingale and that its square minus the stopped clock $s\wedge A_{{\tau }_n}$ is a martingale. These clock stopping times tend to infinity because $A_{\infty }=\infty$. Therefore $\beta$ is a continuous local martingale with ${\beta }_0=0$ and quadratic variation $[\beta {]}_s=s$. The Lévy characterization of Brownian motion identifies it as a ${\mathcal{G}}_s$-Brownian motion.

Strict monotonicity and continuity give $A_{T_s}=s$ and $T_{A_t}=t$, soThe inverse-clock argument proves the asserted representation; the random clock need not be independent of the Brownian motion it produces.

For $t>1$, $B_{t^2}$ is not measurable with respect to the original Brownian filtration ${\mathcal{F}}_t$: its conditional variance is $t^2-t>0$. Thus it is not even adapted on the full half-line. It also fails the martingale condition before time one. With $s=1/4$, $t=3/4$,The appropriate filtration isBrownian independent increments make $M_t=B_{t^2}$ a continuous ${\mathcal{G}}_t$-martingale, with quadratic variation $[M{]}_t=t^2$. Define the stochastic integralThe value of the integrand at zero is immaterial; it is square-integrable against the bracket because ${\int }_0^t(2s{)}^{-1}d(s^2)=t$. Consequently $[W{]}_t=t$, and the Lévy characterization of Brownian motion makes $W$ a ${\mathcal{G}}_t$-Brownian motion. Associativity of stochastic integrals gives the Brownian motion under a quadratic time change representationThe function $f$ is continuous at zero. One can first verify the inverse-integrand identity on $[\epsilon ,t]$ and then let $\epsilon \downarrow 0$ using the Itô isometry, since $\mathbb{E}{M}_{\epsilon }^2={\epsilon }^2$.

Because $W$ is continuous and adapted, the Borel function $\mathrm{sgn}(W)$ is predictable. Its value at zero is $-1$, so its square is exactly one everywhere. Consequently $A$ is a zero-starting continuous local martingale withThe Lévy characterization of Brownian motion therefore makes $A$ a Brownian motion in the same filtration. Here that characterization is the deterministic-clock case of the conditional-exponential argument in question 2.

Apply the Itô formula to $V=W^2$:Since $\sqrt{V_t}=|W_t|$ and $|W_t|\mathrm{sgn}(W_t)=W_t$, including at zero, substitution yieldsThus $W^2$ is a squared Bessel process of dimension one.

Use the common filtration in which the two driving processes are independent Brownian motions and the solutions are adapted. Let $Z=X+Y$, and define predictable bounded weightswith the ratios defined only on $\{Z>0\}$. The nonnegative solutions have $X=Y=0$ when $Z=0$. SetIndependent Brownian drivers have zero cross-variation. Since $h_1^2+h_2^2=1$ even on the zero set, $[B{]}_t=t$, and the Lévy characterization of Brownian motion makes $B$ a Brownian motion. Moreover $\sqrt{Z}{h}_1=\sqrt{X}$ and $\sqrt{Z}{h}_2=\sqrt{Y}$ everywhere. Adding the two original equations therefore proves additivity of independently driven squared Bessel processes:The unit-vector fill on $\{Z=0\}$ is essential: simply dividing by $\sqrt{Z}$ there would leave the Brownian construction undefined.