How CIR++ Handles Negative Short Rates and Why a Shift Is Redundant
Summary
The document compares two ways to accommodate negative short rates in Cox–Ingersoll–Ross models. A displaced CIR shifts the rate by a constant so the shifted process follows the standard square-root model. CIR++ instead adds a deterministic time-dependent function to a CIR factor, with that function calibrated to reproduce the initial discount curve. The answer argues that a separate displacement in CIR++ adds no independent flexibility: its constant can be absorbed into the deterministic shift, which can already make the modeled rate negative.
A separate reply describes changing the level inside the square-root term to impose a chosen lower bound on rates, and notes an analogous idea for SABR. These are brief conceptual answers rather than a full derivation or empirical comparison. They do not address calibration details, parameter identifiability, or how the alternatives affect derivative prices and model behavior, so implementation choices require further analysis.
Key ideas
- CIR models use square-root diffusion, which constrains the rate level under the standard specification.
- A deterministic shift in CIR++ can fit the initial discount curve and permit negative modeled short rates.
- A constant displacement in CIR++ can be folded into its deterministic shift function.
- Changing the level under the square root is presented as another way to specify a lower rate floor.
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Full text
# Extensions of CIR
# Extensions of CIR
I could need some advice on extensions of the CIR model.
The standard CIR reads
$dr(t)=\kappa(\theta-r(t))dt + \sigma \sqrt{r(t)} dW(t)$.
A possible extension, if we would like the short-rate to also include negative values, could be a displaced version, so that $r(t)+\alpha$, where $\alpha>0$, follows a CIR model.
Further, to fit the initial term structure one could also consider the CIR++ (can be seen in Brigo et al) which is that
$r(t)=x(t)+\phi(t)$,
where $x$ is CIR and $\phi(t)$ is deterministic and chosen to fit the initial term structure.
My question is if it would make sense to consider a displaced CIR++, that is that $r(t)+\alpha=x(t)+\phi(t)$. My immediate thought is that the $\alpha$ does not provide any additional value for the model, and that the $\phi$-function already makes it possible for the short-rate to be negative?
## Answer by M. Jeunesse (score 3)
https://quant.stackexchange.com/a/25367
You are right. In the CIR++, $\alpha$ parameter is absorbed into $\phi$. With the CIR++, $\phi(t)$ will allow you to have to have negative rates. You will calibrate your $\phi$ to fit the discount factors.
The shifted idea is the one used to handle negative rates problem in caplet, swaption...
## Answer by Kiwiakos (score -1)
https://quant.stackexchange.com/a/25369
A simple shifting trick is to put $r(t)-f$ instead of $r(t)$ under the square root in your expression. Then $f$ is the new, possibly negative, interest rate floor. If, for example, $f=-100bp$ then the process is defined for all $r(t)>-100bp$.
Same for Sabr, where instead of square root you have another exponent.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.