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Informal Proof: Itô Process is a Martingale ⇒ Itô Process is Driftless

Let us define our Itô process \(\{X_t\}_{t \geq 0}\) with the standard SDE:

where \(a(t, X_t)\) and \(b(t, X_t)\) are the drift and diffusion terms, and \(W_t\) is a standard Brownian motion process. Fix \(t, s \geq 0\). Because we are given \(X_t\) is a martingale, we have that \(\mathbb{E}[X_{t + s} \mid \mathcal{F}_t] = X_t\) or:

Showing that \(\mathbb{E}[\int_{t}^{t + s} b(X_s, s) dW_s \mid \mathcal{F}_t] = 0\) is a standard calculation of evaluating a stochastic integral as a limit of stochastic integrals of elementary processes, which are converging to \(b(X_s, s)\). See Example 2 on how this works. Furthermore, this result is unsurprising as \(\mathbb{E}[\Delta W] = 0\) and \(\Delta W\) is independent of \(b(X_s, s)\) during this same time period.

So we have that \(0 = \mathbb{E}[\int_{t}^{t + s} a(X_s, s) ds \mid \mathcal{F}_t]\). Brushing aside some technicalities, this also gives us \(0 = \int_{t}^{t + s} \mathbb{E}[a(X_s, s) \mid \mathcal{F}_t] \ ds\). From here, we can cleverly use the first Fundamental Theorem of Calculus to get rid of the expectation:

But as we know \(\int_{t}^{t + s} \mathbb{E}[a(X_s, s) \mid \mathcal{F}_t] ds = 0\) and so the above limit is equal to 0 \(\implies a(X_t, t) = 0\). Furthermore \(t\) was arbitrary and so \(\forall t, a(X_t, t) = 0 \implies X_t\) is driftless.

Note that proving the opposite direction is simpler, at least intuitively: