# Log-normal Volatility Approximation

In a comment to this question, it is mentioned that, under the log-normal distribution, \begin{align*} vol(k) \approx vol(atm) \times \sqrt{\frac{atm}{k}}. \end{align*} Here, $k$ is the strike, $atm$ is the at-the-money strike, and $vol(k)$ is the implied volatility corresponding to strike $k$. I have difficulty to derive this approximation. Any suggestion is appreciated.

• @dm63: any suggestion? – Gordon Jul 21 '16 at 14:57
• Gordon, I think Mark Joshi can answer your question without any formula – user16651 Jul 21 '16 at 17:26
• Thanks @BehrouzMaleki. This question (quant.stackexchange.com/questions/25407/…) and this question (quant.stackexchange.com/questions/7761/…) provide a few clues. But there is such approximation as in the question. – Gordon Jul 21 '16 at 17:43
• From memory I think this is addressed in Blyth and Uglum "Rates of Skew" in Risk Magazine from about 10 years ago. – dm63 Jul 21 '16 at 19:31
• The intuition is that if there is a constant normalized local vol s, then the local lognormal vol is initially s/A, where A is the current at the money level. The local lognormal vol around the strike is s/K. So on a path that starts at the current market price and ends around the strike, the geometric average local lognormal vol is s/sqrt(AK). Since the current atm lognormal vol , sigma= s/A, the above is equal to sigma*sqrt(A/K). The key issue to be proved is that the average over all paths is the same as the specific path mentioned. – dm63 Jul 21 '16 at 20:11