Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 29 (Stochastic Calculus and Applications)

The previsible sigma-algebra $\mathcal{P}$ on $\Omega \times [0,\infty )$ is generated by $E\times \{0\}$ with $E\in {\mathcal{F}}_0$ and by $E\times (s,t]$ with $E\in {\mathcal{F}}_s$, $0\le s<t$. Equivalently it is the smallest sigma-algebra making every left-continuous adapted process measurable. A previsible process is a process measurable for $\mathcal{P}$.

A bounded simple predictable process has finitely many time intervals and bounded coefficients known at their left endpoints. One may use deterministic endpoints, or the customary more general elementary predictable process with stopping-time intervals:A bounded ${\mathcal{F}}_0$-measurable value at time zero may be added; it does not affect integration against a continuous martingale. A process with finite support is included by adding a zero coefficient on its terminal interval. Both versions of simple processes satisfy the arguments below. The elementary stochastic integral isWhen a stopping time is infinite, use the terminal value $M_{\infty }$, which exists in $L^2$ and almost surely for an L2-bounded continuous martingale. Refining intervals leaves this definition unchanged.

For a continuous local martingale $M$, its quadratic variation $[M]$ is the unique continuous adapted nondecreasing process starting at zero such that $(M_t-M_0{)}^2-[M{]}_t$ is a local martingale. It is predictable because it is continuous and adapted. To recover it from the path, use refining deterministic grids, for example $t_k^{(n)}=k{2}^{-n}$, and formOnly finitely many increments are nonzero on a compact time interval. ThenIn particular each fixed-time sum converges in probability, and a suitable subsequence converges uniformly on compact intervals almost surely. The partial last grid interval is included in this path approximation.

Use the representation from part (a), and put ${\Delta }_{j}M=M_{T_{j+1}}-M_{T_j}$, $U_j=h_j{\Delta }_{j}M$ and $N=H\cdot M$. The optional stopping theorem in the form needed here says that for a uniformly integrable martingale $L$ and stopping times $\sigma \le \tau$, including terminal infinite times, $\mathbb{E}[L_{\tau }\mid {\mathcal{F}}_{\sigma }]=L_{\sigma }$. An L2-bounded continuous martingale is uniformly integrable and has the required terminal value.

Consequently $\mathbb{E}[{\Delta }_{j}M\mid {\mathcal{F}}_{T_j}]=0$. Conditioning on ${\mathcal{F}}_t$ and distinguishing whether $T_j\le t$ givesThus $N_t=\mathbb{E}[{\sum }_j{U}_j\mid {\mathcal{F}}_t]$ is a martingale. Its paths are continuous, since each stopped increment is continuous and zero until its left endpoint.

For $i<j$, $U_i$ is ${\mathcal{F}}_{T_j}$-measurable, so the same conditional identity yields $\mathbb{E}(U_i{U}_j)=0$. These martingale-difference orthogonality relations also hold for the unweighted increments. ThereforeHere $N_{\infty }={\sum }_j{U}_j\in L^2$. By conditional Jensen, $\mathbb{E}{N}_t^2\le \mathbb{E}{N}_{\infty }^2$ for every $t$, soThis proves the norm bound without presupposing the Itô isometry requested next.

First justify optional sampling for the square-minus-bracket process. The Doob L2 maximal inequality gives $\mathbb{E}{\sup }_{t\ge 0}|M_t{|}^2\le 4\mathbb{E}|M_{\infty }{|}^2<\infty$. Stop the local martingale $M^2-[M]$ using a localizing sequence and bounded bracket levels. The stopped expectation identity, followed by Fatou, bounds $\mathbb{E}[M{]}_{\infty }$ by the expected squared path supremum. Hence $M^2-[M]$ has an integrable absolute supremum, and localization with conditional dominated convergence makes it a uniformly integrable martingale. The optional stopping theorem stated in part (b) is therefore applicable to both $M$ and $M^2-[M]$.

For ordered stopping times $\sigma \le \tau$, expand the squared increment and condition on ${\mathcal{F}}_{\sigma }$. The mean increment is zero, soThis is the conditional bracket isometry for stopped martingale increments. Apply it with $\sigma =T_j$, $\tau =T_{j+1}$ and multiply by the bounded ${\mathcal{F}}_{T_j}$-measurable variable $h_j^2$. The orthogonality proved in part (b) gives the Itô isometry

To identify the whole bracket process, let $N=H\cdot M$ and $C_t={\int }_0^t{H}_s^{2}d[M{]}_s$. For deterministic $s<t$ and $E\in {\mathcal{F}}_s$, apply the proved isometry to the simple integrand $1_E{1}_{(s,t]}H$. It yieldsThus $\mathbb{E}[(N_t-N_s{)}^2\mid {\mathcal{F}}_s]=\mathbb{E}[C_t-C_s\mid {\mathcal{F}}_s]$. Since $N$ is a martingale, the cross term $2N_s(N_t-N_s)$ has conditional mean zero. Consequently $N^2-C$ is a martingale. The process $C$ is continuous, adapted, nondecreasing and starts at zero, so uniqueness in the definition of quadratic variation provesThis deduction uses only the proved elementary isometry, not a previously assumed formula for the quadratic variation of stochastic integrals.

The intended setting is $X=M+A$ with $A$ continuous and adapted. These conditions are needed for the requested continuous-semimartingale conclusion; finite variation alone does not supply them. For example, $M=0$, $A_t=1_{\{t\ge 1\}}$ satisfies the printed deterministic bound and starts at zero, but $f_n(X)$ has a jump for sufficiently large $n$. We use the continuous adapted interpretation throughout this question.

Define $f_n(x)={\int }_0^x\varphi (ny)dy$. Then $f_n\in C^2$, $|f_n^{\prime }|\le 1$ and $f_n^{\prime \prime }(x)=n{\varphi }^{\prime }(nx)\ge 0$. The Itô formula gives the canonical continuous semimartingale decompositionThe initial value is zero. The first term is a continuous martingale, since $M$ is bounded and the integrand is bounded. The second is continuous adapted finite variation, hence predictable. This decomposition is unique because a continuous finite-variation local martingale is constant. The phrase Doob-Meyer here must mean this semimartingale decomposition: $V^{(n)}$ need not be nondecreasing, and the hypotheses do not make $f_n(X)$ a submartingale as required by the Doob-Meyer decomposition theorem. Indeed $f_n^{\prime }(0)=-1$, so the bounded deterministic drift $A_t=t\wedge \delta$ makes $f_n(A_t)$ initially decrease, excluding the submartingale property.

The sign process is predictable because it is a Borel function of the continuous adapted, and therefore predictable, process $X$. Its paths need not themselves be left-continuous. Pointwise, including at $x=0$,For a fixed horizon $T$, $\mathbb{E}[M{]}_T=\mathbb{E}{M}_T^2\le K^2$. The dominated convergence theorem for the finite measure $\mathbb{E}d[M]$ on $[0,T]$ givesThe Itô isometry, extended from simple predictable integrands by their density, and the Doob L2 maximal inequality now implyHence

Fix a sample path on the probability-one event where $X$ is continuous and the total-variation process of $A$ satisfies $V_{\infty }\le K$. The signed measure $dA$ has absolute variation measure $dV$. ThusThe integrand tends to zero at every time, including times when $X_s=0$, and is bounded by $2$. Since $dV$ is a finite measure on this path, the dominated convergence theorem makes the right-hand side tend to zero. ThereforeIn fact the convergence is uniform over the entire time axis under the given total-variation bound. This pathwise estimate establishes one common full-probability event, rather than separate exceptional events for each time.

For $x\le 0$, $f_n(x)=|x|$; for $x>0$, only the interval $(0,1/n)$ contributes an error to the integral of the derivative. HenceBy the Itô formula, the continuous processesare nondecreasing, since $f_n^{\prime \prime }\ge 0$. Parts (a) and (b), together with the uniform approximation of absolute value, show that $C^{(n)}$ converges uniformly on compacts in probability toThe limit is continuous: the martingale integral is continuous, and the bounded-integrand finite-variation integral is continuous because $A$ is continuous. For any rational $s<t$, convergence in probability and $C_t^{(n)}-C_s^{(n)}\ge 0$ imply $Z_t-Z_s\ge 0$ almost surely. Intersect these countably many events and use continuity to extend to all real $s<t$. ThusSince $|X|={\mathrm{sgn}}_{-}(X)\cdot M+({\mathrm{sgn}}_{-}(X)\cdot A+Z)$ is a continuous local martingale plus continuous finite variation, $|X|$ is a continuous semimartingale. This is the smooth convex approximation proof of the Tanaka formula. With the specified sign value $-1$ at zero, $Z$ is the right local time at zero, rather than an unspecified symmetric normalization.

For an arbitrary continuous semimartingale starting at zero, use its continuous decomposition $X=M+A$, with both parts starting at zero, and letThe continuity and local finiteness of variation imply ${\rho }_k\uparrow \infty$ almost surely. Up to ${\rho }_k$ the required deterministic bound holds, and the stopped local martingale is bounded, hence a true $L^2$-bounded martingale. Apply the previous argument to $X^{{\rho }_k}$ for each integer $k$. Stopping the defining integrals gives precisely $Z_{t\wedge {\rho }_k}$. On the intersection of their full-probability events, each stopped $Z$ is nondecreasing; every finite interval is eventually contained before ${\rho }_k$, so $Z$ is nondecreasing globally. Therefore

The Dambis-Dubins-Schwarz theorem states that if $N$ is a continuous local martingale with $N_0=0$ and $[N{]}_{\infty }=\infty$ almost surely, then for ${\sigma }_r=\inf \{t:[N{]}_t>r\}$, $W_r=N_{{\sigma }_r}$ is Brownian motion in ${\mathcal{F}}_{{\sigma }_r}$ and $N_t=W_{[N{]}_t}$. If the terminal bracket is finite, Brownian motion can be continued beyond it on an extension of the probability space. For a nonzero initial value, apply the theorem to $N-N_0$.

Apply time change to the vector $(M^1-M_0^1,M^2-M_0^2)$ with the common bracket $A$. The time-change theorem for continuous local martingales says that at inverse bracket times the processes remain continuous local martingales in the time-changed filtration and their quadratic covariations are composed with those times. Since $A$ is continuous and unbounded, $A_{{\sigma }_r}=r$, givingThe Lévy characterization of multidimensional Brownian motion states that a zero-starting continuous vector local martingale with bracket matrix $rI$ is standard multidimensional Brownian motion. Thus $\tilde{B}$ is standard two-dimensional Brownian motion; this step establishes the joint property, not merely the one-dimensional property of each coordinate.

The printed inverse is ${\tau }_r=\inf \{t:A_t\ge r\}$. For $r>0$ it is a stopping time, since $\{{\tau }_r\le t\}=\{A_t\ge r\}$, and ${\tau }_0=0$. The martingale is constant on each flat interval of $A$, as follows from its Dambis-Dubins-Schwarz representation. Hence $M_{{\tau }_r}^i=M_{{\sigma }_r}^i$, including any initial flat interval. Moreover ${\mathcal{F}}_{{\tau }_r}\subseteq {\mathcal{F}}_{{\sigma }_r}$. The common Brownian process is adapted to the smaller filtration, and its increments are independent of the larger past, so they are independent of the smaller past as well. Therefore the common-clock time change of orthogonal local martingales givesIf Brownian motion is understood to start at zero, the centered vector $B_r-B_0$ is the standard Brownian motion. Retaining the initial vector is the convention required in part (c).

For $X=e^{\beta }\cos \theta$ and $Y=e^{\beta }\sin \theta$, apply the two-variable Itô formula. Independence gives $[\beta ,\theta ]=0$, while each individual bracket is $t$. The second derivatives in the two variables cancel, yieldingTheir integrands are continuous adapted and locally bounded, so both coordinates are continuous local martingales. The quadratic variation of a stochastic integral gives

Here is a full proof that the clock diverges. Let $S_0=0$, and successively defineBy recurrence of one-dimensional Brownian motion, all these times are finite almost surely. By the Strong Markov property at the returns $S_j$, the variables $D_j=(U_j-S_j)\wedge 1$ are independent identically distributed, lie in $(0,1]$ almost surely, and have a strictly positive mean. The strong law of large numbers gives ${\sum }_j{D}_j=\infty$ almost surely. On $[S_j,U_j]$, ${\beta }_s\ge -1$, soThe segments are disjoint, and their divergent duration also ensures they cannot accumulate at a finite time. This proves Divergence of a driftless Brownian exponential clock and

The clock $A_t={\int }_0^t{e}^{2{\beta }_s}ds$ is continuous, strictly increasing and unbounded. Define ${\tau }_r=\inf \{t:A_t\ge r\}$. Part (a), with the initial vector retained, proves thatEvery finite $r$ has finite ${\tau }_r$, and $|{\mathcal{B}}_r|=e^{{\beta }_{{\tau }_r}}>0$. On the common event where the clock is a continuous bijection of $[0,\infty )$, this holds simultaneously for all times. Thus the origin is a polar point for planar Brownian motion started at $(1,0)$:To prove unbounded-time visits near zero, let $t_n=\inf \{t\ge n:{\beta }_t=-n\}$. Brownian recurrence makes each $t_n$ finite almost surely, and $t_n\ge n\to \infty$. Set $r_n=A_{t_n}$. Since $A_t\to \infty$, $r_n\to \infty$, whileConsequentlyFor every fixed neighbourhood of zero there are visits at arbitrarily large times. Since $\mathcal{B}$ has the law of planar Brownian motion with this initial value, both conclusions hold for that Brownian law. This complex exponential construction of planar Brownian motion proves the two properties together without confusing recurrence near a point with hitting the point.

Put $U=M-\frac{1}{2}[M]$. Since the bracket is continuous finite variation, $[U]=[M]$. The Itô formula applied to $Z=e^U$ givesThereforeWith the printed exponential formula, $Z_0=e^{M_0}$. The usual unit-initial-value stochastic exponential is $\exp (M_t-M_0-\frac{1}{2}[M{]}_t)$; the two coincide when $M_0=0$, as in part (c).

For any strictly positive solution of this stochastic equation, its bracket is $d[Z{]}_t=Z_t^{2}d[M{]}_t$. Applying Itô to the logarithm, after localizing away from zero if necessary, givesThe same identity holds for $Z^{\prime }$, so $\log Z_t^{\prime }-\log Z_t$ is constant in time and is zero initially. Their continuous versions agree at all times on one full-probability event. Thus pathwise uniqueness for a multiplicative martingale equation givesThis is indistinguishability of stochastic processes, not just equality almost surely at each separately chosen time.

For $0\le s\le t$ and $E\in {\mathcal{F}}_s$, the definition of the Radon-Nikodym derivative givesThe density is nonnegative and integrable; with the probability normalization used here its expectation is one. Henceso $Z$ is a martingale. When $Z_t=h(X_t,t)$, the Itô formula under Wiener measure givesThe first term is a continuous local martingale, by localization on compact space-time sets. Since $Z$ is a martingale, uniqueness of the continuous semimartingale decomposition makes ${\int }_0^{t}g(X_s,s)ds=0$ identically. The integrand is continuous in time along every continuous path, so $g(X_t,t)=0$ for every $t$, outside one common null event.

For each fixed $t>0$, $X_t$ has a Gaussian distribution with strictly positive density everywhere on $\mathbb{R}$. Since $g(\cdot ,t)$ is continuous, $g(X_t,t)=0$ almost surely implies $g(x,t)=0$ for every real $x$. ThereforeIf the stated smoothness includes time zero, continuity extends the equation there. This is the backward heat equation. The full-support argument is needed to pass from an identity along Brownian paths to an equation at every spatial point.

Define $k(x,t)=h_x(x,t)/h(x,t)={\partial }_x\log h(x,t)$. Positivity and smoothness of $h$ make $k$ continuous and bounded on compact space-time sets. Brownian paths remain in compact spatial sets on each finite horizon, so ${\int }_0^{T}k(X_s,s{)}^{2}ds<\infty$ almost surely for every $T$. Thusis a well-defined continuous local martingale with $M_0=0$. The backward heat equation from part (b) removes the drift in Itô's formula and yieldsBy the allowed fact $Z_0=1$ and the uniqueness in part (a),

The finite-horizon form of the Girsanov theorem used here says: if $M_0=0$ is a continuous local martingale and its stochastic exponential $Z$ is a strictly positive true martingale of mean one on $[0,T]$, then under $d\mathbb{Q}{|}_{{\mathcal{F}}_T}=Z_{T}d\mathbb{P}{|}_{{\mathcal{F}}_T}$ every continuous $\mathbb{P}$-local martingale $N$ becomes the $\mathbb{Q}$-local martingale $N-[N,M]$. Quadratic variation is unchanged by this locally equivalent change of measure.

All the assumptions hold. The integrand defining $M$ is locally square integrable as checked above; $Z$ is strictly positive; and it is already a true martingale because it is the density process. On each finite horizon $Z_t=\mathbb{E}[Z_T\mid {\mathcal{F}}_t]$, so it is uniformly integrable on that horizon. There is no need to establish a separate Novikov condition, nor to assume equivalence on the infinite-time sigma-algebra.

Take $N=X$. Its covariation with $M$ is $[X,M{]}_t={\int }_0^{t}k(X_s,s)ds$. Hence $B_t=X_t-{\int }_0^{t}k(X_s,s)ds$ is a continuous $\mathbb{Q}$-local martingale starting at zero with $[B{]}_t=t$. The Lévy characterization of Brownian motion makes it Brownian under $\mathbb{Q}$. Since this argument holds on every finite horizon for the same process $B$,This is the Brownian change of measure by a positive space-time harmonic function, a Doob h-transform with drift ${\partial }_x\log h$.

Apply the Itô formula to $Y_t=e^{B_t+t/2}$. The time derivative contributes $Y_{t}dt/2$, and the quadratic-variation correction contributes another $Y_{t}dt/2$. ThusBoth the drift $b(y)=y$ and diffusion coefficient $\sigma (y)=y$ are globally Lipschitz. The global existence theorem for stochastic differential equations with Lipschitz coefficients gives a unique strong stochastic solution for a fixed driving Brownian motion and initial value; in particular pathwise uniqueness holds. The explicit solution is strictly positive.

Write $N_t={\int }_0^t{Y}_{s}d{B}_s$. This is a continuous local martingale withThe strong law for Brownian motion gives $B_t/t\to 0$ almost surely, so $\log Y_t/t\to 1/2$, $Y_t\to \infty$, and $H_t\to \infty$. The Dambis-Dubins-Schwarz theorem therefore supplies a Brownian motion $\beta$ in the inverse-clock filtration with $N_t={\beta }_{H_t}$ for all $t$. Finally $d{H}_s=Y_s^{2}ds$, hence $Y_{s}ds=Y_s^{-1}d{H}_s$. Integrating the stochastic equation givesThe infinite clock lifetime verifies that this Brownian motion is defined for every nonnegative clock time without an additional extension.

Since $H$ is continuous, strictly increasing and unbounded, $J_t=\inf \{s>0:H_s>t\}$ is its continuous inverse: $H_{J_t}=t$, $J_{H_s}=s$ and $J_0=0$. These inverse times are stopping times. Put $X_t=Y_{J_t}$, which is continuous and strictly positive and is adapted to the inverse-clock filtration ${\mathcal{G}}_t={\mathcal{F}}_{J_t}$.

Evaluate the identity from part (a) at $J_t$. The Lebesgue-Stieltjes change of variable $r=H_s$ givesThereforeThe process $\beta$ is Brownian motion in this same filtration by the construction in part (a). The drift integral is finite on each compact interval because the positive continuous process $X$ has a positive minimum there. This is the dimension-three case of the Exponential Brownian-to-Bessel time change.

Put $R_t=|W_t|$. First verify that the origin is not reached, so that differentiating the radius is legitimate. On ${\mathbb{R}}^3\setminus \{0\}$ the function $F(w)=|w{|}^{-1}$ is harmonic. For $T_{\epsilon }=\inf \{t:R_t=\epsilon \}$ with $0<\epsilon <1$, the Itô formula makes $F(W_{t\wedge T_{\epsilon }})$ a martingale on each finite horizon: its stochastic integrand has norm at most ${\epsilon }^{-2}$. Since $F(W_0)=1$,Let $t\to \infty$, then $\epsilon \downarrow 0$. Any path hitting zero must first hit every positive $\epsilon$, so $\mathbb{P}(\exists t:R_t=0)=0$.

For $r(w)=|w|$, ${\partial }_{i}r=w_i/r$ and $\Delta r=2/r$ in three dimensions. Itô's formula, initially localized away from zero, yieldsThe continuous local martingale ${\gamma }_t={\sum }_i{\int }_0^t(W_s^i/R_s)d{W}_s^i$ has bracket $[\gamma {]}_t={\int }_0^t{\sum }_i(W_s^i/R_s{)}^{2}ds=t$ and starts at zero. By the Lévy characterization of Brownian motion, $\gamma$ is Brownian motion. ThusThis proves that the radial process is a three-dimensional Bessel process solving the same equation as $X$ in part (b). The permitted uniqueness in distribution identifies their laws as processes.

The strong law in part (a) gives $Y_s\to \infty$, and the inverse clock satisfies $J_t\to \infty$. Hence $X_t=Y_{J_t}\to \infty$. Equality of process laws transfers this almost sure path property to $R$.

For the further logarithmic rate, fix $0<\eta <1/2$. Almost surely, for all sufficiently large $s$, $|B_s|\le \eta s$. With a finite random lower cutoff $S$, for $u>S$,Taking logarithms and dividing by $u$ gives lower and upper asymptotic bounds $1-2\eta$ and $1+2\eta$. Letting $\eta \downarrow 0$ along a countable sequence proves the logarithmic growth of an exponential Brownian clock with positive drift, $\log H_u/u\to 1$. Also $\log Y_u/u=B_u/u+1/2\to 1/2$. Since $t=H_{J_t}$,Again transfer the full path event by uniqueness in distribution. We obtain the logarithmic escape rate of three-dimensional Brownian motion

For an independent way to identify the value once existence of the almost sure limit is granted, use Brownian scaling. At each fixed $t$, $W_t/\sqrt{t}$ has the law of $G+(1/\sqrt{t},0,0)$, where $G$ is standard three-dimensional Gaussian. Its norm converges in distribution to the strictly positive finite variable $|G|$, so $\log (|W_t|/\sqrt{t})$ is tight. ThereforeAn existing almost sure limit must agree with this probability limit. This guesses and identifies the value through scaling, while the inverse-clock argument proves that the almost sure limit actually exists.

The diffusion generator is $\mathcal{L}f=\frac{1}{2}{\sigma }^2{f}^{\prime \prime }+b{f}^{\prime }$. On a compact subinterval of $I$, the coefficient $s^{\prime }\sigma$ is bounded, and the Itô formula up to its exit time $\rho$ givesbecause $\mathcal{L}s=0$. Increasing compact subintervals proves the local-martingale property up to $T$. The notation $s(X_{t\wedge T})$ at an exit endpoint is understood with its continuous endpoint extension when needed; stopping on $[a,b]\subset I$ below is always well-defined. This is the martingale property of a scale function of a one-dimensional diffusion.

For the hitting ratio there are two necessary qualifications: $s(b)\ne s(a)$ and $\tau =T_a\wedge T_b<\infty$ almost surely. A usual scale function is strictly monotone, which ensures the first condition, but the printed hypotheses alone ensure neither. Under these conditions, $s(X_{t\wedge \tau })$ is a bounded local martingale and hence a uniformly integrable martingale. The optional stopping theorem givesContinuity prevents simultaneous hitting of distinct endpoints, and almost sure exit makes these two probabilities sum to one. Solving yields the boundary hitting probability from a diffusion scale function