Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 202 (Stochastic Calculus with Applications to Finance)

A distribution function is nondecreasing and right-continuous. Every nondecreasing function has finite left limits, so $F$ is càdlàg. Along each partition its increments are nonnegative and telescope, givingFor each $n$, the set $A^{(n)}=\{s\in [0,n]:\Delta F(s)\ge 2^{-n}\}$ is finite because the sum of its positive jumps is at most $F(n)-F(0)$. Every jump belongs to some $A^{(n)}$, so $A_F={\bigcup }_n{A}^{(n)}$ is countable.

For a finite partition, the identity $y^2-x^2=2y(y-x)-(y-x{)}^2$ givesThe first sum tends to the Lebesgue-Stieltjes integral $2{\int }_0^{t}FdF$. In the second, intervals containing no prescribed large jump contribute at most their largest increment times $F(t)-F(0)$; first retain finitely many jumps above a threshold and then let the threshold vanish. The limit is therefore ${\sum }_{s\in A_F\cap (0,t]}|\Delta F(s){|}^2$, proving

Refine a dyadic partition by inserting $s$ and $t$. The triangle inequality shows that its variation over $[s,t]$ is at least $|f(t)-f(s)|$. Passing to the defining limit givesConsequently the càdlàg functionsare nondecreasing: for $s\le t$, the displayed inequality makes both increments nonnegative. Thus they are distribution functions in the Stieltjes sense and $f=F-G$; this is the Jordan decomposition of a function of bounded variation.

Now $A_f\subseteq A_F\cup A_G$, so $A_f$ is countable by part (a). For any finite subset $J\subseteq A_f\cap (0,t]$, partitions isolating its points and the triangle inequality giveTaking the supremum over finite $J$ proves ${\sum }_{s\in A_f\cap (0,t]}|\Delta f(s)|<\infty$.

Apply part (a)'s square formula, extended by the Jordan decomposition of a function of bounded variation from nondecreasing functions to arbitrary càdlàg functions of bounded variation, to $f+g$, $f$, and $g$. Sincelinearity of the Lebesgue-Stieltjes integral leaves ${\int }_0^{t}fdg+{\int }_0^{t}gdf$. At each time $s$, polarization of the jump correction givesOnly common jump times contribute, and hence

The bounded continuous local martingale $X$ is a square-integrable martingale. Since $M^{(n)}$ is a discrete predictable transform of $X$, it has mean zero. The identity $A_t^{(n)}=X_t^2-M_t^{(n)}$ therefore givesuniformly in $t$ and $n$.

The Itô isometry for the elementary predictable integrand in $M^{(n)}$ gives

Because $A_t^{(n)}=X_t^2-M_t^{(n)}$, the inequality $(x-y{)}^2\le 2x^2+2y^2$ and part (ii) yield

By the stated Cauchy property and completeness of ${\mathcal{M}}^2$, there is a square-integrable continuous martingale $M$ such thatSet $A_t=X_t^2-M_t$. This process is continuous and adapted, andby the Doob L2 maximal inequality. The process $A$ is the quadratic variation $[X]$.

If $A_t=0$ almost surely, then $X_t^2=M_t$ is a nonnegative martingale starting from zero. A nonnegative random variable of expectation zero vanishes almost surely, so $X_t=0$ almost surely for each $t$. Applying this on the nonnegative rational times and using path continuity shows that $X_t=0$ simultaneously for every $t\ge 0$ almost surely.

Let ${\tau }_m=\inf \{s:|X_s|\ge m\}$. The stopped process $X^{{\tau }_m}$ is bounded, so part (a) gives a continuous adapted quadratic variation $A^{[m]}$. These processes agree before the smaller stopping time, because their dyadic sums agree there and the limits are unique in probability. They therefore paste to a continuous adapted process $A$ with $A_{s\wedge {\tau }_m}=A_s^{[m]}$.

For fixed $t$ and $\epsilon >0$,The second term tends to zero by part (a), while continuity of $X$ on $[0,t]$ makes $\mathbb{P}({\tau }_m\le t)\to 0$. This proves convergence uniformly on compact intervals in probability.

Fix $t$. Uniform continuity of the sample path on $[0,t]$ givesSince $p<2$,The assumed pathwise boundedness of ${\sup }_n{B}_t^{(n)}$ makes the right-hand side tend to zero almost surely. Part (b) also gives $A_t^{(n)}\to A_t$ in probability, so uniqueness of limits in probability yields $A_t=0$ almost surely. The vanishing-quadratic-variation result from part (a), after localization, makes $X$ identically zero. Consequently every $B_t^{(n)}$ is zero and ${\sup }_n{B}_t^{(n)}=0$ almost surely.

Write the $d$-dimensional continuous semimartingale as $X=X_0+M+A$, where $M$ is a continuous local martingale and $A$ is a continuous finite-variation process. For $f\in C^2({\mathbb{R}}^d)$, Itô formula states

The $d$-dimensional Lévy characterization of Brownian motion says that a continuous local martingale $X$ with $X_0=0$ is a standard $d$-dimensional Brownian motion exactly whenfor all $i,j$ and $t$.

One direction follows directly from independent Gaussian increments. Conversely, fix $\theta \in {\mathbb{R}}^d$. Applying Itô formula and the bracket assumption shows thatis a complex local martingale. After stopping $X$ on leaving large balls it is bounded, so optional sampling and then dominated convergence give, for $s<t$,This is the characteristic function of $N(0,(t-s)I_d)$ and is deterministic. Thus each increment is Gaussian with the required covariance and independent of the past. Together with continuity, these are precisely the defining properties of standard $d$-dimensional Brownian motion.

Put ${\theta }_t=a{t}^{a-1}$. When $a>1/2$,The integrand is deterministic, so the Novikov condition holds. Define an equivalent probability measure $Q$ byThe Cameron-Martin-Girsanov theorem makesa $Q$-Brownian motion on $[0,T]$.

The given limit inferior says that almost surely there is a random $\delta >0$ such that $t^{-a}{W}_t>-1/2$ whenever $0<t\le \delta$. Thereforeon that interval. The first entrance timeis a stopping time by continuity and satisfies $\tau >0$ almost surely. Before $\tau$ one has $S_t>t^a/2$, and continuity gives $S_{\tau }\ge {\tau }^a/2>0$. Hence $S_t>0$ for every $0<t\le \tau$.

Suppose an equivalent measure $Q$ made $S$ a local martingale. Stop at the positive time furnished by part (i), chosen so that $S_{t\wedge \tau }\ge 0$. A nonnegative local martingale is a supermartingale, and this one starts from $S_0=0$. Hence $S_{t\wedge \tau }=0$ $Q$-almost surely for every deterministic $t$.

But $Q$ and $P$ have the same null events, while $\tau >0$ almost surely. Some positive rational $q$ therefore satisfies $Q(q<\tau )>0$, and on that event part (i) gives $S_q>0$, a contradiction. No such equivalent measure exists.

A strong solution of a stochastic differential equation is an adapted process $X$ on a prescribed filtered probability space carrying a prescribed Brownian motion $W$, satisfyingalmost surely. A weak solution of a stochastic differential equation consists of a filtered probability space, a Brownian motion, and an adapted process on that space satisfying the same integral equation; the space and driving Brownian motion are part of what may be chosen.

Pathwise uniqueness means that two solutions on the same filtered probability space, driven by the same Brownian motion and having the same initial value almost surely, are indistinguishable. Uniqueness in law means that any two weak solutions with the same initial distribution induce the same probability law on path space.

Let $X$ and $Y$ be two solutions with the same initial value and Brownian motion, and put $Z=X-Y$. Itô formula givesStop when either process or the stochastic integral becomes large. Taking expectations, using the assumed one-sided Lipschitz bound, and then removing the localization givesThe Gronwall inequality yields $\mathbb{E}{Z}_t^2=0$. Thus $X_t=Y_t$ almost surely at every rational time, and path continuity makes the two processes indistinguishable. This proves pathwise uniqueness.

Fix $t$ and apply Itô formula to $M_s=U(t-s,X_s)$ for $0\le s\le t$. Its drift isby the Kolmogorov backward equation. HenceLocalization makes this a martingale, and boundedness of $U$ permits passage to the limit. Conditioning the identity $M_t=U(0,X_t)$ on $X_0$ givesThis is the required special case of the Feynman-Kac formula, proved directly.

The stochastic-integral identity obtained in part (i), evaluated at $s=t$, isSince $d{X}_s-b(X_s)ds=\sigma (X_s)d{W}_s$, this has the requested form with the previsible process

For $T_n=T\wedge n$, Itô formula and the differential equation show thatup to $T$. After localization this is a martingale, and boundedness of $V$ permits optional stopping. Thus, conditionally on $X_0$,On $\{T<\infty \}$, path continuity gives $X_T\in \{\ell ,r\}$ and hence $V(X_T)=1$. On $\{T=\infty \}$, boundedness of $V$ makes $e^{-\lambda n}V(X_n)\to 0$. The dominated convergence theorem therefore yieldson $\{\ell <X_0<r\}$, with the stated convention when $T=\infty$.