From Ito's Correction to an Exponential Martingale
Exponentials turn Brownian motion into positive processes that often appear to be martingales. The decisive point is the correction in Ito's formula: the quadratic-variation term cancels the deterministic drift placed in the exponent. That cancellation produces a local martingale, but a separate argument is needed to establish a true martingale. For constant coefficients, conditional Gaussian expectations give that argument directly.
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. For a fixed finite horizon \(T\), an Ito integral \(\int_0^t H_s\,dB_s\) is used with a progressively measurable integrand \(H\) satisfying \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\); this is the class \(\mathcal{L}^2\). An integral whose integrand is only locally square-integrable is understood after stopping at times that make it square-integrable.
For the general expression, the Ito integral is defined locally when its integrand is locally square-integrable; in the constant-kernel results below, the relevant integrands are in \(\mathcal{L}^2\) on every finite horizon. A useful sufficient condition for the general stochastic exponential to be a true martingale on \([0,T]\) is Novikov's condition: \(\mathbb{E}\exp\!\left(\tfrac12\int_0^T\theta_s^2\,ds\right)<\infty\). Novikov's theorem is a standard result; its proof belongs with the later material on Girsanov's theorem. The condition is sufficient, not necessary. We will verify the constant-kernel case directly instead.
The Constant-Kernel Exponential
Proof. Take \(f(t,x)=\exp(\theta x-\tfrac12\theta^2t)\). Its derivatives are \(f_t=-\tfrac12\theta^2f\), \(f_x=\theta f\), and \(f_{xx}=\theta^2f\). Ito's formula, including its second-derivative correction, gives
Thus Ito's formula gives a local martingale. For the true martingale property, use the independent Brownian increment \(B_t-B_s\), which is independent of \(\mathcal{F}_s\) and has distribution \(N(0,t-s)\). The moment-generating function of this normal variable yields
This conditional expectation is well-defined: the same Gaussian calculation gives \(\mathbb{E}[Z_t]=1\), so \(Z_t\) is integrable. For the second moment, \(Z_t^2=\exp(2\theta B_t-\theta^2t)\), and the normal moment-generating function gives \(\mathbb{E}[Z_t^2]=e^{2\theta^2t-\theta^2t}=e^{\theta^2t}\). When \(\theta=0\), the formula itself gives \(Z_t=1\), and all the claims still hold. \(\square\)
Worked Example: Checking a Constant Exponential
Set \(\theta=\tfrac32\) and \(t=2\). Compute the first and second moments of \(Z_t\), and verify the conditional martingale identity from time \(s=1\).
The exponent is \(\tfrac32 B_t-\tfrac12(\tfrac94)t\). At \(t=2\), this is \(\tfrac32B_2-\tfrac94\). The theorem gives
For \(s=1\), the increment \(B_2-B_1\) is independent of \(\mathcal{F}_1\) and has variance \(1\). Therefore its exponential factor has conditional expectation \(e^{(1/2)(9/4)}e^{-9/8}=e^{9/8}e^{-9/8}=1\), so \(\mathbb{E}[Z_2\mid\mathcal{F}_1]=Z_1\). Check: taking expectations of this conditional identity gives \(\mathbb{E}[Z_2]=\mathbb{E}[Z_1]=1\), in agreement with the direct first-moment calculation.
Geometric Brownian Motion and Its Moments
Let \(S_0\ge0\), and let \(\mu,\sigma\in\mathbb{R}\) be constants. The geometric Brownian motion (GBM) with drift \(\mu\) and diffusion coefficient \(\sigma\) is the process solving \(dS_t=\mu S_t\,dt+\sigma S_t\,dB_t\), with initial value \(S_0\). Here the drift is the coefficient of \(dt\), and the diffusion coefficient is the coefficient of \(dB_t\). Its explicit candidate is
Proof. First suppose \(S_0>0\), and apply Ito's formula to \(f(t,x)=S_0\exp((\mu-\tfrac12\sigma^2)t+\sigma x)\). The derivatives are \(f_t=(\mu-\tfrac12\sigma^2)f\), \(f_x=\sigma f\), and \(f_{xx}=\sigma^2f\). Hence
The Gaussian moment-generating function gives, for each real \(p\), \(\mathbb{E}[S_t^p]=S_0^p\exp((p\mu+\tfrac12p(p-1)\sigma^2)t)\) when \(S_0>0\). In particular, \(p=1\) and \(p=2\) give the stated moments. For \(0\le s\le t\), independence of \(B_t-B_s\) from \(\mathcal{F}_s\) gives \(\mathbb{E}[S_t\mid\mathcal{F}_s]=S_s e^{\mu(t-s)}\). The conditional expectation exists because the first moment is finite. If \(\mu=0\), this equals \(S_s\), proving the martingale property. If \(\mu\ne0\) and \(S_0>0\), then for any \(t>s\), \(e^{\mu(t-s)}\ne1\), so \(S_s e^{\mu(t-s)}\ne S_s\) almost surely; the martingale condition fails. If \(S_0=0\), the explicit solution is identically zero, including when \(\sigma=0\), and its conditional expectations are zero. \(\square\)
The moment formulas can also be checked using Ito's formula on powers of the process. For \(f(x)=x^2\), the derivatives are \(f'(x)=2x\) and \(f''(x)=2\); the correction term is therefore \(\tfrac12 f''(S_t)(dS_t)^2=\sigma^2S_t^2\,dt\). The resulting identity is
Its stochastic integral has zero mean: the integrand \(2\sigma S_t^2\) is progressively measurable and in \(\mathcal{L}^2\) on \([0,T]\), since \(\mathbb{E}\int_0^T4\sigma^2S_s^4\,ds<\infty\) by the Gaussian fourth-moment formula above. The Ito isometry then gives a square-integrable martingale with zero expectation. Taking expectations in the integral identity yields the moment equation corresponding to the displayed second-moment formula. This provides an Ito-based check independent of directly evaluating the second moment at the final time.
Worked Example: First and Second Moments at a Fixed Time
Take \(S_0=4\), \(\mu=\tfrac14\), \(\sigma=\tfrac12\), and \(t=2\). Find the mean, second moment, and variance.
The first-moment exponent is \(\mu t=(\tfrac14)2=\tfrac12\), so \(\mathbb{E}[S_2]=4e^{1/2}\). The second-moment exponent is \((2\mu+\sigma^2)t=(\tfrac12+\tfrac14)2=\tfrac32\), so \(\mathbb{E}[S_2^2]=4^2e^{3/2}=16e^{3/2}\). Subtracting the squared mean gives
Check: the general variance formula is \(S_0^2e^{2\mu t}(e^{\sigma^2t}-1)\). Substitution gives \(16e^{(1/2)2}(e^{(1/4)2}-1)=16e(e^{1/2}-1)\), exactly the same result.
When Is the Process a Martingale?
The condition \(\mu=0\) concerns the process under the stated probability measure \(\mathbb{P}\). It is not a statement about a risk-neutral measure \(\mathbb{Q}\), nor does it assert that a discounted asset price is a martingale. If a constant interest rate \(r\) is introduced, then the conditional expectation of the discounted process is \(\mathbb{E}[e^{-rt}S_t\mid\mathcal{F}_s]=e^{-rs}S_s e^{(\mu-r)(t-s)}\). For \(S_0>0\), this is a martingale under \(\mathbb{P}\) exactly when \(\mu=r\). A change from \(\mathbb{P}\) to \(\mathbb{Q}\) is a separate operation and is not needed for these moment calculations.
Worked Example: A GBM That Is Not a Martingale
Let \(S_0=3\), \(\mu=\tfrac13\), and \(\sigma=\tfrac23\). Compare the conditional expectation from time \(s=1\) to \(t=2\) with \(S_1\).
The conditional-moment formula gives \(\mathbb{E}[S_2\mid\mathcal{F}_1]=S_1e^{(1/3)(2-1)}=S_1e^{1/3}\). Since \(S_1=3\exp((\tfrac13-\tfrac12\cdot\tfrac49)+\tfrac23B_1)=3\exp(\tfrac19+\tfrac23B_1)>0\), the factor \(e^{1/3}\) is strictly greater than \(1\), and the conditional expectation is not \(S_1\). Thus the process is not a martingale. Its mean at time \(2\), computed independently from the first-moment formula, is \(3e^{(1/3)2}=3e^{2/3}\). Check: \(\mathbb{E}[S_1e^{1/3}]=e^{1/3}(3e^{1/3})=3e^{2/3}\), agreeing with that unconditional mean.
Common Pitfall: Omitting the Ito Correction
A tempting but incorrect calculation is to differentiate \(S_0e^{\mu t+\sigma B_t}\) and conclude that it solves \(dS_t=\mu S_t\,dt+\sigma S_t\,dB_t\). The missing term is the second derivative correction: for \(f(t,x)=S_0e^{\mu t+\sigma x}\), \(f_t=\mu f\), \(f_x=\sigma f\), and \(f_{xx}=\sigma^2f\), so Ito's formula gives drift \((\mu+\tfrac12\sigma^2)f\), not \(\mu f\). The correct exponent must subtract \(\tfrac12\sigma^2t\).
A second error is to infer that every positive local martingale is a true martingale. The local-martingale conclusion from Ito's formula alone does not establish integrability and the conditional-expectation identity. In the constant-kernel exponential, those properties were verified explicitly using the normal moment-generating function. For more general kernels, conditions such as Novikov's condition provide a sufficient route, but they are not automatic.
| Process or condition | Conclusion | Reason |
|---|---|---|
| \(\exp(\theta B_t-\tfrac12\theta^2t)\), constant \(\theta\) | True martingale | Conditional Gaussian expectation equals its current value |
| GBM with \(S_0>0\) under \(\mathbb{P}\) | Martingale exactly when \(\mu=0\) | Conditional mean is \(S_se^{\mu(t-s)}\) |
| GBM with \(S_0=0\) | Zero martingale for all coefficients | The explicit solution is identically zero |
| General stochastic exponential satisfying Novikov | True martingale on the stated finite horizon | Novikov's condition is sufficient |
Check Your Understanding
Use the conditional-expectation identities and moment formulas above to justify each answer.
- For \(\theta=-2\), compute \(\mathbb{E}[Z_3]\) and \(\mathbb{E}[Z_3^2]\).
- For GBM with \(S_0=2\), \(\mu=0\), and \(\sigma=1\), find the first and second moments at time \(t\).
- Explain why the Ito formula for \(S_t^2\) contains the term \(\sigma^2S_t^2\,dt\).
- For \(S_0>0\), use the conditional expectation to decide whether GBM with \(\mu=-1\) is a martingale on a nontrivial time interval.
- State Novikov's condition and explain why it is sufficient rather than necessary for the conclusion it guarantees.