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There is an initial-value qualification in the printed conclusion. As written, nothing requires $A_0=0$. For example, $M\equiv 0$ and $A\equiv 1$ satisfy the hypotheses, because $M^2-A\equiv -1$ is a local martingale, but $M_t$ is not $N(0,1)$. The correct variance is $A_t-A_0$; the printed formula holds under the usual normalization $A_0=0$.

Indeed, $M^2-[M]$ is a continuous local martingale. Subtracting it from $M^2-A$ shows that $[M]-A$ is a continuous local martingale with finite variation, hence is its initial constant $-A_0$. ThusIn particular, $C$ is deterministic, continuous, nonnegative and increasing, regardless of the apparent allowance of a general finite-variation process in the premise.

For a real parameter $\theta$, the Itô formula gives the complex local martingaleOn every deterministic interval $[0,T]$, its modulus is at most $e^{{\theta }^2{C}_T/2}$, so its real and imaginary parts are bounded local martingales, hence true martingales. Taking expectations and using $Z_0=1$ yieldsBy the uniqueness theorem for characteristic functions,A zero variance means the point mass at zero. This proves the requested normalized version by a direct characteristic function calculation.

For $g(x)=\log |x|$ on ${\mathbb{R}}^2\setminus \{0\}$,The Itô formula, stopped in a compact annulus, therefore givesThis proves the local martingale property up to a possible hitting of the origin. For the annulus exit time $T$, $M_{t\wedge T}$ is bounded between $\log r$ and $\log R$, so it is a true uniformly integrable martingale.

The exit is finite almost surely. Indeed $|B_t{|}^2-2t$ is a martingale, so stopping at $T\wedge t$ gives $2\mathbb{E}(T\wedge t)=\mathbb{E}|B_{T\wedge t}{|}^2-1\le R^2-1$. Letting $t\to \infty$ proves $\mathbb{E}T<\infty$. Bounded convergence in the logarithmic martingale now yieldsIf $p$ is the probability of reaching radius $r$ before radius $R$, continuity givesHitting zero before reaching radius $R$ would require reaching every positive radius $r$ first. Its probability is bounded by $p$, which tends to zero as $r\downarrow 0$. Any finite-time path hitting zero is bounded before that time and hence hits zero before exiting some integer-radius disk. A countable union over those disks proves $B_t\ne 0$ for every $t\ge 0$, almost surely. This is planar Brownian motion avoids a fixed point; it also makes $M$ a globally defined continuous local martingale.

To decide whether it is a true martingale, compute its expectation directly. Condition on $B_0$ if necessary and use rotation to take $B_0=(1,0)$ and write $B_t=B_0+W_t$. Conditional on $\rho =|W_t|$, the angle of $W_t$ is uniform. The angular average of a logarithmic potential isFor $\rho <1$ this follows by taking the real part of the convergent power series for $\log (1+\rho e^{i\theta })$: each nonconstant Fourier term integrates to zero. For $\rho >1$ factor out $\rho$ and apply the same argument to $1/\rho$. The equality at one follows by the integrable limiting logarithmic singularity. All interchanges are valid: the shifted Gaussian density is bounded near the origin, where ${\int }_0^{1}r|\log r|dr<\infty$, and its Gaussian tail controls logarithmic growth.

The radius $\rho$ has density $(\rho /t)e^{-{\rho }^2/(2t)}$. Consequently, for $t>0$,Since $M_0=0$, $M$ is not a martingale, despite integrability at each deterministic time. This is the logarithmic radius of planar Brownian motion example.

The optional hint is consistent with this conclusion. The upper-radius exit $T(2,0)$ is finite and never occurs at zero. After that exit the stopped square is $(\log 2{)}^2$; before it, a radius in $[1/2,2]$ also gives a logarithmic square no larger than $(\log 2{)}^2$, while a radius below $1/2$ contributes precisely the extra indicated term. This proves the hinted inequality without assuming that the unstopped process is a true martingale.

The multidimensional Itô formula for continuous semimartingales states that a $C^2$ transformation has first-order stochastic and finite-variation integrals plus one-half its Hessian contracted with the quadratic covariations. A continuous finite-variation process has zero quadratic variation and zero quadratic covariation with a continuous local martingale. Thus in the present caseThe first integrand is continuous adapted, hence predictable, and locally bounded by stopping the two coordinate paths on exiting compact sets. The standard construction of the Itô integral then makes this integral a zero-starting continuous local martingale. The remaining integrals are pathwise Lebesgue-Stieltjes integrals. On each compact time interval their integrands are bounded along the continuous paths, so their total variations are bounded by a constant times $V_t(A)+[M{]}_t$, which is finite. Therefore the normalized terms areThe process $A^f$ is continuous adapted of finite variation and $A_0^f=0$. When $f(M_0,A_0)$ is integrable, $M^f$ is a local martingale in the definition of part (a). Uniqueness follows from part (a): the difference of two normalized decompositions is both a continuous local martingale and a continuous finite-variation process starting at zero, hence vanishes.

An initial-value qualification is necessary under that definition. The printed assumptions do not themselves guarantee $\mathbb{E}|f(M_0,A_0)|<\infty$. Take $M=0$, $A_t=G$ with $G$ a standard normal random variable known at time zero, and $f(x,a)=e^{a^2}$. All path assumptions hold, but $\mathbb{E}{e}^{G^2}=\infty$. Any decomposition with $A_0^f=0$ would force the nonintegrable value $M_0^f=e^{G^2}$, contradicting the local-martingale definition. Without an integrability hypothesis, the always-valid continuous semimartingale decomposition is $X=X_0+N+A^f$, with $N$ the zero-starting integral above. Alternatively one may use the convention that a local martingale permits an arbitrary finite initial value and localizes its centered increments. With deterministic initial data, as in part (c), there is no issue. This is initial-value integrability in normalized semimartingale decompositions.

Apply the two-variable Itô formula to $e^w\cos b$. The independent Brownian motions satisfy $[B,W]=0$, while the two diagonal second derivatives sum to zero. ConsequentlyThus $Y$ is a continuous local martingale, but it is not Brownian motion. Brownian motion from any fixed starting point has quadratic variation $t$. If the displayed bracket were identically $t$, differentiating its continuous density would force $e^{2W_t}=1$ for every $t$, hence $W_t=0$ identically, an event of probability zero. This is the cosine-exponential Brownian local martingale. Its initial value $Y_0=1$ is retained in the next part.

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.

Let $N$ be the zero-starting continuous local martingale, and let $V_t$ be the total variation of its path on $[0,t]$. Stop when $|N|$ or $V$ reaches $n$, and also at time $n$. The stopped process is bounded, hence a true square-integrable martingale by the bounded local martingale criterion, and its total variation is bounded by $n$.

For a deterministic partition $0=t_0<\cdots <t_m=t$, martingale increments are orthogonal in $L^2$, soThe sum inside the expectation is at most the largest absolute increment times the stopped total variation. As the mesh tends to zero, path continuity makes the largest increment tend to zero. The product is bounded by $2n^2$, so the dominated convergence theorem makes the displayed expectation zero. Thus the stopped process vanishes at every fixed time. Use rational times, continuity and ${\tau }_n\uparrow \infty$ to concludeThis proves that a continuous finite-variation local martingale is constant.

Write a normalized semimartingale decomposition as $X=X_0+N+A$, with both $N_0=A_0=0$, $N$ a continuous local martingale and $A$ of finite variation. The difference of two such martingale parts is also the difference of two finite-variation parts, and hence vanishes by the result just proved. The other parts coincide too. The normalized continuous semimartingale decomposition is unique. Fixing the initial constants is necessary: without normalization a constant can be transferred between the two parts.

A useful finite-horizon form of the Girsanov theorem is as follows. Let $W$ be a Brownian motion and $\eta$ an adapted, progressively measurable process with ${\int }_0^T{\eta }_s^{2}ds<\infty$ almost surely. Suppose its stochastic exponentialis a true martingale with mean one. The Novikov condition $\mathbb{E}\exp (\frac{1}{2}{\int }_0^T{\eta }_s^{2}ds)<\infty$ is a sufficient condition. Under the equivalent probability measure $dQ=Z_{T}dP$, the process $W_t^Q=W_t+{\int }_0^t{\eta }_{s}ds$ is a Brownian motion.

For the proof sketch, the Itô formula gives $dZ=-Z\eta dW$. Integration by parts cancels the drift in $d(Z{W}^Q)$, since $Z\eta dt$ from $d{W}^Q$ cancels $d[Z,W^Q]=-Z\eta dt$. After localization, the Bayes formula for conditional expectation makes $W^Q$ a continuous local martingale under $Q$. Its quadratic variation is still $t$, because adding a finite-variation process does not change quadratic variation. One can see the Brownian conclusion directly: for real $z$, the Itô formula makes $\exp (iz{W}_t^Q+z^{2}t/2)$ a local martingale. Its modulus is bounded on the fixed horizon, so it is a true martingale. ThusThis is the conditional characteristic function of a centered normal distribution with variance $t-s$, independent of the past. Continuity and these independent normal increments give the defining properties of Brownian motion.

Now put ${\varphi }_t(x)=(2\pi t{)}^{-1/2}{e}^{-x^2/(2t)}$ and $\Phi (z)={\int }_{-\infty }^z{\varphi }_1(x)dx$, the standard normal distribution function. At zero drift the endpoint density of paths staying below $a$ isIndeed, reflect the portion of a path after its first hitting time of $a$. By continuity and the independent symmetric Brownian motion increments after this stopping time, reflection preserves its probability law and maps a crossing path ending at $x<a$ to a path ending at $2a-x>a$. This subtracts exactly ${\varphi }_t(2a-x)$ from the unrestricted endpoint density; it is the Brownian reflection principle.

Taking the constant drift tilt $dQ/dP=e^{\mu W_t-{\mu }^{2}t/2}$ in the Girsanov theorem makes $W$ under $Q$ have the same law as $W+\mu s$ under $P$. Therefore the desired finite-horizon maximum of Brownian motion with drift has probabilityCompleting squares gives $e^{\mu x-{\mu }^{2}t/2}{\varphi }_t(x)={\varphi }_t(x-\mu t)$ and $e^{\mu x-{\mu }^{2}t/2}{\varphi }_t(2a-x)=e^{2\mu a}{\varphi }_t(x-2a-\mu t)$. HenceThe maximum has no atom at a positive level, so strict and weak barrier inequalities agree. This also follows from continuity of the displayed distribution function. At $\mu =0$ the formula reduces to $2\Phi (a/\sqrt{t})-1$.

For the up-and-out power claim, write $T=t_0>0$ and use the risk-neutral measure for the Black-Scholes model, under whichwith $\sigma >0$. The barrier survives exactly when ${\sup }_{t\le T}(W_t+m_{0}t)<A$. Its risk-neutral pricing value isThe factor inside the expectation is a mean-one Exponential martingale for Brownian motion. Under the probability tilted by it, $W_t-2\sigma t$ is a Brownian motion, so the barrier process has drift $m=m_0+2\sigma =(\rho +3{\sigma }^2/2)/\sigma$. Applying the preceding survival formula yieldsAs $c\to \infty$, the bracket tends to one and the price becomes the discounted second moment $S_0^2{e}^{(\rho +{\sigma }^2)T}$. As $c\downarrow S_0$, it tends to zero. If $\sigma =0$, use the deterministic path instead: the price is $S_0^2{e}^{\rho T}$ when $S_0\max (1,e^{\rho T})<c$, and zero otherwise. At $T=0$, with $c>S_0$, the price is simply $S_0^2$.

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.

Relative to a filtration satisfying the usual conditions, a local martingale is an adapted càdlàg process $X$ for which there exist increasing stopping times $S_n\uparrow \infty$ almost surely such that every stopped process $X_t^{S_n}=X_{t\wedge S_n}$ is an integrable martingale. Thus, for $s\le t$,The sequence is a localizing sequence. A continuous local martingale additionally has continuous sample paths. The usual definition entails integrability of the initial value; local martingales need not be integrable at later times without stopping.

The Dambis-Dubins-Schwarz theorem states: let $M$ be a continuous local martingale with $M_0=0$ and $[M{]}_{\infty }=\infty$ almost surely. SetThen $W_u=M_{{\tau }_u}$ is standard Brownian motion relative to $({\mathcal{G}}_u)$ andStrict increase of the bracket is not required in the general theorem: $M$ is constant on intervals on which its bracket is constant. Continuity of $M$ is required. A nonzero initial value is handled by applying the theorem to $M-M_0$ and adding it back. If the terminal bracket is finite, the finite-lifetime extension of the Dambis-Dubins-Schwarz theorem supplies the Brownian motion up to that bracket time and, if necessary, an independent continuation on an enlarged probability space.

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.

For an elementary previsible process $H={\sum }_{j=0}^{r-1}{H}_j{1}_{(t_j,t_{j+1}]}$, with bounded ${\mathcal{F}}_{t_j}$-measurable coefficients, defineThe Itô isometry isMore generally the squared $L^2$ distance between the integrals of $H$ and $K$ is the integral of $(H-K{)}^2$ on the right. The integral is a zero-starting continuous square-integrable martingale.

Elementary predictable processes are dense in $L^2(M)$. Given $H$ in this space, choose elementary $H_n$ converging in its norm and define the Itô integral by the $L^2$ limit of $H_n\cdot M$. The isometry proves that this is independent of the approximating sequence and remains valid for the limit. The Doob L2 maximal inequality bounds the expected squared path supremum of the difference by four times its terminal second moment; hence the convergence also gives a continuous martingale version, with uniform convergence on compacts in probability.

For a continuous local martingale $M$ and locally bounded process $H$ that is previsible, choose increasing stopping times ${\tau }_n\uparrow \infty$ such that $M^{{\tau }_n}$ is L2-bounded, $|H|\le K_n$ up to ${\tau }_n$, and ${\tau }_n\le n$. This can be done by combining localizers with first exits of $M$ from bounded intervals and the local bounds of $H$. Then $H{1}_{[0,{\tau }_n]}$ is in the integrand space for $M^{{\tau }_n}$. Define the integral on each stopped interval by the preceding construction. The stopping property of elementary integrals, extended by the isometry, makes these definitions agree on overlaps. Pasting givesThis is a localization construction; global square integrability of the resulting process is not asserted.

For every zero-starting continuous local martingale $M$, there is a unique, up to indistinguishability, continuous adapted increasing process $[M]$, with $[M{]}_0=0$, such thatThis is its quadratic variation. It also satisfies, for deterministic partitions whose mesh tends to zero,with convergence uniformly on compact time intervals in probability. The theorem includes existence and finiteness on each compact interval; monotonicity is understood pathwise outside one null set.

Uniqueness follows because the difference of two candidate increasing continuous processes is both a finite-variation process and a continuous local martingale, hence is identically zero from its zero initial value. For a continuous semimartingale $X=X_0+M+A$, its quadratic variation is $[X]=[M]$, since continuous finite-variation terms have zero quadratic variation. Polarization defines the quadratic covariation by $[M,N]=([M+N]-[M-N])/4$.

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 càdlàg semimartingales, the semimartingale integration by parts formula isFor continuous processes the left limits can be replaced by the values. The bracket includes the jump products in the general formula.

For a continuous ${\mathbb{R}}^d$-valued semimartingale $X$ and $f\in C^2({\mathbb{R}}^d)$, the Itô formula isThe formula is localized when the derivatives are unbounded; continuity of $X$ makes them bounded after stopping on compact sets.

To prove the one-dimensional formula for polynomials without assuming that formula, use induction. It is immediate for constants and $X$. SupposeWrite $X=X_0+M+A$ with $M$ a continuous local martingale and $A$ finite variation. The induction hypothesis identifies the martingale part of $X^k$ as $k{X}^{k-1}\cdot M$. The stochastic-integral covariation identity, obtained by polarization from the quadratic-variation identity, givesApply semimartingale integration by parts to $X^{k}X$. Its first-order terms combine to $(k+1)X^{k}dX$, and its bracket terms combine toThis proves the induction step. Taking linear combinations proves the polynomial Itô formula from integration by parts for every polynomial $f$.

The initial-value convention is essential. A continuous finite-variation local martingale is constant, so the conclusion is $M_t=M_0$ up to indistinguishability of stochastic processes. The stated zero conclusion uses the usual zero-start convention $M_0=0$; without it, $M_t\equiv 1$ is a counterexample.

Put $N=M-M_0$. Let $V_t$ be its total-variation process, and stop when either $|N|$ or $V$ reaches $n$. The resulting process $N^{{\tau }_n}$ is a bounded continuous local martingale, hence a true martingale, and its total variation on every time interval is bounded by $n$. Fix $t$ and take deterministic partitions $0=t_0<\cdots <t_k=t$ whose mesh tends to zero. Orthogonality of martingale increments givesPathwise,by uniform continuity of a continuous path on $[0,t]$. The sums are bounded by $2n^2$, so the dominated convergence theorem makes their expectations tend to zero. Hence $N_t^{{\tau }_n}=0$ almost surely. Take the countable intersection over rational $t$ and integer $n$, use path continuity, and then let $n\to \infty$. The stopping times tend to infinity because the original path and its finite variation are bounded on every compact time interval. Thus

As with quadratic variation, the increasing process must be normalized by $A_0=0$. Suppose $A,A^{\prime }$ have this initial value and both $M^2-A$ and $M^2-A^{\prime }$ are continuous local martingales. Since $M^2$ and these differences are continuous, both $A$ and $A^{\prime }$ are continuous. Their difference is a finite-variation process, andis a continuous local martingale. The result proved in part (a) makes this difference constant. Its initial value is zero, so $A$ and $A^{\prime }$ are indistinguishable. This is the uniqueness of an increasing square compensator; it uses no existence argument.

If no initial-value normalization is imposed, the literal uniqueness assertion is false: whenever $A$ works, $A+c$ works for every deterministic constant $c$, since subtracting a constant preserves the local martingale property. The exact unnormalized conclusion is $A_t-A_t^{\prime }=A_0-A_0^{\prime }$ for all $t$, up to indistinguishability of stochastic processes.