Defining the Stochastic Exponential

From a Brownian Integral to a Stochastic Exponential

Exponentials convert additive quantities into multiplicative ones. For an ordinary differentiable path \(x_t\), the chain rule gives \(d(e^{x_t})=e^{x_t}\,dx_t\). For a Brownian stochastic integral, that formula misses a term: Brownian motion has quadratic variation, and Ito's formula contributes a second-derivative correction. The stochastic exponential is the exponential with precisely the correction needed to solve a natural linear stochastic differential equation.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\) relative to the filtration. Fix a finite horizon \(T\). Throughout, let \(\theta_t\) be progressively measurable and satisfy \(\mathbb{E}\int_0^T\theta_s^2\,ds<\infty\). Thus \(\theta\) belongs to the Ito-integrand class \(\mathcal{L}^2\), and \(M_t=\int_0^t\theta_s\,dB_s\) is a well-defined Ito integral for \(0\le t\le T\). All expectations and stochastic integrals here are under \(\mathbb{P}\); defining this process alone does not change measure or establish a density for a new measure.

Definition: The stochastic exponential generated by \(\theta\) is the process \(\mathcal{E}(M)_t=\exp\!\left(M_t-\tfrac12[M]_t\right)\), where \(M_t=\int_0^t\theta_s\,dB_s\) and \([M]_t\) is its quadratic variation. For this Brownian integral, \([M]_t=\int_0^t\theta_s^2\,ds\). Equivalently, \(\mathcal{E}(M)_t=\exp\!\left(\int_0^t\theta_s\,dB_s-\tfrac12\int_0^t\theta_s^2\,ds\right)\). The initial value is \(\mathcal{E}(M)_0=1\).

The integrals in the definition are Ito integrals, not Stratonovich integrals. The correction uses the Ito multiplication rules \(dB_t\,dB_t=dt\), \(dB_t\,dt=0\), and \(dt\,dt=0\). In particular, the \(-\tfrac12\int_0^t\theta_s^2\,ds\) term is not optional: it compensates for the quadratic variation of the stochastic part of the exponent.

Why This Formula Solves the Linear Equation

Theorem: Under the stated \(\mathcal{L}^2\) assumption, \(Z_t=\mathcal{E}(M)_t\) is positive, continuous, has \(Z_0=1\), and satisfies \(dZ_t=Z_t\theta_t\,dB_t\). It is a local martingale. The formula is the unique continuous solution, in the locally defined Ito sense, of this equation with initial value \(1\).

Proof. Set \(A_t=\int_0^t\theta_s^2\,ds\), so \(M_t=\int_0^t\theta_s\,dB_s\), \([M]_t=A_t\), and \(dA_t=\theta_t^2\,dt\). Apply Ito's formula to \(F(x,a)=e^{x-a/2}\) at \((M_t,A_t)\). Its derivatives are \(F_x=F\), \(F_a=-\tfrac12F\), and \(F_{xx}=F\). The Ito correction is \(\tfrac12F_{xx}\,d[M]_t=\tfrac12F\,\theta_t^2\,dt\); the finite-variation coordinate \(A\) contributes no second-order term. Therefore

dZ_t =F_x\,dM_t+F_a\,dA_t+\tfrac12F_{xx}\,d[M]_t =Z_t\theta_t\,dB_t-\tfrac12Z_t\theta_t^2\,dt+\tfrac12Z_t\theta_t^2\,dt =Z_t\theta_t\,dB_t.

Since an exponential is strictly positive and \(M_0=A_0=0\), \(Z_t>0\) and \(Z_0=e^0=1\). To justify the local-martingale assertion, stop when either \(Z\) or \(A\) first reaches \(n\), for integers \(n\ge2\). Up to this stopping time, \(Z\le n\) and \(A\le n\), so the stopped integrand \(Z_t\theta_t\) has expected squared integral at most \(n^3\). The stopped Ito integral is therefore a square-integrable martingale, and the stopped \(Z\) is a martingale. On the finite horizon, \(A\) and the continuous process \(Z\) are finite, so these stopping times increase to the horizon almost surely. This proves that \(Z\) is a local martingale; it does not by itself prove that \(Z\) is a true martingale.

For uniqueness, let \(Y\) be any other continuous solution with \(Y_0=1\), with its stochastic integrals defined locally by stopping so their integrands are in \(\mathcal{L}^2\). Since \(Z>0\), apply Ito's formula to \(1/Z\). The derivatives of \(g(z)=z^{-1}\) are \(g'(z)=-z^{-2}\) and \(g''(z)=2z^{-3}\). Using \(d[Z]_t=Z_t^2\theta_t^2\,dt\) gives \(d(Z_t^{-1})=-Z_t^{-1}\theta_t\,dB_t+Z_t^{-1}\theta_t^2\,dt\). The product rule for \(Y/Z\) has covariation term \(d[Y,Z^{-1}]_t=-Y_tZ_t^{-1}\theta_t^2\,dt\). Thus its stochastic terms and its two drift terms cancel, giving \(d(Y_t/Z_t)=0\). Since \(Y_0/Z_0=1\), \(Y_t/Z_t=1\), so \(Y=Z\). \(\square\)

For an initial value \(z_0\) other than \(1\), the unique solution is \(z_0\mathcal{E}(M)_t\), by multiplying the equation and its initial condition by \(z_0\). The same formula also gives a useful interval identity: for \(0\le s\le t\), \(Z_t/Z_s=\exp\!\left(\int_s^t\theta_u\,dB_u-\tfrac12\int_s^t\theta_u^2\,du\right)\). This follows by subtracting the exponents at \(s\) and \(t\), using additivity of the Ito integral and of the ordinary time integral.

Worked Forms of the Definition

Worked Example: Constant Integrand

Take \(T=1\) and \(\theta_t=2\) for \(0\le t\le1\). This deterministic integrand is progressively measurable and \(\mathbb{E}\int_0^1\theta_s^2\,ds=4\), so it is in \(\mathcal{L}^2\). Its Ito integral and accumulated squared integrand are \(\int_0^t2\,dB_s=2B_t\) and \(\int_0^t 2^2\,ds=4t\). Hence

Z_t=\exp(2B_t-2t),\qquad 0\le t\le1.

Check: \(Z_0=\exp(2B_0)=1\). At \(t=1\), \(B_1\) has distribution \(N(0,1)\), whose moment-generating function gives \(\mathbb{E}[e^{2B_1}]=e^{2}\). Thus \(\mathbb{E}[Z_1]=e^{-2}e^2=1\). This endpoint calculation is a check for this example; the definition alone does not assert that every stochastic exponential is a true martingale.

Worked Example: A Simple Adapted Integrand

On \([0,2]\), let \(\theta_t=1\) for \(0\le t\le1\), and let \(\theta_t=2\mathbf{1}_{\{B_1>0\}}\) for \(1<t\le2\). The coefficient on the second interval is \(\mathcal{F}_1\)-measurable, so this is an adapted simple integrand. Since \(B_1\sim N(0,1)\), \(\mathbb{P}(B_1>0)=\tfrac12\), and \(\mathbb{E}\int_0^2\theta_s^2\,ds=1+4\cdot\tfrac12=3\); hence it is in \(\mathcal{L}^2\). Write \(I=\mathbf{1}_{\{B_1>0\}}\). The stochastic integral at time \(2\) is \(B_1+2I(B_2-B_1)\), and the squared-integrand integral is \(1+4I\). Therefore

Z_2=\exp\!\left(B_1-\tfrac12+2I(B_2-B_1)-2I\right).

On the event \(B_1\le0\), \(I=0\), so \(Z_2=e^{B_1-1/2}\). On the event \(B_1>0\), \(I=1\), so \(Z_2=e^{B_1-1/2+2(B_2-B_1)-2}\). The event \(B_1=0\) is included in the first case; it has probability zero, but the formula is defined there too.

Check: At time \(1\), the formula gives \(Z_1=e^{B_1-1/2}\). The increment \(B_2-B_1\) is independent of \(\mathcal{F}_1\) and has distribution \(N(0,1)\). If \(I=0\), the second-interval factor is \(1\). If \(I=1\), its conditional mean is \(\mathbb{E}[e^{2(B_2-B_1)-2}\mid\mathcal{F}_1]=e^{2^2/2-2}=1\). Thus \(\mathbb{E}[Z_2\mid\mathcal{F}_1]=Z_1\), confirming the formula's interval factor in both cases.

Worked Example: A Random, Non-Simple Integrand

Take \(\theta_t=B_t\) on \([0,1]\). Brownian motion is progressively measurable, and \(\mathbb{E}\int_0^1 B_s^2\,ds=\int_0^1s\,ds=\tfrac12\), so this integrand belongs to \(\mathcal{L}^2\). To compute its Ito integral, apply Ito's formula to \(f(x)=x^2\), for which \(f'(x)=2x\), \(f''(x)=2\), and the correction is \(\tfrac12 f''(B_t)\,dt=dt\). Consequently, \(d(B_t^2)=2B_t\,dB_t+dt\), and integration from \(0\) to \(t\) gives \(\int_0^tB_s\,dB_s=\tfrac12(B_t^2-t)\). The accumulated square is \(\int_0^tB_s^2\,ds\). The resulting stochastic exponential is

Z_t=\exp\!\left(\tfrac12(B_t^2-t)-\tfrac12\int_0^tB_s^2\,ds\right).

Check: Set \(L_t=\tfrac12(B_t^2-t)-\tfrac12\int_0^tB_s^2\,ds\), so \(Z_t=e^{L_t}\). The preceding Ito calculation gives \(dL_t=B_t\,dB_t-\tfrac12B_t^2\,dt\) and \(d[L]_t=B_t^2\,dt\). Apply Ito's formula to \(h(x)=e^x\), with \(h'(x)=h''(x)=e^x\): the correction is \(\tfrac12e^{L_t}B_t^2\,dt\), which cancels the drift \(-\tfrac12e^{L_t}B_t^2\,dt\). Hence \(dZ_t=Z_tB_t\,dB_t\), as required. Also \(L_0=0\), so \(Z_0=1\).

A Common Mistake: Omitting the Correction

A tempting but incorrect guess for the equation \(dZ_t=Z_t\,dB_t\), \(Z_0=1\), is \(Y_t=e^{B_t}\), obtained by applying the ordinary chain rule. The error is visible by applying Ito's formula explicitly: for \(h(x)=e^x\), \(h'(x)=e^x\) and \(h''(x)=e^x\), so

d(e^{B_t})=e^{B_t}\,dB_t+\tfrac12e^{B_t}\,dt.

The extra drift means this candidate does not solve the stated equation. The correction gives \(Z_t=e^{B_t-t/2}\): with \(L_t=B_t-t/2\), \(dL_t=dB_t-\tfrac12dt\) and \(d[L]_t=dt\). Ito's formula for \(e^{L_t}\) gives \(dZ_t=Z_t(dB_t-\tfrac12dt)+\tfrac12Z_tdt=Z_t\,dB_t\). The correction is therefore fixed by the Ito quadratic variation, not by an arbitrary convention.

More generally, the formula proves positivity and the local-martingale property, but positivity does not imply that \(Z\) is a true martingale. A local martingale is a process that becomes a martingale after suitable stopping; a true martingale must satisfy the martingale integrability and conditional-expectation identities without such stopping. Establishing when the stochastic exponential is a true martingale, and when it can serve as a probability density, requires additional conditions. No measure change is part of the definition given here.

Check Your Understanding

Use the definition and the Ito correction to answer the following questions.

For constant \(\theta_t=c\) on \([0,T]\), write the stochastic exponential and evaluate its initial value.
For a simple integrand equal to \(\xi\) on \((s,t]\), where \(\xi\) is \(\mathcal{F}_s\)-measurable, identify its contribution to both integrals in the exponent.
Explain why the stochastic exponential is positive but need not be a true martingale solely on the basis of its definition.
Apply Ito's formula to \(e^{B_t}\) and identify the term that prevents it from solving \(dZ_t=Z_t\,dB_t\).
For \(\theta_t=B_t\), state the integrability calculation that verifies \(\theta\in\mathcal{L}^2\) on \([0,1]\).

Defining the Stochastic Exponential

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