The Stochastic Exponential

In ordinary calculus, the function that grows in proportion to its own value, $dz = z\,dx$, is simply $z_t = e^{x_t}$. Once $x_t$ is a stochastic process driven by Brownian motion rather than a smooth deterministic path, that identity breaks — Itô's lemma adds a correction term whenever you differentiate a function of a random process, so naively writing $Z_t = e^{X_t}$ no longer solves $dZ_t = Z_t\,dX_t$. The process that actually solves this equation is called the stochastic exponential (or Doléans-Dade exponential) of $X_t$, written $\mathcal{E}(X)_t$.

For $X_t$ a continuous semimartingale with quadratic variation $[X]_t$, the stochastic exponential is:

$$\mathcal{E}(X)_t = \exp\left( X_t - \tfrac{1}{2}[X]_t \right).$$

In plain English: you still exponentiate, but you first subtract off half the accumulated quadratic variation — a built-in correction for the extra randomness that Itô's lemma introduces. This correction is exactly what makes geometric Brownian motion's solution $S_t = S_0 \exp\big( (\mu - \tfrac{1}{2}\sigma^2)t + \sigma W_t \big)$ have that "$-\tfrac12\sigma^2 t$" drift adjustment that confuses almost everyone the first time they see it: it's not an arbitrary tweak, it's the stochastic exponential of the driving process $X_t = \mu t + \sigma W_t$, whose quadratic variation is $[X]_t = \sigma^2 t$.

Worked check: if $X_t = \sigma W_t$ (pure Brownian motion scaled by $\sigma$), its quadratic variation is $[X]_t = \sigma^2 t$, so $\mathcal{E}(X)_t = \exp(\sigma W_t - \tfrac12\sigma^2 t)$ — exactly the martingale exponential used to price options under the risk-neutral measure via Girsanov's theorem, and it satisfies $\mathbb{E}[\mathcal{E}(X)_t] = 1$ for every $t$, a defining property of a true martingale that $e^{X_t}$ alone would not have.