Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 202 (Stochastic Calculus and Applications)

Let $({\tau }_n)$ be a localizing sequence for $M$. For $s\le t$, the optional sampling theorem givesBoth sides converge almost surely to $M_t$ and $M_s$, and $|M_{u\wedge {\tau }_n}|\le Z$. Conditional dominated convergence therefore yields $\mathbb{E}[M_t\mid {\mathcal{F}}_s]=M_s$, so $M$ is a martingale. Moreover, the family $(M_t{)}_{t\ge 0}$ is dominated by the integrable random variable $Z$, hence is uniformly integrable. Thus $M$ is a uniformly integrable martingale.

Put $A_t={\int }_0^t{H}_s^{2}ds$. It is continuous and tends to infinity almost surely, so $A_{T_{\sigma }}={\sigma }^2$. Define the right-continuous inverse $T(u)=\inf \{t:A_t>u\}$ andThe time-change theorem for local martingales shows that $W$ is a continuous local martingale in the time-changed filtration, andFor completeness, this proves the required case of the Dambis-Dubins-Schwarz theorem: for every $\theta \in \mathbb{R}$, Itô formula makes $\exp (i\theta W_u+{\theta }^{2}u/2)$ a local martingale; stopping and conditioning show that $W_v-W_u$ has conditional characteristic function $e^{-{\theta }^2(v-u)/2}$. Hence the increments are independent centered normal variables with the Brownian variances, and continuity makes $W$ a Brownian motion. ConsequentlyThus $X_{\sigma }$ is centered Gaussian with variance ${\sigma }^2$.

For $\theta >0$, the Exponential martingale for Brownian motion and the Doob maximal inequality for a nonnegative submartingale giveTaking $\theta =a/s$ gives $e^{-a^2/(2s)}$. Apply the same argument to $-B$ and use the union bound:

Itô formula givesso $M$ is a positive local martingale. On every finite interval $[0,T]$, the hypotheses implyThe Gaussian tail bound of the Brownian maximum has finite exponential moments of every subquadratic power. Therefore Novikov condition holds on $[0,T]$, and the stochastic exponential $M$ is a true martingale there. Since $T$ was arbitrary, $M$ is a true martingale.

Write ${\Delta }_{i}M=M_{t_i}-M_{t_{i-1}}$ and $Q={\sum }_i({\Delta }_{i}M{)}^2$. The identityshows, using the martingale-difference orthogonality, thatAlso $\mathbb{E}[Q]=\mathbb{E}[M_t^2]-\mathbb{E}[M_0^2]\le C^2$. Hence $\mathbb{E}[Q^2]\le 10C^4$, which in particular proves the requested bound

Because $\Vert b^{\prime }{\Vert }_{\infty }<\infty$, the function $b$ is Lipschitz continuous. the Picard-Lindelof theorem, proved by iteration onconverges uniformly on every compact interval. The usual factorial estimate proves convergence for arbitrary interval length, and the Gronwall inequality proves uniqueness. Thus there is a unique global continuous solution.

For $w=W$, induction shows that $X_t^{(n)}$ is ${\mathcal{F}}_t$-measurable: its value uses only $(W_s{)}_{s\le t}$ and earlier iterates up to time $t$. The pointwise limit $X_t$ is therefore ${\mathcal{F}}_t$-measurable. Hence $X$ is adapted to $({\mathcal{F}}_t)$.

Let ${\tau }_n$ be the first exit from a compact interval on which $H<n$, chosen so that ${\tau }_n\uparrow T$, and let $m=\inf H>-\infty$. By Itô formula,The assumption gives the drift bound $\frac{1}{2}{H}^{\prime \prime }-(H^{\prime }{)}^2\le C$. After stopping,If $h_n=\inf \{H(y):y\text{ lies outside the }n\text{th compact interval}\}$, then $h_n\to \infty$ because $H$ is coercive, andThus $\mathbb{P}(T\le t)=0$ for every $t$, and $T=\infty$ almost surely.

For an SDE driven by Brownian motion, a strong solution of a stochastic differential equation is adapted to the completed filtration generated by a prescribed Brownian motion and satisfies the equation on that space. A weak solution of a stochastic differential equation consists of a probability space, filtration, Brownian motion, and adapted solution satisfying the equation. Uniqueness in law means that any two weak solutions with the same initial law have the same law as processes. Pathwise uniqueness means that two solutions on the same filtered space, driven by the same Brownian motion and having the same initial value, are indistinguishable.

Since $B_{T_a}=0$, applying Itô formula to $B_t^3$ after $T_a$ givesBefore $T_a$, both sides vanish. Since $T_a$ is a stopping time determined by $B$, this is a strong solution of a stochastic differential equation.

Taking $a=0$ gives $X_t=B_t^3$, whereas any $a>0$ gives a solution that remains zero until $T_a$; these differ with positive probability while using the same Brownian motion and initial value. Therefore pathwise uniqueness fails.

One form of the Feynman-Kac formula is the following. For bounded sufficiently regular $f,V$ andone hasConversely, a bounded classical solution has this representation. To prove it, fix $t$ and apply Itô formula toThe PDE cancels its drift. The remaining stochastic integral is a martingale, so taking expectations at $s=0,t$ gives the representation; the converse follows by the same calculation and uniqueness for the parabolic boundary-value problem.

Apply the Feynman-Kac formula with $V(x)=\sigma x$ and terminal function $1$. The ansatz $u(t,x)=e^{A(t)x+C(t)}$ givesThus $A(t)=-\sigma t$ and $C(t)={\sigma }^2{t}^3/6$, soEquivalently, the Integral of Brownian motion is Gaussian with mean $xt$ and variance $t^3/3$.

For every real $\lambda$, the quadratic variation of $\lambda (M-M_s)+(N-N_s)$ on $[s,t]$ is nonnegative:Its discriminant is therefore nonpositive. This gives the pathwise Kunita-Watanabe inequality

Let $W$ be a Brownian motion, set $X=W$, and defineThis is a continuous local martingale with quadratic variation $t$, hence is Brownian by the Lévy characterization of Brownian motion. Since ${\mathrm{sign}}^2=1$,which gives a weak solution.

Suppose a strong solution existed. It has quadratic variation $t$, so $X$ itself is Brownian. The supplied Tanaka formula gives $|X_t|=B_t+L_t$, and $L$ is adapted to the completed filtration of $|X|$. Hence $B$ and $|X|$ generate the same completed filtration. Strongness would make $X$, and therefore $\mathrm{sign}(X_t)$, measurable with respect to the history of $|X|$. But a Brownian excursion has an independent symmetric sign; in particular, conditionally on the reflected Brownian path, the sign at a fixed nonzero time is not measurable. This contradiction proves that no strong solution exists.

The solution is the geometric Brownian motionIts infinitesimal generator isFor $\gamma =1-2\beta /{\sigma }^2\ne 0$, one has $L(x^{\gamma })=0$. Optional stopping of $X_{t\wedge T_r\wedge T_R}^{\gamma }$ and the boundary values therefore give

A simple predictable process has the formwhere each bounded ${\xi }_i$ is ${\mathcal{F}}_{t_i}$-measurable. DefineIndependent centered Brownian increments show directly by conditioning that this is a martingale. The same conditional expansion, using $\mathbb{E}[(B_v-B_u{)}^2\mid {\mathcal{F}}_u]=v-u$, shows that

Since $B_t={\int }_0^{t}1d{B}_s$, the Itô isometry and polarization give

With $A_t={\int }_0^t\mu (s)ds$, the product rule givesso $X_t{e}^{-A_t}$ is a local martingale under $\mathbb{P}$.

Set $\theta (t)=\mu (t)/\sigma (t)$. This is bounded and compactly supported, so Novikov condition holds anddefines a probability measure $d\mathbb{Q}=Z_{\infty }d\mathbb{P}$. By the Girsanov theorem, $W_t=B_t+{\int }_0^t\theta (s)ds$ is Brownian under $\mathbb{Q}$, andThus $X$ is a local martingale under $\mathbb{Q}$.

A continuous finite-variation path has zero quadratic variation. Hence a continuous finite-variation martingale $M$ satisfies $⟨M⟩=0$. After localizing to make it square-integrable,Letting the localization level tend to infinity shows that $M_t=M_0$ for every $t$ almost surely; continuity makes the equality simultaneous in $t$.

Write $Z_t={\int }_0^t{e}^{-B_s}d{\tilde{B}}_s$, so $X_t=e^{B_t}{Z}_t$. Independence gives $⟨B,\tilde{B}⟩=0$, and Itô formula yieldsThe martingale part has quadratic variation $(1+X_t^2)dt$, so on an enlarged description it equals $\sqrt{1+X_t^2}d{W}_t$. Thus $X$ is a weak solution ofOn the other hand, another application of Itô's formula givesBoth coefficients are Lipschitz continuous, so uniqueness in law gives $X$ and $Y$ the same law.

The variation-of-constants formula gives, when $a\ne 0$,Therefore, with $m=\min (s,t)$,If $a=0$, then $X_t=x+bt+\sigma B_t$ and