Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 202 (Stochastic Calculus and Applications)

Let $V(t)$ be the total variation of $f$ on $[0,t]$. The Jordan decomposition of a function of bounded variation isFor $s<t$, the inequality $V(t)-V(s)\ge |f(t)-f(s)|$ shows that both increments are nonnegative, so $g,h$ are nondecreasing and $f=g-h$. Right-continuity of the finite variation function $f$ implies right-continuity of $V$, and hence of $g,h$.

It suffices by part a to treat a nondecreasing right-continuous integrator, whose increments define a finite Lebesgue–Stieltjes measure ${\mu }_f$ on $[0,1]$. Let ${\alpha }_n$ be the left-endpoint step approximation on the dyadic intervals. The displayed sum is exactlyThe continuous function $\alpha$ is uniformly continuous on the compact interval, so $\Vert {\alpha }_n-\alpha {\Vert }_{\infty }\to 0$. ThereforeApply this separately to $g$ and $h$ to obtain the asserted Lebesgue-Stieltjes integral limit.

For any partition of $[0,t]$,The final sum is at most the largest increment of $f$ times its total variation. It tends to zero because $f$ is uniformly continuous. Part b then gives the integration-by-parts identityThus the formula stated in the question holds when $f(0)=0$; for a general initial value the necessary endpoint correction is $-f(0{)}^2$.

The elementary discrete integration-by-parts identity isBy the supplied fact, $M^{(n)}\to M$ in $L^2$ of the uniform norm. DefineThenwhich proves i, whileis an $L^2$-bounded martingale, proving ii.

Choose a subsequence converging uniformly almost surely. For $s<t$, all complete dyadic increments between $s$ and $t$ contribute nonnegative squares; only the two boundary increments can affect monotonicity, and they vanish uniformly by continuity of $X$. Passing to the limit gives $[X{]}_s\le [X{]}_t$. Thus $[X]$ is nondecreasing and is the quadratic variation of $X$.

Choose stopping times $T_m\uparrow \infty$ such that $X^{T_m}$ is a bounded martingale. Part a constructs $[X^{T_m}]$. Uniqueness in the identityshows consistency on overlapping stopped intervals, so define $[X{]}_{t\wedge T_m}=[X^{T_m}{]}_t$. The stopped dyadic sums converge uniformly on every compact interval in probability, andis a local martingale. This localization constructs the quadratic variation of every continuous local martingale.

Letting $t\to \infty$ in the Burkholder-Davis-Gundy inequalities with exponent two gives absolute constants $c,C>0$ such thatSince $[X]$ is nondecreasing, $\mathbb{E}[X{]}_{\infty }={\sup }_t\mathbb{E}[X{]}_t$. Thus one of the two quantities in the question is finite exactly when the other is.

Apply Itô formula to $\varphi (x)=x\log x$. Since ${\varphi }^{\prime \prime }(x)=1/x$,Boundedness of $\log M$ makes the stochastic integral a true martingale of mean zero. Taking expectations proves

The heat-semigroup form is $U(t,x)=P_{1-t}(f^2)(x)$. For $s<t$, independence and additivity of Brownian increments givewhere ${\tilde{B}}_{1-s}$ is an independent $N(0,1-s)$ increment. Thus $M_t=U(t,W_t)$ is a martingale.

The backward heat equation and Itô formula giveMoreover $M_1=f(W_1{)}^2$ and $M_0=\mathbb{E}f(W_1{)}^2$. Substitution into part a yields

The supplied derivative identity and the Cauchy-Schwarz inequality giveThereforeThe sum $W_t+B_{1-t}$ is standard normal for every $t$, so the right side is $2\mathbb{E}{f}^{\prime }(W_1{)}^2$. Part c proves the Gaussian logarithmic Sobolev inequality

Continuity gives $X_{{\tau }_a}=a$ on $\{{\tau }_a<\infty \}$. The stopped process $X^{{\tau }_a}$ is bounded by $a$ and is therefore a true martingale. HenceThe second term tends to zero by bounded convergence because $X_t\to 0$ and it is bounded by $a$. Thus

The Dambis-Dubins-Schwarz theorem says that there is Brownian motion $W$ such thatWhen $[M]$ is strictly increasing, define its inverseand set $W_s=M_{T_s}$. Optional sampling shows that $W$ is a continuous local martingale, while time change gives $[W{]}_s=s$. The Lévy characterization of Brownian motion makes $W$ Brownian, and inverse time change gives the displayed representation.

Set $s=[M{]}_t$. By hypothesis $s\to \infty$, and part b givesSince $W_s/s\to 0$ almost surely, the expression tends to $-\infty$ almost surely.

The Doléans-Dade exponentialis a positive continuous local martingale with $X_0=1$, and part c shows $X_t\to 0$. Apply part a with $a=e^y$:

A weak solution consists of a filtered probability space carrying a Brownian motion $W$ and an adapted continuous process $X$ satisfyingThe probability space and Brownian motion are part of the unknown solution.

A strong solution is adapted to the augmented filtration generated by a prescribed Brownian motion and initial condition; equivalently, it is constructed measurably from that given noise.

Uniqueness in law means that any two weak solutions with the same initial distribution have the same probability distribution on path space.

Pathwise uniqueness means that two solutions on the same filtered space, driven by the same Brownian motion and with the same initial value, are indistinguishable.

Define the scale function of a one-dimensional diffusionThen $g^{\prime }>0$, so $g$ is strictly increasing, andBy Itô formula,so $Y=g(X)$ is a local martingale.

LetDifferentiating and using the scale equation givesSince $b$ is bounded, $h$ has global Lipschitz continuity. Thushas a pathwise unique strong solution by the standard Lipschitz existence-and-uniqueness theorem for a stochastic differential equation. Applying the deterministic inverse $g^{-1}$ gives a strong solution $X$, and uniqueness of $Y$ gives pathwise uniqueness of $X$.

An $x$-admissible strategy is a predictable vector of holdings $\varphi =({\varphi }^0,\dots ,{\varphi }^d)$ that is integrable against the asset prices, is self-financing, has initial wealthand whose wealth process $V_t={\sum }_i{\varphi }_t^i{S}_t^i$ remains nonnegative.

An arbitrage is a zero-initial-wealth admissible self-financing strategy with $V_T\ge 0$ almost surely and $\mathbb{P}(V_T>0)>0$ for some finite horizon $T$.

Write ${\mu }_t=({\mu }_t^{(1)},\dots ,{\mu }_t^{(d)}{)}^T$ and define the market price of riskIt is bounded by hypothesis. The stochastic exponentialis a true martingale by the Novikov condition. Define the equivalent measure $Q$ by $dQ=Z_{T}dP$. The Girsanov theorem makesa Brownian motion under $Q$. After discounting by the bank account, every risky price has zero drift and is a $Q$-local martingale.

The discounted wealth of an admissible self-financing strategy is a nonnegative local martingale and hence a supermartingale. If an arbitrage existed, its zero initial value would imply nonpositive expected terminal discounted wealth under $Q$, while that wealth is nonnegative and positive with positive $Q$-probability. This contradiction proves that the market has no arbitrage; it is the needed direction of the equivalent local martingale measure criterion.

Let $X$ be a positive strict local martingale solvingand fix $T>0$. Use the bank account $S^0=1$ and two risky assetsBoth discounted prices are nonnegative local martingales under the physical measure itself, so the same supermartingale argument as in part c rules out arbitrage. At maturity,But strictness means $\mathbb{E}[X_T\mid {\mathcal{F}}_t]<X_t$ for some earlier $t$ on a set of positive probability, so the two prices are not indistinguishable before $T$. This no-arbitrage market violates the Law of One Price.