The Exponential Defined by an SDE
For an ordinary differentiable function \(x(t)\), the equation \(z'(t)=z(t)x'(t)\), \(z(0)=1\), has solution \(z(t)=e^{x(t)-x(0)}\). For a stochastic driver, the same differential notation specifies a different equation: the integral is a semimartingale integral, and the driver may have nonzero quadratic variation. The solution is called the Doleans-Dade exponential, or stochastic exponential. It is defined by the equation it solves; its explicit formula for continuous semimartingales is a separate result.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t),\mathbb{P})\) carrying a Brownian motion \(B_t\). Let \(X\) be a continuous semimartingale with decomposition \(X_t=X_0+M_t+A_t\), where \(M_0=A_0=0\), \(M\) is a continuous local martingale, and \(A\) is a continuous finite-variation process. A continuous adapted integrand is progressively measurable. Integrals against \(X\) below are semimartingale integrals, defined locally; against \(B\), an Ito integral \(\int_0^t H_s\,dB_s\) is used for progressively measurable \(H\) with \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\), or locally when this condition holds after stopping.
The coefficient multiplying \(dX_t\) is \(Z_t\) itself. This makes the equation linear, but it does not justify treating \(X\) as a differentiable function. In particular, if \(X\) has a Brownian component, its quadratic variation contributes to the ordinary-exponential comparison.
Existence and Uniqueness
Proof. Write \(C_t=[M]_t+\operatorname{Var}(A)_t\), where \([M]\) is the quadratic variation and \(\operatorname{Var}(A)_t\) is the total variation of \(A\) on \([0,t]\). Both are continuous increasing processes, and \(C_t\) is finite on every compact time interval almost surely. Stop first when \(C\) reaches an integer \(n\). On this stopped time interval, \(C\le n\). Divide it at the successive stopping times when \(C\) reaches levels separated by \(\delta=1/16\). Continuity of \(C\) ensures that its increase on each resulting interval \([\tau,\rho]\) is at most \(\delta\); only finitely many such intervals are needed before the stop.
On one interval, suppose the initial value \(\xi\), measurable with respect to \(\mathcal{F}_\tau\), is square-integrable. Consider adapted continuous processes \(Y\) with finite norm \(\|Y\|_2=(\mathbb{E}\sup_{\tau\le t\le\rho}|Y_t|^2)^{1/2}\). Define $$ \Phi(Y)_t=\xi+\int_\tau^t Y_s\,dM_s+\int_\tau^t Y_s\,dA_s,\qquad \tau\le t\le\rho. $$ These integrals are well-defined: the martingale integral is square-integrable by localization and the Ito isometry, and the finite-variation integral is pathwise defined; the estimates below give the required square-integrability. For any two such processes \(Y,\widetilde Y\), the Burkholder-Davis-Gundy inequality and the bound \([M]_\rho-[M]_\tau\le\delta\) give $$ \left\|\sup_{\tau\le t\le\rho}\left|\int_\tau^t(Y_s-\widetilde Y_s)\,dM_s\right|\right\|_{L^2} \le 2\sqrt{\delta}\,\|Y-\widetilde Y\|_2. $$ The total variation bound gives $$ \left\|\sup_{\tau\le t\le\rho}\left|\int_\tau^t(Y_s-\widetilde Y_s)\,dA_s\right|\right\|_{L^2} \le\delta\,\|Y-\widetilde Y\|_2. $$ By the triangle inequality, \(\|\Phi(Y)-\Phi(\widetilde Y)\|_2\le(2\sqrt{\delta}+\delta)\|Y-\widetilde Y\|_2=\tfrac{9}{16}\|Y-\widetilde Y\|_2\). The space of adapted continuous processes with this norm is complete, and the Banach fixed-point theorem therefore gives a unique fixed point on the interval. The same estimates show that \(\Phi\) maps the space into itself.
Starting with \(\xi=1\), solve on the first interval and use its terminal value as the initial value on the next. There are finitely many intervals before each stop \(C=n\), so concatenation constructs a solution up to that stop. These stopped solutions agree on their common time intervals by the uniqueness argument that follows. Since \(C\) is finite on compact time intervals almost surely, the stopping times \(C=n\) eventually exceed any fixed time, and the compatible solutions define an adapted continuous solution for all finite times.
For uniqueness, let \(Z\) and \(\widetilde Z\) be two solutions with the same initial value. Their difference \(D=Z-\widetilde Z\) satisfies \(D_t=\int_0^t D_s\,dX_s\). Suppose it is zero at the start \(\tau\) of one of the small clock intervals. For each integer \(m\ge1\), stop at the first time \(\eta_m\) in that interval when \(|D|\ge m\), or at the interval's endpoint if this does not occur. The stopped difference is bounded by \(m\), so \(\mathbb{E}\sup_{\tau\le t\le\rho}|D_{t\wedge\eta_m}|^2<\infty\). Applying the same two integral estimates to its stopped equation yields $$ \left(\mathbb{E}\sup_{\tau\le t\le\rho}|D_{t\wedge\eta_m}|^2\right)^{1/2} \le \frac{9}{16}\left(\mathbb{E}\sup_{\tau\le t\le\rho}|D_{t\wedge\eta_m}|^2\right)^{1/2}. $$ Because the quantity is finite and \(9/16<1\), it must be zero. Thus the stopped difference vanishes; it cannot reach level \(m>0\), so the difference vanishes throughout the interval. Repeating over the intervals proves equality up to every stop \(C=n\), and then at every finite time. Continuity gives indistinguishability. The localization is essential: without stopping \(D\), the expectation of its squared supremum need not be finite, and the contraction inequality would not imply it is zero. \(\square\)
The contraction estimate uses the Burkholder-Davis-Gundy inequality, a standard maximal inequality for martingales; its proof is beyond this tutorial. The proof uses it only after stopping so the quantities being estimated are finite. It does not assert that every solution is a true martingale: existence and uniqueness concern the SDE, not the martingale property.
Worked Examples
Worked Example: A Deterministic Driver
Let \(X_t=3t\). Find the solution of \(dZ_t=Z_t\,dX_t\), \(Z_0=1\), and compare it with the ordinary exponential.
Here \(dX_t=3\,dt\), so the equation is the ordinary integral equation \(Z_t=1+\int_0^t3Z_s\,ds\). The candidate \(Z_t=e^{3t}\) gives $$ 1+\int_0^t3e^{3s}\,ds=1+(e^{3t}-1)=e^{3t}. $$ Thus it satisfies the equation, and uniqueness follows from the theorem. The ordinary exponential \(e^{X_t-X_0}=e^{3t}\) agrees in this finite-variation example.
Check: At \(t=0\), \(Z_0=e^0=1\). Differentiating the candidate gives \(dZ_t=3e^{3t}dt=Z_t\,dX_t\), confirming the equation directly.
Worked Example: Brownian Motion as Driver
Let \(X_t=B_t\). Find the solution and test whether the ordinary exponential \(e^{B_t}\) solves the same SDE.
Set \(f(t,x)=e^{x-t/2}\). Its derivatives are \(f_t=-\tfrac12 f\), \(f_x=f\), and \(f_{xx}=f\). Ito's formula, including its second-derivative correction, gives $$ d f(t,B_t)=\left(f_t+\tfrac12 f_{xx}\right)dt+f_x\,dB_t =\left(-\tfrac12 f+\tfrac12 f\right)dt+f\,dB_t =f\,dB_t. $$ Since \(f(0,B_0)=1\), the solution is \(Z_t=e^{B_t-t/2}\). Its Brownian integrand is progressively measurable and satisfies, for finite \(T\), $$ \mathbb{E}\int_0^T Z_s^2\,ds=\int_0^T e^{-s}\mathbb{E}[e^{2B_s}]\,ds =\int_0^T e^s\,ds=e^T-1<\infty. $$ Here \(\mathbb{E}[e^{2B_s}]=e^{2s}\), from the moment-generating function of \(N(0,s)\).
The ordinary exponential has a different differential. For \(g(t,x)=e^x\), \(g_t=0\), \(g_x=g\), and \(g_{xx}=g\); hence Ito's formula gives \(d(e^{B_t})=e^{B_t}\,dB_t+\tfrac12e^{B_t}\,dt\). The extra drift means it does not solve \(dZ=Z\,dB\).
Check: Independent Brownian increments imply, for \(s\le t\), \(\mathbb{E}[Z_t\mid\mathcal{F}_s]=Z_s\mathbb{E}[e^{B_t-B_s-(t-s)/2}]=Z_s\), since a centered normal variable of variance \(t-s\) has exponential moment \(e^{(t-s)/2}\) at argument \(1\). Also \(\mathbb{E}Z_t=1\). Thus this solution is a true martingale, not merely a local martingale.
Worked Example: Brownian Motion with Drift
Let \(X_t=B_t+t\), so \(dX_t=dB_t+dt\). Find the solution and compare it with \(e^{X_t-X_0}\).
Take \(f(t,x)=e^{x+t/2}\). Its derivatives are \(f_t=\tfrac12 f\), \(f_x=f\), and \(f_{xx}=f\). Ito's formula gives $$ df(t,B_t)=\left(\tfrac12 f+\tfrac12 f\right)dt+f\,dB_t =f\,dt+f\,dB_t=f\,dX_t. $$ The initial value is \(f(0,B_0)=1\), so \(Z_t=e^{B_t+t/2}\) solves the defining SDE. For the Ito integral, \(Z\) is progressively measurable and $$ \mathbb{E}\int_0^T Z_s^2\,ds =\int_0^T e^s\mathbb{E}[e^{2B_s}]\,ds =\int_0^T e^{3s}\,ds=\frac{e^{3T}-1}{3}<\infty. $$ Uniqueness follows from the theorem.
The ordinary exponential is \(e^{X_t-X_0}=e^{B_t+t}\). For \(h(t,x)=e^{x+t}\), the derivatives are \(h_t=h\), \(h_x=h\), and \(h_{xx}=h\). Therefore $$ d h(t,B_t)=\left(h+\tfrac12h\right)dt+h\,dB_t =h\,dB_t+\tfrac32h\,dt, $$ whereas the required equation for this candidate would have drift \(h\,dt\) and diffusion \(h\,dB_t\). It fails because its drift is too large by \(\tfrac12h\,dt\).
Check: If the Brownian component is removed, so \(X_t=t\), the same defining equation reduces to \(dZ_t=Z_tdt\), whose solution is \(e^t\). The discrepancy in the Brownian example comes from the Ito correction, not from the drift coefficient \(1\) in \(X\).
What the Definition Does—and Does Not—Say
The Doleans-Dade exponential is first an SDE solution. Its existence and uniqueness do not, by themselves, say that it is positive, a martingale, or equal to \(e^{X_t-X_0}\). The Brownian example shows that the ordinary exponential can have an extra drift, while the drifted Brownian example shows that the discrepancy remains even when \(X\) has both a drift and a diffusion coefficient. For an Ito integral against Brownian motion, the integrability condition \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\) ensures a square-integrable integral; in the examples this was checked explicitly. For a general continuous semimartingale, local semimartingale integrability, rather than that Brownian \(L^2\) condition alone, is the relevant notion.
Common mistake. A tempting but incorrect calculation is to write \(d(e^{B_t})=e^{B_t}dB_t\) by applying the ordinary chain rule. That omits the second derivative term. The correct calculation uses \(f(x)=e^x\), with \(f'=e^x\) and \(f''=e^x\), and yields $$ d(e^{B_t})=e^{B_t}dB_t+\tfrac12e^{B_t}dt. $$ The correction is nonzero because \(dB_t\,dB_t=dt\); in Ito's multiplication rules, \(dB_t\,dt=0\) and \(dt\,dt=0\). Thus the ordinary chain rule cannot be used for a Brownian driver.
In later applications, the defining SDE is useful even before one writes an explicit expression for its solution. It identifies the process through its infinitesimal relation to a driver, and the uniqueness result ensures that any candidate verified by substitution is the same process. Whether that process is a true martingale requires a separate integrability or expectation argument; it is not part of the definition.
Check Your Understanding
For each question, distinguish the defining SDE from the ordinary exponential and state any integrability or conditional-expectation condition you use.
- For \(X_t=2t\), verify directly that \(Z_t=e^{2t}\) satisfies \(dZ_t=Z_t\,dX_t\) and \(Z_0=1\).
- Apply Ito's formula to \(e^{2B_t}\). What drift term prevents it from solving \(dZ_t=2Z_t\,dB_t\)?
- For \(X_t=B_t+2t\), determine a candidate of the form \(e^{B_t+ct}\), then verify it using Ito's formula.
- Why must the uniqueness proof stop the difference of two solutions before taking an expected squared supremum?
- Explain why verifying that a process solves the defining SDE does not, on its own, prove that it is a true martingale.