Why the Exponential Needs a Correction
For an ordinary differentiable path \(x(t)\), the chain rule gives \(d(e^{x(t)})=e^{x(t)}\,dx(t)\). For a continuous semimartingale, Itô’s formula adds a second-order term involving quadratic variation. The Doléans-Dade exponential is the exponential process adjusted so that it solves the stochastic equation \(dZ_t=Z_t\,dX_t\). Its explicit formula therefore contains a correction by half the quadratic variation.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). Let \(X\) be a continuous semimartingale adapted to this filtration. Its quadratic variation is denoted by \([X]_t\), with \([X]_0=0\). Semimartingale integrals against \(X\) are understood locally: for a decomposition of \(X\) into a continuous local martingale and a continuous finite-variation process, the integrand must be integrable against both parts. A continuous adapted integrand is locally bounded, so it is locally integrable against a continuous semimartingale. When an integral is specifically with respect to \(B_t\), we use the Itô integral: its integrand is progressively measurable and in \(\mathcal{L}^2\), meaning \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\) for each finite \(T\).
The expression is written in terms of \(X-X_0\) because the exponential starts at one even if \(X_0\) is not zero. The correction uses quadratic variation, not ordinary variance, and it is the source of the difference between the Itô exponential and the ordinary chain rule.
The Explicit Formula
Proof. Set \(U_t=X_t-X_0\), so that \(U_0=0\) and \([U]_t=[X]_t\). Apply Itô’s formula to the two-variable function \(f(u,v)=\exp(u-v/2)\) at \((U_t,[U]_t)\). Its derivatives are \(f_u=f\), \(f_{uu}=f\), \(f_v=-f/2\), and \(f_{uv}=f_{vv}=0\). The process \([U]\) has finite variation, so its quadratic variation and its covariation with \(U\) are zero. Thus the Itô formula, including its second-order correction, gives $$ \begin{aligned} d f(U_t,[U]_t) &= f_u\,dU_t+f_v\,d[U]_t+\frac12 f_{uu}\,d[U]_t\\ &= f(U_t,[U]_t)\,dU_t -\frac12 f(U_t,[U]_t)\,d[U]_t +\frac12 f(U_t,[U]_t)\,d[U]_t\\ &= f(U_t,[U]_t)\,dX_t. \end{aligned} $$ Also \(f(U_0,[U]_0)=1\), and \(f(U_t,[U]_t)>0\). Hence this process satisfies the defining equation for \(\mathcal{E}(X-X_0)\). By uniqueness for that equation, it is the Doléans-Dade exponential. \(\square\)
The cancellation in the proof is exactly the Itô correction: \(dX_t\,dX_t=d[X]_t\), while \(dX_t\,dt=0\) and \(dt\,dt=0\). These are multiplication rules for Itô differentials, not ordinary algebraic identities between finite increments. In Stratonovich notation the chain rule is restored: for a continuous semimartingale \(U\), the Stratonovich integral satisfies \(\int H_s\circ dU_s=\int H_s\,dU_s+\tfrac12[H,U]\). In particular, \(d(e^{U_t})=e^{U_t}\circ dU_t\); the displayed Doléans-Dade formula is the corresponding Itô form.
Worked Example: Brownian Motion
Take \(X_t=B_t\). Since \([B]_t=t\), the theorem gives $$ Z_t=\mathcal{E}(B)_t=\exp(B_t-t/2). $$ Here the Brownian Itô integrand \(Z_s\) is progressively measurable and belongs to \(\mathcal{L}^2\) on every finite interval: \(Z_s^2=\exp(2B_s-s)\), so the normal moment-generating function gives \(\mathbb{E}[Z_s^2]=e^s\), and \(\mathbb{E}\int_0^T Z_s^2\,ds=\int_0^T e^s\,ds=e^T-1<\infty\).
To check the stochastic equation directly, use \(g(t,b)=\exp(b-t/2)\). Its derivatives are \(g_t=-g/2\), \(g_b=g\), and \(g_{bb}=g\). Itô’s formula gives \(dg=g_b\,dB_t+g_t\,dt+\tfrac12g_{bb}\,dt=g\,dB_t\), with the time terms cancelling. The initial value is \(g(0,0)=1\).
Check: For \(0\le s\le t\), independent Brownian increments give \(\mathbb{E}[Z_t\mid\mathcal{F}_s]=Z_s\,\mathbb{E}[\exp(B_t-B_s-(t-s)/2)]=Z_s\), since \(B_t-B_s\) is normal with variance \(t-s\) and its exponential moment is \(e^{(t-s)/2}\). Also \(\mathbb{E}[Z_t]=1\). Thus this example is a true martingale, not merely a local martingale.
Worked Example: Drifted Brownian Motion
Let \(X_t=\mu t+\sigma B_t\), with constants \(\mu\in\mathbb{R}\) and \(\sigma\in\mathbb{R}\). Its finite-variation drift has zero quadratic variation and zero covariation with \(B\), so \([X]_t=\sigma^2t\). Therefore $$ Z_t=\mathcal{E}(X)_t =\exp\!\left((\mu-\sigma^2/2)t+\sigma B_t\right). $$ For finite \(T\), this is an admissible Brownian integrand in \(\mathcal{L}^2\): \(\mathbb{E}[Z_s^2]=\exp((2\mu+\sigma^2)s)\), and the integral of this continuous function over \([0,T]\) is finite.
For a direct check, set \(h(t,b)=\exp((\mu-\sigma^2/2)t+\sigma b)\). Its derivatives are \(h_t=(\mu-\sigma^2/2)h\), \(h_b=\sigma h\), and \(h_{bb}=\sigma^2h\). Itô’s formula yields $$ dh=h_b\,dB_t+h_t\,dt+\frac12h_{bb}\,dt =\sigma h\,dB_t+\mu h\,dt =h\,dX_t, $$ and \(h(0,0)=1\). The half-variance term in the exponent is necessary: without it, the Itô correction would leave an extra \(\sigma^2h\,dt/2\).
Check: The normal moment-generating function gives \(\mathbb{E}[Z_t]=\exp((\mu-\sigma^2/2)t)\exp(\sigma^2t/2)=e^{\mu t}\). Conditioning on \(\mathcal{F}_s\) gives \(\mathbb{E}[Z_t\mid\mathcal{F}_s]=Z_s e^{\mu(t-s)}\). Consequently \(Z\) is a martingale on a time interval of positive length exactly when \(\mu=0\); it is not a martingale merely because it is positive or has the displayed exponential form.
Finite Variation and the Ordinary Exponential
Proof. A continuous finite-variation process has zero quadratic variation, so \([A]_t=0\). Substitution into the explicit formula gives \(\mathcal{E}(A-A_0)_t=\exp(A_t-A_0)\). The ordinary chain rule also verifies the defining equation: \(d\exp(A_t-A_0)=\exp(A_t-A_0)\,dA_t\), since \(A\) has finite variation. The initial value is one, and uniqueness of the defining equation identifies this solution with the Doléans-Dade exponential. \(\square\)
Worked Example: A Deterministic Finite-Variation Path
Let \(A_t=t^2\), for \(t\ge0\). This is a continuous finite-variation path with \([A]_t=0\), so $$ \mathcal{E}(A)_t=\exp(t^2). $$ The ordinary derivative is \(\frac{d}{dt}\exp(t^2)=2t\exp(t^2)\), while \(dA_t=2t\,dt\). Hence \(d\mathcal{E}(A)_t=\mathcal{E}(A)_t\,dA_t\), as required.
Check: At \(t=1\), the formula gives \(\mathcal{E}(A)_1=e\); differentiating gives derivative \(2e\), and the defining equation gives \(\mathcal{E}(A)_1\cdot 2=2e\). Both calculations agree. No half-variance correction appears because the path has zero quadratic variation.
Martingales and a Common Mistake
If \(X-X_0=M\) is a continuous local martingale, then the defining equation reads \(Z_t=1+\int_0^t Z_s\,dM_s\). The integral is a local martingale, so \(Z\) is a local martingale. It is positive, and a positive local martingale is a supermartingale by the standard positive-local-martingale theorem. It need not be a true martingale: the local property alone does not supply the integrability and conditional-expectation equality required of a martingale. Novikov’s condition is one useful sufficient condition on a finite horizon \(T\): if \(M_0=0\) and \(\mathbb{E}\exp(\tfrac12[M]_T)<\infty\), then \(\mathcal{E}(M)\) is a true martingale on \([0,T]\). This is the standard Novikov theorem; its proof is beyond this derivation. It is a sufficient condition, not a necessary one.
A frequent error is to apply the ordinary chain rule to Brownian motion and claim that its exponential is \(e^{B_t}\). In fact, for \(q(t,b)=e^b\), the derivatives are \(q_t=0\), \(q_b=e^b\), and \(q_{bb}=e^b\), so Itô’s formula gives \(d(e^{B_t})=e^{B_t}\,dB_t+\tfrac12e^{B_t}\,dt\). The unwanted drift shows that this is not the solution of \(dZ_t=Z_t\,dB_t\). The correction \(e^{B_t-t/2}\) removes that drift, as the first worked example verifies.
The formula is useful whenever an equation has the multiplicative form \(dZ_t=Z_t\,dX_t\): it gives the process explicitly and shows immediately how the continuous martingale fluctuations affect it. For a decomposition \(X=M+A\) into continuous local martingale and finite-variation parts, the finite-variation part enters as \(A_t-A_0\), while the correction is \(-[M]_t/2\), because \([X]=[M]\). This separates the effect of drift from the quadratic variation of the fluctuations.
Check Your Understanding
Use the defining equation and the quadratic-variation correction to justify each answer.
- What is \(\mathcal{E}(X-X_0)_t\) when \(X\) is a continuous semimartingale, and which term accounts for the Itô correction?
- For \(X_t=3B_t\), compute \([X]_t\) and write its Doléans-Dade exponential.
- For \(X_t=2t+B_t\), compute \(\mathbb{E}[\mathcal{E}(X)_t]\). Is this process a martingale?
- Why does a continuous finite-variation process have no quadratic-variation correction in its exponential formula?
- State Novikov’s condition and explain why it is sufficient information for a true-martingale conclusion, whereas positivity alone is not.