Doléans-Dade exponential ($\mathcal{E}(X)$)

For a continuous semimartingale $X$, its Doléans-Dade exponential isIt solves $d{Z}_t=Z_{t}d{X}_t$ with $Z_0=1$.

The Itô formula gives $d\log Z=dM-\frac{1}{2}d[M]$ for a strictly positive solution. Therefore equal initial values give indistinguishable solutions. Conversely the displayed exponential satisfies the equation. This is uniqueness for a fixed continuous local martingale driver and does not require the exponential to be a true martingale.

For a strictly positive continuous semimartingale $Z$, its stochastic logarithm is $\int Z^{-1}dZ$. When $Z$ is a continuous local martingale starting at one, this is a local martingale $L$ and the Itô formula gives $Z=\exp (L-⟨L⟩/2)$. Unlike the ordinary logarithm, it has no compensating finite-variation term.

Write a positive continuous local martingale as $M=\exp (X-⟨X⟩/2)$ using its stochastic logarithm. If $M_t\to 0$, its logarithmic bracket must diverge. Otherwise the finite-bracket convergence lemma makes $X$ converge finitely and the exponential has a positive limit.

This is an exponential-moment criterion for the stochastic exponential to be a true martingale. In the infinite-horizon stopped-moment form, a zero-starting convergent continuous local martingale satisfying ${\sup }_T\mathbb{E}{e}^{M_T/2}<\infty$ has a uniformly integrable martingale $\mathcal{E}(M)$. The half-threshold for the exponential-martingale Hölder bound first controls strict scalings. The terminal scaling inequality for stochastic exponentials and terminal expectation criterion for a nonnegative local martingale then include the endpoint scaling.

For a convergent continuous local martingale and $r>1$, factor $\mathcal{E}(X)$ into $\mathcal{E}(rX{)}^{1/r^2}$ and an ordinary exponential. Apply the Holder inequality and then the concave Jensen inequality to the fractional power $2/(r+1)$. The displayed lower bound is informative when the terminal exponential moment is finite.

For a zero-starting continuous local martingale $X$, the stochastic exponential satisfiesThe Holder inequality and the supermartingale property of the first exponential bound its stopped $p$th moment by $(\mathbb{E}{e}^{C(p,q)X_T}{)}^{(q-1)/q}$. The subtraction in the coefficient is $\sqrt{pq}-1$, not a square root of $pq-1$.

For fixed $q>1$, the coefficient in the Hölder factorization of stochastic exponentials exceeds $\sqrt{q}/(\sqrt{q}+1)>1/2$. Taking $q=1+\epsilon$ and $p=1+{\epsilon }^2$ approaches one-half. Thus a uniform stopped exponential moment at coefficient one-half bounds some $p$th moment of each strict scaling $\mathcal{E}(aM)$, $0<a<1$.

If a continuous local martingale $M$ satisfiesthen its stochastic exponential is a true martingale through time $T$.

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.

Write $M_t=B_{[M{]}_t}$ by the Dambis-Dubins-Schwarz theorem and stop the exponential Brownian martingale when $B_s-s$ first reaches $b<0$. The stopped exponential has mean one. Its boundary contribution before $[M{]}_T$ is at mostwhich vanishes as $b\to -\infty$. The remaining term converges monotonically to $\mathcal{E}(M{)}_t$, proving that its expectation is one.

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.

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 a reference measure where $X$ is Brownian motion started at zero, let $T$ be its exit time from $(-1,1)$. The displayed density is the stochastic exponential with integrand $X{1}_{[0,T]}$, by the Itô formula. It is bounded by $e^{1/2}$, hence defines a uniformly integrable density process with terminal density $Z_T$. Under that measure change, $X_t-{\int }_0^{t\wedge T}{X}_{s}ds$ is Brownian motion. At exit, $Z_T\ge e^{1/2-T}$, implying $\mathbb{P}(T\le t)\ge e^{1/2-t}\tilde{\mathbb{P}}(T\le t)$. This is a terminal density on the whole probability space, not merely an unspecified finite-horizon change of measure.

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.

The density-product proof of the Girsanov theorem makes drift-corrected zero-starting increments a local martingale under the new measure. Adding the initial random variable requires its integrability under that measure. This is automatic if the density is one at time zero, because the measures agree on the initial sigma-algebra. Without this normalization it can fail: take $P(n)=2^{-n}$, $Q(n)=c/n^2$, constant $M_t=\log (Q(n)/P(n))$ and constant $X_t=n$. These are P-martingales and the prescribed exponential density is valid, but $X_0$ is not Q-integrable.

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.