Heston Variance, the Feller Condition, and the Ornstein–Uhlenbeck Analogy
Summary
The post examines a mistaken inference from an older version of a reference page: that Heston volatility should be negative half the time and that the Feller condition must always fail. The correction separates an Ornstein–Uhlenbeck process from volatility itself. Squaring an OU process and then taking the principal square root yields its absolute value, which is nonnegative; the signed OU state is not the volatility.
Applying Ito's lemma to the square of that particular OU process produces a CIR variance process with parameter restrictions that violate the Feller condition. But this construction covers only a restricted subset of CIR parameter choices because the Heston variance process has three parameters while the OU construction has two. General Heston models need not use that special parameterization, so the violation is not universal. The discussion is a conceptual correction, tied to the cited page version, rather than a full treatment of boundary behavior or calibration.
Key ideas
- The principal square root of the square of an OU state is its absolute value, not the signed state.
- Squaring the specified OU process yields a restricted CIR parameterization.
- That special parameterization violates the Feller condition under the stated mapping.
- The OU construction does not represent the full three-parameter space of Heston variance models.
- The document notes that its trigger was a page version that has since changed.
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Full text
# Negative volatility and violated Feller condition in Heston model
# Negative volatility and violated Feller condition in Heston model
### EDIT: this question is about the Wikipedia page on april 14, 2025. This page has since been changed.
I think I am missing something, because the equations for the Heston model imply that the volatility will be negative half of the time, and also that the Feller condition is always violated. What is wrong with my reasoning below?
Starting from Wikipedia, the volatility $\sqrt{\nu_t}$ follows that distribution
$$d\sqrt{\nu_t} = -\beta\sqrt{\nu_t}dt + \delta dW_t.$$ This is an Ornstein-Uhlenbeck process with a (long-term) mean of 0. Therefore, the volatility $\sqrt{\nu_t}$ will be negative half of the time.
Next we can apply Ito's lemma to calculate $d\nu_t$. Let $f(\sqrt{\nu_t})=\nu_t$, so $f'(\sqrt{\nu_t})=2\sqrt{\nu_t}$ and $f''(\sqrt{\nu_t})=2$. This gives us
$$d\nu_t = \left(-\beta\sqrt{\nu_t}\cdot 2\sqrt{\nu_t}+\frac{\delta^2}{2}2\right) dt + \delta \cdot 2\sqrt{\nu_t}dW_t.$$ $$d\nu_t = \left(-2\beta\nu_t+\delta^2\right) dt + 2\delta \sqrt{\nu_t}dW_t.$$
Wikipedia tells us that the variance $\nu_t$ is given by $$d\nu_t = \kappa\left(\theta-\nu_t\right) dt + \xi \sqrt{\nu_t}dW_t.$$ So the parameters are $\kappa=2\beta$, $\theta=\delta^2/(2\beta)$, $\xi = 2\delta$. It follows that $2\kappa\theta = 2\delta^2$, while $\xi^2=4\delta^2$. Hence $$2\kappa\theta<\xi^2.$$ So the Feller condition is always violated.
## Answer by Andrea (score 6, accepted)
https://quant.stackexchange.com/a/82341
To answer the first point: the volatility $\sqrt{\nu_t}$ will be negative half of the time, look at
$\sqrt{(-3)^2}=3$
If the OU process is -3, then its square is 9 and the square root is 3 (not -3).
Don't get confused: $\sqrt{x^2} \ne x$, but $\sqrt{x^2} = |x|$.
For the 2nd point, since the CIR process for the Heston model has 3 parameters, and the OU only 2, the equivalence cannot work in all cases.
The CIR process is the square of a OU process only with the choice of parameters you have mentioned. For which the Feller condition is violated.
But, real Heston models use the full space of the 3 parameters: no OU equivalence, on the other hand the Feller condition is not violated.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.