The ATM implied volatility is important in SABR when calibrating the model. Let's consider the ATM vol (for a european call option): $$\sigma = \frac{\alpha}{f^{1-\beta}} \left[ 1+ \left(\frac{(1-\beta)^2}{24}\frac{\alpha^2}{f^{2-2\beta}}+\frac{1}{4}\frac{\rho \beta v}{f^{1-\beta}}+\frac{1}{24} (2-3\rho^2)v^2 \right)T \right]$$ where $v$ is rest is obvious and same notation is used in Hagans original paper.

However, it commonly mentioned in literature that this volality can be estimated by first term: $$\sigma \approx \frac{\alpha}{f^{1-\beta}}$$

How does one proof/show that claim?


  • $\begingroup$ It is pretty obviuos when you look at the terms and some become really small. I can also demonstrate it graphically. But I still want a better argument that this hold for all values of the parameters $\endgroup$ – Kim Sep 14 '18 at 6:41

In short , this claim does not hold under all circumstances.

There are a few ways to break down such approximation.

  1. The options under consideration have very long expiry, i.e. $T$ is very large

  2. As expiration date approaches, the volatility smile becomes more pronounced, i.e. $v$ becomes relatively large.

  3. Under extreme market condition, the magnitude of $\rho$ and $v$ become significant resulting a non-trivial expansion term.

However, in most of the cases, the expansion terms are way smaller than 1, thus I would argue such approximation is legit for most of time.


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