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The scalar stochastic differential equation $d{X}_t=\tanh X_{t}dt+d{W}_t$ has a pathwise unique global strong solution, since its coefficients are globally Lipschitz. The positive martingale $e^{t/2}/\cosh X_t$ gives a Girsanov theorem change of measure under which $X_t-X_0$ is Brownian motion. For $X_0=x$, its transition density isThis is a Doob h-transform with $h=\cosh$, and a mixture of $N(x+t,t)$ and $N(x-t,t)$ with weights $e^x/(2\cosh x)$ and $e^{-x}/(2\cosh x)$.

Let $Z=\exp (X-⟨X⟩/2)$ and $dQ=Z_{T}dP$. The Girsanov theorem first makes $N=M-⟨M,X⟩$ a continuous local Q-martingale. Boundedness of $Y$ gives $⟨X{⟩}_T\le C^{2}T$ and ${\mathbb{E}}_P{Z}_T^2\le e^{C^{2}T}$: write $Z_T^2=\mathcal{E}(2X{)}_T{e}^{⟨X{⟩}_T}$ and apply Novikov's condition to $2X$.

The quadratic covariation bound gives ${\sup }_{s\le T}|⟨M,X{⟩}_s|\le C\sqrt{T}⟨M{⟩}_T^{1/2}$. For a square-integrable martingale $M$, both its maximal square moment and its expected terminal bracket are finite. Cauchy-Schwarz inequality with the density $Z_T$ therefore yields ${\mathbb{E}}_Q{\sup }_{s\le T}|N_s|<\infty$. The dominated limit of its stopped conditional identities proves the true martingale property. Mere P-integrability of $M_T$ does not suffice.

Girsanov's theorem describes how a change of probability measure changes the drift of a semimartingale. If the appropriate stochastic exponential is a true martingale, weighting by it turns $W_t+{\int }_0^t{\theta }_{s}ds$ into a Brownian motion under the new measure.

Here is the Brownian change-of-drift form of the Girsanov theorem. Let $W$ be a standard Brownian motion on $[0,T]$ and $\theta$ predictable with ${\int }_0^T{\theta }_s^{2}ds<\infty$ almost surely. Suppose the stochastic exponentialis a true martingale with $\mathbb{E}{Z}_T=1$. Define $dQ=Z_{T}dP$. Then $Q$ is equivalent to $P$ andThe same statement holds in $d$ dimensions with scalar products, squared norms and a vector drift. Novikov's condition $\mathbb{E}\exp (\frac{1}{2}{\int }_0^T{\theta }_s^{2}ds)<\infty$ is a standard sufficient condition for the true-martingale hypothesis. Bounded $\theta$ is enough for all applications below.

For the proof, Itô formula gives $d{Z}_t=-Z_t{\theta }_{t}d{W}_t$. The Itô product rule gives the cancellationThe two drift terms cancel because $d[Z,W{]}_t=-Z_t{\theta }_{t}dt$. Stop $W^Q$ when its absolute value reaches $n$, at a time ${\tau }_n$ capped by $T$. The same product calculation makes $Z_t{W}_{t\wedge {\tau }_n}^Q$ a local martingale under $P$. Its absolute value is at most $n{Z}_t$; the density martingale is uniformly integrable over this finite horizon, so the stopped product is a true martingale. Bayes' conditional identity then givesAs $n$ increases, these stopping times exhaust the horizon by continuity. Thus $W^Q$ is a continuous local martingale under $Q$. Adding a finite-variation drift does not change quadratic variation, so $[W^Q{]}_t=t$.

To prove the Brownian conclusion rather than just assert it, for every real $u$ apply Itô formula to $F_t=\exp (iu{W}_t^Q+u^{2}t/2)$. Its differential is $iu{F}_{t}d{W}_t^Q$, and its modulus is bounded on the fixed horizon, so it is a true martingale. ConsequentlyThe conditional characteristic function identifies a centered normal increment with variance $t-s$, independent of the past. Together with continuity and $W_0^Q=0$, this proves the Brownian assertion. For vector $u$, the same argument uses $\Vert u{\Vert }^2$ and gives the multivariate version. If $|\theta |\le K$, stopping $Z$ and applying Itô formula to $Z^2$ gives a uniform second-moment bound $\mathbb{E}{Z}_{t\wedge \tau }^2\le e^{K^{2}t}$ by Gronwall inequality; this proves uniform integrability of the stopped densities and the needed true-martingale property in the bounded case.

In the Black-Scholes model with its usual augmented Brownian filtration, write $d{S}_t=S_t(\alpha dt+\sigma d{W}_t)$, $B_t=e^{\rho t}$, with $\sigma >0$. Take $\theta =(\alpha -\rho )/\sigma$, the market price of risk. Under $Q$, the preceding change of drift givesso $S_t/B_t$ is a martingale. The Martingale representation theorem then makes integrable Brownian-market claims attainable and gives their risk-neutral pricing as discounted conditional expectations. The physical drift $\alpha$ has disappeared from the price dynamics.

For the joint terminal-value and maximum law, first use the Brownian reflection principle at level $a>0$. Reflecting a path after its first hit of $a$ preserves its law as a Brownian motion and sends a terminal value $y<a$ to $2a-y>a$; the symmetry and strong Markov property of the post-hit increments justify this map. Thus, with ${\phi }_t(y)=(2\pi t{)}^{-1/2}{e}^{-y^2/(2t)}$,Take the constant-drift exponential density $e^{\mu W_t-{\mu }^{2}t/2}$. By Girsanov theorem, the coordinate process under this tilted measure has the law of $W_s+\mu s$ under $P$. Consequently, for $t>0$ and $x\le a$,Completing the squares givesIntegrating provesHere $\Phi$ is the standard normal distribution function. The first variable is the terminal Brownian value, distinct from its running maximum.

Set $T=t_0$, $a=\log (b/S_0)/\sigma >0$ and $\mu =(\rho -{\sigma }^2/2)/\sigma$. Under the risk-neutral measure, $\log (S_s/S_0)/\sigma =W_s^Q+\mu s$. Taking $x=a$ in the joint law gives the probability of not hitting the upper level. Complementing it and discounting the unit payment at maturity gives the one-touch option priceThe exponential weight is also $(b/S_0{)}^{2\rho /{\sigma }^2-1}$. The payment occurs at $T$, even if the barrier was hit earlier, so its discount factor is $e^{-\rho T}$, not a hitting-time discount factor.

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.

Apply Itô's formula to $M_t=e^{N_t-⟨N{⟩}_t/2}$. The quadratic-variation correction cancels the finite-variation term:Localization of the integrand proves that $M$ is a nonnegative continuous local martingale.

For completeness, Conditional Fatou lemma says that for nonnegative random variables $Y_n$,Choose a localizing sequence ${\tau }_n\uparrow \infty$ such that each stopped $M$ is a martingale. Fatou first gives $\mathbb{E}{M}_t\le 1$. For $s<t$, its conditional form then givesThus $M$ is a supermartingale.

A sufficient condition for it to be a true martingale on $[0,T]$ is Novikov's conditionUnder this condition the stochastic exponential is a uniformly integrable martingale on that finite interval. Requiring it for every finite $T$ gives the martingale property at all finite times; it does not automatically give uniform integrability as $t\to \infty$.

The bounded integrand makes $X_s={\int }_0^s{Y}_{u}d{W}_u$ a square-integrable martingale, with quadratic variationThusNovikov's condition applies to the stochastic exponential $Z_s=\exp (X_s-⟨X{⟩}_s/2)$. Hence $Z$ is a true martingale with $\mathbb{E}{Z}_s=Z_0=1$. It is strictly positive because its exponent is finite almost surely.