Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 34 (Stochastic Calculus and Applications)

Use the standard càdlàg martingale versions and usual conditions for a filtration where optional sampling and semimartingale decompositions are invoked.

Use the predictable sigma-algebra generated by $A\times \{0\}$, $A\in {\mathcal{F}}_0$, and $A\times (s,t]$, $A\in {\mathcal{F}}_s$. A previsible process is a measurable function for this sigma-algebra. Set $J_t=1_{\{S<t\le T\}}$, with $J_0=0$. The events $\{S<t\}$ and $\{T<t\}$ are in ${\mathcal{F}}_t$, because they are countable unions of events $\{S\le q\}$ and $\{T\le q\}$ for rational $q<t$. Thus $J$ is an adapted process. Each path is left-continuous, including at the two endpoints, and is zero if $S=T$.

For a direct measurability argument, letEach summand is measurable for the predictable sigma-algebra, since its coefficient is ${\mathcal{F}}_{k{2}^{-n}}$-measurable. At every $t>0$ the sampling time approaches $t$ strictly from below, so left-continuity gives $J_t^{(n)}\to J_t$. ThereforeThe original PDF has a right-closed interval. The open right endpoint in the converted TeX is a transcription error; replacing $(S,T]$ by $(S,T)$ is not valid for arbitrary stopping times.

An elementary predictable process with stopping-time intervals has the form $H={\sum }_{j=0}^{r-1}{h}_j{1}_{({\tau }_j,{\tau }_{j+1}]}$, where the stopping times are ordered and $h_j$ is bounded and ${\mathcal{F}}_{{\tau }_j}$-measurable. Deterministic endpoints give the usual simple predictable process. Define the zero-starting stochastic integral byCommon refinement and telescoping show that this definition is independent of the representation. Use $M_{\infty }$ for an infinite endpoint. The L2-bounded martingale convergence theorem gives convergence of $M_t$ in $L^2$ to $M_{\infty }$ and $M_t=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_t]$.

Put $D_j=M_{{\tau }_{j+1}}-M_{{\tau }_j}$. The optional sampling theorem gives $\mathbb{E}[D_j\mid {\mathcal{F}}_{{\tau }_j}]=0$. For $j<k$, the variable $h_j{D}_j{h}_k$ is ${\mathcal{F}}_{{\tau }_k}$-measurable; hence $\mathbb{E}[h_j{D}_j{h}_k{D}_k]=0$. Applying the same theorem to the assumed martingale $M^2-[M]$ givesConsequently the Itô isometry follows:The unbounded stopping times cause no optional-sampling gap. The $L^2$ terminal representation makes $M$ uniformly integrable and makes the family of stopped $M^2$ uniformly integrable, by conditional Jensen applied to $M_{\infty }^2$. Also $\mathbb{E}[M{]}_{\infty }=\mathbb{E}{M}_{\infty }^2-\mathbb{E}{M}_0^2<\infty$, from monotone convergence theorem and the assumed square-compensator identity. Thus $M^2-[M]$ has the needed uniform integrability, and truncating stopping times and passing in $L^1$ is legitimate. Bounded coefficients are part of the simple-integrand convention; more general coefficients require the displayed square-integrability condition.

The stopping-interval indicator is an elementary predictable process with stopping-time intervals. Its stochastic integral isIt is zero before $S$, equals $M_t-M_S$ between $S$ and $T$, and equals $M_T-M_S$ after $T$. For a martingale with jumps, the jump at $S$ is excluded and the jump at $T$ is included, exactly as required by the half-open interval.

Use the shifted filtration ${\mathcal{G}}_s={\mathcal{F}}_{T+s}$. Each $T+s$ is a stopping time, and the optional sampling theorem implies that $M_s^{(T)}=M_{T+s}-M_T$ is a ${\mathcal{G}}_s$-martingale. It is $L^2$-bounded, for example by $\mathbb{E}|M_{T+s}-M_T{|}^2\le 4\mathbb{E}{M}_{\infty }^2$. Its quadratic variation is $[M^{(T)}{]}_s=[M{]}_{T+s}-[M{]}_T$. The shifted process $H_{T+s}$ is continuous and ${\mathcal{G}}_s$-adapted, hence predictable. Its boundedness and the integrability of the bracket make the right-hand stochastic integral well-defined in $L^2$.

For a deterministic partition $0=s_0<\cdots <s_r=t$, use left-endpoint coefficients $H_{T+s_j}$. Both integrals have the same approximating sumOn the original clock this is the integral of the simple process on the stopping-time intervals $(T+s_j,T+s_{j+1}]$; on the shifted clock it is a deterministic-partition integral. As the mesh tends to zero, continuity of $H$ gives pathwise uniform approximation on the random compact interval $[T,T+t]$. The Itô isometry and domination by a constant times the integrable $[M{]}_{\infty }$ give $L^2$ convergence of both sums. Thus stopping-time shift of a stochastic integral yieldsThe left integral includes increments strictly after $T$, so no jump at the starting time is added.

For a continuous path $m$ of finite variation on $[0,t]$, every partition satisfiesUniform continuity makes the right side tend to zero as the mesh tends to zero. The quadratic variation of the given continuous local martingale $M$ is therefore identically zero.

Stop $M$ when its magnitude first reaches an integer $n$, so the stopped process is a bounded, square-integrable martingale. The square identity, or the Itô isometry, gives $\mathbb{E}{M}_{t\wedge {\tau }_n}^2=\mathbb{E}[M{]}_{t\wedge {\tau }_n}=0$. Hence it is zero at each rational time almost surely, and continuity extends this simultaneously to all times. Let $n$ tend to infinity. This provesThis is the continuous finite-variation local martingale is constant principle with initial value zero.

For a continuous semimartingale $X$, its quadratic variation is the continuous increasing zero-starting process obtained as the limit, uniformly on compact time intervals in probability, of sums of squared increments along deterministic partitions whose mesh tends to zero. If $X=X_0+N+V$ is its continuous local-martingale/finite-variation decomposition, then $[X]=[N]$; finite-variation terms and their cross sums vanish.

Write $D=d\mathbb{Q}/d\mathbb{P}$. If $\mathbb{P}(E_n)\to 0$, then for every $R>0$,First send $n$ to infinity and then $R$ to infinity. Thus convergence in probability under $\mathbb{P}$ implies convergence in probability under $\mathbb{Q}$, also for a supremum on a compact time interval. The same squared-increment sums converge under $\mathbb{Q}$ to the $\mathbb{P}$ version of $[X]$. Since $X$ is also a semimartingale under $\mathbb{Q}$, these sums converge to its $\mathbb{Q}$ quadratic variation. Uniqueness of limits in probability gives equality at every fixed time; continuity and countably many rational times giveAbsolute continuity suffices; equivalence is not required. This is quadratic variation under an absolutely continuous measure change.

Use just the parameters $1$ and $-1$. Positivity and adaptation of their local martingales giveThus both processes are adapted. A positive continuous local martingale has a semimartingale logarithm, by Itô formula after localization away from zero. Since $X=\log Z^{(1)}+A/2$ and $A$ has finite variation, $X$ is a continuous semimartingale. Write $X=N+V$, where $N$ is a zero-starting continuous local martingale and $V$ is continuous finite variation, also starting from zero.

Apply Itô formula to each exponential:Because $Z^{(\theta )}$ is itself a local martingale, the finite-variation term is zero by part (a). Divide its finite signed measure by the strictly positive $Z^{(\theta )}$. For $\theta =1,-1$, respectively,Adding and subtracting show $V=0$ and $[N]=A$. Therefore the exponential criterion for a continuous local martingale and its bracket givesThe semimartingale property was established before applying Itô's formula to $X$; it was not assumed from the outset.

For $h(x,y)=\frac{1}{2}\log (x^2+y^2)$ away from the origin,The Itô formula for planar Brownian motion therefore has no drift term before the radius reaches $r$. Since the initial logarithm is zero,The integrand's squared norm is $|B_s{|}^{-2}\le r^{-2}$ on this stopped interval, so this is a square-integrable martingale on every finite horizon, in particular a local martingale. One may first stop in an outer annulus to apply the formula on a compact smooth domain and then remove that localization.

Let $\tau =S_r\wedge T_R$. The stopped logarithm lies between $\log r$ and $\log R$, so the bounded local martingale criterion makes it a true bounded martingale. Also $\tau$ is finite almost surely. Indeed $|B_t{|}^2-2t$ is a martingale, and applying its stopped identity through $t\wedge T_R$ givesmonotone convergence theorem implies $\mathbb{E}{T}_R\le (R^2-1)/2<\infty$, hence $\tau \le T_R<\infty$ almost surely. Boundedness permits dominated convergence theorem in the stopped logarithm as $t\to \infty$. ConsequentlyThe standard results used are Itô's formula, the bounded-local-martingale criterion, the stopped martingale identity at bounded stopping times, and dominated/monotone convergence theorem. At exit the radius is either $r$ or $R$ by continuity. Solving $p\log r+(1-p)\log R=0$ also gives the planar Brownian annulus hitting probability

For $k\ge 2$, substitute $r_k=k^{-k}$ and $R=k$ into the exit formula:Let ${\tau }_0$ be the first hit of the origin. On any path with ${\tau }_0<\infty$, continuity makes the radius bounded on $[0,{\tau }_0]$. For all sufficiently large $k$, that path reaches radius $r_k$ before ${\tau }_0$ and has not yet reached radius $k$, so it belongs to $\{S_{r_k}\le T_k\}$. Thus $\{{\tau }_0<\infty \}$ is contained in the liminf of these events. Fatou's inequality for their indicators gives probability at most ${\mathrm{liminf}}_{k}1/(k+1)=0$ for that liminf. Hence the polar point for planar Brownian motion conclusion isThe argument controls one event for all times, rather than merely proving zero hitting probability at each fixed time.

On a compact time interval a continuous path avoiding the origin has a strictly positive minimum radius. Thus the times $S_{1/n}$ tend to infinity almost surely. Part (a) localizes $M_t=\log |B_t|$, so $M$ is a continuous local martingale.

To show that it is not a martingale, compute its expectation at a positive deterministic time. In distribution $B_t=(1,0)+W_t$, where $W_t$ is a centered isotropic Gaussian vector. Conditional on $\rho =|W_t|$, its angle is uniform. The angular average of a logarithmic potential isFor $\rho <1$ this follows by averaging the real part of the power series for $\log (1+\rho e^{i\theta })$; for $\rho >1$ factor out $\rho$ and apply the same argument to ${\rho }^{-1}$. The value at $1$ follows by the integrable logarithmic singularity, or is irrelevant for the continuously distributed radius.

The radius has density $(\rho /t)e^{-{\rho }^2/(2t)}$. The logarithm of $|B_t|$ is absolutely integrable: near the origin its Gaussian density is bounded and ${\int }_0^{1}r|\log r|dr<\infty$; away from the origin its positive part is bounded by a constant plus $|B_t|$. Therefore conditioning is legitimate, and integration by parts yieldsSince $M_0=0$, this contradicts preservation of expectation by a martingale. Thus the logarithmic radius of planar Brownian motion is a strict local martingale. It is not a nonnegative local martingale; its negative values allow its expectation to increase.

The stochastic exponential of $M$ isItô's formula gives $d\mathcal{E}(M{)}_t=\mathcal{E}(M{)}_{t}d{M}_t$, so it is a positive continuous local martingale starting from one. For $p>1$,Every nonnegative local martingale is a supermartingale. Hence, under $[M{]}_{\infty }\le C$,The bound applies also to stopped versions. Localize the exponential to true martingales. Their uniform $L^p$ bound permits $L^1$ passage to the limit, proving that the unstopped exponential is a true martingale of mean one. The same bound makes its entire time-indexed family uniformly integrable. This proves the bounded-bracket criterion for a stochastic exponential, without needing an additional exponential-moment assumption.

One continuous Girsanov theorem formulation is as follows. Let $L$ be a zero-starting continuous local martingale such that $Z=\mathcal{E}(L)$ is a strictly positive uniformly integrable martingale, and define $d\mathbb{Q}=Z_{\infty }d\mathbb{P}$. For every continuous $\mathbb{P}$-local martingale $N$,On a finite horizon the same assertion holds with terminal density $Z_T$, provided $Z$ is a true martingale through $T$. In particular, if $L_t={\int }_0^t{\theta }_{s}d{B}_s$, thenis a Brownian motion under $\mathbb{Q}$, on the specified horizon. Its quadratic variation remains $t$ under this absolutely continuous measure (equivalent on every finite-horizon sigma-algebra), and the Lévy characterization of Brownian motion identifies it as Brownian. The density sign and drift subtraction are paired: choosing the exponential of $-\int \theta dB$ instead produces $B+\int \theta ds$.

Use the predictable processThe value at zero is irrelevant to the Brownian integral. The numerator is bounded and ${\sigma }_s\ge \delta >0$, so $\theta$ is bounded and $[L{]}_{\infty }={\int }_0^T{\theta }_s^{2}ds$ has a deterministic bound. Part (a) supplies a uniformly integrable density martingale. Define a measure on the entire given sigma-algebra byIts density is positive and has expectation one, so $\mathbb{Q}$ is a probability measure equivalent to $\mathbb{P}$. The Girsanov theorem makes ${\tilde{B}}_t=B_t-{\int }_0^{t\wedge T}({\nu }_s-{\mu }_s)/{\sigma }_{s}ds$ a Brownian motion under $\mathbb{Q}$. For $t\le T$, substitute $dB=d\tilde{B}+\theta dt$:Continuity of $\sigma$ gives pathwise local square integrability, even though it is not assumed bounded. Equivalence preserves this property. This is finite-horizon drift replacement by a change of measure.

Before $\tau$, put $a_s=\sigma (Y_s)-\sigma (X_s)$ and $d_s=b_2(Y_s)-b_1(X_s)$, so $d{Z}_s=a_{s}d{B}_s+d_{s}ds$. For $Z_s\ge 0$, the Lipschitz and drift-order hypotheses give $|a_s|\le K{Z}_s$ and $d_s\ge b_1(Y_s)-b_1(X_s)\ge -L{Z}_s$. The derivatives of $f_{ϵ}(z)=(z+ϵ{)}^{-1}$ are $f_{ϵ}^{\prime }=-(z+ϵ{)}^{-2}$ and $f_{ϵ}^{\prime \prime }=2(z+ϵ{)}^{-3}$. Itô's formula givesThe drift integrand is bounded above byThe stochastic integrand has magnitude at most $K{Z}_s/(Z_s+ϵ{)}^2\le K/(4ϵ)$, so its integral on a finite horizon is a square-integrable martingale with mean zero. Global Lipschitz coefficients give finite second moments of the SDE solutions on finite horizons, which also justify the drift expectation; alternatively localize first and use boundedness of $f_{ϵ}$ and Fatou. Since $Z_{t\wedge \tau }\ge 0$, taking expectations and replacing the stopped integral by the larger positive full-time bound yields the reciprocal barrier proof of scalar diffusion comparison:

The Gronwall inequality applied to part (a) givesContinuity implies $Z_{\tau }=0$ on $\{\tau <\infty \}$. Thus on $\{\tau \le t\}$ the stopped reciprocal equals $1/ϵ$. Positivity givesLet $ϵ\downarrow 0$, using $z_0>0$, to obtain $\mathbb{P}(\tau \le t)=0$. Taking the union over integer $t$ proves $\tau =\infty$ almost surely. Hence the strict order preservation for scalar Lipschitz diffusions conclusion iswhich is stronger than the requested weak inequality. The single-event argument prevents a gap between fixed-time ordering and simultaneous ordering at all times.

For the diffusion operatora continuous adapted process solves the diffusion martingale problem when, for every $f\in C_b^2({\mathbb{R}}^d)$,Here $C_b^2$ means that the function and its derivatives through order two are bounded and continuous. An equivalent compact-support/local convention is often used for the general martingale problem; the bounded-coefficient setting lets us use the true-martingale convention explicitly. The matrix $a$ is interpreted as a covariance matrix in the diffusion formulation, hence is symmetric nonnegative; the displayed operator in any case uses only its symmetric part.

A well-posed martingale problem has existence and uniqueness in law on continuous path space for each specified starting point, or for each initial law in the initial-law formulation. This is uniqueness of the process law, not an assertion of pathwise uniqueness on a prescribed Brownian probability space.

Take $f\in C_b^2$. Boundedness of $a,b$ and the derivatives gives a finite bound $\Vert Lf{\Vert }_{\infty }$. On $0\le t\le T$ the assumed local martingale satisfiesIt is therefore a bounded local martingale on this time interval. To check the true martingale property directly, stop at a localizing sequence, apply its conditional martingale identity, and pass to the limit by dominated convergence theorem using this deterministic bound. Since $T$ is arbitrary,The hypothesis for every $C^2$ test function contains the required $C_b^2$ class. No integrability assumption on the unbounded test functions is needed.

The sampling rule in the PDF is with replacement. Thus conditional on the current proportion $y=j/n$, the next count is $W\sim \mathrm{Bin}(n,y)$. Put $\Delta =W/n-y$ and $q=y(1-y)$. Direct binomial moments giveThe fourth centered moment of a binomial distribution implies $\mathbb{E}|\Delta {|}^3\le (\mathbb{E}{\Delta }^2\mathbb{E}{\Delta }^4{)}^{1/2}\le C{n}^{-3/2}$. The accelerated generator of the Wright–Fisher binomial sampling chain isFor $f\in C^3([0,1])$, Taylor's formula uniformly yieldsThe initial condition is $Z_0^n=\lfloor xn\rfloor$, so $X_0^n\to x$. The endpoints are absorbing.

We give the path-space argument, rather than only a formal generator calculation. Regard $X^n$ as a càdlàg process in $D([0,1],[0,1])$ with the Skorokhod $J_1$ topology. The chain proportion is a bounded martingale, with predictable square compensator $V_t^n=n^{-1}{\sum }_{k<\lfloor nt\rfloor }{X}_{k/n}^n(1-X_{k/n}^n)$. For bounded stopping times ${\tau }_n\le {\sigma }_n\le {\tau }_n+{\delta }_n$, optional sampling gives $\mathbb{E}|X_{{\sigma }_n}^n-X_{{\tau }_n}^n{|}^2=\mathbb{E}(V_{{\sigma }_n}^n-V_{{\tau }_n}^n)\le ({\delta }_n+n^{-1})/4$. This bound uses the deterministic number of grid jumps in an interval of length at most ${\delta }_n$. Compact containment is automatic. The Aldous tightness criterion therefore proves tightness.

Furthermore, the fourth-moment bound and a union bound giveEvery subsequential limit is continuous. For each smooth $f$, the discrete compensated processis a martingale. Generator convergence and continuity of any limiting path turn the sum into ${\int }_0^t\frac{1}{2}{X}_s(1-X_s)f^{\prime \prime }(X_s)ds$. These compensated processes are uniformly bounded on $[0,1]$, so their martingale identities pass to the limit against bounded continuous functions of past coordinates. A Monotone class theorem argument extends the identities to the natural filtration. Smooth approximation in the $C^2$ norm extends the test class. Thus every limit solves the martingale problem for $Lf(y)=\frac{1}{2}y(1-y)f^{\prime \prime }(y)$ on $[0,1]$.

To identify this problem with the stated SDE, tests equal to $y$ and $y^2$ on $[0,1]$ (using bounded smooth extensions) give a continuous martingale with bracket ${\int }_0^t{X}_s(1-X_s)ds$. On an enlargement carrying independent Brownian motion $W$, setIts bracket is $t$, and the Lévy characterization of Brownian motion makes it Brownian. The zero-coefficient portion of $dX$ has zero bracket, so $d{X}_t=\sqrt{X_t(1-X_t)}d{B}_t$. Conversely Itô's formula shows that any such SDE solution solves the limiting martingale problem. The allowed uniqueness of the SDE, understood at least as uniqueness in law, therefore makes every subsequential limit have the same law. ConsequentlyThe limit is the Wright–Fisher diffusion. Linear interpolation has the same weak limit in the uniform topology because the maximum jump tends to zero. The results applied are Aldous tightness, the vanishing-jump continuous-limit criterion, passage of uniformly bounded martingale identities under weak convergence, Lévy characterization, and uniqueness in law; the estimates above verify their hypotheses.