Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 33 (Stochastic Calculus and Applications)

A continuous adapted process $M$ is a local martingale if there are stopping times ${\tau }_n\uparrow \infty$ almost surely such that each stopped process $M_t^{{\tau }_n}=M_{t\wedge {\tau }_n}$ is a martingale. In this definition $M_0$ is integrable. A continuous adapted process $A$ has finite variation if, almost surely, its paths have finite total variation of a function on each compact time interval:The permitted fact that this total-variation process is continuous and adapted makes its level-hitting times stopping times.

Suppose $M=A$, and put $N=M-M_0$. DefineThese times increase to infinity almost surely. The optional stopping theorem applied to the localized martingale shows that $N^{{\sigma }_n}$ is a martingale; it is also bounded in absolute value by $n$, with total variation at most $n$. Fix $n,t$ and write $L=N^{{\sigma }_n}$. For a deterministic partition of an interval $0=t_0<\cdots <t_r=t$, orthogonality of square-integrable martingale increments givesPathwise the sum on the right is bounded by $n{\max }_j|L_{t_j}-L_{t_{j-1}}|$. For partitions with mesh tending to zero, continuity makes this converge to zero. It is bounded by $2n^2$, so dominated convergence gives $\mathbb{E}{L}_t^2=0$. First use rational $t$, then continuity, to obtain $L\equiv 0$ almost surely. Letting $n\to \infty$ provesThis proves continuous finite-variation local martingale is constant without assuming integrability of the unstopped total variation. Continuity matters: compensated jump martingales can have finite variation without being constant.

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.

Here $[M{]}_t=t$ and $d{A}_t=dt$, so the formula in part (b) givesIf $g$ vanishes on $\mathbb{R}\times [0,\infty )$, this integral is zero. Conversely suppose $A_t^f=0$ for all $t$ almost surely. Its integrand is continuous along each path; the fundamental theorem of calculus implies $g(B_t,t)=0$ for all $t$ on that event. For any fixed $t>0$, the normal distribution of $B_t$ has a strictly positive density at every real point. Since $g(\cdot ,t)$ is continuous, a nonzero value at any spatial point would give a nonzero value on an open interval of positive Brownian probability, a contradiction. Thus $g(x,t)=0$ for every $x$ and every positive $t$. Continuity extends this to $t=0$.

Consequently the exact condition isThese are space-time harmonic functions along Brownian motion, satisfying the backward heat equation. The second coordinate never takes negative values, so no equation is imposed for $y<0$, beyond compatibility with the assumed global $C^2$ regularity. The normalized finite-variation term is unique by part (a), so it cannot be removed by choosing a different decomposition.

Subtract $M_0$, which does not affect either the integral or its quadratic variation. Let $N=M-M_0$. Choose a localizing sequence ${\tau }_n$ for $N$. By local boundedness of the previsible process $H$, there are increasing stopping times ${\rho }_n\uparrow \infty$ and finite deterministic constants $K_n$ with $|H_s|\le K_n$ on $0<s\le {\rho }_n$. PutContinuity and finiteness on compact intervals imply ${\sigma }_n\uparrow \infty$ almost surely. The stopped process $N^{{\sigma }_n}$ is a bounded square-integrable martingale, its bracket is bounded by $n$, and its stopped integrand is bounded by $K_n$. The indicator $1_{\{s\le {\sigma }_n\}}$ is predictable, so the stopped stochastic integral identity is valid.

Fix $n$ and abbreviateThe Itô isometry gives $\mathbb{E}{J}_t^2=\mathbb{E}{Q}_t\le K_n^{2}n$, so $J$ is a square-integrable martingale. For $F\in {\mathcal{F}}_s$, apply the same isometry to the predictable integrand $1_F{1}_{(s,t]}H$, stopped at ${\sigma }_n$. This yieldsAlso $\mathbb{E}[J_t-J_s\mid {\mathcal{F}}_s]=0$. Expanding $J_t^2-J_s^2$ therefore proves that $J^2-Q$ is a true martingale. The process $Q$ is continuous adapted increasing and starts at zero. By the characterization of continuous martingale quadratic variation as its increasing square compensator, and uniqueness of an increasing square compensator, we have $[J]=Q$.

Stochastic integration and quadratic variation commute with stopping; for the latter this follows either from squared-increment sums or from the same square-compensator characterization. Hence the desired identity holds up to every ${\sigma }_n$. Since these times increase to infinity, they exhaust every finite time interval almost surely, and the continuous identities patch to giveThis is the localized isometry proof of stochastic-integral quadratic variation. No unstopped second-moment or bracket-integrability assumption has been made.

The one-dimensional Lévy characterization of Brownian motion states: a continuous adapted process $L$, with $L_0=0$, is standard Brownian motion relative to its filtration if and only if it is a local martingale and $[L{]}_t=t$ for all $t$ almost surely.

For necessity, the centered independent Gaussian increments of Brownian motion give both the martingale property and the martingale $L_t^2-t$. The uniqueness characterization of quadratic variation then gives $[L{]}_t=t$.

For sufficiency fix $\theta \in \mathbb{R}$. The Itô formula givesThe second-order term cancels the time derivative because $d[L{]}_t=dt$. Thus $E$ is a complex local martingale. Its modulus is the deterministic quantity $e^{{\theta }^{2}t/2}$, bounded on every finite horizon, so its real and imaginary parts are true martingales by the bounded local martingale criterion. For $s<t$,Uniqueness of characteristic functions says that the increment has normal distribution $N(0,t-s)$ independently of ${\mathcal{F}}_s$. Indeed multiplying the identity by any bounded ${\mathcal{F}}_s$-measurable variable gives the product characteristic-function identity expressing independence. Iterating over deterministic times gives independent increments. Together with continuity and $L_0=0$, these are precisely the defining properties of standard Brownian motion. If the starting value is a fixed $x$, the same argument identifies $L-x$ as standard Brownian motion.

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.

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.

Put $C_u={\int }_0^u{e}^{2W_s}ds$. Its paths are continuous and strictly increasing. We also need $C_{\infty }=\infty$ to ensure that the printed inverse is finite for every $t$.

Here is a duration argument for Divergence of a driftless Brownian exponential clock. Start at zero and choose successive return times $T_j$ to zero, each at least one time unit after the preceding return. Recurrence of one-dimensional Brownian motion makes all these times finite. By the Strong Markov property, each unit interval after $T_j$ has the same fixed positive probability that the Brownian path stays above $-1$. The intervals are disjoint, and repeated strong Markov conditioning makes these trials independent. Infinitely many succeed almost surely, by the Borel-Cantelli lemma. Each successful interval contributes at least $e^{-2}$ to $C_{\infty }$, proving divergence.

Therefore ${\tau }_t$ is the unique finite inverse of $C$, with $C_{{\tau }_t}=t$. It is an increasing family of stopping times, since $\{{\tau }_t\le u\}=\{C_u\ge t\}$. The continuous-local-martingale time-change theorem states that inverse-bracket time change preserves the local-martingale property in the filtration ${\mathcal{G}}_t={\mathcal{F}}_{{\tau }_t}$ and transforms the bracket by composition. Applied to $Y$, this givesThe Lévy characterization of Brownian motion now proves that $Z$ is Brownian motion starting at one; $Z-1$ is standard Brownian motion, in the time-changed filtration. If the term Brownian motion is reserved for a process starting at zero, $Z$ itself fails only that normalization. The substantive Brownian property, including independent Gaussian increments in the new time, holds.

For the first equation use an integrating factor. The Itô product rule gives $d(e^{\lambda t}{X}_t)=\sigma e^{\lambda t}d{B}_t$, soThe Gaussianity of deterministic Brownian stochastic integrals follows by approximating its deterministic integrand by step functions and using the Itô isometry. HenceThe variance is positive also when $\lambda <0$, since numerator and denominator then both have negative sign. This is the explicit Ornstein-Uhlenbeck solution with initial value one.

For the second equation, take independent standard Brownian motions $U,V$ and define $R_t=\sqrt{(1+U_t{)}^2+V_t^2}$. The associated planar Brownian motion does not hit the origin. To check this rather than silently differentiate at zero, stop its radius upon hitting $\epsilon$ or $L$, with $0<\epsilon <1<L$. The function $\log r$ is harmonic on that annulus, so the optional stopping theorem givesLet $\epsilon \downarrow 0$: hitting zero before $T_L$ has probability zero. Taking integer $L\to \infty$ covers every possible finite hitting time, since continuous paths are bounded on compact intervals. Thus planar Brownian motion avoids a fixed point and $R$ stays positive.

The radial Itô formula is now legitimate and givesThe sum of the two martingale terms has bracket $t$ and starts at zero, so it is a standard Brownian motion $\beta$ by Lévy characterization of Brownian motion. Thus $dR=d\beta +(2R{)}^{-1}dt$ with $R_0=1$. The law-determination assumption permitted in the question identifies the law of $Y$ with that of this dimension-two Bessel process. For independent standard normal random variables $G_1,G_2$,This is the two-dimensional Bessel transition law and answers the requested distribution without needing a special-function density.

The original PDF specifies a bounded solution class $C_b^2$, rather than the $C_0^2$ transcribed in the TeX. Write $\varphi (x)=\min (1,\max (0,-x))$ and fix a time $t>0$. For Brownian motion starting at $x$, considerThe Itô formula and ordinary product rule give a drift proportional to $-u_t+\frac{1}{2}{u}_{xx}-\lambda \varphi u$, which vanishes by the equation. Therefore $F$ is a local martingale. It is bounded on this horizon because $u$ is bounded and $\varphi \ge 0$, so the bounded local martingale criterion makes it a true martingale. Its endpoint values are $F_0=u(t,x)$ and $F_t=e^{-\lambda {\int }_0^t\varphi (B_r)dr}$, using $u(0,\cdot )=1$. ConsequentlyThis proves the required Feynman-Kac formula with a bounded potential directly, and also gives uniqueness within the stated bounded solution class.

Let $I={\int }_0^t\varphi (B_s)ds$. By dominated convergence, $e^{-\lambda I}\to 1_{\{I=0\}}$ in expectation. Continuity of $B$ and $\varphi$ shows that $I=0$ exactly when $B_s\ge 0$ throughout $[0,t]$: a negative value at any time produces a nontrivial interval with positive integrand. For $x<0$ such an interval occurs immediately, so the limiting probability is zero. For $x>0$, the Brownian reflection principle and the atom-free distribution of the running minimum giveFor $x=0$ this probability is zero: the reflection formula for levels $-\epsilon$, followed by $\epsilon \downarrow 0$, shows that Brownian motion goes below zero before $t$ almost surely. ThusThis is the soft killing limit for Brownian nonnegative survival. The assertion is pointwise for positive time, as requested; no claim of uniform convergence across the initial-time boundary is needed.

The PDF defines multiplication by an indicator: $X_t=U{1}_{\{V\ge e^{-t}\}}$. PutThen $T>0$ almost surely, $\{T\le t\}=\{X_t>0\}\in {\mathcal{F}}_t$, and $T$ is a stopping time. It has exponential distribution of rate one and is independent of $U$. The path stays at zero until $T$, jumps to $U$ at $T$, and remains there.

Let $p=\mathbb{P}(U\in B)$ and $N_t=1_{\{T\le t,U\in B\}}$. We verify the martingale property directly. On $\{T\le s\}$ both increments $N_t-N_s$ and $(T\wedge t)-(T\wedge s)$ vanish. On $\{T>s\}$, every observation up to $s$ is zero, so the past reveals only that survival event. Conditional on it, $R=T-s$ is rate-one exponential by memorylessness and remains independent of the mark $U$. With $h=t-s$,whileSubtracting $p$ times the second identity from the first gives zero. The process is adapted and integrable, since its absolute value is bounded by $1+pt$ on a finite horizon. ThereforeOn $\{T\le t\}$ we have $X_t=U$, so this is exactly the requested process. This first-principles argument uses the natural filtration, which does not disclose the future mark or lifetime. It describes a single marked exponential jump process.

At a positive state $x$, the future is constant. At zero, the residual holding time is independent rate-one exponential and the next state is an independent uniform mark. Thus the conditional future given the natural filtration depends only on the present state: $X$ is a Markov jump process. For any bounded Borel $g$, its transition semigroup isThe Lévy kernel of a Markov jump process, written in destination-state convention, isIn displacement convention it has the same expression $K(x,dz)=1_{\{x=0\}}{1}_{(0,1)}(z)dz$, since its only nonzero jump rate is from zero. The Markov jump-process generator is $Lg(x)=1_{\{x=0\}}({\int }_0^{1}g(y)dy-g(0))$.

We state the generator martingale theorem used here: for a nonexplosive Markov jump process and a bounded $g$ with bounded $Lg$, the process $g(X_t)-g(X_0)-{\int }_0^{t}Lg(X_{s-})ds$ is a true martingale. In this one-jump model it also follows directly from part (a), first for mark indicators and then by bounded-function approximation.

Take $g_{\theta }(x)=e^{\theta x}$, and putHere $L{g}_{\theta }(x)/g_{\theta }(x)=r_{\theta }{1}_{\{x=0\}}$. Therefore choose the previsible processIt is continuous adapted and thus predictable. To verify the exponential, write $d{g}_{\theta }(X_t)=d{N}_t+L{g}_{\theta }(X_{t-})dt$ with the generator martingale $N$. The semimartingale integration by parts formula, using the continuous finite-variation factor $e^{-Y_t}$, givesThus $Z_t=e^{\theta X_t-Y_t}$ is a local martingale. On each fixed horizon it is bounded by $e^{|\theta |+|r_{\theta }|t}$, so it is a true martingale with $Z_0=1$. This is the exponential martingale from a jump generator; for $\theta =0$ it reduces to the constant process one.

Let $(A_t,B_t)$ be a Markov jump process on integer populations, with rateswhen $a,b>0$, where $\kappa >0$. Make both axes absorbing, so the fight ends when one population reaches zero. Start at $(N,b_N)$, with $b_N=\lfloor \alpha N\rfloor$; when $\alpha N$ is integral this is the printed initial condition exactly. The harmless rounding convention makes the model defined for every integer $N$.

Use the embedded jump chain $(A_j,B_j)$ rather than physical time. Its probabilities of the two losses are respectively $b/(a+b)$ and $a/(a+b)$, independently of $\kappa$. It reaches an axis after at most $N+b_N\le 2N$ steps; hold it constant thereafter. PutBefore absorption its two increments are $-2a+1$ and $2b-1$, soAfter absorption set ${\mu }_j=0$. Consequentlyis a discrete-time martingale. All populations are at most $N$, so each martingale difference is bounded in absolute value by $2N+2\le 4N$. The total drift up to step $2N$ is at most $2N$. This is the squared-population near-invariant for stochastic attrition.

The two-sided Azuma-Hoeffding inequality states that a martingale with $m$ differences bounded by $C$ satisfies $\mathbb{P}(|M_m|\ge z)\le 2\exp (-z^2/(2m{C}^2))$. Its exponential estimate follows by applying the conditional Hoeffding lemma to each centered difference, iterating, and optimizing the exponential Markov bound. With $m=2N$ and $C=4N$ it gives

Let $c=\sqrt{1-{\alpha }^2}>0$, and let $X^N=A_{2N}/N$ be the final proportion. If the first population survives, then the second is zero and $D_{2N}/N^2=(X^N{)}^2$. Hence $|X^N-c|>\delta$ impliesIf the first population is extinguished, $X^N=0$ and $D_{2N}\le 0$. This can satisfy the indicated deviation event only when $\delta <c$, and then $|D_{2N}-c^2{N}^2|\ge c^2{N}^2>c\delta N^2$. Thus the same implication holds in both cases.

The initial rounding error obeys $|D_0-c^2{N}^2|=|{\alpha }^2{N}^2-b_N^2|\le 2N$. Combining it with the drift bound shows that the deviation event forcesfor all sufficiently large $N$. Set $\eta =c\delta /2$ in the preceding concentration inequality. We concludefor sufficiently large $N$, and thereforeThe symmetric stochastic attrition process thus has exponential concentration of terminal stochastic attrition survivors. The deterministic equations conserve $a^2-b^2$ and predict the same surviving fraction, but the martingale argument supplies the stronger exponential probability estimate required here.