Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 31 / 6 / Solution

Here is a finite-horizon Girsanov theorem statement with the sign fixed explicitly. Let $B$ be a Brownian motion under $P$, and let $\theta$ be predictable with ${\int }_0^T{\theta }_s^{2}ds<\infty$ almost surely. Ifis a true martingale with $\mathbb{E}{L}_T=1$, then the measure $Q$ defined by $dQ=L_{T}dP$ makes $B_t-{\int }_0^t{\theta }_{s}ds$ a Brownian motion up to $T$. The sufficient Novikov condition is $\mathbb{E}\exp (\frac{1}{2}{\int }_0^T{\theta }_s^{2}ds)<\infty$. Merely being a local martingale is not a sufficient density condition.

For a deterministic absolutely continuous function $h$ with $h(0)=0$ and derivative $g\in L^2[0,T]$, take $\theta =g$. Novikov's condition holds because its exponential is deterministic and finite. Under $Q$, the canonical process has the law of a Brownian motion plus $h$. Thus for every bounded measurable path functional $F$,This derives the Cameron-Martin theorem density formula from the Girsanov theorem, rather than assuming the two measures have the same drift convention. Replacing $h$ by $-h$ changes the sign of the stochastic integral.

The question's path space is the whole half-line, not just one finite horizon. When $g\in L^2(0,\infty )$, the density martingale satisfiesIt has a uniformly integrable limit $L_{\infty }=\exp ({\int }_0^{\infty }{g}_{s}d{B}_s-\Vert g{\Vert }_2^2/2)$ of mean one, strictly positive almost surely. Integrals here converge in $L^2$ and almost surely. The finite-time identities then identify $L_{\infty }$ as the density on the full path sigma-algebra, since finite-time cylinder events generate it. This is Cameron-Martin shifts on infinite Wiener path space. Finite-horizon absolute continuity of measures alone does not imply infinite-horizon absolute continuity of measures.