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Can square-root variance reach zero15 min read
Feller Condition in CIR and Heston Models Explained
Understand the Feller condition, zero-boundary behavior, nonnegativity versus strict positivity, and numerical consequences for CIR and Heston models
Prepared by Mark · Primary sources below
Direct answer
For dX_t = κ(θ-X_t)dt + σ√X_t dW_t, the Feller condition 2κθ ≥ σ² keeps zero unattainable from a positive start. If it fails, the exact CIR process remains nonnegative but can reach zero; this is distinct from numerical negatives
Square-root diffusion links scale to uncertainty
The CIR process uses drift κ(θ-X_t) and diffusion σ√X_t for a nonnegative state such as a short rate or variance
Kappa controls mean-reversion speed, theta the long-run level, and sigma the state volatility
As X approaches zero, random shock size shrinks while the positive drift κθ pushes the process away from the boundary
The Feller inequality compares drift and noise
The standard boundary condition is 2κθ ≥ σ², assuming positive parameters and a positive initial state
It compares the inward drift at zero with diffusion intensity, rather than testing whether a calibration fits option prices
Some texts use a Feller index 2κθ/σ² or dimension 4κθ/σ², so threshold conventions must be checked
Passing the condition gives strict positivity
When 2κθ ≥ σ² and X_0 > 0, zero is unattainable for the ideal continuous-time CIR process
The state stays strictly positive at finite times, avoiding contact with the degenerate square-root boundary
Strict positivity is stronger than nonnegativity and should not be reduced to a vague claim that the model is stable
Failure permits contact, not negative states
When 0 < 2κθ < σ², zero is attainable, but the standard CIR solution remains nonnegative
With positive θ, the zero boundary is instantaneously reflecting rather than a permanently absorbing state
Therefore a failed Feller condition does not by itself make the stochastic differential equation invalid or negative
Heston applies the same test to variance
Heston variance follows dv_t = κ(θ-v_t)dt + ξ√v_t dW_t, replacing σ with vol-of-vol ξ
Its common condition is 2κθ ≥ ξ²; failure allows variance to touch zero and can sharpen boundary sensitivity
A calibrated Heston model can still be used when the condition fails, but interpretation and numerical treatment need extra care
Discretization creates a separate problem
A naive Euler step can become negative even when the exact process is nonnegative and the Feller condition holds
Full truncation, implicit methods, quadratic-exponential schemes, or exact-transition sampling handle the boundary differently
Clipping negative values changes the simulated model, so convergence, bias, and Greek stability should be tested
Calibration needs boundary-aware diagnostics
Report the ratio 2κθ/ξ², not only whether it passes, and compare it across dates, measures, and parameter uncertainty
Physical and risk-neutral parameters need not match because variance risk premia can change mean-reversion dynamics
Test near-zero paths, time-step sensitivity, transition moments, option errors, and hedge outputs before trusting a fitted model
Common questions
What is the Feller condition for a CIR process?
For dX = κ(θ-X)dt + σ√X dW, the standard strict-positivity condition is 2κθ ≥ σ²
What happens if the Feller condition fails?
The exact process can reach zero but stays nonnegative under the standard CIR specification with positive parameters
Does Heston require the Feller condition?
Not for every pricing use, but failure permits variance to touch zero and raises analytical and numerical boundary concerns
Why can a simulation produce negative variance anyway?
A discrete scheme such as naive Euler may violate the exact process boundary, so the simulation method needs separate validation
Sources and further reading
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