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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?

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  • $\begingroup$ 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 Aug 17 '18 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 Sep 3 '18 at 20:04
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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) $$ \frac{\chi(\zeta)}{\zeta} \approx 1 + \frac{\rho}{2}\zeta. $$

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