Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 29 / 4

Put $U=M-\frac{1}{2}[M]$. Since the bracket is continuous finite variation, $[U]=[M]$. The Itô formula applied to $Z=e^U$ givesThereforeWith the printed exponential formula, $Z_0=e^{M_0}$. The usual unit-initial-value stochastic exponential is $\exp (M_t-M_0-\frac{1}{2}[M{]}_t)$; the two coincide when $M_0=0$, as in part (c).

For any strictly positive solution of this stochastic equation, its bracket is $d[Z{]}_t=Z_t^{2}d[M{]}_t$. Applying Itô to the logarithm, after localizing away from zero if necessary, givesThe same identity holds for $Z^{\prime }$, so $\log Z_t^{\prime }-\log Z_t$ is constant in time and is zero initially. Their continuous versions agree at all times on one full-probability event. Thus pathwise uniqueness for a multiplicative martingale equation givesThis is indistinguishability of stochastic processes, not just equality almost surely at each separately chosen time.

For $0\le s\le t$ and $E\in {\mathcal{F}}_s$, the definition of the Radon-Nikodym derivative givesThe density is nonnegative and integrable; with the probability normalization used here its expectation is one. Henceso $Z$ is a martingale. When $Z_t=h(X_t,t)$, the Itô formula under Wiener measure givesThe first term is a continuous local martingale, by localization on compact space-time sets. Since $Z$ is a martingale, uniqueness of the continuous semimartingale decomposition makes ${\int }_0^{t}g(X_s,s)ds=0$ identically. The integrand is continuous in time along every continuous path, so $g(X_t,t)=0$ for every $t$, outside one common null event.

For each fixed $t>0$, $X_t$ has a Gaussian distribution with strictly positive density everywhere on $\mathbb{R}$. Since $g(\cdot ,t)$ is continuous, $g(X_t,t)=0$ almost surely implies $g(x,t)=0$ for every real $x$. ThereforeIf the stated smoothness includes time zero, continuity extends the equation there. This is the backward heat equation. The full-support argument is needed to pass from an identity along Brownian paths to an equation at every spatial point.

Define $k(x,t)=h_x(x,t)/h(x,t)={\partial }_x\log h(x,t)$. Positivity and smoothness of $h$ make $k$ continuous and bounded on compact space-time sets. Brownian paths remain in compact spatial sets on each finite horizon, so ${\int }_0^{T}k(X_s,s{)}^{2}ds<\infty$ almost surely for every $T$. Thusis a well-defined continuous local martingale with $M_0=0$. The backward heat equation from part (b) removes the drift in Itô's formula and yieldsBy the allowed fact $Z_0=1$ and the uniqueness in part (a),

The finite-horizon form of the Girsanov theorem used here says: if $M_0=0$ is a continuous local martingale and its stochastic exponential $Z$ is a strictly positive true martingale of mean one on $[0,T]$, then under $d\mathbb{Q}{|}_{{\mathcal{F}}_T}=Z_{T}d\mathbb{P}{|}_{{\mathcal{F}}_T}$ every continuous $\mathbb{P}$-local martingale $N$ becomes the $\mathbb{Q}$-local martingale $N-[N,M]$. Quadratic variation is unchanged by this locally equivalent change of measure.

All the assumptions hold. The integrand defining $M$ is locally square integrable as checked above; $Z$ is strictly positive; and it is already a true martingale because it is the density process. On each finite horizon $Z_t=\mathbb{E}[Z_T\mid {\mathcal{F}}_t]$, so it is uniformly integrable on that horizon. There is no need to establish a separate Novikov condition, nor to assume equivalence on the infinite-time sigma-algebra.

Take $N=X$. Its covariation with $M$ is $[X,M{]}_t={\int }_0^{t}k(X_s,s)ds$. Hence $B_t=X_t-{\int }_0^{t}k(X_s,s)ds$ is a continuous $\mathbb{Q}$-local martingale starting at zero with $[B{]}_t=t$. The Lévy characterization of Brownian motion makes it Brownian under $\mathbb{Q}$. Since this argument holds on every finite horizon for the same process $B$,This is the Brownian change of measure by a positive space-time harmonic function, a Doob h-transform with drift ${\partial }_x\log h$.