The Stochastic Exponential

From a Linear SDE to an Exponential

For an ordinary differential equation $dz_t=a(t)z_t\,dt$, taking a logarithm turns multiplication by $z_t$ into addition, and integration gives $z_t=z_0\exp(\int_0^t a(s)\,ds)$. A Brownian-driven equation has a similar exponential solution, but the ordinary chain rule misses a term: Brownian motion has quadratic variation, so Ito's formula contributes a second-derivative correction. That correction determines the minus one-half term in the exponent.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$. Fix a finite horizon $T$, and let $\theta_t$ be progressively measurable with $\mathbb{E}\int_0^T\theta_s^2\,ds<\infty$. Thus $\theta$ belongs to the class $\mathcal{L}^2$ on $[0,T]$, and the Ito integral $\int_0^t\theta_s\,dB_s$ is defined for $t\le T$. Assume this integrability on every finite horizon. The quadratic variation of $B$ is $[B]_t=t$; in the Ito multiplication rules, $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$.

Definition: For $M_t=\int_0^t\theta_s\,dB_s$, its stochastic exponential is $\mathcal{E}(M)_t=\exp\big(M_t-\tfrac12[M]_t\big)$. Here $M$ is a continuous local martingale and $[M]_t=\int_0^t\theta_s^2\,ds$, so $\mathcal{E}(M)_t=\exp\big(\int_0^t\theta_s\,dB_s-\tfrac12\int_0^t\theta_s^2\,ds\big)$.

The stochastic exponential is positive and finite at every finite time, almost surely. The integrability assumption makes both integrals in its exponent finite on each finite horizon. The theorem below identifies it as the solution of the linear SDE, and also specifies what kind of solution and uniqueness are obtained.

Derivation and Verification

Theorem: Under the assumptions above, for any deterministic $z_0\in\mathbb{R}$, the equation $dZ_t=Z_t\theta_t\,dB_t$, $Z_0=z_0$, has the continuous adapted solution $$ Z_t=z_0\exp\left(\int_0^t\theta_s\,dB_s-\frac12\int_0^t\theta_s^2\,ds\right). $$ It is pathwise unique: any two continuous adapted solutions on the same filtered probability space, driven by the same $B$ and using the same $\theta$, are indistinguishable. In particular, when $z_0=1$, the solution is $\mathcal{E}(M)_t$. The construction on the given space is a strong solution; pathwise uniqueness is stronger than merely requiring solutions to have the same law (uniqueness in law).

Proof. First take $z_0=1$, and set $Y_t=\int_0^t\theta_s\,dB_s-\tfrac12\int_0^t\theta_s^2\,ds$, so that the proposed solution is $Z_t=e^{Y_t}$. The function is $f(y)=e^y$, with $f'(y)=e^y$ and $f''(y)=e^y$. The finite-variation part of $Y$ is $-\tfrac12\int_0^t\theta_s^2\,ds$, and its quadratic variation is $[Y]_t=\int_0^t\theta_s^2\,ds$. Ito's formula, including its second-derivative correction, gives $$ dZ_t=f'(Y_t)\,dY_t+\frac12f''(Y_t)\,d[Y]_t =Z_t\left(\theta_t\,dB_t-\frac12\theta_t^2\,dt\right) +\frac12Z_t\theta_t^2\,dt =Z_t\theta_t\,dB_t. $$ Also $Z_0=1$, proving existence for this initial value. Multiplication by $z_0$ gives a solution for any $z_0$, including $z_0=0$.

For uniqueness, let $Z$ denote the proposed solution with initial value $1$, and let $X$ be any other solution with the same initial value. Since $Z_t>0$, define $Y_t=1/Z_t$. To compute its differential, use $f(z)=1/z$, $f'(z)=-z^{-2}$, and $f''(z)=2z^{-3}$. From $d[Z]_t=Z_t^2\theta_t^2\,dt$, Ito's formula yields $$ dY_t=-Z_t^{-2}\,dZ_t+\frac12(2Z_t^{-3})\,d[Z]_t =-\theta_tY_t\,dB_t+\theta_t^2Y_t\,dt. $$ The Ito product rule for $X_tY_t$ has covariation term $d[X,Y]_t=-\theta_t^2X_tY_t\,dt$. Therefore $$ d(X_tY_t)=Y_t\,dX_t+X_t\,dY_t+d[X,Y]_t=0, $$ because the two stochastic terms cancel and the two drift terms cancel. Since $X_0Y_0=1$, it follows that $X_tY_t=1$ and $X_t=Z_t$. The same argument applies to arbitrary initial value by comparing with the candidate $z_0Z$; if $z_0=0$, multiplication by $Y$ shows that every solution remains zero. All these calculations can be stopped when the processes or $\int_0^t\theta_s^2\,ds$ reach a finite bound, so the Ito product rule applies to locally square-integrable integrands; letting the bounds increase gives the asserted equality on every finite horizon. $\square$

The verification also explains the correction. The exponent has differential $dY_t=\theta_t\,dB_t-\tfrac12\theta_t^2\,dt$. Its Ito correction is $\tfrac12 e^{Y_t}\theta_t^2\,dt$, which exactly cancels the negative drift. Omitting the correction would leave a drift term and would not solve the stated SDE.

Worked Forms of the Exponential

Worked Example: Constant Integrand

Let $\theta_t=2$ on $[0,1]$ and $Z_0=1$. This integrand is progressively measurable and $\mathbb{E}\int_0^1\theta_s^2\,ds=4<\infty$. The two terms in the exponent are $$ \int_0^t 2\,dB_s=2B_t,\qquad \frac12\int_0^t 2^2\,ds=\frac12\cdot4t=2t. $$ Thus $Z_t=e^{2B_t-2t}$. To check it directly, use $f(t,x)=e^{2x-2t}$. Its derivatives are $f_t=-2f$, $f_x=2f$, and $f_{xx}=4f$. Ito's formula gives $$ dZ_t=f_t\,dt+f_x\,dB_t+\frac12f_{xx}\,dt =-2Z_t\,dt+2Z_t\,dB_t+2Z_t\,dt =2Z_t\,dB_t. $$ This is the required equation. The initial value check is $Z_0=e^0=1$; the drift terms cancel exactly.

Worked Example: A Piecewise Constant Integrand

On $[0,1]$, let $\theta_t=1$ for $0\le t\le\tfrac12$, and $\theta_t=2$ for $\tfrac12<t\le1$, with $Z_0=1$. The value at the single switching time does not change either integral. The integrand is deterministic and square-integrable. At $t=1$, $$ \int_0^1\theta_s\,dB_s=B_{1/2}+2(B_1-B_{1/2})=2B_1-B_{1/2}, $$ and $$ \frac12\int_0^1\theta_s^2\,ds =\frac12\left(1^2\cdot\frac12+2^2\cdot\frac12\right) =\frac12\left(\frac12+2\right)=\frac54. $$ Consequently $Z_1=\exp(2B_1-B_{1/2}-\tfrac54)$. On the first half-interval the formula is $Z_t=e^{B_t-t/2}$, whose Ito differential is $Z_t\,dB_t$. On the second half it is $$ Z_t=Z_{1/2}\exp\left(2(B_t-B_{1/2})-2(t-\tfrac12)\right). $$ The constant factor $Z_{1/2}$ is fixed at the switching time. For the exponential factor, the time derivative is $-2$ times the factor, the first space derivative is $2$ times the factor, and the second space derivative is $4$ times the factor. The Ito drift is $-2+ \tfrac12\cdot4=0$, leaving $dZ_t=2Z_t\,dB_t$ on that interval. The exponent at $t=1$ agrees with the general formula, including the total correction $\tfrac54$.

Worked Example: An Integrand Depending on Brownian Motion

On $[0,T]$, take $\theta_t=B_t$. This is progressively measurable, and $\mathbb{E}\int_0^T B_s^2\,ds=\int_0^T s\,ds=T^2/2<\infty$, so the Ito integral is defined in $\mathcal{L}^2$. Applying Ito's formula to $g(x)=x^2$, with $g'(x)=2x$ and $g''(x)=2$, gives $d(B_t^2)=2B_t\,dB_t+dt$. Hence $$ \int_0^t B_s\,dB_s=\frac12(B_t^2-t). $$ The stochastic exponential therefore has the explicit form $$ Z_t=\exp\left(\frac12(B_t^2-t)-\frac12\int_0^tB_s^2\,ds\right). $$ For a direct check, denote the exponent by $U_t$. The displayed Ito identity gives $dU_t=B_t\,dB_t-\tfrac12B_t^2\,dt$, and $d[U]_t=B_t^2\,dt$. For $f(u)=e^u$, $f'(u)=f''(u)=e^u$, so Ito's formula gives $d(e^{U_t})=e^{U_t}(B_t\,dB_t-\tfrac12B_t^2\,dt)+\tfrac12e^{U_t}B_t^2\,dt=Z_tB_t\,dB_t$. Also $U_0=0$, so $Z_0=1$. This verifies both the formula and the initial condition.

Local Martingale Does Not Mean Martingale

The equation $dZ_t=Z_t\theta_t\,dB_t$ expresses $Z$ as a stochastic integral plus its initial value. However, the integrand $Z_t\theta_t$ is not automatically in $\mathcal{L}^2$ on the whole time interval: the assumptions on $\theta$ alone do not give $\mathbb{E}\int_0^T Z_t^2\theta_t^2\,dt<\infty$. The appropriate general conclusion is a local martingale.

Proposition: Under the theorem's assumptions with $Z_0=1$, $Z_t=\mathcal{E}(M)_t$ is a positive local martingale and a supermartingale. These conclusions do not assert that $Z$ is a true martingale on $[0,T]$.

Proof. Let $A_t=\int_0^t\theta_s^2\,ds$, which is finite and continuous almost surely on $[0,T]$. For integers $n>1$, stop at the first time $Z_t\ge n$ or $A_t\ge n$, and denote this stopping time by $\tau_n$ (if neither event occurs by $T$, use $T$). The process is positive and continuous, so before this stop $Z_t\le n$, and the stopped quadratic energy satisfies $\int_0^{T\wedge\tau_n}Z_s^2\theta_s^2\,ds\le n^2n=n^3$. Thus the stopped Ito integral is square-integrable, has conditional mean-zero increments by the Ito integral martingale property, and $Z_{t\wedge\tau_n}=1+\int_0^{t\wedge\tau_n}Z_s\theta_s\,dB_s$ is a martingale. As $n$ increases, $\tau_n$ reaches $T$ eventually almost surely, so this proves that $Z$ is a local martingale.

For $0\le s\le t\le T$, conditional Fatou's lemma applied to the nonnegative variables $Z_{t\wedge\tau_n}$ gives $$ \mathbb{E}[Z_t\mid\mathcal{F}_s] \le\liminf_{n\to\infty}\mathbb{E}[Z_{t\wedge\tau_n}\mid\mathcal{F}_s] =\liminf_{n\to\infty}Z_{s\wedge\tau_n} =Z_s. $$ This proves the supermartingale property. The stopping argument proves a martingale property for each stopped process, but it does not provide a uniform integrability bound as $n$ increases. Consequently it cannot be used to pass equality of expectations from the stopped processes to $Z$; a positive local martingale need not be a true martingale. Additional conditions, such as an appropriate exponential-integrability criterion, are needed for a general martingale conclusion. $\square$

This distinction matters whenever the stochastic exponential is used as a candidate density process: positivity and the local martingale property alone do not guarantee that its expectation remains one. In this tutorial no martingale or change-of-measure conclusion is inferred from the formula. The ordinary chain rule is another common pitfall: using $d(e^{Y_t})=e^{Y_t}dY_t$ would retain the drift $-\tfrac12Z_t\theta_t^2dt$ and give the wrong SDE. Ito's correction, determined by $dB_t\,dB_t=dt$, is essential.

Check Your Understanding

Use Ito's formula and the stated integrability assumptions to check your reasoning.

For a progressively measurable $\theta$ with $\mathbb{E}\int_0^T\theta_s^2\,ds<\infty$, write the stochastic exponential solving $dZ_t=Z_t\theta_t\,dB_t$ with initial value $z_0$.
For constant $\theta_t=c$, identify the drift from differentiating $e^{cB_t-\frac12c^2t}$ with Ito's formula and show how it cancels.
Explain why the process $Z_t=\exp(\int_0^t B_s\,dB_s-\frac12\int_0^t B_s^2\,ds)$ has differential $dZ_t=Z_tB_t\,dB_t$.
In the uniqueness argument, why is the quadratic covariation term in $d(X_tY_t)$ necessary?
What does the localization argument establish about $Z$, and why does it not by itself establish that $Z$ is a true martingale?

What You'll Learn