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Fix a deterministic $a\ge 0$. By the Strong Markov property at the finite Brownian first-passage time $H_a$, the process ${\tilde{B}}_s=B_{H_a+s}-a$ is a standard Brownian motion started at zero. Such a process takes positive values arbitrarily soon almost surely: for every $\epsilon >0$, the Brownian reflection principle makes its maximum on $[0,\epsilon ]$ have the distribution of $|B_{\epsilon }|$, so the probability that the maximum is zero is zero. Intersecting these events for $\epsilon =1/n$ gives an infimum of positive-crossing times equal to zero. ThereforeThe infimum defining the strict crossing time need not itself be a time when $B_t>a$; continuity gives $B_{T_a}=a$. The quantifier here is fixed-level almost sure equality, not indistinguishability of stochastic processes as the level varies.

Let $A={\max }_{0\le t\le 1}{B}_t$, a random level. The Brownian reflection principle gives $A\stackrel{d}{=}|B_1|$, so $A>0$ almost surely. Also $A-B_1\stackrel{d}{=}|B_1|$: this follows by applying reflection to the reversed increment process $(B_1-B_{1-s}{)}_{0\le s\le 1}$, which has the law of a standard Brownian motion. Thus $B_1<A$ almost surely as well.

By continuity and compactness, $A$ is attained at some time strictly between zero and one, so $H_A<1$. It is not exceeded anywhere on $[0,1]$; since $B_1<A$, continuity even excludes an exceedance in a positive interval just after time one. Consequently $T_A>1$. Both conclusions hold on a single probability-one event, giving the fixed-level versus simultaneous Brownian passage-time equality distinction:Thus the simultaneous assertion is false. Equality for each deterministic level, and even simultaneously for all rational levels, cannot be extended to all real levels by an uncountable intersection.

Use the usual completed right-continuous Brownian filtration. The strict passage process $(T_a{)}_{a\ge 0}$ is finite at every level on one probability-one event: finiteness at all integer levels from (a) suffices by monotonicity. Also $T_0=0$ almost surely by (b).

For each fixed $a$, $T_a$ is a stopping time with $B_{T_a}=a$. The Strong Markov property shows that the passage times above this level depend on a new independent standard Brownian motion. Thus for $0\le a<b$, $T_b-T_a$ is independent of the past at $T_a$ and has the same law as $T_{b-a}$. Iteration gives independent increments and stationary increments in the level parameter. From (a) and fixed-level equality,For $\lambda ,\eta >0$, this also proves continuity in probability at zero, sinceFinally, with $M_t={\sup }_{s\le t}{B}_s$, we have $T_a=\inf \{t:M_t>a\}$. This strict generalized inverse of a nondecreasing function is right-continuous: if $a_n\downarrow a$, then for any $t>T_a$ we have $M_t>a$, and eventually $M_t>a_n$, which forces $T_{a_n}\le t$. Monotonicity gives the reverse bound. Monotonicity and local finiteness also give finite left limits. Hence $T$ is the Brownian first-passage subordinator, with càdlàg paths.

Now condition on this clock, which is independent of $W$. For a deterministic partition $0=a_0<a_1<\cdots <a_k$, write $\Delta T_j=T_{a_j}-T_{a_{j-1}}$ and $\Delta X_j=W_{T_{a_j}}-W_{T_{a_{j-1}}}$. Conditional on the clock these are independent centered Gaussian increments, with respective variances $\Delta T_j$. ConsequentlyFactorization and dependence only on interval lengths prove independent increments and stationary increments for $X$. For small $h$, $T_h\to 0$ in probability, and independence and continuity of $W$ imply $W_{T_h}\to 0$ in probability. For example, bound its deviation probability by $\mathbb{P}(T_h>\eta )+\mathbb{P}({\sup }_{t\le \eta }|W_t|>\epsilon )$ and then let $h\downarrow 0$ and $\eta \downarrow 0$. Stationary increments give stochastic continuity at every deterministic level. Composition of the continuous path of $W$ with the nondecreasing càdlàg clock gives càdlàg paths for $X$, and $X_0=0$. ThusThis is subordination of a Lévy process. Using strict passage times ensures the required right-continuous path choice, despite their fixed-level equality with the non-strict times.

The relevant fact is uniform convergence on compacts in probability under an absolutely continuous measure change. Write $Z=d\tilde{\mathbb{P}}/d\mathbb{P}$. For any event $E$ and $K>0$,The last term tends to zero as $K\to \infty$ because $Z$ is integrable. Consequently, $\mathbb{P}(E_n)\to 0$ implies $\tilde{\mathbb{P}}(E_n)\to 0$. Apply this to $E_n=\{{\sup }_{t\le T}|C_t^n-[M,N{]}_t|>\epsilon \}$ for each finite $T$ and $\epsilon >0$. The given dyadic sums therefore have the same uniform convergence on compacts in probability limit under the new probability measure.

This proves quadratic covariation under an absolutely continuous measure change:Taking $N=M$ proves the corresponding quadratic variation assertion directly; it has not been assumed. The ceiling in the printed sum includes one grid interval extending beyond $t$. This does not affect the limit: on $[0,T]$ the resulting extra product is bounded by the product of the two path oscillations on mesh $2^{-n}$, which tends to zero by uniform continuity on $[0,T+1]$.

The bracket under the new probability measure must be understood as the quadratic covariation of continuous semimartingales. Semimartingale stability under an absolutely continuous measure change ensures this class is preserved; the local martingale property need not be. For example, on a finite horizon, weighting Brownian motion by $\exp (\mu B_T-{\mu }^{2}T/2)$ makes $B_t-\mu t$ a Brownian motion by the Girsanov theorem, so $B$ has nonzero drift under the new measure even though its quadratic variation is still $t$. The PDF additionally assumes that $M,N$ remain local martingales under the new probability measure, so its bracket is also defined directly by the local martingale characterization.

Specify an initial value or initial probability distribution. A strong solution of a stochastic differential equation uses a prescribed Brownian motion on a prescribed filtered probability space. It is continuous, adapted to the completed filtration generated by that Brownian motion and the initial value, and satisfiesalmost surely for every $t$, with ${\int }_0^t(|b(X_s)|+\sigma (X_s{)}^2)ds<\infty$ almost surely. The given initial variable is independent of future Brownian motion increments. Some courses allow a larger prescribed filtration in their definition of a strong solution of a stochastic differential equation; the Brownian-generated convention states explicitly what absence of extra randomness means.

A weak solution of a stochastic differential equation consists of a filtered probability space, a Brownian motion relative to its filtration, and a continuous adapted process $X$ satisfying the same integrability and integral equation. Here the space and noise are part of the unknown; the filtration may contain randomness beyond the initial variable and the driving path.

Pathwise uniqueness means that on any common filtered probability space, any two solutions with the same driving Brownian motion and the same initial value are indistinguishable. Uniqueness in law means that all weak solutions with the same initial probability distribution have the same law as path-valued random variables, even on different spaces. Pathwise uniqueness compares shared-noise paths; uniqueness in law compares path distributions. Neither uniqueness notion by itself asserts existence.

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.

Define a probability measure on the same full measurable space byThe Radon-Nikodym derivative is positive and has expectation one, so this is a probability measure and even equivalent to $\mathbb{P}$, in particular ${\mathbb{P}}^T\ll \mathbb{P}$. For $t\le T$ its density on ${\mathcal{F}}_t$ is $\mathbb{E}[Z_T\mid {\mathcal{F}}_t]=Z_t$.

The finite-horizon Girsanov theorem says that if $W$ is a Brownian motion and $\mathcal{E}(\int hdW)$ is a true martingale through $T$ defining the density, then $W_t-{\int }_0^t{h}_{s}ds$ is a Brownian motion under that density for $t\le T$. Here $h_s=\mu$, and the stochastic exponential is exactly $Z$. The previous part verifies the necessary martingale property; alternatively the Novikov condition is $e^{{\mu }^{2}T/2}<\infty$. Hence $B_t-\mu t$ is a standard Brownian motion on $[0,T]$ relative to the same filtration under ${\mathbb{P}}^T$.

No absolutely continuous probability measure can remove this nonzero drift on the whole half-line. By the strong law for Brownian motion, the eventhas $\mathbb{P}$-probability one. If ${\mathbb{P}}^{\infty }\ll \mathbb{P}$, absolute continuity of measures gives ${\mathbb{P}}^{\infty }(A)=1$. But if $W_t=B_t-\mu t$ is a Brownian motion under ${\mathbb{P}}^{\infty }$, the same strong law for Brownian motion gives $B_t/t=W_t/t+\mu \to \mu$ with ${\mathbb{P}}^{\infty }$-probability one. Since $\mu >0$, these two limiting events are disjoint, a contradiction.

This is infinite-horizon singularity of Brownian motion with constant drift. The finite-horizon densities are consistent on ${\mathcal{F}}_t$ but cannot define a globally absolutely continuous measure with the requested property. Indeed,because $t^{-1}\log Z_t\to -{\mu }^2/2$. Hence $Z$ lacks uniform integrability on $[0,\infty )$. A drifted law does exist on canonical path space, but it is mutually singular with Wiener measure on the infinite-horizon path sigma-algebra.

The printed assumptions omit the orthogonality needed for the claimed local martingales. Merely placing two Brownian motions on the same probability space does not make them independent. Interpret both as Brownian motions relative to a common filtration, and put $C_t=[B,\vartheta {]}_t$. For $X=e^B\cos \vartheta$ and $Y=e^B\sin \vartheta$, the two-variable Itô formula givesThe pure second derivatives cancel because $[B{]}_t=[\vartheta {]}_t=t$, while the mixed partial derivatives are $X_{b\vartheta }=-Y$ and $Y_{b\vartheta }=X$. Thus the mixed quadratic covariation term cannot be omitted.

For a counterexample allowed by the PDF, take $\vartheta =B$. Then $C_t=t$ and $X$ has drift $-Ydt$, while $Y$ has drift $Xdt$. In particular $X_0=1$ makes the latter drift nonzero near zero, so $Y$ is not a local martingale. Nor is $X$: $Y$ cannot be identically zero, so its accumulated continuous drift is not identically zero. Uniqueness of the continuous semimartingale decomposition excludes cancellation by another local martingale term.

The corrected hypothesis is $[B,\vartheta ]\equiv 0$, for example independent Brownian motions in their joint filtration. Conversely, if both were local martingales, uniqueness of the continuous semimartingale decomposition would force $YdC=XdC=0$; since $X^2+Y^2=e^{2B}>0$, this forces $dC=0$. Thus the correction is exactly the condition needed for both local martingales. Under this hypothesis the drift terms vanish and the intended stochastic differential equations are $dX=XdB-Yd\vartheta$ and $dY=YdB+Xd\vartheta$. The continuous integrands are locally bounded, so localization makes these Itô integrals continuous local martingales. Subsequent intended conclusions are proved under this explicitly stated correction.