Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 202 (Stochastic Calculus and Applications)

A semimartingale is an adapted process of the form $X=X_0+M+A$, where $M$ is a local martingale and $A$ is an adapted finite-variation process. A sequence of processes $X^n$ converges to $X$ in uniform convergence on compacts in probability, abbreviated ucp, when for every $T<\infty$ and $\delta >0$,

For $f_{ϵ}(x)=({ϵ}^2+x^2{)}^{1/2}$, direct differentiation givesSince the quadratic variation of standard Brownian motion is $[B{]}_t=t$, the Itô formula gives

The deterministic boundholds for every real number $x$. Thereforefor every $T$, so $f_{ϵ}(B)$ converges to $|B|$ uniformly on every compact time interval, and hence ucp.

Set $g_{ϵ}(x)=x/\sqrt{{ϵ}^2+x^2}$ and $g(x)=\mathrm{sgn}(x)$, with $g(0)=0$. For every $s>0$, the normal distribution of $B_s$ has no atom at zero, so $g_{ϵ}(B_s)\to g(B_s)$ almost surely. Since $|g_{ϵ}-g|\le 2$, the dominated convergence theorem and Tonelli theorem giveThe Itô isometry followed by the Doob L2 maximal inequality now yieldsThus the stochastic integrals converge ucp to ${\int }_0^t\mathrm{sgn}(B_s)d{B}_s$.

Rearranging part b expresses the last term asParts c and d, together with stability of ucp convergence under addition, show that

Each $A^{ϵ}$ is continuous, adapted, and increasing. From ucp convergence one can choose a subsequence that converges uniformly almost surely on every compact interval. Its limit $A$ is therefore also continuous and increasing, hence a finite-variation process. The stochastic integral $M_t={\int }_0^t\mathrm{sgn}(B_s)d{B}_s$ is a continuous local martingale, and part e gives the semimartingale decompositionConsequently $|B|$ is a semimartingale. In fact, comparison with the Tanaka formula identifies $A_t$ as the local time of a semimartingale $L_t^0(B)$.

The predictable sigma-algebra $\mathcal{P}$ on $\Omega \times [0,\infty )$ is generated by the sets $A\times \{0\}$ with $A\in {\mathcal{F}}_0$ and $A\times (s,t]$ with $0\le s<t$ and $A\in {\mathcal{F}}_s$. A process is previsible precisely when it is $\mathcal{P}$-measurable.

A simple predictable process has the formwhere $0=t_0<\cdots <t_m<\infty$, $H_0$ is bounded and ${\mathcal{F}}_0$-measurable, and each $H_k$ is bounded and ${\mathcal{F}}_{t_k}$-measurable. Finite linear combinations of indicators of predictable rectangles are of this form after refining the finitely many time partitions.

Those rectangles form a semiring of sets generating $\mathcal{P}$. The collection of sets whose indicators can be approximated in $L^2(\nu )$ by simple predictable processes is a monotone class: for an increasing sequence, truncate the union and use the finiteness of $\nu$; complements and finite disjoint unions are handled by linearity. The Monotone class theorem therefore puts every $\mathcal{P}$-measurable indicator in the closure. Ordinary measurable simple functions are dense in $L^2(\mathcal{P},\nu )$, so simple predictable processes are dense there as well.

Every simple predictable process is left-continuous and adapted: on $(t_k,t_{k+1}]$ its value is already ${\mathcal{F}}_{t_k}$-measurable. Hence any sigma-algebra making all left-continuous adapted processes measurable contains the generators of $\mathcal{P}$.

Conversely, let $X$ be left-continuous and adapted. For $n\ge 1$, defineOn each bounded time interval this is a simple predictable process, and left continuity gives $X_t^{(n)}\to X_t$ for every $t$. Thus $X$ is $\mathcal{P}$-measurable. This proves that $\mathcal{P}$ is exactly the smallest sigma-algebra making every left-continuous adapted process measurable.

If $X:(\Omega \times [0,\infty ),\mathcal{P})\to (\mathbb{R},\mathcal{B}(\mathbb{R}))$ is measurable and $f$ is Borel measurable, then $f\circ X$ is measurable by composition of measurable functions. Hence applying a Borel function pointwise to a predictable process again gives a predictable process.

A deterministic càdlàg function is Borel measurable. The sets $\Omega \times (s,t]$ and $\Omega \times \{0\}$ show that $\mathcal{B}([0,\infty ))$ is contained in $\mathcal{P}$. Therefore every deterministic càdlàg function, regarded as a process constant in $\omega$, is predictable.

Let $Y$ be a nonconstant Bernoulli random variable, let ${\mathcal{F}}_t$ be trivial for $t<1$, and let ${\mathcal{F}}_t=\sigma (Y)$ for $t\ge 1$. This is a right-continuous filtration after completion. The processis adapted and càdlàg. If it were predictable, its value $X_1=Y$ at the deterministic time $1$ would be measurable with respect to the left-limit sigma-algebra ${\mathcal{F}}_{1-}=\sigma ({\bigcup }_{t<1}{\mathcal{F}}_t)$, which is trivial. That contradicts the choice of $Y$, so $X$ is not predictable.

For a continuous local martingale $M$ with $M_0=0$, its stochastic exponential isThe Itô formula gives $d\mathcal{E}(M{)}_t=\mathcal{E}(M{)}_{t}d{M}_t$, so $\mathcal{E}(M)$ is a positive continuous local martingale. Every nonnegative local martingale is a supermartingale, because localization and the Conditional Fatou lemma turn the localized martingale equality into the supermartingale inequality.

The exponential process $Z_t=\exp (-B_t-t/2)$ is a martingale. Under the Girsanov theorem change of measure $d\mathbb{Q}{|}_{{\mathcal{F}}_t}=Z_t,d\mathbb{P}{|}_{{\mathcal{F}}_t}$, the process $W_t=B_t+t$ is Brownian motion. Its first hitting time $T_a$ of $a>0$ is finite $\mathbb{Q}$-almost surely. On $\{T_a\le t\}$, the optional sampling theorem givesbecause $B_{T_a}=a-T_a$. Letting $t\to \infty$ and applying the monotone convergence theorem yieldsThis is the Critical exponential moment of a drifted Brownian hitting time.

By the Reflection invariance of Brownian motion, $-B$ is Brownian motion, and the first time $B_t-t$ reaches $b<0$ is the first time $-B_t+t$ reaches $-b>0$. Part b therefore gives

Now $L_t=\exp (B_t-t/2)$ is the exponential Brownian martingale. At $S_b$, the identity $B_{S_b}=b+S_b$ gives $L_{S_b}=e^{b+S_b/2}$, so $\mathbb{E}{L}_{S_b}=1$. The nonnegative stopped martingale $(L_{t\wedge S_b})$ therefore loses no mass at infinity and is uniformly integrable. The optional sampling theorem at any stopping time $R$ gives

Put $A_t=[M{]}_t$ and use the Dambis-Dubins-Schwarz theorem to write $M_t=B_{A_t}$, enlarging the space if necessary after the terminal clock value. In the time-changed filtration, $A_t$ is a stopping time. For $b<0$, let $S_b=\inf \{s:B_s-s=b\}$. Applying part c with $R=A_t$ givesOn $\{S_b\le A_t\}$, the first integrand is at most $e^b{e}^{A_t/2}$. The assumed Novikov condition therefore impliesas $b\to -\infty$, uniformly for $t\le T$. Meanwhile $mathbf{1}_{\{A_t<S_b\}}\uparrow 1$, so the monotone convergence theorem gives $\mathbb{E}\mathcal{E}(M{)}_t=1$. A nonnegative local martingale with constant expectation is a martingale. Thus $\mathcal{E}(M^T)$ is a martingale, proving the Novikov condition.

The strictly increasing continuous clock $[M]$ has a finite continuous inverse ${\tau }_s$. The time change of a continuous process therefore preserves continuity and adaptedness. Moreover,The optional time-change theorem makes $M_{{\tau }_s}$ a continuous local martingale for ${\tilde{\mathcal{F}}}_s={\mathcal{F}}_{{\tau }_s}$, while composition with the increasing map $s\mapsto {\tau }_s$ preserves the finite variation of $A$. Hence $\tilde{X}$ is a continuous semimartingale in the time-changed filtration.

For $Z_t=e^{B_t+at}$, the Itô formula givesThus its local-martingale part is $M_t={\int }_0^t{Z}_s,d{B}_s$, its finite-variation part is $(a+1/2){\int }_0^t{Z}_{s}ds$, and its quadratic variation clock is

Let $C_t={\int }_0^t{Z}_s^{2}ds$ and ${\tau }_u=\inf \{t:C_t>u\}$. By the Dambis-Dubins-Schwarz theorem, $W_u=M_{{\tau }_u}$ is Brownian motion. Setting $Y_u=Z_{{\tau }_u}$ and using $du=Z_t^{2}dt$ transforms the decomposition in part b intoComparison with the Bessel process equation $d{Y}_u=d{W}_u+(d-1)(2Y_u{)}^{-1}du$ givesThus an exponential Brownian motion with drift becomes a Bessel process under its quadratic-variation time change; this is an Exponential Brownian-to-Bessel time change.

The lifetime of the time-changed process isBy the strong law for Brownian motion, $B_t/t\to 0$ almost surely. If $a<0$, the exponent is eventually at most $at<0$, so the integral is finite. If $a>0$, it is eventually positive and grows linearly; if $a=0$, the recurrence of one-dimensional Brownian motion makes $B$ spend infinite total time in, for example, $[-1,1]$, so the integral is infinite. Consequently

For $Y=X^{\alpha }$, the Itô formula and the Bessel equation giveUse the clockand its inverse. The Dambis-Dubins-Schwarz theorem turns the first term into Brownian motion, while division of the drift by the clock rate givesHence $\tilde{Y}$ is a Bessel process of dimensionThis is the Power time change of a Bessel process.

The drift $v/2$ is continuously differentiable and hence locally Lipschitz on the open interval $(0,1)$, while the diffusion coefficient is the constant one. The local existence and pathwise uniqueness theorem for a stochastic differential equation therefore gives a unique strong solution up to its first exit from every compact subinterval. These solutions agree by pathwise uniqueness, producing a unique maximal local solution of a stochastic differential equation whose lifetime is

Since $V^{\prime }=v$, the fundamental theorem of calculus givesApplying the Itô formula before the lifetime, the two drift terms cancel:Thus $\psi (X)$ is a continuous local martingale. The increasing function $\psi$ is the scale function of a one-dimensional diffusion.

Because $v\in L^1(0,1)$, its indefinite integral $V$ is bounded, so $e^{-V}$ is bounded above and away from zero. Hence $\psi$ extends continuously and strictly increasingly to $[0,1]$. The quadratic-variation clock of $\psi (X)$ iswhose rate is bounded above and away from zero before exit. The Dambis-Dubins-Schwarz theorem therefore identifies $\psi (X)$, up to an equivalent time change, with Brownian motion in the bounded interval $(0,\psi (1))$; in particular, $\mathcal{T}<\infty$ almost surely.

The bounded stopped local martingale $\psi (X_{t\wedge \mathcal{T}})$ is a martingale. If $q={\mathbb{P}}_x(X_{\mathcal{T}}=0)$, the optional sampling theorem givesThereforeThis is the boundary hitting probability from a diffusion scale function.

Let ${\mathcal{T}}_n$ be the first exit from $[1/n,1-1/n]$. On that compact interval the drift $v/2$ is bounded. The Girsanov theorem therefore gives, through every fixed time $t$, a probability measure equivalent to the original one under which the stopped process $X_{\cdot \wedge {\mathcal{T}}_n}$ has Brownian increments before ${\mathcal{T}}_n$.

If $A\subset (0,1)$ has Lebesgue measure zero, the normal distribution of Brownian motion givesSince $\{t<\mathcal{T}\}={\bigcup }_n\{t<{\mathcal{T}}_n\}$, countable subadditivity gives $\mathbb{P}(X_t\in A,t<\mathcal{T})=0$. Thus the killed law at time $t$ is absolutely continuous with respect to Lebesgue measure on $(0,1)$.

The killed transition operator isThe infinitesimal generator of the diffusion isThe Markov property and the Chapman-Kolmogorov equation give, for each fixed $y$,Dividing by $h$, letting $h\downarrow 0$, and using the generator definition together with the assumed $C^{1,2}$ regularity gives the pointwise Kolmogorov backward equation

For Brownian motion started at $x\in D$, the Itô formula and $\Delta u=0$ show thatis a local martingale. Since $D$ is bounded and $u$ is continuous on the compact set $\overline{D}$, the stopped process is bounded and hence a true martingale. Brownian motion exits every bounded domain almost surely, so $t\wedge \tau \to \tau$. The dominated convergence theorem, continuity at the boundary, and $u=f$ on $\partial D$ give the Brownian representation of the Dirichlet problem

The assertion is false without a boundedness or uniform integrability condition. Take the upper half-planeand $u(x,y)=y$. This function is harmonic and continuous on $\overline{D}$, with boundary value $f=0$. The second coordinate of planar Brownian motion is one-dimensional Brownian motion, so its first hitting time $\tau$ of zero is finite almost surely. NeverthelessThe stopped local martingale $u(B_{t\wedge \tau })$ is not uniformly integrable, which is exactly why optional stopping fails in the limit.

For each $x$, the bounded local martingale $u(x+B_t)$ is a true martingale. Taking expectations givesfor every $t>0$. Thus $u$ equals its convolution with every heat kernel. The convolution is smooth, so the originally Borel function $u$ is smooth. Differentiating the heat semigroup identity at $t=0$ gives $\Delta u=0$, so $u$ is a bounded harmonic function on the plane. The Harmonic Liouville theorem now implies that $u$ is constant.