Mixed Problems on Ito Martingales and Geometric Brownian Motion

Combining Ito Calculations with Martingale Tests

In mixed problems, a single Ito calculation may produce several different conclusions: a process may be a martingale, a transformed process may have a known moment, or a continuous martingale may satisfy the conditions of Levy's characterization. The key is to keep the drift calculation, integrability check, and quadratic-variation test separate. A zero drift term alone does not establish that a process is a true martingale.

Throughout, work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\) relative to the filtration. Ito integrals below are with respect to this Brownian motion. Their integrands are adapted and progressively measurable, and when a true square-integrable Ito integral on \([0,T]\) is used, it belongs to \(\mathcal{L}^2\): \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). The Ito rules are \(dB_t\,dB_t=dt\), \(dB_t\,dt=0\), and \(dt\,dt=0\); they encode quadratic variation and do not constitute an ordinary chain rule.

Exponential Processes: Cancel the Drift, Then Check the Martingale

Proof. Set \(f(t,x)=\exp(\lambda x-\tfrac12\lambda^2t)\). Its derivatives are \(f_t=-\tfrac12\lambda^2f\), \(f_x=\lambda f\), and \(f_{xx}=\lambda^2f\). Ito's formula, including the second-derivative correction, gives

This calculation gives a local martingale; to verify a true martingale, use the explicit expression and the independent normal increment \(B_t-B_s\), which is independent of \(\mathcal{F}_s\) and has law \(N(0,t-s)\). The normal moment-generating function gives \(\mathbb{E}[\exp(\lambda(B_t-B_s))]=\exp(\tfrac12\lambda^2(t-s))\). Therefore

The same calculation at \(s=0\) gives \(\mathbb{E}[Z_t]=1\), so \(Z_t\) is integrable. This proves the martingale property, not just the local martingale property. \(\square\)

Geometric Brownian Motion: Moments from Ito's Formula

Let \(s>0\), and let \(\mu,\sigma\in\mathbb{R}\) be constants. Geometric Brownian motion is the positive process \(S_t=s\exp((\mu-\tfrac12\sigma^2)t+\sigma B_t)\). It satisfies \(dS_t=\mu S_t\,dt+\sigma S_t\,dB_t\): the drift coefficient is \(\mu S_t\), while the diffusion coefficient is \(\sigma S_t\). When \(\sigma=0\), the process is deterministic; the moment formula below includes that case.

Proof. Apply Ito's formula to \(f(x)=x^p\) on \(x>0\). Its derivatives are \(f'(x)=px^{p-1}\) and \(f''(x)=p(p-1)x^{p-2}\). Since \(S_t>0\), these derivatives are defined along the process, and the formula gives

Write \(c=p\mu+\tfrac12p(p-1)\sigma^2\) and \(Y_t=e^{-ct}S_t^p\). Applying the product rule (the deterministic factor has zero quadratic variation and zero covariation with \(S^p\)) cancels the drift and yields \(dY_t=p\sigma Y_t\,dB_t\). To justify taking expectations, \(Y\)'s stochastic-integral integrand \(p\sigma Y_t\) is square-integrable on each finite interval: the explicit exponential expression for \(S_t\) and the normal moment-generating function show that \(\mathbb{E}[S_t^{2p}]\) is a finite exponential function of \(t\), whose integral on \([0,T]\) is finite. Thus \(\int_0^t p\sigma Y_u\,dB_u\) is an \(\mathcal{L}^2\) Ito integral. It is a true martingale, since its conditional expectation increments are zero. Consequently \(\mathbb{E}[Y_t]=Y_0=s^p\), proving the moment formula. The same conditional-expectation argument gives \(\mathbb{E}[Y_t\mid\mathcal{F}_u]=Y_u\) for \(u\le t\). \(\square\)

A Martingale and Quadratic-Variation Test for Brownian Motion

Levy's characterization is a way to identify Brownian motion without deriving its distribution directly. In the form used here, it states: if \(M_t\) is a continuous local martingale with \(M_0=0\) and quadratic variation \([M]_t=t\), then \(M_t\) is a Brownian motion relative to the given filtration. This is a standard result from the earlier material on Levy's characterization; the result is invoked here, not reproved.

Common Pitfall: A Vanishing Drift Is Not the Whole Argument

A frequent error is to apply the ordinary chain rule to an exponential of Brownian motion. For \(f(t,x)=e^x\), the derivatives are \(f_t=0\), \(f_x=e^x\), and \(f_{xx}=e^x\). Ito's formula therefore gives

The incorrect calculation \(d(e^{B_t})=e^{B_t}\,dB_t\) omits the \(\tfrac12f_{xx}\,dt\) correction. It would falsely label \(e^{B_t}\) a martingale. In fact, \(\mathbb{E}[e^{B_t}]=e^{t/2}\), which is not constant for \(t>0\). Multiplying by \(e^{-t/2}\) cancels this drift, and the exponential-martingale calculation above verifies the resulting process is a true martingale.

A second distinction is equally important: a local martingale need not be a true martingale. A zero drift in an Ito differential gives a local-martingale conclusion in general. To claim a true martingale, verify an appropriate condition, such as square-integrability of the Ito integrand on finite intervals or a direct conditional-expectation identity. In the constant-parameter exponential example, the conditional expectation was computed directly; in the sign-integral example, square-integrability was checked before invoking Levy's characterization.