SABR Normal Volatility when F = K

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?

• 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. Aug 17 '18 at 20:10
• 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. Sep 3 '18 at 20:04

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) $$\frac{\chi(\zeta)}{\zeta} \approx 1 + \frac{\rho}{2}\zeta.$$