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For a zero-start continuous local martingale $M$ with $[M{]}_{\infty }\le C$ deterministically, its stochastic exponential is a uniformly integrable martingale. For each $p>1$, ${\sup }_t\mathbb{E}\mathcal{E}(M{)}_t^p\le e^{p(p-1)C/2}$. Its terminal value is strictly positive, so it defines an equivalent infinite-horizon change of measure by the Girsanov theorem.

Under $P$, let $X$ be standard Brownian motion started at zero and let $T$ be its first exit from $(-1,1)$. The Itô formula makesthe stochastic exponential with integrand $X{1}_{[0,T]}$. Since $Z_t\le e^{1/2}$, it is a uniformly integrable martingale and defines $dQ=Z_{T}dP$ on the whole original sigma-algebra. The Girsanov theorem makes $X_t-{\int }_0^{t\wedge T}{X}_{s}ds$ a Brownian motion under $Q$.

At exit, $|X_T|=1$ and ${\int }_0^T{X}_s^{2}ds\le T$, so $e^{1/2-T}\le Z_T\le e^{1/2}$. On $\{T\le t\}$, integrate these likelihood bounds in $Q(T\le t)={\mathbb{E}}_P[Z_T{1}_{\{T\le t\}}]$ to obtainA lower bound on $Z_T=dQ/dP$ gives an upper bound on $P$ in terms of $Q$; reversing that direction gives an invalid comparison.

Let $W$ be a P-Brownian motion on $[0,1]$ and tilt by the bounded constant drift integrand one, so $dQ=e^{W_1-1/2}dP$. Then $W_s^Q=W_s-s$ is Q-Brownian motion, either by Girsanov theorem or direct Gaussian exponential tilting. Set $a=1/2$,The density of $W_a$ under P is ${\pi }^{-1/2}{e}^{-x^2}$, so ${\mathbb{E}}_{P}H={\pi }^{-1/2}\int (1+x^2{)}^{-1}dx<\infty$. Independence of future Brownian increments makes $M$ a continuous true P-martingale. Under Q, that density is multiplied by $e^{x-1/4}$, so ${\mathbb{E}}_{Q}H=\infty$.

For $s>a$, its compensated local martingale isThe future Q-Brownian increment is independent of the positive variable $H$ and has positive finite mean absolute value. Hence the displayed variable has infinite expected absolute value under Q and is not a true martingale. This shows why the general Girsanov theorem promises a local martingale, while a true-martingale conclusion needs an additional integrability condition such as Girsanov compensation with square-integrable input.

If $h>0$ is smooth and $Z_t=h(X_t,t)$ is the unit-mean density martingale of a locally absolutely continuous probability measure relative to Wiener measure, then $h_t+\frac{1}{2}{h}_{xx}=0$. The Itô formula gives $Z=\mathcal{E}(\int (h_x/h)(X_s,s)d{X}_s)$ when $Z_0=1$. The true-martingale hypothesis is already supplied by the density process. On each finite horizon the Girsanov theorem makes $X-\int (h_x/h)(X_s,s)ds$ Brownian under the new measure. Positivity provides local equivalence; global infinite-horizon equivalence and the Novikov condition are not additional requirements.

For $0<x<M$, a three-dimensional Bessel process stopped at $M$ has the law of Brownian motion conditioned to hit $M$ before zero and stopped at $M$. On $(\epsilon ,M)$, the Girsanov theorem and the Itô formula give terminal density $X_{\tau }/x$. Its value is $M/x$ at the upper exit and $\epsilon /x$ at the lower exit. Letting $\epsilon \downarrow 0$ in bounded stopped-path functionals gives density $(M/x)1_{\{T_M<T_0\}}$ relative to the corresponding stopped Wiener measure. The exit probability is $x/M$ by the optional stopping theorem, so the density is normalized. On the unstopped full path space the same density defines the conditioned Brownian law; pushing it forward by the stopping map gives the stopped Bessel law. The already-stopped law is not absolutely continuous relative to unstopped Wiener measure, since eventually constant paths have Wiener probability zero.

The scalar stochastic differential equation $d{X}_t=\tanh X_{t}dt+d{W}_t$, $X_0=x$, has a global strong stochastic solution and pathwise uniqueness because its coefficients are globally Lipschitz continuous. The Itô formula gives the normalized positive martingaleBoundedness of $\tanh$ verifies Novikov's condition. Under $dQ=Z_{T}dP$, the Girsanov theorem makes $X_t-x$ a Brownian motion through time $T$. The factor $\cosh x$ ensures $Z_0=1$.

The transition density under the original measure isThis is a Doob h-transform with $h=\cosh$. Expanding the hyperbolic cosine gives a mixture of $N(x+t,t)$ and $N(x-t,t)$ with weights $e^x/(2\cosh x)$ and $e^{-x}/(2\cosh x)$, showing directly that the density integrates to one.

For a positive level $a$, this density describes the first-passage time of $B_t+ct$. The Girsanov theorem multiplies the zero-drift density by the stopped likelihood $e^{ca-c^{2}t/2}$. Its total mass is one for nonnegative drift and $e^{2ac}$ for negative drift; the latter has remaining mass at infinite hitting time.

For $a>0$ and $t>0$, the distribution function is $\Phi ((a-\nu t)/\sqrt{t})-e^{2a\nu }\Phi ((-a-\nu t)/\sqrt{t})$. The Brownian reflection principle gives the zero-drift terminal density below the barrier as ${\varphi }_t(y)-{\varphi }_t(2a-y)$ for $y<a$. Weight it by $e^{\nu y-{\nu }^{2}t/2}$ using the Girsanov theorem and complete squares in the two Gaussian integrals.

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.

For bounded predictable $\sigma$ on $[0,T]$, the stochastic exponential $Z_s=\exp (\frac{1}{2}{\int }_0^s{\sigma }_{u}d{W}_u-\frac{1}{8}{\int }_0^s{\sigma }_u^{2}du)$ is a positive martingale by the Novikov condition. Under $dQ/dP=Z_T$, the Bayes formula for conditional expectation givesIndeed, the square-root stock ratio is $(Z_T/Z_t)\exp (-\frac{1}{8}{\int }_t^T{\sigma }_u^{2}du)$. This exact identity gives the bounds on square-root stock implied volatility by bounding integrated variance pathwise. The Girsanov theorem gives $W^Q=W-\frac{1}{2}\int \sigma du$. The measure is an auxiliary pricing identity for this payoff, not a claim that it is an equivalent martingale measure for the original stock.

The Brownian transition semigroup identity proves that this positive density process has mean one for $t<1$. The backward heat equation and Itô's formula give $d{M}_t=M_t(y-B_t)(1-t{)}^{-1}d{B}_t$. On every fixed horizon $T<1$, Girsanov theorem under $dQ=M_{T}dP$ makes $B_t-{\int }_0^t(y-B_s)(1-s{)}^{-1}ds$ Brownian motion. Thus the original coordinate acquires the bridge drift coefficient toward $y$ at time one, without defining an equivalent endpoint density at the singular time.

The laws on continuous paths of a standard Brownian motion and a Brownian motion with drift $\mu \ne 0$ are mutually singular measures on the whole half-line. The event ${\lim }_{t\to \infty }\omega (t)/t=0$ has full measure for the former and zero measure for the latter by the strong law for Brownian motion. On every fixed finite horizon, their laws are equivalent by the Girsanov theorem. The finite-horizon densities $e^{\mu B_t-{\mu }^{2}t/2}$ tend to zero almost surely while retaining expectation one, so they lack uniform integrability over all time.

For an integrable density $Z=dQ/dP$, $Z\ge 0$ and ${\mathbb{E}}_{P}Z=1$, its density process is $Z_t={\mathbb{E}}_P[Z\mid {\mathcal{F}}_t]$. If $Z_0=1$, then $P$ and $Q$ agree on ${\mathcal{F}}_0$, so every ${\mathcal{F}}_0$-measurable random variable integrable under $P$ remains integrable under $Q$. This covers the normalized stochastic exponential densities used in the Brownian Girsanov theorem.

Without that initial normalization the assertion can fail. On the positive integers with the full sigma-algebra already known at time zero, take $P(n)=2^{-n}$ and $Q(n)=c/n^2$, where $c=({\sum }_{n\ge 1}{n}^{-2}{)}^{-1}$. The constant-in-time process $X_t=n$ is a martingale under $P$, but ${\mathbb{E}}_Q{X}_0=c{\sum }_{n\ge 1}1/n=\infty$. The density process $Z_t=Q(n)/P(n)$ is constant in time with random $Z_0$ and mean one. It is not the Doléans-Dade exponential of a constant process, which would start at one and remain one. Thus drift-corrected zero-starting increments and integrability of an added initial variable are separate issues.

For standard Brownian motion and $\alpha ,t\ge 0$, $\beta >0$, the title gives the joint Laplace transform of the terminal square and the integrated square of Brownian motion. At $\beta =0$ the answer is $(1+2\alpha t{)}^{-1/2}$. To prove the formula, use the stochastic exponentialOn every finite interval $[0,T]$, $0<Z_t\le e^{\beta T/2}$, so the bounded local martingale criterion makes it a martingale. The Girsanov theorem changes $B$ into the zero-started Ornstein-Uhlenbeck process $dB=dW-\beta Bdt$ under the terminal density $Z_t$. Its terminal variance is $v=(1-e^{-2\beta t})/(2\beta )$. Multiplying $Z_t$ by $\exp [(\beta /2-\alpha )B_t^2-\beta t/2]$ recovers the integrand on the left of the title. The Gaussian quadratic exponential moment then gives $e^{-\beta t/2}[1-(\beta -2\alpha )v{]}^{-1/2}$, which simplifies to the title formula. At $\alpha =0$ this reduces to the Laplace transform of the integrated square of Brownian motion after Brownian scaling.

For constant real $a,b$ and $\sigma >0$, the stochastic exponential with integrand ${\theta }_t=(a-b)X_t/\sigma$ defines an equivalent change of measure on every finite horizon. The Girsanov theorem then changes the drift of this linear stochastic differential equation from $aX$ to $bX$.

A localization argument verifies the necessary unit expectation. Stop at ${\tau }_m=\inf \{t:|X_t|\ge m\}$. The stopped integrand is bounded, so its exponential density defines a probability ${\mathbb{Q}}_m$. Under this probability the stopped process has drift $bX$; the usual second-moment maximal estimate and Gronwall inequality give ${\mathbb{E}}_{{\mathbb{Q}}_m}[{\sup }_{u\le T}|X_{u\wedge {\tau }_m}{|}^2]\le C_T$ independent of $m$. Thus ${\mathbb{Q}}_m({\tau }_m\le T)\le C_T/m^2$. On $\{{\tau }_m>T\}$, the stopped and unstopped densities agree, so the expectation of the unstopped density is at least ${\lim }_m{\mathbb{Q}}_m({\tau }_m>T)=1$. A nonnegative local martingale has expectation at most one, proving equality. The density is strictly positive because each finite-horizon continuous path has finite $\int {\theta }^{2}dt$. This establishes equivalence without asserting that the stronger Novikov condition necessarily holds for every horizon.

Set ${\theta }_t=(\mu -r)S_t/\sigma$. The Girsanov theorem uses the densityFor these nonexplosive linear stochastic differential equations, the linear drift change of measure ensures that $Z$ is a true martingale on each finite horizon. This matters: a stochastic exponential being a local martingale alone would not prove that its terminal expectation is one. Its strict positivity gives an equivalent risk-neutral measure $\mathbb{Q}$, under whichThe discounted asset is then a martingale. The Feynman-Kac formula, equivalently risk-neutral valuation of the replicable claim, givesHere the payoff must meet the usual integrability and replication assumptions; the bounded digital payoff below does.

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.

The entire term structure is naturally indexed by observation time and maturity. Work under a risk-neutral measure $Q$ and let $f(t,T)$ be the instantaneous forward rate, with $0\le t\le T$. Define the short rate $r_t=f(t,t)$, the bank account $B_t=\exp ({\int }_0^t{r}_{s}ds)$ and the zero-coupon bond priceAn arbitrary Gaussian mean and covariance need not make these bond prices consistent. Absence of arbitrage requires the discounted traded bonds to be martingales under a suitable equivalent measure. We derive the constraint explicitly.

Consider the Gaussian forward-rate field $f(t,T)=m(t,T)+X(t,T)$ with deterministic mean and centered Gaussian random field satisfyingThe maturity kernel is symmetric, and its increments in the first parameter must be positive semidefinite kernels. Assume continuity, separability and sufficient integrability for the following mean-square integrals and maturity derivatives. This covariance gives independent observation-time increments, while permitting correlated fluctuations across all maturities. The filtration contains the history of the whole observed forward curve.

For a fixed maturity $T$, collect the past short rate and the present curve into the integrated Gaussian forward-rate processThen $P(t,T)/B_t=e^{-A(t,T)-Y(t,T)}$. Its centered Gaussian variance isFor $s<t$ and an observed field coordinate $X(z,U)$ with $z\le s$, the covariance of $Y(t,T)-Y(s,T)$ with it is zero: the integrand isThe uncorrelated jointly Gaussian variables are independent principle therefore makes the increment independent of the earlier field filtration. Its variance is $v(t,T)-v(s,T)$, since the same covariance calculation gives $\mathrm{Cov}(Y(t,T),Y(s,T))=v(s,T)$. Conditional Gaussian exponential expectation now givesHence the exact bond-martingale condition isDifferentiating in maturity, using symmetry of the covariance, gives the Gaussian forward-rate covariance drift restrictionThis is necessary and sufficient. For sufficiency, the derivatives of the half-variance identity agree by the displayed restriction. Its integration constant also agrees at $T=t$, becauseThus the full identity follows for every $T\ge t$, and the conditional expectation calculation proves a true martingale, not merely zero formal drift. The initial mean $f(0,T)=-{\partial }_T\log P(0,T)$ fits the observed initial curve exactly; thereafter the mean adjustment is fixed by the covariance.

If $c(t;T,U)={\int }_0^t{k}_s(T,U)ds$ is time-differentiable, the forward drift is $a(t,T)={\int }_t^T{k}_t(T,u)du$. Writingwith deterministic finite- or Hilbert-space factor loadings gives the Heath-Jarrow-Morton modelWith $\Sigma (t,T)={\int }_t^T\sigma (t,u)du$, Itô formula independently verifies bond drift $r_t-{\int }_t^{T}a(t,u)du+\Vert \Sigma (t,T){\Vert }^2/2=r_t$. Finite-factor models give low-rank maturity correlations; a genuine random field or an infinite factor space permits a much richer covariance structure. Gaussian forward rates and short rates may be negative, although the exponential bond prices stay positive. Under a physical measure, a Girsanov theorem market-risk-premium change adds $⟨\sigma ,\lambda ⟩$ to the forward drift; deterministic premia preserve Gaussianity, whereas arbitrary adapted premia need not.

For a concrete two-parameter example, let $X$ be a standard Brownian sheet, so $c(t;T,U)=t\min (T,U)$. The restriction givesIndeed ${\int }_0^T\min (t,u)udu=t^3/3+t(T^2-t^2)/2$. This demonstrates directly how a Gaussian random field determines the compensating forward drift.

For the one-factor constant-volatility case, $\sigma (t,T)=\eta$, the Constant-coefficient Ho-Lee model hasandThese follow by integrating the explicit curve, not by assigning the short-rate drift independently. Exponentially decaying loadings $\eta e^{-\kappa (T-t)}$ instead give a Gaussian mean-reverting Hull-White model, with bond-rate loading $(1-e^{-\kappa (T-t)})/\kappa$.

A practical advantage is an explicit Gaussian bond-option formula. For a call of strike $K>0$ exercised at $T$ on a bond maturing at $U>T$, use the T-forward measure with numéraire $P(t,T)$. The forward bond ratio $F_t=P(t,U)/P(t,T)$ has a lognormal distribution under that measure, with remaining integrated varianceIts martingale property makes the conditional log mean $\log F_t-q/2$. Evaluating the positive-part lognormal integral givesFor $q=0$, the limit is $(P(t,U)-KP(t,T){)}^{+}$. The forward measure accounts for the stochastic short rate; discounting by one deterministic interest rate would generally be wrong. Calibration selects a positive-semidefinite covariance structure fitting observed rate co-movements and option prices, with the no-arbitrage mean restriction imposed afterward. With more independent factors than the available traded bond exposures span, the market is incomplete and additional claims need a specified pricing measure or risk-premium model. The Gaussian field is therefore a flexible curve model with explicit consistency and pricing equations, rather than a claim that all possible rate risks are hedgeable.

Let ${\varphi }_T(y)=(2\pi T{)}^{-1/2}{e}^{-y^2/(2T)}$ be the normal distribution density. The Brownian reflection principle gives the killed endpoint densityIndeed, reflection after the first hit of $a$ maps paths ending below $a$ that have hit the barrier to paths ending at $2a-y$; symmetry gives ${\varphi }_T(2a-y)={\varphi }_T(y-2a)$.

By the Girsanov theorem, weighting the path law by $\exp (\nu W_T-{\nu }^{2}T/2)$ makes the coordinate process a Brownian motion with drift $\nu$. Completing the two squares yieldsIntegration up to $x\le a$ proves the joint endpoint and maximum law for drifted Brownian motion:Here $\Phi$ is the standard normal distribution function. Setting $x=a$ also gives a continuous barrier distribution, so equality at the barrier has zero probability.

For the up-and-in option, use the risk-neutral measure of the Black-Scholes model with interest rate $\rho$ and spot volatility $\sigma >0$. ThenAbove $a$, the barrier has certainly been reached. Below $a$, subtracting the killed density from the unrestricted endpoint density leaves $e^{2a\nu }{\varphi }_T(y-2a-\nu T)$. Thus the discounted expected value of its payoff isIn the second integral put $z=y-2a$, and defineThen $S_0{e}^{\sigma y}=S_0{e}^{\sigma z}/\kappa$, $z\le -a$ means $S_0{e}^{\sigma z}\le \kappa b$, and $e^{2a\nu }={\kappa }^{-\nu /\sigma }$. Both integrals now use the unrestricted terminal stock distribution. This proves that the reflected European payoff for an up-and-in option isTheir initial risk-neutral pricing values agree for payoffs with finite absolute expectations in these integrals. The identity concerns prices; the two terminal payoffs need not agree along individual paths.

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.