Looking at the papers

  1. Arbitrage free SABR (Hagan)
  2. Managing Smile Risk (Hagan)
  3. Explicit SABR Calibration through simple expansions (Floch)

all 3 papers have similar forms for expression for implied volaility (Normal) but differ quite a little. But that is not my main question.

When F = K, the implied normal volaility breaks down when i tried to implement them, either divide by 0 or 0/0.

Can i get some help on this?

  • $\begingroup$ You may need to write out the implied volatility formulas in the mentioned papers for review, as we have no clue how did your 0/0 happen. For $K=F$, the formula should be much simpler. $\endgroup$
    – Gordon
    Commented Aug 17, 2018 at 20:10
  • $\begingroup$ For the "Managing Smile Risk" paper for example, I guess you are talking about equation (A.59a). In this case you see that the first term $\left(\frac{\varepsilon\alpha(f-K)}{\int_K^f \frac{df'}{C(f')}}.\left(\frac{\xi}{x(\xi)}\right)\right)$ become an indeterminate form when $f \rightarrow K$. You can compute this limit (a Taylor expansion will do the job) which is the preferable way, or you can take a numerical limit instead ($f = K+0.0001$ for example) for your computations. $\endgroup$
    – loyd.f
    Commented Sep 3, 2018 at 20:04

1 Answer 1


This is indeed an important issue if you use SABR in production.

If I am correct, you'll need this term $$ \chi(\zeta) = \log \left( \frac{\sqrt{1-2\rho\zeta+\zeta^2}-\rho+\zeta}{1-\rho} \right) $$ and $\zeta=0$ if $F_0=K$.

When $\zeta$ is VERY small (e.g., $|\zeta|<10^{-8}$), you can use the Taylor expansion $$\sqrt{1+\varepsilon} \approx 1+\varepsilon/2-\varepsilon^2/8 \quad \text{and}\quad \log(1+\varepsilon)\approx \varepsilon - \varepsilon^2/2$$ to get (make sure to keep both $\zeta$ and $\zeta^2$ terms) $$ \frac{\chi(\zeta)}{\zeta} \approx 1 + \frac{\rho}{2}\zeta. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.