Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 202 (Stochastic Calculus and Applications)

Let $\mathcal{P}$ be the predictable sigma-algebra on $\Omega \times [0,\infty )$, generated by $A\times \{0\}$ with $A\in {\mathcal{F}}_0$ and $A\times (s,u]$ with $A\in {\mathcal{F}}_s$. A stochastic process $H$ is previsible precisely when $(\omega ,t)\mapsto H_t(\omega )$ is $\mathcal{P}$-measurable.

Every left-continuous adapted process is previsible. Indeed, on intervals $(k{2}^{-n},(k+1)2^{-n}]$, put $H_t^{(n)}=H_{k{2}^{-n}}$, and put $H_0^{(n)}=H_0$. Each approximation is a predictable process, since its coefficient is ${\mathcal{F}}_{k{2}^{-n}}$-measurable; left continuity gives $H_t^{(n)}\to H_t$ for every $t$, so measurability passes to the limit.

An adapted càdlàg process need not be previsible. For a concrete example, let $\xi$ take values $0,1$ with equal probabilities, let ${\mathcal{F}}_t$ be trivial for $t<1$ and ${\mathcal{F}}_t=\sigma (\xi )$ for $t\ge 1$, and set $H_t=\xi 1_{[1,\infty )}(t)$. This is an adapted càdlàg process, but $H_1=\xi$ is not ${\mathcal{F}}_{1-}$-measurable, where ${\mathcal{F}}_{1-}=\sigma ({\bigcup }_{s<1}{\mathcal{F}}_s)$. Sections at time $1$ of the generators of $\mathcal{P}$ all belong to ${\mathcal{F}}_{1-}$, so every predictable process has an ${\mathcal{F}}_{1-}$-measurable value there; this proves the counterexample directly.

Use the standard convention that the finite-variation process $X$ has càdlàg paths and that the stochastic integral starts from zero. Write $V(X)$ for its total-variation process. The necessary admissibility condition is ${\int }_{(0,t]}|H_s|d{V}_s(X)<\infty$ almost surely for each finite $t$; it holds, for example, for locally bounded $H$. Define the pathwise Lebesgue-Stieltjes integralAn atom at $s>0$ contributes $H_s\Delta X_s$, so this convention also fixes the treatment of jumps.

Fix $t$. For a generator $H_s=1_A{1}_{(a,b]}(s)$ of the predictable sigma-algebra, with $A\in {\mathcal{F}}_a$, the Lebesgue-Stieltjes integral isIt vanishes when $t\le a$ and is ${\mathcal{F}}_t$-measurable otherwise. The total-variation process is adapted: for a càdlàg path its value at $t$ is the supremum over a countable collection of partitions of an interval using rational interior points and endpoint $t$. Thus the positive and negative Lebesgue-Stieltjes measures $d{X}^{\pm }=(dV(X)\pm dX)/2$ have ${\mathcal{F}}_t$-measurable masses on the generator intervals.

The Monotone class theorem, applied separately to these positive random measures on $[0,t]$, extends this measurability from the generators of the predictable sigma-algebra to all nonnegative predictable processes. Truncation and positive/negative decomposition then cover every admissible $H$. Consequently $H\cdot X$ is adapted.

The integrability condition is essential. Taken literally, an arbitrary previsible process cannot always be integrated against a finite-variation process: with $X_t=t$ and deterministic $H_s=s^{-1}$ for $s>0$, $H_0=0$, the proposed value at every $t>0$ diverges. The usual càdlàg convention is also what identifies $dX$ by $dX((a,b])=X_b-X_a$.

For deterministic step functions, define directly from Brownian motion incrementswith $s_0$ the left endpoint of the first interval and $f=0$ outside the listed intervals. The printed endpoint list omits $s_0$; the intended convention is $0\le s_0<s_1<\cdots <s_n$, usually $s_0=0$.

Given finitely many observation times, refine the deterministic partition of an interval to include all observation times and all $s_i$. Every $Z_{t_j}$ is a linear combination of the same finite family of independent centered normal Brownian motion increments. Thus every linear combination of $(Z_{t_1},\dots ,Z_{t_m})$ is normal, including possibly a degenerate constant: $Z$ is a Gaussian process. By independence, only increments in the common interval $[0,s\wedge t]$ contribute to the covariance, givingThe same calculation for two step functions $f,g$ proves, rather than assumes, the Itô isometryFor $f\in L^2({\mathbb{R}}_{+})$, choose deterministic step functions $f_n\to f$ in the L2 space. This identity makes $Z_t(f_n)$ a Cauchy sequence in the L2 space; the Riesz-Fischer theorem defines $Z_t(f)$ by mean-square convergence, independently of the approximation. The covariance formula passes to the limit by the Cauchy-Schwarz inequality. For finitely many times and coefficients $a_j$, the characteristic functions converge toso the extension remains a Gaussian process.

For completeness, the construction has a continuous version. Each approximating $Z(f_n)$ is a martingale, directly by the centered independent increments of Brownian motion; the Doob L2 maximal inequality givesTo prove the inequality used here without assuming Itô integral properties, a continuous martingale $N$ satisfies $\lambda \mathbb{P}(N_T^{\ast }\ge \lambda )\le \mathbb{E}(|N_T|1_{\{N_T^{\ast }\ge \lambda \}})$, by conditioning at the first crossing on finite deterministic grids and then using continuity; integrating in $\lambda$ and using the Cauchy-Schwarz inequality yields $\Vert N_T^{\ast }{\Vert }_2\le 2\Vert N_T{\Vert }_2$. A sufficiently fast subsequence therefore converges uniformly almost surely on every compact time interval, giving the continuous version of the stochastic integral.

For a deterministic partition of an interval $0=t_0<\cdots <t_m=t$, put $Q_{\pi }={\sum }_j{\Delta }_{j}B{\Delta }_j\tilde{B}$. Independence of the two Brownian motions and their independent increments givesThe defining sums for quadratic covariation therefore converge to zero in the L2 space and in probability. Hence, with the continuous versions of the quadratic covariation,Equality holds simultaneously for all $t$ by checking rational times and using continuity. For continuous local martingales, the angle-bracket notation here agrees with quadratic covariation.

The stochastic process $W$ starts from zero and has continuous paths. Its increments on disjoint intervals are independent, since the corresponding pairs of Brownian motion increments are independent, andThus $W$ is a standard Brownian motion, including $\rho =\pm 1$. Here the intended filtration is the joint natural filtration, or another filtration relative to which the pair is a two-dimensional Brownian motion; independence of their path laws alone would not justify arbitrary extra information in a larger filtration.

By bilinearity of quadratic covariation and part (i),

The Itô formula applied to $F(t,x)=e^{t/2}\cos x$ gives exact cancellation of the drift:On every finite interval $[0,T]$, the Itô integral is square-integrable since $\mathbb{E}{\int }_0^T{e}^s{\sin }^2{B}_{s}ds\le e^T-1$. Therefore $e^{t/2}\cos B_t$ is a true martingale, and hence a local martingale.

For the other stochastic process, $d(B_t-t^2)=d{B}_t-2tdt$. If $B_t-t^2$ were a local martingale, its difference with $B$ would make the continuous finite-variation process $-t^2$ a local martingale, contradicting the fact that a continuous finite-variation local martingale is constant. Also $\mathbb{E}(B_t-t^2)=-t^2$ is not constant. Thus $B_t-t^2$ is neither a local martingale nor a martingale.

On the right-hand side, ${\mathbb{E}}_x$ means that the reference Brownian motion starts at $B_0=x$; on the left it means the stochastic differential equation starts at $X_0=x$. Since $h$ is smooth and compactly supported, $h^{\prime }$ is bounded and globally Lipschitz continuous. Define the stochastic exponentialThe Novikov condition holds on every finite horizon $T$, because $\mathbb{E}\exp (\frac{1}{2}{\int }_0^T{h}^{\prime }(B_s{)}^{2}ds)\le \exp (T\Vert h^{\prime }{\Vert }_{\infty }^2/2)$. Thus $D$ is a true martingale. Under $d\mathbb{Q}=D_{T}d\mathbb{P}$, the Girsanov theorem makes ${\hat{B}}_t=B_t-x-{\int }_0^t{h}^{\prime }(B_s)ds$ a standard Brownian motion for $t\le T$. Consequently $B$ under $\mathbb{Q}$ solves the prescribed stochastic differential equation; the existence and pathwise uniqueness theorem for a stochastic differential equation identifies its law with that of $X$.

The Itô formula gives ${\int }_0^t{h}^{\prime }(B_s)d{B}_s=h(B_t)-h(x)-\frac{1}{2}{\int }_0^t{h}^{\prime \prime }(B_s)ds$. Hence, for any bounded measurable $f$,The finite-horizon argument applies separately to every $t\ge 0$.

Let $X,Y$ be two continuous solutions on the same filtered probability space, driven by the same Brownian motion and with $X_0=Y_0$ almost surely. Write $D=X-Y$ and choose a common Lipschitz constant $L$ for $b,\sigma$. Introduce the stopping time ${\tau }_n=\inf \{t\ge 0:|X_t|+|Y_t|\ge n\}$. On $\{{\tau }_n=0\}$ the stopped difference is zero; before ${\tau }_n$, both coefficients and the difference are bounded.

The Itô formula and the zero expectation of the stopped square-integrable Itô integral yieldThe Gronwall inequality implies $\mathbb{E}{D}_{t\wedge {\tau }_n}^2=0$. Continuous solution paths are bounded on each compact time interval, so ${\tau }_n\uparrow \infty$ almost surely. Letting $n\to \infty$ proves $X_t=Y_t$ almost surely at each fixed time, and a countable dense set of times together with continuity gives indistinguishability of stochastic processes. Therefore the stochastic differential equation has pathwise uniqueness.

The Itô formula verifies the geometric Brownian motion solutionThis is valid for negative $x$ as well as positive $x$; $x=0$ gives the zero solution. Pathwise uniqueness follows from the Lipschitz continuity of the linear coefficients. The moment-generating function of a normal distribution gives $\mathbb{E}{e}^{k\sigma B_t}=e^{k^2{\sigma }^{2}t/2}$, and thereforeThese moments are finite on every finite horizon. If the convention $0\in \mathbb{N}$ is used, the zeroth moment is $1$.

Multiplication by the integrating factor $e^{t/2}$ and the Itô product rule give the Ornstein-Uhlenbeck processThe two summands are independent centered normal random variables, because $X_0$ is independent of the Brownian motion and the integrand is deterministic. By the Itô isometry, their variances sum to $e^{-t}+{\int }_0^t{e}^{-(t-u)}du=1$. ThusFor $t\ge s$, split the Itô integral at $s$ to obtain $X_t=e^{-(t-s)/2}{X}_s+{\int }_s^t{e}^{-(t-u)/2}d{B}_u$, where the latter term is centered and independent of $X_s$. ConsequentlyIn particular this initialization makes the Gaussian process a stationary process.

Use the standard definition of a local martingale, including an integrable initial value, and let $T_n\uparrow \infty$ be a localizing sequence. If $X\ge 0$, the Fatou lemma gives $\mathbb{E}{X}_t\le {\mathrm{liminf}}_n\mathbb{E}{X}_{t\wedge T_n}=\mathbb{E}{X}_0<\infty$. For $s\le t$, the Conditional Fatou lemma givesFor clarity, the Conditional Fatou lemma itself follows by setting $U_m={\inf }_{n\ge m}{U}_n$ for nonnegative $U_n$, applying the conditional version of the monotone convergence theorem to $U_m\uparrow \mathrm{liminf}{U}_n$, and observing $\mathbb{E}(U_m\mid {\mathcal{F}}_s)\le {\inf }_{n\ge m}\mathbb{E}(U_n\mid {\mathcal{F}}_s)$. Hence a nonnegative local martingale is a supermartingale.

If $|X_t|\le C$ simultaneously for all $t$, with a deterministic $C<\infty$, the dominated convergence theorem applied to the stopped martingale identity gives, for each $A\in {\mathcal{F}}_s$,This is exactly the conditional expectation identity, so a bounded local martingale is a true martingale. The same proof works for deterministic bounds on every finite time interval; merely having a finite random pathwise bound supplies no dominating integrable random variable and is a different convention.

Let $U_t=(1,0,0)+(B_t^1,B_t^2,B_t^3)$, with three independent standard Brownian motions, and set $R_t=|U_t|$ and $M_t=R_t^{-1}$. The radial stochastic process $R$ is a three-dimensional Bessel process, which never hits zero by the Hitting-zero classification for a Bessel process. In three dimensions $u\mapsto |u{|}^{-1}$ is a harmonic function away from zero: direct differentiation gives $\Delta (|u{|}^{-1})=0$.

For the stopping times $T_n=\inf \{t:R_t\notin (1/n,n)\}$, $n\ge 2$, the Itô formula givesThe coefficients are bounded by $n^2$, so each stopped Itô integral is a true martingale on every finite horizon. The stopping times form a localizing sequence; as permitted, its divergence need not be proved here. Thus $M$ is a nonnegative continuous local martingale.

To see that $M$ is not a true martingale, integrate the three-dimensional multivariate normal density of $U_t$ in spherical coordinates. The angular integral yieldswhere $\Phi$ is the standard normal distribution function. Therefore the reciprocal of a three-dimensional Bessel process is a strict local martingale.

Put $\Delta B_h=B_{t+h}-B_t$ and $R_h={\int }_t^{t+h}(H_s-H_t)d{B}_s$. The constant-in-time coefficient $H_t$ on $(t,t+h]$ is previsible, so the ratio minus $H_t$ is $R_h/\Delta B_h$. Since $|H|\le C$, continuity and the dominated convergence theorem implyThe Itô isometry gives $\mathbb{E}{R}_h^2\le h{\eta }_h$. For $\epsilon ,\delta >0$, use the Chebyshev inequality and the normal distribution of $\Delta B_h/\sqrt{h}$ to obtainFirst let $h\downarrow 0$, then $\delta \downarrow 0$; the normal distribution has no atom at zero. This provesNo independence between $R_h$ and the denominator is needed.

The first estimate suggested by the hint is valid: Brownian scaling and the Gamma function give $\mathbb{E}|\Delta B_h{|}^{-1/2}=h^{-1/4}{2}^{-1/4}\Gamma (1/4)/\sqrt{\pi }$. The second hint requires care about the placement of its expectation: with $V_h={\int }_t^{t+h}(H_s-H_t{)}^{2}ds$, the interpretation $\mathbb{E}|R_h{|}^{1/2}\le (\mathbb{E}{V}_h{)}^{1/4}$ is correct by the Jensen inequality and the Itô isometry, whereas the interpretation $\mathbb{E}|R_h{|}^{1/2}\le \mathbb{E}(V_h^{1/4})$ is false with constant $1$. For the latter interpretation, the Burkholder-Davis-Gundy inequality supplies a constant $C_{1/2}$ instead. One can see the obstruction with a Brownian motion stopped on exiting $(-1,1)$: if $\tau$ is that Brownian exit time, then $|B_{\tau }|=1$, $\mathbb{E}\tau =1$, and strict Jensen inequality gives $\mathbb{E}{\tau }^{1/4}<1=\mathbb{E}|B_{\tau }{|}^{1/2}$.

This obstruction also survives the continuity condition on $H$. On a long finite interval approximate $1_{\{s<\tau \}}$ by a continuous adapted process equal to $1$ until $\tau$ and decreasing linearly to $0$ during the next $ϵ$ units of time, and take its initial value to be $0$ by inserting a short continuous ramp before starting the exit experiment. The Itô isometry controls these approximations, and letting the interval length tend to infinity recovers the strict inequality above; hence some bounded continuous adapted example also violates the latter interpretation with constant $1$. The direct proof avoids relying on this ambiguous hint.

Let $A=⟨M⟩$ be the quadratic variation and stop at $T_b=\inf \{u\ge 0:A_u>b\}$. Its continuity and monotonicity imply $A_{u\wedge T_b}\le b$ and $A_{T_b}=b$ whenever $T_b<\infty$. The standard martingale isometry after stopping by bounded quadratic variation gives $\mathbb{E}(M_{t\wedge T_b}^2)=\mathbb{E}{A}_{t\wedge T_b}\le b$; in particular the stopped local martingale is square-integrable. This criterion follows by localization and the Doob L2 maximal inequality, which bound the second moment of the stopped supremum by $4b$ and then permit removal of localization.

On $\{A_t\le b\}$ one has $T_b\ge t$, so the original and stopped paths agree through time $t$. Therefore the Doob L2 maximal inequality and the Markov inequality giveThus

Take the stopping timesContinuity gives $|M_{t\wedge S_n}|\le n$, with no overshoot at the first crossing. Stopping preserves the local martingale property, and the bounded case proved in question 4(i) makes each $M^{S_n}$ a continuous true martingale.

Every continuous path is bounded on each compact time interval, so $S_n\uparrow \infty$ almost surely. Hence this is also a localizing sequence consisting of bounded stopped martingales.

For the dyadic power variation of a continuous local martingale, write ${\Delta }_i^{n}M=M_{i{2}^{-n}}-M_{(i-1)2^{-n}}$, ${\delta }_n={\max }_i|{\Delta }_i^{n}M|$ and $Q_n={\sum }_i({\Delta }_i^{n}M{)}^2$. Uniform continuity on $[0,1]$ gives ${\delta }_n\to 0$ almost surely, while the deterministic-partition characterization of quadratic variation gives $Q_n\to ⟨M{⟩}_1$ in probability. In particular $(Q_n)$ is bounded in probability.

If $p>2$, then $Z_n\le {\delta }_n^{p-2}{Q}_n$, and a factor tending to zero in probability times a sequence bounded in probability tends to zero in probability. ThusIf $1<p<2$, the reverse comparison is $Q_n\le {\delta }_n^{2-p}{Z}_n$. The assumed finite $\mathrm{limsup}{Z}_n$ makes $Z_n$ eventually bounded almost surely, so $Q_n\to 0$ almost surely. By uniqueness of a limit in probability, $⟨M{⟩}_1=0$ almost surely. Its monotonicity gives $⟨M{⟩}_t=0$ for $t\le 1$.

To finish without assuming any integrability of $M$, apply the bound from part (i) to both $M$ and $-M$, with $t=1$, and then let $b\downarrow 0$. For each $a>0$ the probability of a positive or negative excursion beyond $a$ is zero. Taking countably many $a\downarrow 0$ proves $M$ is indistinguishable from zero on $[0,1]$. The same argument in fact works for every $0<p<2$ under the stated pathwise boundedness assumption.

The usual quadratic variation convention includes $⟨M{⟩}_0=0$. If $A,C$ are two continuous finite-variation processes with $A_0=C_0=0$ such that $M^2-A$ and $M^2-C$ are continuous local martingales, then $A-C$ is a continuous local martingale of finite variation. It must be constant, and its initial value is zero, so $A=C$ up to indistinguishability of stochastic processes.

Here is a proof of the fact used. Subtract the initial value of a continuous local martingale $N$ of finite variation, and stop so that both $|N|$ and its total-variation process are bounded by deterministic constants. The stopped local martingale is a bounded true martingale, so orthogonality of its increments gives, on any deterministic partition of an interval of $[0,t]$,Continuity and the dominated convergence theorem justify the last limit. Remove stopping and use continuity on a countable dense set of times to conclude $N\equiv 0$.

As literally printed, uniqueness is false without normalization. If $A$ works then $A+c$ also works for any deterministic constant $c$, since subtracting a constant preserves the local martingale property; this is a genuine omitted convention.

With normalization, existence always holds for a continuous local martingale. Its quadratic variation $[M]$, obtained as the limit in probability of squared increments, is continuous, increasing, starts at zero and satisfies $M^2-[M]$ is a continuous local martingale, so taking $⟨M⟩=[M]$ supplies the required finite-variation process.

Interpret the Stratonovich integral in its standard extension to continuous semimartingale integrands:This extension is needed because $f^{\prime }(X)$ is generally a semimartingale rather than a local martingale. Also the introductory prose in the PDF reverses integrand and integrator relative to its displayed definition; the displayed definition fixes the convention used here.

Apply the Itô formula to $f^{\prime }(X)$, using $f\in C^3$:The continuous finite-variation process in the second term has zero quadratic covariation with $X$, while the quadratic covariation rule for an Itô integral givesA second application of the Itô formula, now to $f(X)$, therefore yieldsAll unbounded coefficients are handled by stopping $X$ on compact spatial intervals, then removing the localizing sequence.

Take $X=Y=B$, a standard Brownian motion with $B_0=0$. The Itô formula and $⟨B{⟩}_t=t$ giveIf this nonnegative stochastic process, which starts from zero, were a local martingale, question 4(i) would make it a supermartingale and force its expectation to be at most zero. But $\mathbb{E}(B_t^2/2)=t/2>0$ for $t>0$. Hence a Stratonovich integral is in general not a local martingale.

The limit is in probability, or more strongly in uniform convergence on compacts in probability; a pathwise limit for every continuous local martingale is not implicit. The original PDF uses $\lceil t{2}^n\rceil$, whereas the supplied TeX changes this to $\lfloor t{2}^n\rfloor$; both versions have the same limit by continuity.

Put $\delta =2^{-n}$ and $m=\lfloor t/\delta \rfloor$. For the completed intervals, the algebraic identityseparates a left-point sum from a quadratic covariation sum.

The results from stochastic calculus used are these: predictable left-step approximations to a continuous adapted integrand converge in their Itô integrals in uniform convergence on compacts in probability; and products of increments of continuous semimartingales on deterministic partitions of an interval with vanishing mesh converge to their quadratic covariation in the same sense. The first follows after localization from the Itô isometry and the Doob L2 maximal inequality, since the integrand approximations converge uniformly on compact intervals; the second is the deterministic-partition characterization of quadratic covariation.

Thus the two sums converge to ${\int }_0^t{Y}_{s}d{X}_s$ and $\frac{1}{2}⟨Y,X{⟩}_t$. For the ceiling convention, at most one additional full interval extends past $t$; on $[0,T+1]$ its absolute contribution is at most ${\sup }_{s\le T+1}|Y_s|$ times the modulus of continuity of $X$ at scale $\delta$, which tends to zero almost surely. Therefore

The space-time Hermite polynomials obtained by scaling the Probabilists' Hermite polynomials are genuine polynomials in both variables, including at $t=0$:This expansion follows by differentiating the defining Gaussian exponential, and shows that $H_n(x,0)=x^n$ extends smoothly through $t=0$. Consequently the Itô formula is applicable from time zero. The given backward heat equation cancels the drift, and the derivative identity givesThe integrand is continuous and adapted, hence previsible, and stopping the Brownian motion and time makes it bounded. Thus $H_n(B_t,t)$ is a local martingale for every $n\ge 0$. The $n=0$ case is the constant $H_0=1$; in fact Gaussian moments make the displayed Itô integral square-integrable on every finite horizon, so these are true martingales as well.

The first three Probabilists' Hermite polynomials are $h_1(x)=x$, $h_2(x)=x^2-1$ and $h_3(x)=x^3-3x$. Hence