Auditing Ito Derivations with Explicit Moment Checks

What a Moment Check Can Detect

An explicit solution to a stochastic differential equation (SDE) can look plausible and still have a missing drift term, a wrong diffusion coefficient, or an incorrect Ito correction. A useful audit is to compute the solution’s mean and variance by a route independent of the proposed formula, then compare those moments with the SDE. This tutorial carries out that check for geometric Brownian motion and the Ornstein–Uhlenbeck process.

All calculations are under \(\mathbb{P}\) on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). An Ito integral \(\int_0^t H_s\,dB_s\) below has a progressively measurable integrand \(H\) satisfying \(\mathbb{E}\int_0^t H_s^2\,ds<\infty\); this class is denoted by \(\mathcal{L}^2\). In that case the integral is a square-integrable martingale, and the Ito isometry gives \(\mathbb{E}[(\int_0^t H_s\,dB_s)^2]=\mathbb{E}\int_0^t H_s^2\,ds\). No risk-neutral measure \(\mathbb{Q}\) is used here.

We will also use the normal moment-generating formula. If \(B_t\sim N(0,t)\), then completing the square in its density gives \(\mathbb{E}[e^{aB_t}]=e^{a^2t/2}\) for every real \(a\). This provides a direct way to calculate moments of an exponential solution without taking expectations of an SDE.

Geometric Brownian Motion: The Independent Moment Calculation

Consider the SDE with initial value \(s>0\), drift \(\mu\in\mathbb{R}\), and diffusion coefficient \(\sigma\in\mathbb{R}\):

The drift coefficient is \(\mu S_t\), while the diffusion coefficient is \(\sigma S_t\). To obtain its candidate explicit solution, apply Ito’s formula to \(f(t,x)=\log x\) for \(x>0\). Its derivatives are \(f_t=0\), \(f_x=1/x\), and \(f_{xx}=-1/x^2\). The second-order term is therefore \(\tfrac12 f_{xx}(\sigma x)^2dt=-\tfrac12\sigma^2dt\), so \(d\log S_t=(\mu-\tfrac12\sigma^2)dt+\sigma\,dB_t\). Integrating and exponentiating gives the candidate below. Equivalently, applying Ito’s formula directly verifies it:

For that direct verification, set \(g(t,b)=s\exp((\mu-\tfrac12\sigma^2)t+\sigma b)\). Then \(g_t=(\mu-\tfrac12\sigma^2)g\), \(g_b=\sigma g\), and \(g_{bb}=\sigma^2g\). Ito’s formula gives \(dg=(g_t+\tfrac12g_{bb})dt+g_b\,dB_t=\mu g\,dt+\sigma g\,dB_t\), as required, and \(g(0,0)=s\). The coefficients are globally Lipschitz and of linear growth, so the standard existence-and-uniqueness theorem for SDEs ensures this solution agrees with the unique strong solution.

Proof. The displayed explicit solution gives \(S_t^k=s^k\exp(k(\mu-\tfrac12\sigma^2)t+k\sigma B_t)\) for \(k=1,2\). Applying the normal moment-generating formula with \(a=k\sigma\) yields \(\mathbb{E}[S_t^k]=s^k\exp(k(\mu-\tfrac12\sigma^2)t+\tfrac12k^2\sigma^2t)\). Substituting \(k=1\) gives \(se^{\mu t}\), and substituting \(k=2\) gives \(s^2e^{(2\mu+\sigma^2)t}\). Subtracting the square of the mean from the second moment gives \(s^2e^{2\mu t}(e^{\sigma^2t}-1)\). These moments are finite, including the fourth moment, which will justify the Ito-integral check below. \(\square\)

A Second Audit Through Ito’s Formula

There is another check for the geometric Brownian motion second moment. Apply Ito’s formula to \(f(x)=x^2\). Its derivatives are \(f'(x)=2x\) and \(f''(x)=2\); the correction term is \(\tfrac12 f''(S_t)(\sigma S_t)^2dt=\sigma^2S_t^2dt\). Thus

Taking expectations is justified here, rather than automatic: the explicit formula has finite fourth moment, so \(\mathbb{E}\int_0^t(2\sigma S_s^2)^2ds=4\sigma^2\int_0^t\mathbb{E}[S_s^4]ds<\infty\). The stochastic integral is consequently a true square-integrable martingale with mean zero. Taking expectations in the integrated identity gives \(m_2(t)=s^2+(2\mu+\sigma^2)\int_0^t m_2(u)\,du\), where \(m_2(t)=\mathbb{E}[S_t^2]\). The explicit moment already derived satisfies this equation: its derivative is \((2\mu+\sigma^2)s^2e^{(2\mu+\sigma^2)t}\) and its initial value is \(s^2\). This agreement audits both the Ito correction and the second-moment exponent.

The equation \(dB_t\,dB_t=dt\) accounts for the positive \(\sigma^2S_t^2dt\) term. The other multiplication rules are \(dB_t\,dt=0\) and \(dt\,dt=0\); they do not remove the Brownian quadratic-variation term. The equality \(dB_t\,dB_t=dt\) is an Ito-calculus rule, not an ordinary chain rule for differentiable paths.

Ornstein–Uhlenbeck: Mean and Variance from the Integral

Consider an Ornstein–Uhlenbeck process with initial value \(x\), mean-reversion rate \(\lambda>0\), and constant diffusion coefficient \(\eta\in\mathbb{R}\):

Here the drift is \(-\lambda X_t\) and the diffusion coefficient is \(\eta\). Multiplying by the deterministic integrating factor \(e^{\lambda t}\), using the Ito product rule, and integrating gives the candidate solution

For fixed \(t\), the integrand is deterministic and \(\int_0^t e^{-2\lambda(t-s)}ds=(1-e^{-2\lambda t})/(2\lambda)<\infty\), so it belongs to \(\mathcal{L}^2\). The integral has mean zero and, by the Ito isometry, variance \(\int_0^t e^{-2\lambda(t-s)}ds\). Therefore

Proof. The stochastic integral in the explicit solution is a square-integrable martingale at its upper time limit, so its expectation is zero. Its covariance with the deterministic constant \(xe^{-\lambda t}\) is zero. Applying the Ito isometry to its variance gives \(\eta^2\int_0^t e^{-2\lambda(t-s)}ds\). Evaluating the integral yields \(\eta^2(1-e^{-2\lambda t})/(2\lambda)\), proving both formulas. \(\square\)

Common Mistake: Dropping the Ito Correction

How to Use the Audit

For an Ito-derived explicit formula, separate two questions. First, does Ito’s formula verify that the candidate satisfies the SDE, including the second-derivative correction? Second, do independent calculations of its mean and variance agree with the drift and diffusion in that SDE? The checks are complementary: substitution detects a pathwise algebra error, while moments can reveal a wrong exponential factor or a stochastic integral whose expectation has been assumed to vanish without adequate integrability.

Do not confuse a local martingale with a true martingale. A stochastic integral with an \(\mathcal{L}^2\) integrand is a true square-integrable martingale, so its expectation and conditional expectation increments are zero. Without that integrability (or another sufficient condition), an Ito integral may only be a local martingale, and taking expectations may be invalid. In the checks above, the needed square-integrability was verified explicitly.