Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 37 (Stochastic Calculus and Applications)

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.

Work first with $M_0=0$. More generally, the assertion needs $M_0\in L^2$ and then applies to $M-M_0$. Without this assumption, an integrable but non-square-integrable initial random variable kept constant in time has zero quadratic variation but is not bounded in L2 space.

Choose increasing stopping timesThey tend to infinity almost surely, because the continuous path and its quadratic variation are finite on every compact time interval. The stopped process is a bounded martingale, and $M_{t\wedge {\tau }_n}^2-[M{]}_{t\wedge {\tau }_n}$ is bounded and a local martingale. Taking expectations givesSuppose $C=\mathbb{E}[M{]}_{\infty }<\infty$. These squared expectations are at most $C$. For each fixed $t$, the stopped values converge almost surely to $M_t$ and are uniformly integrable, since they are uniformly bounded in L2 space. They therefore converge in $L^1$. Passing in the stopped conditional-expectation identity, using contraction of conditional expectation in $L^1$, gives $\mathbb{E}(M_t\mid {\mathcal{F}}_s)=M_s$ for $s\le t$. Thus $M$ is a true martingale. The Fatou lemma gives $\mathbb{E}{M}_t^2\le C$, uniformly in $t$, so it is an L2-bounded continuous martingale.

Conversely, suppose $M$ is a true martingale with ${\sup }_t\mathbb{E}{M}_t^2\le C$. For fixed $t$, bounded-time optional sampling theorem gives $M_{t\wedge {\tau }_n}=\mathbb{E}(M_t\mid {\mathcal{F}}_{t\wedge {\tau }_n})$. Conditional Jensen inequality then yieldsThe brackets increase to $[M{]}_t$. Applying the monotone convergence theorem first in $n$ and then in $t\uparrow \infty$ proves $\mathbb{E}[M{]}_{\infty }\le C<\infty$.

The exact identities and terminal limits can also be justified. On each fixed horizon $t$, the Doob L2 maximal inequality givesThis is an integrable dominating variable for $M_{t\wedge {\tau }_n}^2$, so the dominated convergence theorem and increasing convergence of the brackets remove the stopping:The L2 martingale convergence theorem supplies $M_{\infty }$ with $M_t\to M_{\infty }$ almost surely and in L2 space. Consequently the squares converge in $L^1$, while $[M{]}_t\uparrow [M{]}_{\infty }$. We obtain the integrable terminal bracket criterion with the precise terminal identityFor $M_0\in L^2$ rather than zero, the final identity becomes $\mathbb{E}{M}_{\infty }^2=\mathbb{E}{M}_0^2+\mathbb{E}[M{]}_{\infty }$.

Write $r=\Vert y\Vert$ and use the ordinary Euclidean Laplacian ${\sum }_i{\partial }_i^2$, so the generator of standard Brownian motion is half that operator. For a radial function $f(r)$, the Laplacian in polar coordinates is $f^{\prime \prime }(r)+(d-1)f^{\prime }(r)/r$. With $f(r)=r^{2-d}$,Thus $h$ is a harmonic function on the punctured space.

We first justify that the singularity is not visited. For $0<\eta <x<R$, stop at $S={\tau }_{\eta }\wedge {\tau }_R$. The Itô formula makes $h(B_{t\wedge S})$ a local martingale, and it is bounded on this annulus, hence a true martingale. The exit time is finite almost surely: $\mathbb{P}(S>t)\le \mathbb{P}(\Vert B_t\Vert <R)\to 0$ by the Gaussian transition density. Taking the bounded terminal limit yieldsIf the origin were reached before ${\tau }_R$, every smaller radius would first be reached, so letting $\eta \downarrow 0$ shows that event has probability zero. Taking a countable sequence $R\to \infty$ proves ${\tau }_0=\infty$ almost surely. Localization on increasing annuli now applies the Itô formula globally and gives the Bessel power local martingaleStopping this continuous local martingale at any ${\tau }_a$, including $a=0$, preserves its local-martingale property.

For $0<a<x$, the stopped path stays at radius at least $a$, so $0<M_t\le a^{2-d}$. The bounded local martingale criterion makes it a true martingale. For $a=x$, Brownian radial oscillations at its starting sphere give ${\tau }_x=0$ almost surely and the stopped process is constant. Thus the same true-martingale conclusion holds for $0<a\le x$.

For $a=0$, the process is a strict local martingale. To prove this, write $B_t=\bar{x}+\sqrt{t}G$ in distribution, with $G$ a standard $d$-dimensional normal vector. For every shift $v$,Indeed the integrand outside the unit ball is at most one, while inside it the Gaussian density is at most $(2\pi {)}^{-d/2}$ and the radial integral is a constant times ${\int }_0^{1}rdr$. Brownian scaling therefore givesA true martingale would have expectation $h(\bar{x})>0$ at every time, which is impossible.

The printed range also permits $a>x$, and this case must not be grouped with the bounded case. Here ${\tau }_a<\infty$ almost surely, by the same Gaussian bound for exit from a bounded ball. Moreover,since the second expectation is at most $\mathbb{E}h(B_t)\to 0$. It too is a strict local martingale. The complete stopped inverse radial power martingale classification is therefore

For $0<a<x<b$, put $S={\tau }_a\wedge {\tau }_b$. The process $\varphi (\Vert B_{t\wedge S}\Vert )$, where $\varphi (r)=r^{2-d}$, is a bounded martingale by part (a). Also $S<\infty$ almost surely: before exit the endpoint $B_t$ lies inside the radius-$b$ ball, whose Gaussian probability tends to zero as $t\to \infty$. Path continuity makes the terminal radius either $a$ or $b$. If $p={\mathbb{P}}_{\bar{x}}({\tau }_a<{\tau }_b)$, the dominated convergence theorem applied to this bounded stopped martingale givesSolving yieldsThe denominator is nonzero because $\varphi$ is strictly decreasing for $d\ge 3$.

As $b\uparrow \infty$, the events $\{{\tau }_a<{\tau }_b\}$ increase to $\{{\tau }_a<\infty \}$. In one direction membership already implies a finite hit of $a$; in the other direction a continuous path up to its finite hit of $a$ has a finite maximum radius, so it lies in the event for every sufficiently large $b$. Continuity from below of probability and $\varphi (b)\to 0$ now give

Set $N_t={\int }_0^t{Z}_s^{-1}d{Z}_s$, the stochastic logarithm of $Z$. On each finite horizon, a strictly positive continuous path has its reciprocal bounded; stopping at levels of $Z$, $Z^{-1}$ and $[Z]$ rigorously localizes this stochastic integral. Thus $N$ is a zero-start continuous local martingale, withThe Itô formula for the ordinary logarithm givesAccordingly, with the exponential convention printed in this question,The initial constant in $M$ contributes no quadratic variation. Under the standard normalized Doléans-Dade exponential convention, which subtracts the driver's initial value in the exponent, the same formula is written $Z=Z_0\mathcal{E}(N)$.

For uniqueness, suppose another continuous driver $L$ gives the printed exponential. Evaluating at zero forces $L_0=\log Z_0$. Applying the Itô formula to its exponential gives $d{Z}_t=Z_{t}d{L}_t$. Since $Z_t>0$, associativity of stochastic integration yields $d{L}_t=Z_t^{-1}d{Z}_t$. Thus $L_t=L_0+N_t=M_t$, up to indistinguishability of stochastic processes.

There is an initial-integrability qualification if “local martingale” is used with the usual integrable-initial-value definition. The displayed $M$ is a continuous local martingale when $\log Z_0\in L^1$, in particular when $Z_0$ is a positive deterministic number. An arbitrary positive integrable $Z_0$ does not ensure this. For example, take an ${\mathcal{F}}_0$-measurable positive integer $W$ with $\mathbb{P}(W=n)=6/({\pi }^2{n}^2)$ and put $Z_t=e^{-W}$ for all $t$. This is a bounded strictly positive martingale, but any representing driver must have $M_0=-W$, which is not integrable. Thus the unqualified claim is false under that definition. For general $Z_0$, the always-valid local-martingale statement is the zero-start representation $Z=Z_0\mathcal{E}(N)$; the printed representation additionally requires the stated integrability or the convention that allows arbitrary finite initial constants.

Let $D=d\mathbb{Q}/d\mathbb{P}$ on $\mathcal{F}$. The Radon-Nikodym derivative is nonnegative and ${\mathbb{E}}_{\mathbb{P}}D=1$. For $A\in {\mathcal{F}}_t$,Uniqueness of the restricted Radon-Nikodym derivative therefore identifiesFor $s\le t$, the tower property of conditional expectation gives ${\mathbb{E}}_{\mathbb{P}}(Z_t\mid {\mathcal{F}}_s)=Z_s$, and ${\mathbb{E}}_{\mathbb{P}}{Z}_t=1$. Thus $Z$ is a nonnegative true martingale, indeed a uniformly integrable martingale, since it consists of conditional expectations of one integrable random variable. This is the Radon-Nikodym density martingale. Under the usual conditions for a filtration, it has the standard càdlàg version; continuity is the additional assumption in part (c), not a consequence of absolute continuity alone.

On the finite horizon $[0,T]$, strict positivity gives equivalence of the restricted probability measures: for $A\in {\mathcal{F}}_T$, $\mathbb{Q}(A)={\mathbb{E}}_{\mathbb{P}}(Z_T{1}_A)$ vanishes if and only if $\mathbb{P}(A)$ vanishes. The standard change-of-measure theorem gives semimartingale invariance under equivalent measures: a process is a semimartingale under $\mathbb{P}$ on this horizon if and only if it is one under $\mathbb{Q}$. The sample-path quadratic variation is unchanged, but the local-martingale and finite-variation parts generally change.

More explicitly, set $N=\int Z^{-1}dZ$, the zero-start stochastic logarithm of the density. For every continuous $\mathbb{P}$-local martingale $L$, the Girsanov theorem states thatis a continuous $\mathbb{Q}$-local martingale. Consequently, if the continuous semimartingale decomposition under $\mathbb{P}$ is $X=X_0+L+V$, its decomposition under $\mathbb{Q}$ isThe covariation correction is a continuous finite-variation process, locally well-defined because $Z$ stays strictly positive. In the Brownian case, if $d{Z}_t=Z_t{\theta }_{t}d{B}_t$, thenis a Brownian motion under $\mathbb{Q}$. The inverse finite-horizon density is $1/Z$, giving the reverse change of measure. These assertions apply on each horizon on which the density is strictly positive; they do not assert equivalence on a larger terminal sigma-algebra where the full density $D$ may vanish.

The Brownian reflection principle gives, for $t>0$,For a standard normal random variable $N^{\prime }$, this is also $\mathbb{P}(|N^{\prime }|\ge x/\sqrt{t})=\mathbb{P}(x^2/(N^{\prime }{)}^2\le t)$. Thus the inverse-square Gaussian law of Brownian first passage isIn particular, ${\tau }_x$ is finite almost surely.

The Brownian motions $X,Y$ are independent, so conditional on ${\tau }_x=s$, the ordinate $Y_{{\tau }_x}$ has the same distribution as $\sqrt{s}N$, with $N$ a standard normal independent of the random time. By Brownian scaling and the hitting-time identity,The last equality uses symmetry of $N$ and independence of $N,N^{\prime }$: inserting the independent sign of $N^{\prime }$ does not change the distribution of the numerator. ThereforeFor completeness, the ratio has density ${\int }_{\mathbb{R}}|v|\phi (cv)\phi (v)dv=1/(\pi (1+c^2))$, where $\phi$ is the standard normal density. The scaled exit ordinate has density $x/(\pi (x^2+y^2))$.

Translate the horizontal coordinate by $x$. Then $X_t+x$ is a standard Brownian motion starting at zero, and continuity identifies $\tau$ with its first-passage time to $x$. The vertical coordinate remains an independent standard Brownian motion starting at zero. Part (a) therefore gives the Cauchy exit law from a Brownian half-plane:The TeX transcription's initial value $-z$ and later $Y_T$ are both corrected here to the PDF's $-x$ and $Y_{\tau }$.

Start the planar Brownian motion $X_t+i{Y}_t$ at $-x$ and stop at $\tau =\inf \{t:X_t=0\}$. The complex exponential is holomorphic with nonzero derivative. By conformal invariance of planar Brownian motion, the process $e^{X_t+i{Y}_t}$, run on its conformal Brownian clockis planar Brownian motion started at $\epsilon =e^{-x}$, until its first exit from the unit disk. The clock is strictly increasing before $\tau$, and $A_{\tau }\le \tau <\infty$ almost surely. It is legitimate to use the exponential although it is not globally injective: the Brownian coordinate brackets depend locally on the nonzero derivative, and the clock construction gives the stopped planar Brownian law. Before $\tau$, $|e^{X_t+i{Y}_t}|<1$; at $\tau$ it equals one. Hence the disk exit time in the new clock is exactly $A_{\tau }$.

The continuous lifted argument of this path, starting from angle zero, is $Y_t$. Its total argument at exit is therefore $Y_{\tau }$, which has distribution $xC$ by part (b). If the winding count records only completed integer turns, it differs from $Y_{\tau }/(2\pi )$ by a remainder $R_{\epsilon }$ with $|R_{\epsilon }|\le 1$. If fractional turns are retained, that remainder is zero. In either convention, since $\log \epsilon =-x$,The first term has distribution $-C/(2\pi )$ for every $x$, and the second tends to zero deterministically in absolute value as $x\to \infty$. Symmetry of the Cauchy distribution removes the minus sign, so Slutsky theorem gives the small-radius Brownian winding law:

A strong solution of a stochastic differential equation is constructed for the prescribed driving Brownian motion on the prescribed stochastic basis, adapted to the augmented filtration generated by that driver and the initial data, and satisfies the integral equation almost surely. For this equation it must be continuous, nonnegative and satisfy $Z_t=z+{\int }_0^t\sqrt{Z_s}d{B}_s$. Pathwise uniqueness means that any two solutions on the same filtered space, with the same initial value and the same driving Brownian motion, are indistinguishable.

The standard global existence theorem for stochastic differential equations with Lipschitz coefficients cannot be applied to the square-root branching diffusion, because $\sigma (x)=\sqrt{x}$ is not Lipschitz at zero:It is locally Lipschitz on $(0,\infty )$, which gives a solution only up to its possible approach to zero by that theorem. Thus the missing hypothesis is the Lipschitz bound at the absorbing boundary, not any failure of continuity. Failure of this sufficient criterion does not establish failure of strong existence or uniqueness; part (c) constructs strong existence directly.

Work on a common stochastic basis on which $B,B^{\prime }$ are independent Brownian motions and the solutions are adapted; independent weak solutions can be put on the product stochastic basis. Independence gives $[B,B^{\prime }]=0$. Put $S=Z+Z^{\prime }\ge 0$ and define the predictable weightsThey are bounded and satisfy $u_t^2+v_t^2=1$ everywhere. DefineThis is a zero-start continuous local martingale, and its quadratic variation is $[\beta {]}_t={\int }_0^t(u_s^2+v_s^2)ds=t$. By the Lévy characterization of Brownian motion, $\beta$ is a Brownian motion in the common filtration.

On $\{S_t>0\}$, $\sqrt{S_t}{u}_t=\sqrt{Z_t}$ and $\sqrt{S_t}{v}_t=\sqrt{Z_t^{\prime }}$. On $\{S_t=0\}$, nonnegativity forces both $Z_t,Z_t^{\prime }$ to be zero, so the same identities hold. Associativity of stochastic integration therefore givesThis proves the addition law for square-root branching diffusions: $Z+Z^{\prime }$ is a weak solution with initial value $z+z^{\prime }$. Specifying the weights at zero is necessary to keep $[\beta {]}_t=t$ after extinction; simply setting both weights to zero would fail to produce a Brownian driver.

For fixed $\epsilon >0$, the derivative of ${\sigma }_{\epsilon }$ is zero on $(-\infty ,\epsilon /2]$, bounded on the compact transition interval $[\epsilon /2,\epsilon ]$, and equals $1/(2\sqrt{x})\le 1/(2\sqrt{\epsilon })$ for $x\ge \epsilon$. Thus ${\sigma }_{\epsilon }$ is globally Lipschitz, with at most linear growth. The global existence theorem for stochastic differential equations with Lipschitz coefficients supplies a global strong stochastic solution and pathwise uniqueness for the modified equation.

The solution stays nonnegative. More precisely, if $z>\epsilon /2$, it cannot cross $\epsilon /2$: on reaching this level the constant continuation solves the equation, and pathwise uniqueness forces that continuation. If $z\le \epsilon /2$, it is constant from the outset. In either case $Z^{\epsilon }$ is a nonnegative local martingale, so $\mathbb{E}{Z}_s^{\epsilon }\le z$.

For ${\epsilon }^{\prime }<\epsilon <z$, both coefficients agree with $\sqrt{x}$ on $[\epsilon ,\infty )$. Stop the two solutions when either first reaches $\epsilon$, and also at a common upper bound. On the resulting compact state interval, the coefficient is Lipschitz. The Itô isometry applied to their difference, followed by the Gronwall inequality, gives zero expected squared difference. Remove the upper bound using the nonexplosion of each modified solution. The two solutions agree until the first of these lower exits, and by continuity both have value $\epsilon$ there, so their exits at that level coincide. In particular,up to indistinguishability. If $\epsilon \ge z$, the first exit time is zero and the same assertion is immediate from their initial values.

Now take a decreasing sequence $0<{\epsilon }_n<z$ with ${\epsilon }_n\downarrow 0$. Write $Z^n=Z^{{\epsilon }_n}$ and ${\tau }_n=T_{{\epsilon }_n}$. Compatibility gives ${\tau }_n\le {\tau }_{n+1}$; indeed, a continuous path cannot reach the smaller level before reaching the larger one. Let $\tau ={\lim }_n{\tau }_n$, a stopping time. On $[0,\tau )$ the compatible solutions define an adapted continuous process $Z$ by setting $Z_t=Z_t^n$ whenever $t\le {\tau }_n$. We still need to justify the finite limiting endpoint and the integral equation there, rather than assume they exist.

For this purpose use the compatible predictable integrandsTheir squares increase pointwise with $n$: on the smaller interval compatibility makes the values equal, and outside it the old integrand is zero. Their limit $H$ is predictable and equals $\sqrt{Z_s}$ before $\tau$, and zero after $\tau$, apart from irrelevant single-time endpoints. For every finite $t$,The monotone convergence theorem gives $\mathbb{E}{\int }_0^t{H}_s^{2}ds\le zt$. Moreover, compatibility makes $(H-H^n{)}^2=H^2-(H^n{)}^2$ away from endpoints, so its expected time-integral tends to zero. The Itô isometry, and the Doob L2 maximal inequality if uniform convergence on a finite time interval is desired, therefore give a continuous stochastic-integral limitBefore ${\tau }_n$, the integrand agrees with $\sqrt{Z^n}$, which is the original square-root coefficient there. Locality of the stochastic integral yields $L_{t\wedge {\tau }_n}=Z_{t\wedge {\tau }_n}^n$. A countable intersection in $n$ and path continuity give this equality simultaneously for all times.

On $\{\tau <\infty \}$, every ${\tau }_n$ is finite and $L_{{\tau }_n}={\epsilon }_n$. Continuity of $L$ thus gives $L_{\tau }=0$. The limiting integrand is zero after $\tau$, so $L$ remains zero afterward. On $\{\tau =\infty \}$, every finite time is covered by one of the compatible positive solutions. Consequently $L$ is nonnegative everywhere and $H_s=\sqrt{L_s}$ for almost every $s$, including after absorption. The cutoff construction of an absorbed square-root diffusion has producedAll modified solutions, stopping times and integrands were constructed from the same prescribed Brownian motion; hence this is a strong stochastic solution. The integral is square-integrable on every finite horizon by the bound $zt$, which also rules out a hidden finite-time explosion in the patching argument.

Put $R_{\epsilon }={\int }_0^{\epsilon }(H_u-H_0)d{B}_u$ and $d_{\epsilon }={\epsilon }^{-1}{\int }_0^{\epsilon }\mathbb{E}|H_u-H_0{|}^{2}du$. Boundedness and path continuity of $H$ make $\mathbb{E}|H_u-H_0{|}^2\to 0$ by the dominated convergence theorem, so $d_{\epsilon }\to 0$. The Itô isometry gives $\mathbb{E}{R}_{\epsilon }^2=\epsilon d_{\epsilon }$.

We must not assume independence between the integral and its Brownian denominator. Instead, the Cauchy-Schwarz inequality givesConcavity, or Jensen inequality, bounds the first factor by $(\mathbb{E}{R}_{\epsilon }^2{)}^{1/8}$. Since $B_{\epsilon }\stackrel{d}{=}\sqrt{\epsilon }N$, the second factor is ${\epsilon }^{-1/8}(\mathbb{E}|N{|}^{-1/2}{)}^{1/2}$. This negative normal moment is finite: the density is bounded near zero, where ${\int }_0^1{u}^{-1/2}du<\infty$, and has Gaussian decay at infinity. In fact $\mathbb{E}|N{|}^{-1/2}=2^{-1/4}\Gamma (1/4)/\sqrt{\pi }$. The powers of $\epsilon$ cancel, yielding the fractional-moment control of Brownian increment ratiosThe ratio may be assigned any value on the null event $B_{\epsilon }=0$.

Fix a deterministic $t>0$ and define ${\hat{B}}_u=B_{t+u}-B_t$, ${\hat{H}}_u=H_{t+u}$ and ${\hat{\mathcal{F}}}_u={\mathcal{F}}_{t+u}$. The increment process is a standard Brownian motion in the shifted filtration, and $\hat{H}$ remains bounded, continuous and adapted. Thus part (a) applies even though the initial value ${\hat{H}}_0=H_t$ is random. For $\epsilon >0$,Here the constant-in-time integrand $H_t$ is ${\mathcal{F}}_t$-measurable, so its stochastic integral equals $H_t(B_{t+\epsilon }-B_t)$. The expectation of the absolute value of the right-hand side to the power $1/4$ tends to zero by part (a). For every $\eta >0$, Markov inequality consequently givesThereforeThe stated convergence in probability holds for each fixed deterministic time.