Feller Condition

In the Heston stochastic volatility model, the variance $v_t$ mean-reverts toward a long-run level at speed $\kappa$, with the size of its random wiggles scaled by a "vol-of-vol" parameter $\xi$ (bigger $\xi$ means variance itself is noisier). The catch is that when variance gets close to zero, the random shocks (proportional to $\sqrt{v_t}$) can shrink faster than the pull back up (proportional to $\kappa$ times the long-run level $\theta$), and in that case the process really can hit zero — variance briefly "touching the floor" before bouncing back. Whether that's possible or not depends on a single condition.

The Feller condition states that variance stays strictly positive (never touches zero) if and only if:

$$2 \kappa \theta \geq \xi^2 .$$

In plain English: the mean-reversion pull toward the long-run variance level, scaled up by two, must be at least as strong as the square of how noisy variance's own volatility is. If $2\kappa\theta \geq \xi^2$ holds, the process is a well-behaved diffusion that never actually reaches zero variance. If it fails, the discretized process used in simulation can go negative unless special care is taken (e.g. truncation or reflection schemes), and the "true" continuous-time process still spends time arbitrarily close to zero even if it doesn't literally hit it.

Worked example: with $\kappa = 2$, $\theta = 0.04$ (long-run variance corresponding to 20% vol), and $\xi = 0.3$, check $2\kappa\theta = 2 \times 2 \times 0.04 = 0.16$ against $\xi^2 = 0.09$. Since $0.16 \geq 0.09$, the Feller condition holds and variance is guaranteed to stay strictly positive. Many Heston calibrations to real equity index options end up violating this condition, because matching the steep observed volatility skew requires a large $\xi$.