Girsanov theorem

The Girsanov theorem describes how an absolutely continuous change of measure alters the drift of a semimartingale. In a Brownian version, let $\theta$ be predictable with ${\int }_0^T{\theta }_s^{2}ds<\infty$ almost surely, and supposeis a true martingale on $[0,T]$. Its initial value is one. Under $dQ=Z_{T}dP$, the process $W_t^Q=W_t+{\int }_0^t{\theta }_{s}ds$ is standard Brownian motion through time $T$. Novikov's condition is one sufficient test for the density assumption.^[ref]

For $d{X}_t=\mu dt+\sigma d{W}_t$ with constant $\sigma >0$, choosing $\theta =\mu /\sigma$ gives $d{X}_t=\sigma d{W}_t^Q$: the same paths have different statistical weights and zero drift under $Q$. Their quadratic variation is unchanged. A positive local martingale density is insufficient unless its expectation remains one, and consistent finite-horizon changes need not give an equivalent infinite-horizon measure.

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.

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 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.

On each finite horizon the displayed Radon-Nikodym derivative changes Wiener measure into the law of $dX=dB-\mu Xdt$. Stop at coordinate magnitude n so the Novikov condition applies. Under each stopped tilted measure the coordinate solves the drifted equation, and the Gronwall inequality with the Doob L2 maximal inequality bounds its maximal second moment uniformly in n. The probability of reaching n therefore tends to zero; this proves that the unstopped density has expectation one. No unrestricted large-time Novikov estimate is needed.

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.

On a fixed finite horizon, equivalent probability measures have the same semimartingales. For a strictly positive continuous Radon-Nikodym density martingale $Z$ and a continuous local martingale $L$ under the original measure, $L-\int Z^{-1}d[L,Z]$ is a local martingale under the new one, by the Girsanov theorem. Thus the finite-variation part of a continuous semimartingale decomposition changes by the opposite covariation correction.

For $dX=\mu dt+\sigma dB$ on $[0,T]$, take $\theta =(\nu -\mu )/\sigma$ and use density $\mathcal{E}(\int \theta dB{)}_T$. If $\mu ,\nu$ are bounded and $\sigma$ is continuous and bounded below by a positive constant, the exponential has bounded bracket and defines an equivalent measure. Under it, $\tilde{B}=B-\int \theta dt$ is Brownian and $dX=\nu dt+\sigma d\tilde{B}$. Set $\theta =0$ after $T$ to define the density on the entire original sigma-algebra and the new Brownian motion on the whole time axis.

For reference Brownian motion $R=x+W$, stopped at $T=t\wedge {\tau }_{(a,b)}$ with $0<a<x<b$, this expression is the Radon-Nikodym derivative of the stopped Bessel process law of dimension $\delta$. The bounded stopped drift verifies the Novikov condition. The comparison laws must have the same initial value and the same stopping rule.

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.

Let $Z$ be a positive uniformly integrable martingale defining $d\tilde{\mathbb{P}}=Z_{\infty }d\mathbb{P}$. If $Y$ is bounded and $ZY$ is a true $\mathbb{P}$-martingale, then $Y$ is a $\tilde{\mathbb{P}}$-martingale. This follows from the conditional-expectation change-of-measure identity.

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.

Let $Z$ be a positive uniformly integrable density martingale. If $Y$ is bounded and $ZY$ is a local martingale under the original measure, the product is a true martingale: its stopped absolute values are dominated by a fixed multiple of the uniformly integrable stopped density. Bayes then makes $Y$ a martingale under the new measure. Stop a locally bounded continuous $Y$ at absolute-value levels to obtain the local version.