Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 202 / 2

On the same measurable space, absolute continuity of measures meansThe Radon-Nikodym theorem then supplies a nonnegative measurable density $Z$ with ${\mathbb{E}}_{\mathbb{P}}Z=1$ and $\tilde{\mathbb{P}}(A)={\mathbb{E}}_{\mathbb{P}}[Z{1}_A]$. The converse follows directly from this formula. Strict positivity of the Radon-Nikodym derivative gives equivalence of the two probability measures; absolute continuity of measures alone need not give equivalence.

For a continuous local martingale $M$, its stochastic exponential isThe Itô formula gives $d{Z}_t=Z_{t}d{M}_t$ and $Z_0=1$, so $Z$ is a strictly positive local martingale. The finite-horizon Girsanov theorem says: if $Z$ is a true martingale on $[0,T]$ and $d\tilde{\mathbb{P}}=Z_{T}d\mathbb{P}$ on ${\mathcal{F}}_T$, then for every continuous local martingale $N$ under $\mathbb{P}$,For an infinite-horizon change of measure, assume $Z$ is a uniformly integrable martingale and use its terminal density $Z_{\infty }$. In particular, if $M=\int \theta dB$, the process $\tilde{B}=B-\int \theta ds$ is Brownian motion under the new probability measure, by the Lévy characterization of Brownian motion.

The quadratic variation cannot change. Adding the continuous finite-variation process $-[N,M]$ leaves $[\tilde{N}]=[N]$. Also the same pathwise partition sums defining $[N]$ have the same limit under an absolutely continuous change of measure; convergence in probability transfers by integrability of the density. Thus the bracket of the unchanged process $N$, now a semimartingale, is also unchanged. The local martingale property itself need not survive. This is quadratic covariation under an absolutely continuous measure change.

The Novikov condition holds on every horizon, sinceTherefore $Z=\mathcal{E}(M)$ is a true martingale and the Girsanov theorem applies through each finite $T$.

The deterministic bound also gives a genuine infinite-horizon density. For any $p>1$, the stochastic exponential identity isA nonnegative local martingale is a supermartingale, soThis $L^p$ bound gives uniform integrability. Also $M$ is a L2-bounded continuous martingale, since localization and the Itô isometry give $\mathbb{E}{M}_t^2=\mathbb{E}[M{]}_t\le C$. The L2 martingale convergence theorem gives a finite limit $M_{\infty }$; hence $Z_{\infty }=\exp (M_{\infty }-[M{]}_{\infty }/2)>0$ and $\mathbb{E}{Z}_{\infty }=1$. Weighting by $Z_{\infty }$ defines an equivalent probability measure on ${\mathcal{F}}_{\infty }$. This is the bounded-bracket criterion for a stochastic exponential.

The reference Brownian motion must start at $x$, and both laws must use the same stopped-path rule. Under the reference probability measure let $R_s=x+W_s$, let $\tau$ be its exit time from $(a,b)$, and put $T=t\wedge \tau$ and $c=(\delta -1)/2$. The Girsanov density for a stopped Bessel process, as a functional of the stopped reference path, isUse the predictable process ${\theta }_s=c{R}_s^{-1}{1}_{\{s\le \tau \}}$, with any harmless choice at the endpoint. On $[0,t]$, $\int {\theta }_s^{2}ds\le c^{2}t/a^2$. The Novikov condition holds; under the density $D$, the Girsanov theorem makes ${\tilde{W}}_s=W_s-{\int }_0^s{\theta }_{u}du$ a Brownian motion. Before exit, $d{R}_s=c{R}_s^{-1}ds+d{\tilde{W}}_s$, which is the required stochastic differential equation. The drift is a locally Lipschitz function on $(0,\infty )$, so pathwise uniqueness and uniqueness in law hold up to this exit time.

For a version involving just the endpoint and an ordinary integral, the Itô formula for $\log R$ givesThusThe stopped density is measurable from the stopped path, so it is also the Radon-Nikodym derivative of the stopped-path laws. The stopping times in this formula are evaluated on the reference path, rather than taken from a different realization of $X$. A zero-start reference Brownian motion would have a different deterministic initial value and could not dominate this law.