Past exam of the stochastic differential equations course of the University of Oxford / 2011 (C10.1a, DMAC 2A83, DMCO 2A83, DMSU 2A83)

Source: released examination bundle, Part C 2007 11.pdf, pages 454–456. All three questions are covered. No mark allocations are printed.

The Bessel process is stopped in the compact interval $[\alpha ,\beta ]$, where its drift and diffusion coefficients are bounded. In particular its stopped stochastic integral has expectation zero on every finite time interval.

For $k>1$, taking expectations in the stopped equation givesSince $Y_t\le \beta$, this bounds $\mathbb{E}(t\wedge \tau )$ uniformly in $t$. For $k<1$, the same argument with $1-\mathbb{E}{Y}_t$ gives $\mathbb{E}(t\wedge \tau )\le 2\beta (1-\alpha )/(1-k)$.

If $k=1$, the stopped process is a stopped Brownian motion. Itô's formula gives $\mathbb{E}(Y_t-1{)}^2=\mathbb{E}(t\wedge \tau )$, which is at most $\max \{(1-\alpha {)}^2,(\beta -1{)}^2\}$. Thus in every case the monotone convergence theorem gives $\mathbb{E}\tau <\infty$, so $\tau <\infty$ almost surely. By path continuity its value at exit is one of the two endpoints. They are disjoint possibilities, hence

For a twice continuously differentiable function, Itô's formula in this stopped diffusion readsFor $f(x)=x^{2-d}$,Setting $d=k$ eliminates the drift. Moreover $f^{\prime }$ is bounded on $[\alpha ,\beta ]$, so the stochastic integral is square-integrable on each finite horizon by the Itô isometry. Hence $f(Y_t)$ is a true martingale, not merely a local one. The process itself is bounded on the stopped interval. For $k=2$, this particular function is the constant one; the logarithmic scale function will supply the nontrivial hitting identity.

Let $p=\mathbb{P}(y_{\tau }=\alpha )$. For $k\ne 2$, the bounded stopped martingale from part (b) converges to $y_{\tau }^{2-k}$, so the dominated convergence theorem givesTherefore the boundary hitting probability from a diffusion scale function isFor $k=2$, applying Itô's formula to $\log Y_t$ gives a bounded stopped martingale instead, yieldingIf $k>2$ and $\beta >1$ is fixed, the first expression tends to zero as $\alpha \downarrow 0$. A continuous path that reaches zero before it reaches $\beta$ must first reach every such $\alpha$ before $\beta$, so that event has probability zero. Any path hitting zero in finite time has a finite maximum up to that time, and therefore belongs to one of these zero-probability events with an integer upper boundary $\beta >1$. Their countable union still has probability zero. Thus the process does not hit zero almost surely. This is the relevant part of the Hitting-zero classification for a Bessel process.

The given Doob L2 maximal inequality givesApply Markov's inequality to the nonnegative variable $(M_t^{\ast }{)}^2$, at the threshold printed in the question:The corresponding threshold for $M_t^{\ast }$ itself is $2^{-(n-1)/2}$.

Set $A_n=\{{\sup }_{s\le t}|M_s^{(n)}|\ge 2^{-(n-1)/2}\}$. Part (a) gives ${\sum }_n\mathbb{P}(A_n)<\infty$. The Borel-Cantelli first lemma implies that, almost surely, only finitely many $A_n$ occur. On the complementary full-probability event, all sufficiently late terms satisfyThese deterministic bounds form a convergent geometric series. The finitely many earlier suprema are finite by continuity on the compact time interval. HenceThe series therefore converges absolutely and uniformly on $[0,t]$ by the Weierstrass M-test, and its sample-path limit is continuous.

Work modulo indistinguishability of stochastic processes, with norm $\Vert M\Vert =(\mathbb{E}|M_t{|}^2{)}^{1/2}$. It is a genuine norm: zero terminal value implies $M_s=\mathbb{E}[M_t\mid {\mathcal{F}}_s]=0$ at every fixed time, and continuity extends equality from countably many rational times to all times simultaneously.

Let $M^j$ be a Cauchy sequence in this norm. Choose a subsequence with $\Vert M^{j_{n+1}}-M^{j_n}\Vert <2^{-n}$. Part (b) applies to these difference martingales, so their telescoping series converges uniformly almost surely. Its limit has continuous paths. At each time $s$, conditional-expectation contractivity givesso the fixed-time limit also exists in $L^2$ and agrees with the pathwise limit. It has an adapted version obtained by taking the finite pointwise subsequence limit when it exists, and zero otherwise. Each such definition is measurable at its own time, and the almost surely uniform limit gives continuous paths. Passing to the $L^2$ limit in $\mathbb{E}[M_v^j\mid {\mathcal{F}}_s]=M_s^j$ shows that the limit is a martingale. The original Cauchy sequence converges to it in terminal norm. Thus the space is complete.

Part (b) supplies what fixed-time $L^2$ completeness alone does not: a single full-probability set on which a subsequence converges uniformly in time, preserving path continuity. This is completeness of continuous square-integrable martingales.

For simple predictable integrands, the Itô integral is a continuous square-integrable martingale and the Itô isometry givesApproximate a general square-integrable predictable integrand by simple ones. The resulting integral martingales are Cauchy in terminal $L^2$ norm. Completeness of continuous square-integrable martingales guarantees a unique continuous martingale limit, independent of the approximation, which defines the integral on the completed integrand space.

The bounded integrand makes $X_s={\int }_0^s{Y}_{u}d{W}_u$ a square-integrable martingale, with quadratic variationThusNovikov's condition applies to the stochastic exponential $Z_s=\exp (X_s-⟨X{⟩}_s/2)$. Hence $Z$ is a true martingale with $\mathbb{E}{Z}_s=Z_0=1$. It is strictly positive because its exponent is finite almost surely.

Define $\mathbb{Q}$ on ${\mathcal{F}}_t$ by density $Z_t$. Nonnegativity and countable additivity of the integral make it a measure, and $\mathbb{Q}(\Omega )={\mathbb{E}}_P{Z}_t=1$. Strict positivity of $Z_t$ also makes the two measures equivalent. The Radon-Nikodym derivative identity gives ${\mathbb{E}}_Q[1_{A}Y]={\mathbb{E}}_P[1_{A}Y{Z}_t]$, first for simple variables and then for every Q-integrable $Y$ by approximation.

Write $U={\mathbb{E}}_Q[Y\mid {\mathcal{F}}_s]$. The density on ${\mathcal{F}}_s$ is $Z_s={\mathbb{E}}_P[Z_t\mid {\mathcal{F}}_s]$. For every $A\in {\mathcal{F}}_s$,Moreover $Z_{s}U$ is P-integrable because ${\mathbb{E}}_P[Z_s|U|]={\mathbb{E}}_Q|U|<\infty$. Uniqueness in the definition of conditional expectation provesThis is conditional expectation under a martingale density.

Boundedness makes each $N_u$ Q-integrable. For $s\le u\le t$, the density formula and conditioning first on ${\mathcal{F}}_u$ giveThe middle equality uses that $N_u$ is ${\mathcal{F}}_u$-measurable and $Z$ is a martingale; the next uses the assumed martingale property of $NZ$. Thus $N$ is a Q-martingale.

The initial value can first be subtracted: it is ${\mathcal{F}}_0$-measurable and has the same integrability under both measures, since the density has conditional expectation one on ${\mathcal{F}}_0$. Put $N=M-⟨M,X⟩$. Since $dZ=ZdX$, the Itô product rule and the quadratic covariation identity $d⟨M,Z⟩=Zd⟨M,X⟩$ giveThis is a local P-martingale. Stop $N$ at increasing bounded levels; the resulting products with $Z$ are true martingales, since they are local martingales dominated at stopping times by a constant times the uniformly integrable density process. Part (c) then shows that the stopped $N$ processes are Q-martingales. ConsequentlyThe unqualified true-martingale assertion printed in this part needs an additional integrability hypothesis. One sufficient version is a square-integrable P-martingale $M$. Indeed $Z_t\in L^2(P)$ by boundedness of $Y$, and the covariation bound givesThe Doob L2 maximal inequality, ${\mathbb{E}}_P⟨M{⟩}_t<\infty$, and Cauchy-Schwarz inequality therefore give ${\mathbb{E}}_Q{\sup }_{s\le t}|N_s|<\infty$. Its local martingale identities pass to the limit, making it a true Q-martingale. This is Girsanov compensation with square-integrable input.

Here is a counterexample to the broader printed assertion. On $[0,1]$, take $X_s=W_s$, so $Y_s\equiv 1$ and $Z_s=e^{W_s-s/2}$. Put $a=1/2$ andUnder P, $W_a$ has density ${\pi }^{-1/2}{e}^{-x^2}$, so ${\mathbb{E}}_{P}H<\infty$. Independence of future Brownian increments proves $M$ is a continuous true P-martingale. Under Q, $W_s^Q=W_s-s$ is Brownian and the density of $W_a$ is multiplied by $e^{x-1/4}$, so ${\mathbb{E}}_{Q}H=\infty$. The compensated process after $a$ isThe future Q-Brownian increment is independent of $H$ and has positive finite expected absolute value. Thus ${\mathbb{E}}_Q|N_s|=\infty$ for every $s>a$, so it cannot be a true martingale. This proves that Bounded Girsanov drift does not preserve true martingales under compensation.

For the requested application, the Brownian driver does satisfy the needed conditions. To distinguish the state from the density-driving integral, call the state $U$. Set $Y_s=b(U_s)/a(U_s)$; the stated coefficient bounds make this integrand bounded. Applying the local result to $M=W$ givesIt is a continuous local Q-martingale with quadratic variation $s$. The Lévy characterization of Brownian motion therefore makes it a Q-Brownian motion. Substituting $d{W}_s=d{\tilde{W}}_s+[b(U_s)/a(U_s)]ds$ into the original state equation givesThe Brownian driver after the change is $\tilde{W}$; writing it as $W$ again in the last printed equation amounts to renaming that driver.