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

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.

There is no density on the full infinite path space; the two laws are mutually singular. Under Wiener measure, $B_t/t\to 0$ almost surely. Under the shifted law, $(B_t+t)/t\to 1$ almost surely, so the same measurable tail event separates the laws.

For completeness the Brownian limit follows from the Gaussian bound at integers and the maximal bound on each interval $[n,n+1]$: for every fixed $\eta >0$, both ${\sum }_n\mathbb{P}(|B_n|>\eta n)$ and ${\sum }_n\mathbb{P}({\sup }_{0\le s\le 1}|B_{n+s}-B_n|>\eta n)$ are finite. The Borel-Cantelli lemma and continuity fill the gaps between integers. Intersect over positive rational $\eta$. This proves the limit and the claimed singularity, which is infinite-horizon singularity of Brownian motion with constant drift.

On every finite horizon there nevertheless is the density $e^{B_T-T/2}$. Its failure to give a global density is exactly the distinction required by the printed half-line path space.

The deterministic shift has derivative $g_s=-1_{[0,1]}(s)$ and finite total energy one. The infinite-horizon Cameron-Martin formula therefore givesThe integral of this density is one by the moment-generating function of $B_1\sim N(0,1)$. It is strictly positive, so the shifted law and Wiener measure are equivalent on the whole half-line. The density depends only on the path through time one; after that time the shift is constant.

There is no density; scaling by two gives a law singular to Wiener measure. On $[0,1]$ form the pathwise dyadic sumsUnder Wiener measure, independent normal increments give $\mathbb{E}{Q}_n=1$ and $\mathrm{var}(Q_n)=2^{1-n}$. Chebyshev's inequality and the Borel-Cantelli lemma show $Q_n\to 1$ almost surely, because the variances are summable. For $2B$, every sum is four times its Brownian counterpart and hence tends to four almost surely. The event that the limit equals one separates the two laws. Unlike the linear-drift case, this obstruction already occurs on a finite interval: it is the Pathwise quadratic variation distinguishes Brownian speeds argument.

The PDF's subscript is $0$, and the Brownian motion starts from zero. Thus $B_0=0$ almost surely and the process is exactly $B$ at every time. Its law is Wiener measure itself, with density $1$. There is no random future coefficient or anticipating transformation in this printed case.