Past exam of the stochastic differential equations course of the University of Oxford / 2025 / 2

For $t>0$, $B_t$ has the normal distribution $N(0,t)$. Evaluate its Gaussian integral:At $t=0$ the defining random variable is one, consistent with the same formula. Thus

Let $Y_t=-\beta {\int }_0^t{B}_{u}d{B}_u-{\int }_0^t{h}_{u}du$. Its quadratic variation is $[Y{]}_t={\beta }^2{\int }_0^t{B}_u^{2}du$. The Itô formula for $Z=e^Y$ givesChooseThe chosen process is continuous and adapted, hence progressively measurable. Its time integral is finite on compact intervals because Brownian paths are continuous. The resulting $Z$ is the stochastic exponential of $-\beta {\int }_0^{\cdot }{B}_{u}d{B}_u$, and is a positive local martingale.

The Itô formula for $B_t^2$ yields ${\int }_0^t{B}_{u}d{B}_u=(B_t^2-t)/2$. Substituting into the expression for $Z_t$ gives the pathwise identityConsequentlyTaking expected values provesThe random variable on the preceding line is bounded by one for $\alpha \ge 0$, so its integrability is immediate.

The expression obtained in part (a)(iii) gives a direct deterministic bound. For every fixed $T<\infty$ and $0\le s\le T$,Thus the bounded local martingale criterion makes $Z$ a true martingale on every finite time interval. More explicitly, if ${\sigma }_n$ is a localizing sequence, the identities $\mathbb{E}[Z_{t\wedge {\sigma }_n}\mid {\mathcal{F}}_s]=Z_{s\wedge {\sigma }_n}$ pass to the limit by dominated convergence, using the displayed bound for $t\le T$. ThereforeThis establishes exactly the finite-horizon property needed for the subsequent change of measure.

The relation between $B$ and $W$ is the linear equation $d{B}_s=d{W}_s-\beta B_{s}ds$ with $B_0=0$. Multiplication by the deterministic integrating factor $e^{\beta s}$ and the Itô product rule giveHenceThe second equality follows by semimartingale integration by parts. It expresses $B$ as an explicit function of the continuous path of $W$, using only an ordinary time integral. This inversion holds before changing measure; $W$ is already a continuous semimartingale under the original measure.

Fix $t<\infty$. Part (b)(i) makes $d{\mathbb{Q}}_t=Z_{t}d\mathbb{P}$ a probability measure on ${\mathcal{F}}_t$, equivalent to $\mathbb{P}$ since $Z_t>0$. Its density process on $[0,t]$ is $Z_s=\mathbb{E}[Z_t\mid {\mathcal{F}}_s]$. The Girsanov theorem, applied to the stochastic exponential with integrand $-\beta B_s$, says thatis standard Brownian motion under ${\mathbb{Q}}_t$. The required local square-integrability follows from path continuity, and the true martingale property was proved in part (b)(i).

By part (b)(ii), under ${\mathbb{Q}}_t$ the process $B$ is therefore the zero-started Ornstein-Uhlenbeck process:This identifies its full law on continuous path space, rather than just its one-time marginal. It is a centered Gaussian process, since it consists of deterministic Itô integrals against Brownian motion. For $r,s\in [0,t]$, the Itô isometry gives its covariance function:In particular, its Ornstein-Uhlenbeck transition distribution at the terminal time is

Put $c=\beta /2-\alpha$. The change of measure and part (a)(iii) yieldFor a centered normal random variable $G$ of variance $v>0$, a Gaussian integral gives the Gaussian quadratic exponential momentHere $c\le \beta /2$ and $\beta v_t<1/2$ for finite positive $t$, so the required positivity holds, including when $c>0$. ThusExpanding the two exponentials in terms of hyperbolic functions givesAt $t=0$ this is one. As $\beta \downarrow 0$, the denominator tends to $1+2\alpha t$, recovering part (a)(i). At $\alpha =0$ it gives the Laplace transform of the integrated square of Brownian motion. The joint terminal-and-integral transform is the Joint Laplace transform of Brownian square and its time integral.