Ito's Formula for Exponential Processes

Why an Exponential Needs a Correction

For a differentiable path $x_t$, the ordinary chain rule gives $d(e^{x_t})=e^{x_t}\,dx_t$. For a Brownian integral, Itô’s formula adds a second-derivative term because the integral has nonzero quadratic variation. Consequently, simply exponentiating a Brownian integral does not produce a process whose differential consists only of a stochastic-integral term. Subtracting half of its quadratic variation in the exponent cancels that extra drift.

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 time horizon $T$. Throughout, $H$ is progressively measurable and belongs to $\mathcal{L}^2$ on $[0,T]$, meaning $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. In particular, the Itô integral $\int_0^t H_s\,dB_s$ is defined for $0\le t\le T$.

Definition: For such an integrand $H$, set $M_t=\int_0^t H_s\,dB_s$ and $A_t=\int_0^t H_s^2\,ds$. The stochastic exponential of $M$, on $[0,T]$, is $$ \mathcal{E}(M)_t=\exp\!\left(M_t-\frac12 A_t\right). $$ The process $A_t$ is the quadratic variation $[M]_t$.

Since $H\in\mathcal{L}^2$, $A_T$ is finite almost surely. The process $M$ has continuous paths, so the exponential is continuous and strictly positive. The correction is $-\tfrac12 A_t$, not an arbitrary normalization: it is exactly the term required by the quadratic-variation contribution in Itô’s formula.

The Exponential Differential

Proposition: If $H$ is progressively measurable and $\mathbb{E}\int_0^T H_s^2\,ds<\infty$, then $Z_t=\mathcal{E}(M)_t$, with $M_t=\int_0^t H_s\,dB_s$, satisfies $$ dZ_t=Z_tH_t\,dB_t,\qquad Z_0=1. $$ The stochastic integral on the right is understood locally: $Z H$ need not have finite expected squared integral on the whole horizon.

Proof. Write $Y_t=M_t-\tfrac12 A_t$, so that $Z_t=e^{Y_t}$. The Itô integral $M$ has quadratic variation $d[M]_t=H_t^2\,dt$. The process $A$ has finite variation, and a continuous finite-variation term has zero quadratic variation and zero covariation with $M$. Therefore $d[Y]_t=H_t^2\,dt$ and $dY_t=H_t\,dB_t-\tfrac12H_t^2\,dt$. Apply Itô’s formula to $g(y)=e^y$: $g'(y)=e^y$ and $g''(y)=e^y$. Including the second-derivative correction explicitly gives $$ dZ_t=g'(Y_t)\,dY_t+\frac12g''(Y_t)\,d[Y]_t =Z_t\left(H_t\,dB_t-\frac12H_t^2\,dt\right)+\frac12Z_tH_t^2\,dt =Z_tH_t\,dB_t. $$ Also $M_0=A_0=0$, hence $Z_0=1$.

To justify the local integral, let $\tau_n$ be the first time $Z$ reaches $n$ or $A$ reaches $n$, capped at $T$. Before $\tau_n$, $Z\le n$ and $A_{\tau_n}\le n$, so $$ \mathbb{E}\int_0^{\tau_n} Z_s^2H_s^2\,ds \le n^2\mathbb{E}[A_{\tau_n}] \le n^3. $$ Thus $Z H\mathbf{1}_{[0,\tau_n]}$ is progressively measurable and square-integrable, and its Itô integral is a square-integrable martingale. The stopped equation is therefore a martingale-integral equation; as the stopping levels increase, it gives the stated local equation. This proves a local martingale conclusion, not by itself a true martingale conclusion: this calculation has not established the unconditional or conditional-expectation identities required of a martingale. $\square$

The cancellation can also be read directly from the Itô multiplication rules. The exponent has a drift $-\tfrac12H_t^2\,dt$, while its quadratic variation contributes $+\tfrac12H_t^2\,dt$ after applying the exponential function. The two drift terms cancel. In differential shorthand, the relevant rule is $dB_t\,dB_t=dt$; the finite-variation products $dB_t\,dt$ and $dt\,dt$ vanish.

Worked Examples

Worked Example: A Constant Integrand

Take $T=1$ and $H_t=2$ for $0\le t\le1$. This integrand is progressively measurable and $\mathbb{E}\int_0^1H_s^2\,ds=4$. Directly, $$ M_t=2B_t,\qquad A_t=4t,\qquad Z_t=e^{2B_t-2t}. $$ The proposition gives $dZ_t=2Z_t\,dB_t$.

Check by direct Itô calculation. Regard the displayed candidate as $f(t,x)=e^{2x-2t}$, evaluated at $x=B_t$. Its derivatives are $f_t=-2f$, $f_x=2f$, and $f_{xx}=4f$. The Itô formula, including its correction, gives $$ df(t,B_t)=f_t\,dt+f_x\,dB_t+\frac12 f_{xx}\,dt =(-2f+2f)\,dt+2f\,dB_t =2f\,dB_t. $$ Since $f(t,B_t)=Z_t$, this recovers the stated equation. At $t=0$, the candidate equals $1$, as required.

Worked Example: A Piecewise Deterministic Integrand

On $[0,1]$, let $H_s=1$ for $0\le s\le\tfrac12$ and $H_s=2$ for $\tfrac12<s\le1$. Its squared-integrability check is $$ \mathbb{E}\int_0^1H_s^2\,ds =\frac12+4\cdot\frac12=\frac52. $$ For $t\le\tfrac12$, $M_t=B_t$ and $A_t=t$, so $Z_t=e^{B_t-t/2}$. For $t\ge\tfrac12$, $$ M_t=B_{1/2}+2(B_t-B_{1/2})=2B_t-B_{1/2},\qquad A_t=\frac12+4\left(t-\frac12\right)=4t-\frac32. $$ Thus $Z_t=\exp(2B_t-B_{1/2}-2t+\tfrac34)$ on the second interval, and $Z_1=\exp(2B_1-B_{1/2}-\tfrac54)$. The differential is $dZ_t=Z_t\,dB_t$ before $\tfrac12$ and $dZ_t=2Z_t\,dB_t$ after $\tfrac12$.

Check the joining point and the equation. At $t=\tfrac12$, the first expression is $e^{B_{1/2}-1/4}$; the second is $e^{2B_{1/2}-B_{1/2}-1+3/4}=e^{B_{1/2}-1/4}$, so there is no jump. On the second interval, condition on the already observed value $B_{1/2}=c$ and use $f(t,x)=e^{2x-c-2t+3/4}$. Its derivatives are $f_t=-2f$, $f_x=2f$, $f_{xx}=4f$. Hence $$ f_t\,dt+f_x\,dB_t+\frac12f_{xx}\,dt =(-2f+2f)\,dt+2f\,dB_t, $$ which verifies the second-interval equation. On the first interval, $f(t,x)=e^{x-t/2}$ has derivatives $f_t=-f/2$, $f_x=f$, $f_{xx}=f$, and the drift is $-f/2+f/2=0$. Both pieces agree with the proposed differential.

Worked Example: An Integrand Chosen from the Observed Past

On $[0,\tfrac32]$, define $I=\mathbf{1}_{\{B_{1/2}>0\}}$ and $H_s=I\mathbf{1}_{(1/2,\,3/2]}(s)$. The event is known at time $\tfrac12$, so this process is progressively measurable. The strict inequality in the event is part of the definition; since $B_{1/2}$ has a continuous normal distribution, the boundary event $B_{1/2}=0$ has probability zero. By symmetry, $\mathbb{P}(I=1)=\tfrac12$, and $$ \mathbb{E}\int_0^{3/2}H_s^2\,ds =\left(1\right)\mathbb{E}[I] =\frac12. $$ For $t\le\tfrac12$, $Z_t=1$. For $t\ge\tfrac12$, $$ M_t=I(B_t-B_{1/2}),\qquad A_t=I(t-\tfrac12),\qquad Z_t=e^{I(B_t-B_{1/2})-\frac12I(t-\frac12)}. $$ The differential is zero before $\tfrac12$ and $dZ_t=I Z_t\,dB_t$ after it.

Check both possible values of the observed indicator. Once time $\tfrac12$ has passed, $I$ is fixed by the observed past. If $I=0$, the candidate is identically $1$, and both sides of the equation are zero. If $I=1$, condition on $B_{1/2}=c$ and set $f(t,x)=e^{x-c-(t-1/2)/2}$. Here $f_t=-f/2$, $f_x=f$, and $f_{xx}=f$. Itô’s formula gives $$ f_t\,dt+f_x\,dB_t+\frac12f_{xx}\,dt =(-f/2+f/2)\,dt+f\,dB_t=f\,dB_t. $$ Thus the equation holds in either case, including at the initial value $Z_{1/2}=1$.

The Converse and the Role of Positivity

Theorem: Suppose $Z_0=1$, $Z$ is a strictly positive continuous adapted process, and, locally on $[0,T]$, it satisfies $dZ_t=Z_tH_t\,dB_t$ for a progressively measurable $H$ with $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. Then $$ Z_t=\exp\!\left(\int_0^t H_s\,dB_s-\frac12\int_0^t H_s^2\,ds\right). $$

Proof. Apply Itô’s formula to $g(z)=\log z$, which is twice differentiable on the positive half-line. Its derivatives are $g'(z)=1/z$ and $g''(z)=-1/z^2$. The quadratic variation of $Z$, from its stated stochastic differential, is $d[Z]_t=Z_t^2H_t^2\,dt$. Therefore the correction term is explicitly $$ d(\log Z_t)=\frac1{Z_t}\,dZ_t+\frac12\left(-\frac1{Z_t^2}\right)d[Z]_t =H_t\,dB_t-\frac12H_t^2\,dt. $$ Integrating from $0$ to $t$ and using $\log Z_0=0$ gives $\log Z_t=\int_0^tH_s\,dB_s-\tfrac12\int_0^tH_s^2\,ds$. Exponentiating proves the representation. In particular, there is at most one strictly positive solution with initial value $1$: any such solution must equal this same expression. $\square$

Positivity is essential to this converse argument because it permits applying $\log Z_t$. The exponential constructed from the formula is strictly positive; the theorem does not assert a corresponding logarithmic representation for a process that can reach zero.

Common Pitfall: Omitting the Correction

A tempting but incorrect calculation is to set $Y_t=\exp(\int_0^tH_s\,dB_s)$ and use the ordinary chain rule to write $dY_t=Y_tH_t\,dB_t$. For $H_t=2$, this would assert that $Y_t=e^{2B_t}$ satisfies $dY_t=2Y_t\,dB_t$. But $f(t,x)=e^{2x}$ has $f_t=0$, $f_x=2f$, and $f_{xx}=4f$. Itô’s formula gives $$ d(e^{2B_t})=2e^{2B_t}\,dB_t+\frac12(4e^{2B_t})\,dt =2e^{2B_t}\,dB_t+2e^{2B_t}\,dt. $$ The missing drift is not negligible. Instead, $e^{2B_t-2t}$ has time derivative $-2f$, which cancels the correction $+\tfrac12(4f)=2f$, leaving only the stochastic term.

The differential equation identifies the stochastic exponential as a positive local martingale. A local martingale is a process that becomes a martingale after suitable stopping; a martingale additionally satisfies the relevant integrability and conditional-expectation identities without that stopping. The derivation here establishes the former property through localization, but does not establish the latter. That distinction matters when an exponential is later used as a density process: positivity and the differential alone do not complete the required martingale verification.

Check Your Understanding

For each question, identify the quadratic-variation correction explicitly.

For $H_t=3$ on $[0,T]$, write $M_t$, $A_t$, and the stochastic exponential, then derive its differential.
Why does the exponential of $\int_0^t H_s\,dB_s$ generally have a drift term under Itô’s formula?
For the piecewise integrand in the second worked example, verify directly that the two expressions for $Z_t$ agree at $t=\tfrac12$.
What does the localization argument establish about $Z$, and what martingale property does it not establish?
Why does the converse representation require strict positivity of $Z$?