Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 25 (Stochastic Calculus and Applications)

Choose an orthonormal basis $(e_j)$ of the separable Hilbert space and independent standard normal random variables $(Z_j)$ on a common probability space. In finite dimension the sums below are finite; in dimension zero take $X(0)=0$. In infinite dimension defineIndependence, centring and unit variance giveBy the Parseval identity for a Hilbertian basis, the coefficient sequence is square-summable. Completeness of $L^2(\Omega )$ therefore supplies a limit, and we defineEvery partial-sum map is linear, and passage to the $L^2$ limit preserves this identity. Thus for each fixed $a,b,g,h$, $X(ag+bh)=aX(g)+bX(h)$ as an $L^2$ identity, hence an almost sure equality. This constructs an isonormal Gaussian process.

The usual meaning is almost-sure linearity for each fixed choice of arguments. A version linear simultaneously for every argument can also be chosen: select a Hamel basis of $H$, choose a measurable representative of $X$ on each basis vector, and extend each sample algebraically by finite sums. For every fixed $h$, this extension equals the constructed $L^2$ variable almost surely, so all its required distributions are unchanged.

For the construction above, independence of the normal random variables makes $S_N(h)$ normal with mean zero and variance $v_N={\sum }_{j\le N}⟨h,e_j{⟩}^2$. Its characteristic function is $\exp (-{\theta }^2{v}_N/2)$. The $L^2$ convergence gives $L^1$ convergence, andSince $v_N\to \Vert h{\Vert }^2$ by the Parseval identity for a Hilbertian basis, the limiting characteristic function identifiesThis includes $h=0$, where the normal distribution is degenerate at zero.

For any $r_1,\dots ,r_n\in \mathbb{R}$, the linearity from part (a) givesalmost surely, and the right side has a normal distribution. This is the defining linear-combination criterion for a multivariate normal distribution; singular covariance matrices are allowed.

Passing to the limit in the inner products of the partial sums givesThe passage to the limit is justified by Cauchy-Schwarz inequality and $L^2$ convergence. Equivalently, the vector's characteristic function isThus both joint normality and the complete covariance matrix follow from the Hilbert-space inner product.

Apply the construction of part (a) to the real Hilbert space $H=L^2([0,\infty ),du)$, and setChoose $X(0)=0$. The interval of length zero represents the zero vector of $H$, so $W_0=0$. All variables are defined on the single probability space already used for the isonormal Gaussian process.

For $0\le s<t$, linearity of the isonormal Gaussian process givesThe squared $L^2$ norm of this indicator is $t-s$, so part (a) yieldsAlso, the covariance formula gives $\mathrm{Cov}(W_s,W_t)=s\wedge t$, the Brownian covariance kernel. The Gaussian statement concerns the signed increment.

Let ${\Delta }_j=W_{t_j}-W_{t_{j-1}}$ for $1\le j\le n$. These increments are jointly normal, because each is obtained by evaluating the isonormal Gaussian process on an interval indicator. Indicators of distinct intervals are orthogonal in $L^2([0,\infty ))$, soBy uncorrelated jointly Gaussian variables are independent, the increments on all these disjoint intervals are independent. Reversing the sign of the first increment, as in the printed list, preserves this independence.

The three requested properties have now been obtained without an existence theorem for Brownian motion. One can also obtain continuous paths: the Gaussian fourth moment gives $\mathbb{E}|W_t-W_s{|}^4=3|t-s{|}^2$. The Kolmogorov continuity theorem therefore supplies a continuous modification on every finite time interval, which can be chosen consistently on the half-line. Modification preserves every finite-dimensional distribution and hence the independent Gaussian increments. This yields Brownian motion itself.

The density $Z=\exp (X(k)-\Vert k{\Vert }^2/2)$ is strictly positive. The Gaussian moment-generating function gives $\mathbb{E}Z=1$, so it defines an equivalent probability measure.

The joint normal distribution of $(X(h),X(k))$, with covariance $⟨h,k⟩$, gives the mixed exponential formulaMultiplying by the normalizing and centring factors therefore yieldsThe characteristic function identifies the answer:This is exponential tilting of an isonormal Gaussian process: the mean shifts by the inner product while its covariance remains unchanged.

If $T\le s$, then $\varphi (W_T)K$ is ${\mathcal{F}}_s$-measurable. Independence and centring of the future Brownian increment make the desired left side zero; the time multiplier on the right is zero as well.

Suppose $T>s$. Conditional on ${\mathcal{F}}_s$, write $W_T=W_s+U$ and $W_t-W_s=V$. The pair $(U,V)$ is jointly normal and independent of ${\mathcal{F}}_s$, withApply the supplied Gaussian integration by parts formula to $z\mapsto \varphi (W_s+z)$, treating the known $W_s$ as its parameter. This givesMultiply by the bounded ${\mathcal{F}}_s$-measurable $K$ and use the defining property of conditional expectation. Thus the required expectation identity holds for every $s<t$, including intervals crossing or lying after $T$.

First take a bounded elementary predictable process $\alpha ={\sum }_j{K}_j{1}_{(s_j,t_j]}$, with each $K_j$ bounded and ${\mathcal{F}}_{s_j}$-measurable and with finite time support. The Itô integral is the corresponding finite sum ${\sum }_j{K}_j(W_{t_j}-W_{s_j})$. Applying part (a) term by term givesSuch elementary predictable processes are dense among predictable processes in $L^2(d\mathbb{P}du)$. The Itô isometry makes the left functional continuous, with boundThe Cauchy-Schwarz inequality makes the right functional continuous, with bound $\Vert {\varphi }^{\prime }{\Vert }_{\infty }\sqrt{T}{\left(\mathbb{E}\int {\alpha }_u^{2}du\right)}^{1/2}$. Approximation therefore proves the same identity for every allowed predictable $\alpha$. The integral over the infinite time interval is the $L^2$ limit of its finite-horizon Itô integrals.

The terminal variable $F=\varphi (W_T)$ is bounded and hence square-integrable. In the completed natural Brownian filtration, the Brownian martingale representation theorem says that any square-integrable ${\mathcal{F}}_T$-measurable variable admits a representationwith $\beta$ predictable and $\mathbb{E}{\int }_0^T{\beta }_u^{2}du<\infty$. Extend $\beta$ by zero after $T$. Thus the requested constant and integrability areExpectation determines $c$ uniquely. If two integrands give the same representation, the Itô isometry gives $\mathbb{E}{\int }_0^{\infty }({\beta }_u-{\tilde{\beta }}_u{)}^{2}du=0$. Consequently $\beta$ is unique up to $d\mathbb{P}du$-almost everywhere equality, rather than pointwise equality at every time. The corresponding integral martingales are indistinguishable.

Multiply the representation in part (c) by ${\int }_0^{\infty }{\alpha }_{u}d{W}_u$ and take expectations. This Itô integral has mean zero, and the bilinear form of the Itô isometry givesEquating this with part (b), and writing both ordinary integrals with the same time variable, yieldsAll terms are integrable by the Cauchy-Schwarz inequality, the assumed square integrability of $\alpha$, and boundedness of ${\varphi }^{\prime }$. This is an orthogonality statement against predictable processes; its second term need not itself be predictable.

Set $Z_t=\mathbb{E}[{\varphi }^{\prime }(W_T)\mid {\mathcal{F}}_t]$. The Brownian martingale representation theorem applied to the bounded terminal variable ${\varphi }^{\prime }(W_T)$ supplies a continuous adapted version of $Z$, so it is predictable. Henceis predictable and satisfies $\mathbb{E}\int {\gamma }_t^{2}dt\le T\Vert {\varphi }^{\prime }{\Vert }_{\infty }^2$. For every square-integrable predictable process $\alpha$, conditioning at each deterministic time and using Fubini theorem givesThus part (d) says $\mathbb{E}\int ({\beta }_t-{\gamma }_t){\alpha }_{t}dt=0$. Taking $\alpha =\beta -\gamma$, which is an allowed predictable square-integrable process, makes its squared norm zero. We concludeThis is the Clark-Ocone formula for a smooth Brownian terminal payoff, with exactly the uniqueness established in part (c).

Put $t_j=j{2}^{-n}$ and ${\Delta }_j=X_{t_{j+1}}-X_{t_j}$. Telescoping $X_{t_{j+1}}^2-X_{t_j}^2=2X_{t_j}{\Delta }_j+{\Delta }_j^2$ givesThe martingale transform summands are orthogonal in $L^2$: for an earlier summand, conditioning on the sigma-algebra at the start of the later increment makes the cross expectation zero. HenceThe martingale increments themselves are also orthogonal. Since $X_0=0$, their variance sum equals $\mathbb{E}{X}_1^2\le C^2$. ThereforeOnly discrete martingale orthogonality is used here; no pre-existing quadratic variation calculation is needed.

The defining relation gives $A_1^{(n)}=X_1^2-2M_1^{(n)}$. Applying $(r-s{)}^2\le 2r^2+2s^2$ and the bound from part (a),Thusuniformly in the dyadic mesh. This controls the approximations to quadratic variation without assuming that their limits already exist.

For $n>m$, define $r_j=\lfloor j{2}^{m-n}\rfloor 2^{-m}$ and $h_j=X_{j{2}^{-n}}-X_{r_j}$. Each $h_j$ is measurable at time $j{2}^{-n}$, so the summands in the given difference formula are orthogonal martingale transforms. ThereforeLetPath continuity on the compact interval gives ${\omega }_m\to 0$ almost surely, and ${\omega }_m\le 2C$. Since $0\le j{2}^{-n}-r_j<2^{-m}$, $|h_j|\le {\omega }_m$. Thus, by the Cauchy-Schwarz inequality and part (b),The last step is the dominated convergence theorem. The bound is uniform in $n>m$, and the other ordering follows by symmetry. Hence the terminal martingale transforms are Cauchy in $L^2$.

For every $t$, subtraction of the definitions givesThe difference of the two supplied continuous martingales is a square-integrable martingale on $[0,1]$; for each fixed mesh its finite sums are bounded. For this dyadic quadratic variation of a bounded continuous martingale, the Doob L2 maximal inequality therefore givesThus the dyadic approximations to quadratic variation are Cauchy for the expected squared uniform norm, as required. No monotonicity in time of the partially completed squared-increment sums is needed.

The drift $b(z)=\tanh z$ has derivative $b^{\prime }(z)=1/{\cosh }^{2}z$, so $|b^{\prime }(z)|\le 1$. Hence it is globally Lipschitz. The diffusion coefficient $\sigma (z)=1$ is globally Lipschitz as well, and both coefficients satisfy a linear growth bound.

The global existence theorem for stochastic differential equations with Lipschitz coefficients states that globally Lipschitz coefficients with linear growth give, for each deterministic initial point, an adapted continuous strong solution of a stochastic differential equation on every finite interval, with pathwise uniqueness and no finite-time explosion. Applying this theorem gives a unique strong solution for every $X_0=x\in \mathbb{R}$, satisfying

Write $f(z)=1/\cosh z$. Differentiating givesTherefore the diffusion operator satisfies $\frac{1}{2}{f}^{\prime \prime }+\tanh z{f}^{\prime }=-\frac{1}{2}f$. Applying the Itô formula to $Y_t=e^{t/2}f(X_t)$ cancels its drift:So $Y$ is a positive local martingale. On every finite interval $[0,R]$, $0<Y_t\le e^{R/2}$. The bounded local martingale criterion makes it a true martingale on that interval. Since $R$ is arbitrary,

For $0\le t\le T$, set $Z_t=\cosh x{Y}_t$. Part (b) gives $Z_0=1$, $\mathbb{E}{Z}_T=1$, and $Z_t>0$. Its stochastic differential isThus $Z$ is the stochastic exponential of $\int \eta dW$. The Novikov condition also holds, since $\exp (\frac{1}{2}{\int }_0^T{\eta }_t^{2}dt)\le e^{T/2}$.

The Girsanov theorem says that under the measure with Radon-Nikodym derivative $Z_T$, the process $W_t-{\int }_0^t{\eta }_{s}ds$ is a Brownian motion up to $T$. Substituting the sign of $\eta$ and the original stochastic differential equation givesThe density is strictly positive, so $\mathbb{Q}$ and $\mathbb{P}$ are equivalent probability measures.

Under $\mathbb{Q}$, $X_T=x+W_T^{\mathbb{Q}}$ has density ${\phi }_T(z-x)=(2\pi T{)}^{-1/2}\exp (-(z-x{)}^2/(2T))$. The reciprocal Radon-Nikodym derivative depends only on $X_T$:Consequently the probability density function under the original measure isIt integrates to one because ${\mathbb{E}}_{\mathbb{Q}}\cosh X_T=e^{T/2}\cosh x$. Completing the square also gives the useful mixture distribution formThus the terminal law is a mixture of $N(x+T,T)$ and $N(x-T,T)$ with the displayed positive weights. The diffusion with hyperbolic tangent drift density also exhibits the Doob h-transform with $h(z)=\cosh z$.

For a deterministic $t\le T$, conditional symmetry makes the conditional characteristic function of $X_T-X_t$ invariant under $\theta \mapsto -\theta$. The bounded real and imaginary parts of the exponential are legitimate test functions. HenceThe right side is a bounded complex martingale. Both sides have continuous versions by the assumptions; equality on rational times and continuity make them indistinguishable. Thus $e^{-2i\theta X_t}{M}_t$ is a martingale on $[0,T]$. Complex martingale assertions mean the corresponding assertions for both real and imaginary parts.

Let $A=⟨X⟩$ be the quadratic variation. We use bilinear quadratic covariation for the complex martingale: $⟨M,X⟩=⟨\mathrm{Re}M,X⟩+i⟨\mathrm{Im}M,X⟩$. The Itô formula givesBy the Itô product rule, the finite-variation part of $d(e^{-2i\theta X}M)$ isPart (a) says the product is a martingale, so uniqueness of the continuous semimartingale decomposition makes this finite-variation part zero. For $\theta \ne 0$, division givesFor $\theta =0$, $M\equiv 1$ and both sides are zero, so the identity holds without exception.

Put $F_t=\exp (-i\theta X_t-\frac{1}{2}{\theta }^2{A}_t)$. The Itô formula givesThe explicit decreasing exponential contributes half the finite-variation term; the other half is the quadratic correction from $e^{-i\theta X_t}$. Now the Itô product rule and part (b) yieldThe finite-variation terms cancel because $-i\theta (i\theta )={\theta }^2$. This proves the product is a local martingale. Moreover $|F_t|=e^{-{\theta }^2{A}_t/2}\le 1$ and $|M_t|\le 1$. The bounded local martingale criterion therefore proves the product is a true martingale.

At the terminal time, $M_T=e^{i\theta X_T}$, so the martingale from part (c) has terminal value $e^{-{\theta }^2⟨X{⟩}_T/2}$. Its initial value is $M_0$, since $X_0=⟨X{⟩}_0=0$. Taking expectations givesHere $M_0$ may be a nonconstant ${\mathcal{F}}_0$-measurable variable; the tower property of conditional expectation still gives $\mathbb{E}{M}_0=\mathbb{E}{e}^{i\theta X_T}$. This characteristic function under conditionally symmetric martingale increments identity relates the characteristic function of the terminal martingale to the Laplace transform of a nonnegative random variable given by its quadratic variation.

Fix $T>0$. The given normal distribution and part (d) implyOne can deduce determinism without any moment assumption on the bracket. Put $Z=e^{-⟨X{⟩}_T}$. Taking $\lambda =1,2$ gives $\mathbb{E}Z=e^{-T}$ and $\mathbb{E}{Z}^2=e^{-2T}$, so $\mathrm{Var}(Z)=0$. Therefore $⟨X{⟩}_T=T$ almost surely. Applying this at every rational time and using continuity of quadratic variation gives $⟨X{⟩}_t=t$ simultaneously for all $t\ge 0$ outside a single null set.

The Lévy characterization of Brownian motion states that a continuous local martingale starting at zero with this bracket is Brownian motion in its filtration. To see the independent-increment conclusion directly, the Itô formula shows that $e^{i\theta X_t+{\theta }^{2}t/2}$ is a martingale on any fixed bounded time interval: it is a local martingale with a deterministic bound on its modulus. ThusThe deterministic conditional characteristic function identifies an $N(0,t-s)$ increment independent of ${\mathcal{F}}_s$. Together with the given path continuity and $X_0=0$, this proves $X$ is Brownian motion.

Let $a(x)=\sigma (x)\sigma (x{)}^{\mathsf{T}}$. The multidimensional Itô formula gives the second-order diffusion generatorIndeed,The integrand is locally bounded along the continuous path after stopping on compact sets, because the coefficients and derivatives are continuous. Thus the final stochastic integral is a local martingale. No global growth or uniqueness assumption on this already-given weak solution of a stochastic differential equation is needed.

Apply the Itô product rule to the deterministic discount factor and $u(X_t)$:The prescribed differential equation makes the drift vanish. ThereforeThis is the discounted generator-eigenfunction martingale underlying the Feynman-Kac formula.

Continuity and adaptedness make the first boundary hit a stopping time. A continuous path starting in the open domain cannot leave it before meeting its boundary. Thus $X_{t\wedge T}\in \mathcal{D}\cup \partial \mathcal{D}$, with the usual interpretation when $T=\infty$. If $|u|\le K$ on this set, thenThe stopped process is a bounded local martingale; the bounded local martingale criterion makes it a martingale and, in fact, a uniformly integrable martingale. The Martingale convergence theorem gives almost sure and $L^1$ convergence as $t\to \infty$.

Its limit can also be identified pathwise. On $\{T<\infty \}$ the stopped process is eventually constant at $e^{-\lambda T}u(X_T)$. On $\{T=\infty \}$ its absolute value is at most $K{e}^{-\lambda t}$ and hence tends to zero. Thus the limit is $e^{-\lambda T}u(X_T)1_{\{T<\infty \}}$.

On the event $T<\infty$, continuity puts $X_T$ on the boundary, where $u=1$. The limit found in part (c) is therefore $e^{-\lambda T}$, defining this expression to be zero when $T=\infty$. Taking expectations and using the bounded convergence justified in part (c) givesThe conclusion does not require almost-sure finiteness of $T$. It is the discounted boundary-hitting representation for a bounded solution of $\mathcal{L}u=\lambda u$.

For the drifted Brownian motion $Y_t=y+B_t+at$ on $(0,\infty )$, the diffusion generator is $\mathcal{L}=\frac{1}{2}\frac{d^2}{d{z}^2}+a\frac{d}{dz}$. Seek a bounded solution of $\mathcal{L}u=\lambda u$ with $u(0)=1$. The exponential ansatz $u(z)=e^{rz}$ givesBecause $\lambda >0$, the plus root is positive and the minus root is negative. Boundedness on $[0,\infty )$ therefore selectsIt satisfies the boundary condition and all hypotheses of part (d). HenceThis is the first-passage Laplace transform for Brownian motion with drift, with $e^{-\lambda \infty }=0$. At $a=0$ it reduces to $e^{-y\sqrt{2\lambda }}$. As $\lambda \downarrow 0$, it gives $\mathbb{P}(S<\infty )=e^{-2y\max (a,0)}$, agreeing with certainty of hitting when the drift points towards zero and a possible escape when it points away.