Past exam of the stochastic analysis and partial differential equations course of the University of Oxford

Source: Oxford 2025 C8.2 paper, parent PDF pages 8–10. The separate three-page extract has the cover on page 1 and the questions on pages 2–3.

Let $Z=(Z_n{)}_{n\ge 0}$ be a Markov chain with transition kernel $K$, and let $N$ be an independent Poisson process of constant rate $r>0$. A Pseudo-Poisson process is $X_t=Z_{N_t}$. On the Banach space $B_b(E)$ of bounded Borel measurable functions, with the supremum norm, writeIts infinitesimal generator, transition semigroup, and resolvent operator are respectivelyand, for $\lambda >0$,The transition semigroup averages $K^n$ against the Poisson distribution of the number of updates. Since $K$ is a Markov operator, $\Vert K\Vert \le 1$; both series converge in operator norm. Updates with $Z_{n+1}=Z_n$ are allowed and do not count as changes of the observed state. The degenerate rate $r=0$ gives $L=0$, $P_t=I$, and $R_{\lambda }=I/\lambda$.

The displayed identity requires a correction to the process construction in the source. In a piecewise-deterministic Markov process, the instantaneous hazard function is evaluated at the current flowing state $X_{u-}$, and the destination is sampled from $K_u(X_{u-},dy)$. Keeping the jump times and destinations of the original Markov jump process $Y$ instead evaluates both quantities at its last reset state $Y_{u-}$, which generally differs from $X_{u-}$. The source PDF confirms that these original jump times and destinations were retained. Consequently its definition does not in general produce the asserted transition kernel, or even a Markov process in the state variable $X$ alone.

For an explicit counterexample, take $a(t,x)=1$, $\alpha (x)=2+\tanh x$, $y_0=0$, and let every $K_t(x,dy)$ be the standard normal distribution, independently of $t,x$. These are smooth coefficients with bounded positive jump rate, so nonexplosion holds. In the literal construction, the first holding time has the exponential distribution of rate $2$. Thus the probability distribution of $X_t$ has an atom of mass $e^{-2t}$ at $t$. Paths with at least one jump contribute no atom there, because the reset has a normal density. The asserted identity instead assigns to this no-jump atom the masswhich differs for every $t>0$.

Here is the full derivation for the intended piecewise-deterministic Markov process. Assume nonexplosion; smoothness alone does not guarantee it. Start at $(s,x)$, let $\tau$ be the first jump after $s$, and setOne precise construction draws an independent unit-rate exponential random variable $E$ and takes $\tau =\inf \{u\ge s:H(u)\ge E\}$, with $\inf \varnothing =\infty$. This implements a hazard function along the flow of a vector field, rather than a fixed exponential holding time at the last reset. Its survival probability and jump-time probability density function areEquivalently, $S^{\prime }=-\alpha (z)S$ with $S(s)=1$. On $\{\tau >t\}$, the terminal state is $z(t)$. Conditional on $\tau =u$, draw the new state $y$ from $K_u(z(u),dy)$ and restart the same construction from $(u,y)$. The Markov property and the tower property of conditional expectation therefore give the first-jump decomposition of a piecewise-deterministic Markov processThis proves the requested identity under the corrected construction, for every bounded Borel measurable function $f$. The two contributions partition according to whether the first jump occurs before $t$; later jumps are already contained in $P_{u,t}$.

For comparison, the literal construction does have a different renewal identity when started at a reset with $X_s=Y_s=x$. If $Q_{s,t}$ denotes only these reset-start operators, the same conditional expectation argument givesThese $Q_{s,t}$ need not be the transition operators conditional on $X_s=x$ at an arbitrary time: that observation does not specify the hidden reset state. This distinction explains exactly why replacing only the between-jump paths is insufficient.

The identity for the killed deterministic flow follows directly from the chain rule and the fundamental theorem of calculus, independently of the construction issue in part (b). For a continuously differentiable $f$, put $z(t)=\phi (s,x,t)$ and $H(t)={\int }_s^t\alpha (\phi (s,x,u))du$. ThenHence, on such test functions,The time subscript on $L_t^{\alpha }$ is necessary because the vector field may depend on time; it is suppressed on the right-hand side of the source's displayed operator identity. The killed deterministic flow transports an observable along the flow of a vector field and multiplies it by the no-jump survival probability.

Use the corrected piecewise-deterministic Markov process from part (b). For a smooth test function $f$, its time-dependent infinitesimal generator isIndeed, over an interval of length $h$, deterministic motion contributes $ha(t,x)f^{\prime }(x)$, while a jump has probability $h\alpha (x)+o(h)$ and changes the conditional mean of $f$ by $K_{t}f(x)-f(x)$. Terms involving both drift and a jump contribute only to the remainder. Thus $P_{t,t+h}f=f+h{L}_{t}f+o(h)$.

The Dynkin formula, or this infinitesimal expansion integrated against the current probability distribution, gives the weak evolution identityTake $f\in C_c^{\infty }(\mathbb{R})$, so integration by parts has no boundary contribution. Assuming the displayed jump integrals are finite, Fubini's theorem givesConsequently the forward equation of a piecewise-deterministic Markov process isHere $A_t=L_t^{\ast }$ is the formal adjoint operator acting on probability density functions. Its terms are transport, loss from $x$, and incoming mass from every $y$. In particular, the incoming rate is $\alpha (y)$, evaluated at the departure state. The operator denoted $A$ in the source is generally time-dependent. Because $\int m_t(y,x)dx=1$, the gain and loss integrals cancel; with vanishing transport flux at infinity, $\frac{d}{dt}\int p_t=0$.

The deterministic initial state in part (b) does not actually yield an everywhere smooth full density. For finite times the no-jump contribution is the positive measurewhere ${\delta }_z$ is the Dirac measure at $z$. Smooth jump destination probability density functions cannot remove this positive atom. The density equation above is therefore conditional on the extra density hypothesis, for example with a suitable random initial probability density function. For the specified fixed initial state, write ${\nu }_t$ for the whole probability measure and use insteadThis weak measure evolution includes both the atom and any continuous component. It becomes the displayed partial differential equation whenever ${\nu }_t(dx)=p_t(x)dx$ with the assumed regularity. It does not describe the literal hidden-reset construction without the correction in part (b).

For the stochastic differential equation $d{X}_s=b(s,X_s)ds+\Sigma (s,X_s)d{B}_s$, the diffusion generator on smooth functions isOne form of the Feynman-Kac formula is the following. Suppose the stochastic differential equation is well posed and nonexplosive, $c\ge 0$ is bounded, the terminal data $g$ and source $h$ are bounded, and $F$ is a classical solution bounded on finite time intervals ofWith the regularity needed for the Itô formula, the parabolic Feynman-Kac formula with a source statesHere ${\mathbb{E}}_{t,x}$ means expectation for the diffusion process started at $X_t=x$. Under hypotheses ensuring existence of a classical solution, this representation also identifies that solution uniquely in the stated bounded class. In the remaining parts, $c=h=0$, so $F(t,x)={\mathbb{E}}_{t,x}g(X_T)$ solves the terminal-value Kolmogorov backward equation $F_t+{\mathcal{L}}_{t}F=0$. Complex terminal data are allowed by applying the real formula to their real part and imaginary part separately.

The diffusion generator is $\mathcal{L}=\mu {\partial }_x+\frac{1}{2}{\sigma }^2{\partial }_{xx}$. By the Feynman-Kac formula, the required conditional characteristic function solvesUse separation of variables in the form $F(t,x)=e^{iux}h(t)$. Since $F_x=iuF$ and $F_{xx}=-u^{2}F$, the temporal factor satisfies the ordinary differential equationThus $h(t)=\exp [(iu\mu -\frac{1}{2}{\sigma }^2{u}^2)(T-t)]$, andThis is bounded in absolute value by one, so it lies in the uniqueness class of the Feynman-Kac formula. As an independent check, integrating the stochastic differential equation gives $X_T=x+\mu (T-t)+\sigma (B_T-B_t)$. The independent Brownian increments have the normal distribution with mean zero and variance $T-t$, yielding exactly the same characteristic function. For $\sigma =0$, the conditional distribution is a point mass at $x+\mu (T-t)$, and the same boxed expression still applies.

Here the diffusion generator is $\mathcal{L}=ax{\partial }_x+\frac{1}{2}{\sigma }^2{\partial }_{xx}$. The Feynman-Kac formula givesTry $F(t,x)=\exp [iu\alpha (t)x+\beta (t)]$, where $\alpha (T)=1$ and $\beta (T)=0$. Substitution and collection of the coefficients of $x$ and the constant term giveThe solutions of these ordinary differential equations, with the terminal conditions imposed, areFor $\Delta =T-t$, define the conditional varianceThen the required conditional characteristic function isIn particular, for $a\ne 0$ this exponent is $iu{e}^{a\Delta }x-\frac{{\sigma }^2{u}^2}{4a}(e^{2a\Delta }-1)$. The variance $v_{\Delta }$ is nonnegative for either sign of $a$, and its limit at $a=0$ is ${\sigma }^2\Delta$. Hence the solution is bounded in absolute value by one and the Feynman-Kac formula identifies it with the required expectation. The value $u=0$ gives one directly, including that case without dividing by $u$.

For an independent check, the integrating factor $e^{-as}$ and the Itô product rule giveThe stochastic integral has a normal distribution because its integrand is deterministic; the Itô isometry gives its variance as $v_{\Delta }$. For $a<0$ this is the Ornstein-Uhlenbeck transition distribution. The formula also covers the unstable case $a>0$, and degenerates to deterministic motion when $\sigma =0$.

For $a<0$, set $v_{\infty }=-{\sigma }^2/(2a)>0$. The candidate invariant probability law of a Markov process isTo prove invariance rather than merely guess a limiting probability distribution, initialize $X_t\sim \pi$ independently of future Brownian increments. Integrating the conditional characteristic function from part (i) against this centered normal distribution givesbecause $v_{\Delta }=v_{\infty }(1-e^{2a\Delta })$. The uniqueness theorem for characteristic functions implies $X_{t+\Delta }\sim \pi$ for every $\Delta \ge 0$. This proves $\pi P_{\Delta }=\pi$, as required. The time-homogeneous Markov property then also makes the process with this Ornstein-Uhlenbeck stationary initialization a strictly stationary process.

Let $E$ be the underlying Banach space. By the Laplace-transform formula for a semigroup resolvent,These are Bochner integrals, defined for every $f\in E$ and $\lambda >0$. The contraction semigroup bound and semigroup property giveInterchanging and regrouping these Bochner integrals is justified by the integrable bound $e^{-\lambda s-\mu t}\Vert f\Vert$, whose integral is $\Vert f\Vert /(\lambda \mu )$. For $\lambda \ne \mu$, set $v=s+t$ and evaluate the inner scalar integral:Since $f$ was arbitrary, the resolvent identity follows:For $\lambda =\mu$, both sides vanish, so that case holds as well.

For $\lambda \ne \mu$, apply the resolvent identity from part (i) in both orders:The two quotients are equal. For $\lambda =\mu$ the assertion is immediate. HenceThus resolvent operators from the same strongly continuous semigroup commute. This conclusion does not require commuting arbitrary bounded linear operators; it uses the fact that both arise from the same semigroup generator.

A second-order nondivergence-form elliptic operator on a domain $D\subseteq {\mathbb{R}}^d$ has the formOnly the symmetric part of $a$ affects the Hessian matrix term, so one may take $a(x)$ symmetric. The operator is a uniformly elliptic operator if there is a constant $\kappa >0$, independent of $x$, such thatA common bounded-coefficient convention additionally requires an upper bound $\sum a_{jk}(x){\xi }_j{\xi }_k\le \Lambda |\xi {|}^2$ with $\Lambda <\infty$. The lower bound says that the principal quadratic form cannot degenerate in any direction or at any location.

For an explicit stochastic example, standard $d$-dimensional Brownian motion has infinitesimal generatorIts principal quadratic form is $|\xi {|}^2/2$, so it is uniformly elliptic with $\kappa =\Lambda =1/2$. More generally, a sufficiently regular stochastic differential equation $d{X}_t=b(X_t)dt+\Sigma (X_t)d{B}_t$ has diffusion generator $b\cdot \nabla +\frac{1}{2}(\Sigma {\Sigma }^{\mathsf{T}}):D^2$; it is uniformly elliptic when $\Sigma {\Sigma }^{\mathsf{T}}$ has a uniformly positive lower bound. A conservative diffusion process has no zeroth-order term, since its infinitesimal generator annihilates constants.

The convergence is in the strong operator topology: for each fixed $f\in E$, one must prove $\Vert \lambda R_{\lambda }f-f\Vert \to 0$. First, the contraction semigroup bound givesThus $\lambda R_{\lambda }$ is a contraction. This is the useful scaling of the contraction estimate in the source's hint. Since ${\int }_0^{\infty }\lambda e^{-\lambda t}dt=1$,Fix $\epsilon >0$. By strong continuity, choose $\delta >0$ so that $\Vert P_{t}f-f\Vert <\epsilon$ for $0\le t\le \delta$. Split the integral there and use $\Vert P_{t}f-f\Vert \le 2\Vert f\Vert$ for the tail:Let $\lambda \to \infty$ and then $\epsilon \downarrow 0$. ThereforeThe distinction from operator norm convergence is essential. For example, the multiplication semigroup on ${\ell }^2(\mathbb{N})$ defined by $(P_{t}f{)}_n=e^{-nt}{f}_n$ is a strongly continuous semigroup of contractions: coordinatewise convergence and dominated convergence applied to ${\sum }_n|f_n{|}^2$ prove strong continuity. Its resolvent operator is $(R_{\lambda }f{)}_n=f_n/(\lambda +n)$, andThus the claimed convergence cannot in general be read as convergence in operator norm.

Take $f$ in the generator domain $D(L)$. The semigroup commutation with its unbounded generator implies $P_{t}f\in D(L)$ andIndeed, applying the bounded operator $P_t$ to the difference quotient at zero gives the right derivative $P_{t}Lf$; strong continuity gives the full orbit derivative for $t>0$. Now integration by parts in the Bochner integral yieldsThe boundary term at infinity vanishes because $\Vert e^{-\lambda t}{P}_{t}f\Vert \le e^{-\lambda t}\Vert f\Vert$. The derivative integral exists as a Bochner integral, since its integrand is bounded in norm by $e^{-\lambda t}\Vert Lf\Vert$.

Multiplying this identity by $\lambda$ givesPart (b) applies to the fixed vector $Lf\in E$, soThe bounded operators $L_{\lambda }={\lambda }^2{R}_{\lambda }-\lambda I$ are the Yosida approximation of the semigroup generator. The proof only applies the unbounded operator $L$ to $f\in D(L)$; the contraction and limit arguments then act on the ordinary vector $Lf$.