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A left-continuous adapted process is a predictable process. Define the elementary predictable approximationAt every $s>0$, its sample time approaches $s$ from the left, including when $s$ is itself a grid point. Thus $H_s^{(n)}\to H_s$ pathwise by left continuity.

Use the continuous semimartingale decomposition $X=X_0+M+A$, with $M$ a continuous local martingale and $A$ a continuous finite-variation process. Localize so that on a fixed horizon $T$, the integrands are bounded by a deterministic constant, $[M{]}_T$ is bounded, and the total-variation process of $A$ at $T$ is bounded. This is possible using local boundedness and the finite pathwise clock and variation.

For the finite-variation integral, the dominated convergence theorem givesalmost surely. For the martingale integral, the Itô isometry and the Doob L2 maximal inequality giveHere the localized bounds justify dominated convergence also in expectation. Remove localization to conclude that the full elementary integrals converge in uniform convergence on compacts in probability to $\int HdX$.

The sum of completed intervals differs from the full elementary integral at time $t$ byIts maximum on $[0,T]$ tends to zero almost surely, by the local bound on $H$ and uniform continuity of $X$. ThereforeThis proves left-endpoint approximation of a continuous semimartingale integral, including the unfinished interval convention.

Use the local convention for the martingale problem, which is appropriate for arbitrary measurable coefficients. A continuous adapted process $X$ solves $\mathbf{M}(a,b)$ if the coefficient integrals are locally finite and, for every $f\in C_c^{\infty }(\mathbb{R})$,is a local martingale. Here locally finite means ${\int }_0^T(|a(X_s)|+|b(X_s)|)ds<\infty$ almost surely for each $T<\infty$. Suitable localization upgrades the identities to true martingale identities. With bounded coefficients and bounded test derivatives, this distinction disappears on finite horizons.

A Markov diffusion generated by $L$ is a time-homogeneous Markov process with continuous paths whose infinitesimal generator on the stated test domain is $L$. Its transition operators satisfy the generator limit, and Dynkin formula gives precisely the preceding martingale problem identities. Some terminology uses L-diffusion for the test-function identities alone; the Markov property is an additional assertion unless uniqueness supplies it.

Suppose now $a={\sigma }^2>0$. Use smooth compactly supported functions agreeing with $x$ and $x^2$ on $[-R,R]$, and stop before exiting this interval. Removing these stops shows thatis a continuous local martingale. Consequently $X$ is a continuous semimartingale. The martingale problem applied to the second cutoff function says thatis a local martingale. The Itô formula for $X^2$ also says thatis a local martingale. Subtracting, the continuous finite-variation process $[N{]}_t-{\int }_0^{t}a(X_s)ds$ is a local martingale and starts at zero. The theorem that a continuous finite-variation local martingale is constant givesDefineThe integrand is predictable and is stochastically integrable, since its integral against the bracket is exactly $t$. Thus $B$ is a continuous local martingale with $B_0=0$ and $[B{]}_t=t$. The Lévy characterization of Brownian motion states that these conditions make $B$ a Brownian motion relative to the given filtration. Associativity of stochastic integrals recovers $N=\int \sigma (X)dB$, and thereforeThe original process is a weak solution of a stochastic differential equation, with its driving Brownian motion constructed on the original space.

Conversely, the Itô formula applied to any weak solution with locally integrable coefficients gives the martingale problem identities. A Markov diffusion with generator $L$ gives them by Dynkin formula. A well-posed martingale problem, with the usual measurable path-space setup, identifies a unique family of diffusion laws and yields the Strong Markov property. Without uniqueness, the test-function identities alone do not assert that every chosen solution is Markov. This specifies both the equivalence with weak equations and the role of uniqueness in identifying a diffusion.

For a continuous path $m$ of finite variation on $[0,t]$, every partition satisfiesUniform continuity makes the right side tend to zero as the mesh tends to zero. The quadratic variation of the given continuous local martingale $M$ is therefore identically zero.

Stop $M$ when its magnitude first reaches an integer $n$, so the stopped process is a bounded, square-integrable martingale. The square identity, or the Itô isometry, gives $\mathbb{E}{M}_{t\wedge {\tau }_n}^2=\mathbb{E}[M{]}_{t\wedge {\tau }_n}=0$. Hence it is zero at each rational time almost surely, and continuity extends this simultaneously to all times. Let $n$ tend to infinity. This provesThis is the continuous finite-variation local martingale is constant principle with initial value zero.

Use just the parameters $1$ and $-1$. Positivity and adaptation of their local martingales giveThus both processes are adapted. A positive continuous local martingale has a semimartingale logarithm, by Itô formula after localization away from zero. Since $X=\log Z^{(1)}+A/2$ and $A$ has finite variation, $X$ is a continuous semimartingale. Write $X=N+V$, where $N$ is a zero-starting continuous local martingale and $V$ is continuous finite variation, also starting from zero.

Apply Itô formula to each exponential:Because $Z^{(\theta )}$ is itself a local martingale, the finite-variation term is zero by part (a). Divide its finite signed measure by the strictly positive $Z^{(\theta )}$. For $\theta =1,-1$, respectively,Adding and subtracting show $V=0$ and $[N]=A$. Therefore the exponential criterion for a continuous local martingale and its bracket givesThe semimartingale property was established before applying Itô's formula to $X$; it was not assumed from the outset.

On a compact time interval a continuous path avoiding the origin has a strictly positive minimum radius. Thus the times $S_{1/n}$ tend to infinity almost surely. Part (a) localizes $M_t=\log |B_t|$, so $M$ is a continuous local martingale.

To show that it is not a martingale, compute its expectation at a positive deterministic time. In distribution $B_t=(1,0)+W_t$, where $W_t$ is a centered isotropic Gaussian vector. Conditional on $\rho =|W_t|$, its angle is uniform. The angular average of a logarithmic potential isFor $\rho <1$ this follows by averaging the real part of the power series for $\log (1+\rho e^{i\theta })$; for $\rho >1$ factor out $\rho$ and apply the same argument to ${\rho }^{-1}$. The value at $1$ follows by the integrable logarithmic singularity, or is irrelevant for the continuously distributed radius.

The radius has density $(\rho /t)e^{-{\rho }^2/(2t)}$. The logarithm of $|B_t|$ is absolutely integrable: near the origin its Gaussian density is bounded and ${\int }_0^{1}r|\log r|dr<\infty$; away from the origin its positive part is bounded by a constant plus $|B_t|$. Therefore conditioning is legitimate, and integration by parts yieldsSince $M_0=0$, this contradicts preservation of expectation by a martingale. Thus the logarithmic radius of planar Brownian motion is a strict local martingale. It is not a nonnegative local martingale; its negative values allow its expectation to increase.

Put $V_t=⟨X{⟩}_t$ and $S_t={\sup }_{s\le t}|X_s|$. Suppose first that $|X|\le K$. The Itô formula for $|x{|}^p$, which is twice continuously differentiable for $p\ge 2$, givesFor $p=2$, the second derivative is interpreted as the constant $2$. The stochastic term has mean zero: its integrand is bounded, and $\mathbb{E}{V}_t<\infty$ makes it a square-integrable martingale. ConsequentlyThis is the first required estimate.

A bounded local martingale is a true martingale. Apply the allowed Doob Lp maximal inequality, and then Hölder's inequality with conjugate exponents $p/(p-2)$ and $p/2$ for $p>2$. With $D_p=(p/(p-1){)}^p$, this yieldsIf the left side is zero there is nothing to prove; otherwise divide by its indicated power and raise to $p/2$. For $p=2$, the same conclusion follows directly from $\mathbb{E}{X}_t^2=\mathbb{E}{V}_t$ and Doob's inequality. A usable constant is thereforeIn particular $C_2=4$. This is the upper maximal moment bound for a continuous local martingale.

For an unbounded continuous local martingale, stop at ${\tau }_n=\inf \{s:|X_s|\ge n\}$. Its stopped path is bounded, and its bracket at $t$ is $V_{t\wedge {\tau }_n}\le V_t$. The proved inequality gives a bound by $C_p\mathbb{E}{V}_t^{p/2}$, independent of $n$. Continuity makes ${\tau }_n\uparrow \infty$, and the stopped maxima increase to $S_t$. Monotone convergence proves the same inequality for the original unbounded process, with the same constant. This localization step also shows that the assumed bracket moments supply all the needed maximal moments.

Use $V=⟨X⟩$. Applying the Itô formula and the Itô product rule givesThe finite-variation terms cancel in the prescribed combination, leavingThus $Y$ is initially a continuous local martingale. It is a true martingale, not merely local. On each fixed horizon $T$,Part (a) with $p=4$, the assumed bracket moments and Cauchy-Schwarz inequality make this bound integrable. The integrable-supremum martingale criterion now applies, by localization and dominated conditional expectations. The same reasoning makes $X^2-V$ a martingale, so $\mathbb{E}{V}_t=\mathbb{E}{X}_t^2=t$.

The zero covariance now means $\mathbb{E}(X_t^2{V}_t)=t^2$. Since $Y_0=0$ and $\mathbb{E}{Y}_t=0$,This is the fourth-moment deficit and bracket variance identity. If equality holds for every $t$, then $V_t=t$ almost surely for each $t$. Take a single probability-one event for all rational times and use continuity to obtain $V_t=t$ simultaneously for all times. The Lévy characterization of Brownian motion then proves $X$ is Brownian motion in its given filtration. A proof of that characterization by conditional characteristic functions is included in Question 2(a).

Let $V_t=⟨M{⟩}_t$, and first suppose $V_{\infty }=\infty$ almost surely in addition to strict increase. The Dambis-Dubins-Schwarz theorem says thatdefine a Brownian motion $B$ in the time-changed filtration and giveEach $T_u$ is a stopping time and is finite. Continuity and strict increase of $V$ make $u\mapsto T_u$ continuous, $V_{T_u}=u$ and $T_{V_t}=t$.

Here are the martingale details behind this inverse-clock proof of the Dambis-Dubins-Schwarz theorem. Stopping a continuous local martingale when its bracket reaches $n$ makes it an L2-bounded continuous martingale. This follows from the stopped Itô isometry or from the $p=2$ estimate proved in Question 1(a), applied after localization. In particular it is uniformly integrable, and optional sampling is valid even at an unbounded stopping time by taking limits. Applying this to $M$ stopped at $T_n$ shows that $(M_{T_{u\wedge n}}{)}_{u\ge 0}$ is a martingale. Thus $B$ is a continuous local martingale. Time-changing $M^2-V$ in the same way shows $B_u^2-u$ is a local martingale, so $⟨B{⟩}_u=u$.

For completeness, the Lévy characterization of Brownian motion follows directly from the Itô formula. If a continuous local martingale $N$, starting at zero, has bracket $u$, thenis a complex local martingale. Its modulus is bounded on each deterministic finite horizon, so it is a true martingale there. ConsequentlyConditional characteristic functions give Gaussian increments independent of the past. Iterating this identity gives independent increments, and continuity completes the Brownian characterization. The identical vector argument proves the Lévy characterization of multidimensional Brownian motion when the bracket matrix is $uI$.

The printed strict-increase hypothesis does not imply $V_{\infty }=\infty$. For example, $M_t={\int }_0^t{e}^{-s}d{W}_s$ has strictly increasing bracket $(1-e^{-2t})/2$. To state the theorem under exactly the printed hypothesis, allow an independent enlargement of the probability space if the terminal clock $L=V_{\infty }$ can be finite.

On $\{L<\infty \}$ the martingale $M$ has a finite terminal limit. Indeed, stopping at each bracket level $n$ gives an L2-bounded continuous martingale which converges; on $\{V_{\infty }<n\}$ the stopped process is the original one. This proves the finite-bracket convergence lemma. Set $T_u=\infty$ when $u\ge L$ and continue $N_u=M_{T_u}$ by that terminal limit. The optional-sampling argument just given makes $N$ a continuous local martingale with bracket $u\wedge L$. Moreover $L$ is a stopping time in ${\mathcal{G}}_u={\mathcal{F}}_{T_u}$.

On a product extension add an independent Brownian motion $\beta$ in clock time, and putThe two summands have zero quadratic covariation, and their brackets are $u\wedge L$ and $(u-L{)}^{+}$. Thus $⟨\tilde{B}{⟩}_u=u$; the proved characterization makes $\tilde{B}$ Brownian. Since $V_t<L$ at every finite $t$ when $L$ is finite, $M_t={\tilde{B}}_{V_t}$ still holds. This is the finite-lifetime extension of the Dambis-Dubins-Schwarz theorem. An infinite clock gives Brownian motion on the original space; a finite clock may require the independent extension.

Since the density martingale is uniformly integrable, $Z_t={\mathbb{E}}_{\mathbb{P}}[Z_{\infty }\mid {\mathcal{F}}_t]$. Equivalence of measures, as stipulated, means $Z_{\infty }>0$ almost surely. Write $L=\int Z^{-1}dZ$ for the stochastic logarithm. The Itô formula gives $\log Z=L-⟨L⟩/2$, and hence the quadratic covariation in the drift correction isIt involves the local-martingale part of $\log Z$; its continuous finite-variation part has zero covariation.

With $Y=X-C$, the Itô product rule now gives an exact cancellation:Thus $ZY$ is a P-local martingale. To transfer this conclusion rigorously, set ${\tau }_n=\inf \{t:|Y_t|\ge n\}$. The same computation for the stopped $Y$ givesThis is again a P-local martingale. Moreover its absolute value is at most $n{Z}_t$. Uniform integrability of $Z$ makes the family over bounded stopping times uniformly integrable, so the product is a true martingale. This is the bounded-process density-product criterion.

The Bayes formula for conditional expectation therefore gives, for $s\le t$,Continuity gives ${\tau }_n\uparrow \infty$, proving $X-⟨\log Z,X⟩$ is a Q-local martingale. This proves the needed Girsanov theorem rather than invoking it.

For the Brownian-filtration conclusion, use this precise Brownian martingale representation theorem: every continuous local martingale in the usual augmentation of the natural Brownian filtration is an Itô integral with a predictable integrand locally square integrable in time. In particularDefineOn each finite horizon a strictly positive continuous $Z$ has a positive pathwise minimum. Thus ${\int }_0^t{\alpha }_s^{2}ds<\infty$ almost surely, andThe already proved measure-change result makes $\hat{W}=W-\int \alpha ds$ a continuous Q-local martingale. Its quadratic variation is $t$, unchanged by its finite-variation correction. The Lévy characterization of Brownian motion provesNo additional exponential-integrability condition is needed, since the equivalent uniformly integrable density process is already given.

Strict positivity lets us define the continuous local martingaleThe integrand is locally bounded because a positive continuous path has positive minimum on every compact time interval. The Itô formula for $\log M$ givesThis is the stochastic exponential representation of a positive continuous local martingale.

If $⟨X{⟩}_{\infty }$ were finite on an event of positive probability, the finite-bracket convergence lemma proved in Question 2(a) would make $X_t$ converge to a finite limit there. The exponential would then have a strictly positive limit, contradicting the assumed $M_t\to 0$. ThereforeThis is the divergent logarithmic clock for a positive local martingale tending to zero.

The scalar Lévy characterization of Brownian motion states that an adapted process $X$ starting at zero is a Brownian motion if and only if it is a continuous local martingale with

For necessity, the centered independent increments make a Brownian motion a martingale. Their conditional second moments show that $X_t^2-t$ is also a martingale. The defining uniqueness of the quadratic variation compensator gives $⟨X{⟩}_t=t$.

For sufficiency, fix $\theta \in \mathbb{R}$ and apply the Itô formula toThe time drift cancels the second-order Itô term, leaving $d{Z}_t=i\theta Z_{t}d{X}_t$. Its real and imaginary parts are local martingales. On any deterministic interval $[0,T]$, $|Z_t|=e^{{\theta }^{2}t/2}\le e^{{\theta }^{2}T/2}$, so the bounded local martingale criterion makes them true martingales. ThereforePart (b) proves the Brownian property. This also supplies a proof of Lévy's characterization of Brownian motion through conditional Fourier transforms.

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.

Write ${\delta }_n=2^{-n}$ and let $Q_n(U)$ denote the printed squared-increment sum for a process $U$, on grid times $k{\delta }_n$. We construct its limit first for bounded martingales, then use localizing sequences and the finite variation part of a semimartingale.

Continuity and monotonicity of the bounded-martingale limit. Let $U$ be a uniformly bounded continuous martingale. The given result supplies a limit $R^U$ in uniform convergence on compacts in probability. Each $Q_n(U)$ is continuous and adapted. A subsequence converges almost surely uniformly on each compact time interval, by choosing summable error probabilities and diagonalizing. Its limit therefore has a continuous version; in the usual completed filtration this version is adapted.

The sums $Q_n(U)$ themselves can decrease between grid points, because the last partial increment is being squared. To establish monotonicity, use instead the quadratic variation from completed grid incrementsThese step processes are nondecreasing, and pathwise uniform continuity giveswhere ${\omega }_U(\delta ;T)=\sup \{|U_t-U_s|:s,t\le T,|t-s|\le \delta \}$. Thus the same almost surely uniform subsequence of $S_n$ converges to $R^U$, proving that $R^U$ is nondecreasing. Also $R_0^U=0$.

Localization of a continuous local martingale. Subtract the initial value, which does not affect increments, so that $M_0=0$. SetContinuity gives ${\tau }_m\uparrow \infty$ almost surely and makes $M^{{\tau }_m}$ bounded. The bounded local martingale criterion makes it a true martingale. Let $R^m$ be its continuous, adapted, nondecreasing limit.

The discrete sums commute exactly with stopping:For $\ell \le m$, uniqueness of the probability limit and stability of uniform convergence on compacts in probability under stopping giveup to indistinguishability. Taking a common null set for the countably many pairs, patch these processes into $R^M$ by setting $R_t^M=R_t^m$ when $t\le {\tau }_m$. Compatibility makes this definition independent of $m$. It is continuous, adapted, and nondecreasing. For each $T,\epsilon >0$,Let $n\to \infty$ and then $m\to \infty$. This proves the localization and patching of quadratic variation.

Adding finite variation. Decompose the continuous semimartingale as $X=X_0+M+A$, where $M$ is a continuous local martingale starting at zero and $A$ is continuous, adapted, and locally of finite variation. For every compact interval, the sum of the absolute increments of $A$ is at most its total variation of a function. Hence, pathwise,The Cauchy-Schwarz inequality bounds the mixed increment sum byThe first factor is bounded in probability, since $Q_n(M)\to R^M$ uniformly on compacts in probability; the second tends to zero almost surely. Thus the mixed term tends to zero in probability. Expanding the square provesThe constructed $R=R^M$ is continuous, nondecreasing, and adapted. This is the quadratic variation of $X$, and expresses the fact that finite-variation terms do not change quadratic variation.

Put $A_t=e^{-W_t+t/2}$ and $J_t={\int }_0^t{A}_s^{-1}d{B}_s$. The Itô formula gives $d{A}_t=-A_{t}d{W}_t+A_{t}dt$. The independence of the two Brownian motions gives $[W,B]=0$, so the product formula yieldsFor $f(x)=\arctan x$, $f^{\prime }(x)=(1+x^2{)}^{-1}$ and $f^{\prime \prime }(x)=-2x(1+x^2{)}^{-2}$. The drift terms cancel in the Itô formula:Define the rotated stochastic integralIt is a continuous local martingale withThe Lévy characterization of Brownian motion makes $Z$ a Brownian motion. Since ${\Theta }_t\in (-\pi /2,\pi /2)$ and $\cos {\Theta }_t=(1+X_t^2{)}^{-1/2}$,This is the arctangent transform of a two-noise affine diffusion.