I found the following explanation in a paper by Grunspan (see attached paper page 6) but have trouble understanding it:

By differentiating Formula (3) with respect to m, it turns out that the Black-Scholes skew $\frac{\partial\sigma_{LN}}{\partial m}$ at the money ($m = 1$) generated by the Bachelier model is $\frac{\partial\sigma_{LN}}{\partial m} = -\frac{1}{2}\frac{\sigma_N}{S}$ ($\sigma_{LN}$ is by definition the implied lognormal volatility). Therefore, the Bachelier model is highly skewed ATM (a slope of $−50\%\times\frac{\sigma_N}{S}$). Another way to explain this feature is that given call prices, when we use the BS model, the function $\sigma_{LN}$ is a decreasing and convex function of $m$, i.e., it generates a skew, while the function $\sigma_N$ is a rather flat function of $m$. Thus, normal volatility is most suited for products such as swaptions for instance.

I am not sure what Formula (3) is, but it might be $\sigma_{LN} = \frac{1}{S}\frac{\ln m}{m-1}\sigma_N$.

My two questions are:

  1. How does he get the formula above, i.e. $\frac{\partial\sigma_{LN}}{\partial m} = -\frac{1}{2}\frac{\sigma_N}{S}$ and more importantly what does this tell us about the two skews?
  2. Doesn't this concern the slope of the Black-Scholes IV, since the slope of the log-normal volatility is equal to that?

Therefore, the Bachelier model is highly skewed ATM (a slope of $−50\%\times\frac{\sigma_N}{S}$).

Here is the paper: Grunspan Paper

  • $\begingroup$ What happens if you price up a load of options with a bachelier constant vol, and then calculate the log normal vol that corresponds? $\endgroup$
    – will
    Oct 25, 2020 at 11:04

1 Answer 1

  1. You can take a derivative $\frac{\partial}{\partial m}\frac{\ln m}{m-1}$ at point $m=1$, so you will get $-\frac{1}{2}$.
  2. Yes, Black-Scholes volatility is log-normal volatility. In other terms it's comparison of Black-Scholes IV and Bachelier IV.

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