Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 29 (Stochastic Calculus and Applications)

Set $h_n=2^{-n}$ and $t_n=h_n\lceil t/h_n\rceil$. Replace the last sampled value $X_{t_n}$ by $X_t$, and call the resulting absolute-increment sum $W_n(t)$. This is a sum over a genuine partition of an interval of $[0,t]$, andby right continuity. This is the dyadic approximation of total variation. Let $V(t)$ be the total variation of a function $X$ on $[0,t]$, allowing $+\infty$. Every $W_n(t)$ is at most $V(t)$. Conversely, take any finite partition of an interval of $[0,t]$. Approximate its interior points from the right by dyadic grid points, retaining the endpoint $t$. For large $n$ these approximating points remain distinct, and right continuity makes their absolute-increment sum converge to that of the original partition. The triangle inequality makes $W_n(t)$ at least this selected sum. Taking the supremum over all partitions provesThe limit is the total-variation function; for locally finite variation it is the total-variation process.

There is a missing hypothesis in the assertion about being càdlàg. A general càdlàg function can have infinite total variation of a function arbitrarily close to zero. For example,is continuous. At the successive points $s_j=(\pi /2+j\pi {)}^{-1}$ its values alternate in sign with magnitude $s_j$. The variation along these points dominates a divergent harmonic series. Thus $V(t)=\infty$ for every $t>0$, but $V(0)=0$, so even extended-valued right continuity fails at zero.

The correct càdlàg conclusion holds when $X$ has finite variation on every compact interval. Indeed $V$ is then finite and nondecreasing, so it has finite left limits. To prove right continuity at $t$, fix $s>t$ and choose a partition of $[t,s]$ whose sum is within $\epsilon$ of its variation. If $u>t$ lies before the first interior partition point, changing the first endpoint from $t$ to $u$ changes that sum by at most $|X_u-X_t|$. Additivity of total variation of a function therefore givesLet $u\downarrow t$ and then $\epsilon \downarrow 0$. This proves right continuity. This proves càdlàg regularity of finite total variation. Under that qualification, the jump formula isIf the variation is infinite, a difference of two infinite variation values is not a defined jump.

Fix $T<\infty$. By uniform continuity of each path of the continuous local martingale on $[0,T+1]$, its modulustends to zero almost surely. For every $s\le T$, comparison of powers of each increment givesThe squared-increment sum converges in probability to $[X{]}_T$, by the dyadic definition of quadratic variation. Including the last full grid interval instead of the last partial interval changes that sum by a quantity bounded by $2{\eta }_n^2$, so the ceiling convention has the same limit. In particular these squared-increment sums are bounded in probability.

Since $p-2>0$, a quantity tending to zero almost surely multiplied by a family bounded in probability tends to zero in probability. HenceThis is uniform convergence on compacts in probability, and proves the superquadratic case of dyadic power variation of a continuous local martingale.

Fix the specified time $t$. With the path modulus ${\eta }_n$ on $[0,t+1]$ as above, comparison of increment powers now givesThe hypothesis makes the second factor eventually bounded almost surely, while the first tends to zero almost surely because $2-p>0$. Thus the squared-increment sums tend to zero almost surely. They also converge in probability to the quadratic variation $[X{]}_t$, so uniqueness of the limit in probability implies $[X{]}_t=0$ almost surely.

The quadratic variation is nondecreasing, hence it vanishes throughout $[0,t]$. To see explicitly that the continuous local martingale must then be constant, localize it to square-integrable stopped martingales. The Itô isometry givesRemove the localization, apply the conclusion at every rational $s\le t$, and use continuity. Since $X_0=0$, this provesThus $X$ is indistinguishable from the zero process on that interval. This is the subquadratic case of dyadic power variation of a continuous local martingale.

No quadratic variation argument is needed. For deterministic $u<v$, the martingale property and square integrability giveConsequentlyApply this identity separately to each consecutive grid increment. The second moments telescope, givingThis uniform bound follows directly from martingale-difference orthogonality, including the final interval specified by the ceiling.

The gradient bound makes $f$ a Lipschitz function, with $|f(x)|\le |f(0)|+K\Vert x\Vert$, so its value at a Gaussian random variable is integrable and square integrable. The future Brownian increment $X_1-X_t$ is independent of ${\mathcal{F}}_t$ and has distribution $N(0,(1-t)I_d)$. Conditioning on the current location therefore givesThis is the Markov property expressed through the Brownian transition semigroup. In particular $M_0=\mu$ and $M_1=f(X_1)$.

Let $u(t,x)=P_{1-t}f(x)$ for $t<1$. Differentiating the Brownian transition semigroup in the spatial variables is justified by the bounded gradient and givesThe heat equation for this semigroup is the backward equation ${\partial }_{t}u+\frac{1}{2}\Delta u=0$. Applying the multidimensional Itô formula yieldsThe finite-variation term vanishes. ThusInitially this calculation holds on $[0,1-\epsilon ]$. The vector of stochastic integral coefficients has norm at most $K$, so the Itô isometry lets the integrals extend to time one. Also $P_{1-t}f(X_t)\to f(X_1)$ in $L^2$ by the Lipschitz function bound and continuity of Brownian motion. The same representation therefore holds on the entire interval.

Let $g_t=P_{1-t}(\nabla f)(X_t)$. The triangle inequality for the conditional average givesThe independent coordinate Brownian motions have zero cross-variation, so the quadratic variation of a stochastic integral givesThe Dambis-Dubins-Schwarz theorem says that a continuous local martingale $N$ starting at zero whose quadratic variation diverges admits a Brownian motion $B$, with the inverse-clock filtration, such that $N_t=B_{[N{]}_t}$. If the total clock is finite, one may enlarge the probability space and append independent Brownian motion after the clock terminates. This gives the same identity up to the original lifetime.

Apply this theorem to $N=M-M_0$. The clock bound implies the event inclusionThe latter event is contained in the union of the upper and lower crossing events. Symmetry of Brownian motion givesThis uses only an event inclusion; independence between the Brownian motion and its random clock is neither assumed nor needed.

The Brownian reflection principle and Brownian scaling givewhere $Z$ has the standard normal distribution. Combine part (c) with the allowed Gaussian tail bound to obtainThis is the Brownian martingale proof of Gaussian concentration. It proves the intended Gaussian concentration inequality. The denominator printed in the initial bound is $2K$, whereas the gradient hypothesis and part (c) give $2K^2$. The stronger printed bound is false in general. For a concrete counterexample, take $d=1$, $K=2$, and $f(x)=2x$. At $r=10$,where the first strict inequality follows by integrating the decreasing normal density over $[5,6]$. The rightmost term is the printed bound. Thus the square on $K$ is a necessary source correction, rather than a change of proof technique.

Use the following form of the Girsanov theorem. If $W$ is Brownian motion, $\vartheta$ is predictable, andis a uniformly integrable martingale with $Z_0=1$, then the probability measure with terminal density $Z_{\infty }$ makes $W_t-{\int }_0^t{\vartheta }_{s}ds$ a Brownian motion. The analogous finite-horizon statement uses the terminal density at that horizon.

Here take $W_t=X_t-x$ and ${\vartheta }_s=1_{\{s<\tau \}}/X_s$. Before $\tau$, $X_s\in [\epsilon ,M]$, so the stochastic integral is well-defined. The assumed uniform integrability of the stopped stochastic exponential givesThus the proposed weighting really defines a probability measure, and the Girsanov theorem gives the Brownian motionunder that measure. In particularAll calculations are first made on this compact stopped interval. The integral involving $1/X$ is naturally defined on $[0,T_0)$; at $T_0$ its quadratic variation may diverge. The density will have a zero limiting extension there, as established below, rather than a finite stochastic logarithm at the hitting time.

For $w\ne 0$ in ${\mathbb{R}}^3$, the radial function $r(w)=\Vert w\Vert$ hasThe multidimensional Itô formula, initially stopped away from zero, givesThe martingale term has quadratic variation $dt$, since ${\sum }_i(W_t^i/R_t{)}^2=1$. The Lévy characterization of Brownian motion says that a continuous local martingale starting at zero with quadratic variation $t$ is a standard Brownian motion. Consequently the radial equation isOn $[\epsilon ,M]$ the drift $1/r$ is Lipschitz continuous. Extend it to a bounded globally Lipschitz function on the real line. The standard existence and pathwise uniqueness theorem for stochastic differential equations with globally Lipschitz coefficients applies to this extension, and gives uniqueness in law up to the first exit from $[\epsilon ,M]$. Both this radial process and the process in part (a) solve the same equation from $x$. Thuswhere each exit time is computed from its own coordinate path.

For completeness, the radial equation is valid globally: the three-dimensional Bessel process does not hit zero. Apply the Itô formula to $1/R$ on the annulus. Its drift is zero, and the stopped process is a bounded martingale. At the first exit,soThe exit time is finite because one coordinate of the underlying Brownian motion reaches $M$ in finite time. Letting $\epsilon \downarrow 0$ and then considering arbitrarily large $M$ proves that zero is never reached. The radial martingale therefore has quadratic variation $t$ on the whole time axis, completing the Lévy characterization of Brownian motion argument.

Under Wiener measure, $d{X}_t=d{W}_t$ andThe definition of the stochastic exponential givesOn the other hand, the Itô formula applied to the logarithm gives preciselyTheir initial values differ by $\log x$, soalmost surely under Wiener measure. In particular,

Write ${\tau }_{\epsilon }=T_{\epsilon }\wedge T_M$. The stopped Bessel process has the same law as the weighted stopped Brownian motion, so for every bounded measurable functional $F$ of a continuous path,The Bessel process hits $M$ in finite time and never hits zero. Its minimum before that hitting time is strictly positive. Thus for sufficiently small $\epsilon$, depending on the path, its stopped path $R^{{\tau }_{\epsilon }}$ equals $R^{T_M}$ exactly. Under Wiener measure, the events $\{T_M<T_{\epsilon }\}$ increase to $\{T_M<T_0\}$, and on that event the stopped paths eventually equal $X^{T_M}$. The remaining term is bounded by $\epsilon \Vert F{\Vert }_{\infty }/x$. The dominated convergence theorem therefore yields the precise limiting identityThe bounded stopped martingale $X_{t\wedge T_0\wedge T_M}$ givesHence the limiting identity saysThis establishes Brownian conditioning by a stopped Bessel density.

The displayed density in the question requires a stopping convention. On the full, unstopped path space, the probability measureis well-defined, but its paths continue as Brownian motion after $T_M$. Its pushforward by the stopping map $\omega \mapsto \omega (\cdot \wedge T_M(\omega ))$ is the law ${\mathbb{Q}}^M$ of the stopped Bessel process. Equivalently, if ${\mathbb{W}}_x^{0,M}$ denotes the law of Brownian motion stopped at $T_0\wedge T_M$, then the actual stopped laws satisfyLiteral absolute continuity of the stopped law with respect to unstopped Wiener measure is false: paths that become permanently constant at $M$ have probability one for the stopped law and zero under Wiener measure. The same density is also valid on the stopped sigma-algebra of the original coordinate space. Thus the corrected stopped-law formulation proves the intended conclusion without equating these different full-path measures.

The conformal invariance of planar Brownian motion is a time-change statement. If $\varphi :D\to D^{\prime }$ is a conformal bijection of planar domains and $B$ is planar Brownian motion started in $D$, let ${\tau }_D$ be its exit time and defineBefore exit this clock is strictly increasing. Its inverse $C_u$ makes $\varphi (B_{C_u})$ a planar Brownian motion started at $\varphi (B_0)$, up to its transformed lifetime $A_{{\tau }_D}$. The image of the stopped trajectory is therefore invariant in law after this change of time. A holomorphic map with nonzero derivative gives the corresponding local statement even without global injectivity.

Here is the proof. Write $\varphi =u+iv$ and $B=(B^1,B^2)$. The real and imaginary parts are harmonic. The Itô formula therefore gives the two continuous local martingalesstopped on compact subsets of $D$. The Cauchy-Riemann equations implyThus both coordinate quadratic variations equal $A_t$ and their quadratic covariation is zero. After inverse time change, the covariance matrix of the two continuous local martingales is $u{I}_2$. The multidimensional Lévy characterization of multidimensional Brownian motion states that a continuous vector local martingale starting at zero with this covariance matrix is standard vector Brownian motion. Apply it after subtracting $\varphi (B_0)$, and exhaust $D$ by compact subsets. This proves the theorem, with killing at the transformed exit lifetime. The clock is the conformal Brownian clock; omitting it would generally give an incorrect speed.

The exponential map is a conformal bijection from the stripto the wedge containing the starting point $1$. The restriction $\alpha +\beta <2\pi$ makes it injective. A planar Brownian motion $U+iV$ started at zero in this strip maps, after the conformal Brownian clock, to the required Brownian motion in the wedge.

Let $\sigma$ be the first time $V$ exits $(-\alpha ,\beta )$. It is finite almost surely, and $V_{t\wedge \sigma }$ is a bounded martingale. The optional stopping theorem givesThe strip exit maps to the corresponding bounding ray. The exponential clock is finite at $\sigma$, because its integrand is continuous on the compact time interval $[0,\sigma ]$. Thus it does not remove or reverse the exit event. We obtainThis is the Brownian wedge-exit probability obtained directly from the strip, using conformal invariance of planar Brownian motion.

Put $L_r={\theta }_{T_r}$. The hitting times $T_r$ are finite almost surely by recurrence of planar Brownian motion, and increase with $r$. At $T_r$, the position is $e^{-r}{e}^{i{L}_r}$. By the Strong Markov property, Brownian scaling, and rotational invariance of planar Brownian motion, the future processis a standard planar Brownian motion started at $1$, independent of ${\mathcal{F}}_{T_r}$. The random rotation does not change its conditional law, since that law is rotationally invariant.

The first time this rescaled process reaches radius $e^{-h}$ corresponds to the original time $T_{r+h}$. Its continuous argument starts at zero and equals the increment of the original winding angle. HenceFor any ordered finite collection of levels, apply this assertion successively. Earlier winding values are measurable at the current hitting time, so all increments are independent and their laws depend only on the level differences. This proves the Lévy process property of winding at logarithmic radial passage levels.

One can also verify stochastic continuity and identify this process. The logarithmic-radius and argument martingales have the common clockThe same Cauchy-Riemann equations calculation as in part (a), applied locally to the logarithm and joined along the continuous argument, gives independent Brownian motions $U,V$ such that $\log R_t=U_{H_t}$ and ${\theta }_t=V_{H_t}$. This clock diverges: if it had a finite limit, convergence of martingales with finite total quadratic variation would make $\log R_t$ converge to a finite value, forcing the clock integral itself to diverge.

Thus, with ${\eta }_r=\inf \{u:U_u=-r\}$, we have $L_r=V_{{\eta }_r}$. The Brownian first-passage Laplace transform is $\mathbb{E}{e}^{-q{\eta }_r}=e^{-r\sqrt{2q}}$. It follows either from the Exponential martingale for Brownian motion stopped at ${\eta }_r\wedge n$, or from the one-dimensional exit equation. Conditioning on the independent process $U$ givesSo this is a symmetric Cauchy process, with level parameter $r$. The formula implies stochastic continuity at zero and, by stationary increments, everywhere. Under the usual definition requiring càdlàg paths, use the right-continuous passage clock ${\eta }_r^{+}=\inf \{u:U_u<-r\}$. It agrees with ${\eta }_r$ for each fixed level almost surely, and $V_{{\eta }_r^{+}}$ is the càdlàg modification. This distinction concerns simultaneous passage levels and does not alter any increment law.

A left-continuous adapted process is a predictable process. Define the elementary predictable approximationAt every $s>0$, its sample time approaches $s$ from the left, including when $s$ is itself a grid point. Thus $H_s^{(n)}\to H_s$ pathwise by left continuity.

Use the continuous semimartingale decomposition $X=X_0+M+A$, with $M$ a continuous local martingale and $A$ a continuous finite-variation process. Localize so that on a fixed horizon $T$, the integrands are bounded by a deterministic constant, $[M{]}_T$ is bounded, and the total-variation process of $A$ at $T$ is bounded. This is possible using local boundedness and the finite pathwise clock and variation.

For the finite-variation integral, the dominated convergence theorem givesalmost surely. For the martingale integral, the Itô isometry and the Doob L2 maximal inequality giveHere the localized bounds justify dominated convergence also in expectation. Remove localization to conclude that the full elementary integrals converge in uniform convergence on compacts in probability to $\int HdX$.

The sum of completed intervals differs from the full elementary integral at time $t$ byIts maximum on $[0,T]$ tends to zero almost surely, by the local bound on $H$ and uniform continuity of $X$. ThereforeThis proves left-endpoint approximation of a continuous semimartingale integral, including the unfinished interval convention.

Define the quadratic covariation of two continuous semimartingales by polarization:The available quadratic variation convergence theorem and the polarization identity imply that the sums of products of simultaneous grid increments converge in uniform convergence on compacts in probability to $[X,Y]$. The quadratic covariation is a continuous finite-variation process, vanishing at time zero.

For a grid ending at $t_n=h_n\lfloor t/h_n\rfloor$, the elementary algebraic identitytelescopes. The first two sums converge by part (a), with the continuous processes $X,Y$ as integrands; continuous paths are left-continuous and locally bounded. The last sum converges to the quadratic covariation by the preceding polarization argument. Finally, $X_{t_n}{Y}_{t_n}\to X_t{Y}_t$ locally uniformly by continuity. HenceThis is the Itô product rule, or stochastic integration by parts. Uniqueness of the limit in uniform convergence on compacts in probability, followed by continuity, makes the identity hold simultaneously for all $t$ outside one null set. The proof used finite-grid algebra, part (a), and the permitted quadratic variation theorem; it did not use the Itô formula.

Use the local convention for the martingale problem, which is appropriate for arbitrary measurable coefficients. A continuous adapted process $X$ solves $\mathbf{M}(a,b)$ if the coefficient integrals are locally finite and, for every $f\in C_c^{\infty }(\mathbb{R})$,is a local martingale. Here locally finite means ${\int }_0^T(|a(X_s)|+|b(X_s)|)ds<\infty$ almost surely for each $T<\infty$. Suitable localization upgrades the identities to true martingale identities. With bounded coefficients and bounded test derivatives, this distinction disappears on finite horizons.

A Markov diffusion generated by $L$ is a time-homogeneous Markov process with continuous paths whose infinitesimal generator on the stated test domain is $L$. Its transition operators satisfy the generator limit, and Dynkin formula gives precisely the preceding martingale problem identities. Some terminology uses L-diffusion for the test-function identities alone; the Markov property is an additional assertion unless uniqueness supplies it.

Suppose now $a={\sigma }^2>0$. Use smooth compactly supported functions agreeing with $x$ and $x^2$ on $[-R,R]$, and stop before exiting this interval. Removing these stops shows thatis a continuous local martingale. Consequently $X$ is a continuous semimartingale. The martingale problem applied to the second cutoff function says thatis a local martingale. The Itô formula for $X^2$ also says thatis a local martingale. Subtracting, the continuous finite-variation process $[N{]}_t-{\int }_0^{t}a(X_s)ds$ is a local martingale and starts at zero. The theorem that a continuous finite-variation local martingale is constant givesDefineThe integrand is predictable and is stochastically integrable, since its integral against the bracket is exactly $t$. Thus $B$ is a continuous local martingale with $B_0=0$ and $[B{]}_t=t$. The Lévy characterization of Brownian motion states that these conditions make $B$ a Brownian motion relative to the given filtration. Associativity of stochastic integrals recovers $N=\int \sigma (X)dB$, and thereforeThe original process is a weak solution of a stochastic differential equation, with its driving Brownian motion constructed on the original space.

Conversely, the Itô formula applied to any weak solution with locally integrable coefficients gives the martingale problem identities. A Markov diffusion with generator $L$ gives them by Dynkin formula. A well-posed martingale problem, with the usual measurable path-space setup, identifies a unique family of diffusion laws and yields the Strong Markov property. Without uniqueness, the test-function identities alone do not assert that every chosen solution is Markov. This specifies both the equivalence with weak equations and the role of uniqueness in identifying a diffusion.

Set $u_0=\alpha /\beta$ and use the accelerated grid time $j/n$. At population $k=n{u}_0+\sqrt{n}x$, the fluctuation increment is $h_n=n^{-1/2}$, $-h_n$, or zero. Its first and second conditional moments, multiplied by $n$, areThus $b_n\to b(x)=-\alpha x$ and $a_n\to a=2{\alpha }^2/\beta$, uniformly on every compact set. The centered conditional variance has the same limit, since the squared conditional mean contributes only $b_n(x{)}^2/n$.

Here is the diffusion approximation theorem being applied. For Markov chains sampled with time step $1/n$, suppose the first and second conditional increment moments multiplied by $n$ converge uniformly on compact sets to continuous coefficients $b,a$; the large-jump conditional second moments satisfy the local Lindeberg condition; the initial laws converge; compact containment holds on finite time intervals; and the limiting martingale problem is well-posed and nonexplosive. Then the interpolated chains converge weakly, uniformly on finite time intervals, to that problem's continuous solution. The step versions converge in the Skorokhod J1 topology. Local moment bounds supply the stopped-process tightness, and the limiting test-function identities identify every subsequential limit by the martingale problem.

We verify the global condition rather than assume it. Before the stopping rule is triggered, $p_k+q_k\le 1$, and $k\ge 0$ implies $x\ge -u_0\sqrt{n}$. ThereforeFor the Lyapunov function $V(x)=x^2$, the exact one-step generator is consequentlyAfter stopping, the process is constant. Applying the elementary stopped supermartingale inequality for $V(X_{j/n}^n)-j/n$, and stopping also on first reaching $|X^n|\ge L$, givesThis is compact containment. At zero population we use the natural absorbing convention $p_0=q_0=0$; the same calculation is valid there.

The invalid-probability threshold is at the positive rootBecause $2{\alpha }^2<\beta$, we have $u_{\ast }>u_0$. The corresponding fluctuation level $\sqrt{n}(u_{\ast }-u_0)$ tends to infinity, so the stopping rule changes none of the compact-set moment limits. The initial fluctuation satisfies $|X_0^n|\le n^{-1/2}$ and hence tends to zero. All jumps have magnitude at most $n^{-1/2}$, so the Lindeberg condition is immediate. The limiting equation has globally Lipschitz coefficients and is well-posed and nonexplosive. The theorem therefore givesThis is the Ornstein-Uhlenbeck fluctuation limit at a stable population equilibrium. The limiting Ornstein-Uhlenbeck process has explicit solutionThe PDF samples the linearly interpolated population at $\lfloor nt\rfloor$, so its displayed process is actually the step version. Using $nt$ instead gives the continuous interpolation; their uniform distance is at most $n^{-1/2}$. Both therefore have the asserted continuous diffusion limit, with the step version interpreted in Skorokhod J1 topology.

Finally, if ${\tau }_n\le 1$, the stopped fluctuation reaches a value greater than $\sqrt{n}(u_{\ast }-u_0)$. The preceding Lyapunov function bound gives directlyThis also shows explicitly why the artificial stopping rule is asymptotically irrelevant.