Related questions to this topic/subject:
In both posts, the authors/respondents recommend using the second derivative to capture the curvature of the volatility skew:
$$\frac{f(x+h)-2f(x)+f(x-h)}{h^2} = f^{''}(K)\approx\frac{f(K_{ATM}+h)-2f(K_{ATM})+f(K_{ATM}-h)}{h^2}$$
I understand that this is merely an approximation as options are offered at discrete intervals, which does not meet the theoretical requirements of the second derivative that:
$$h \rightarrow 0$$
Furthermore, there could be instances where $f^{''}(K)$ could be convex (concave) with positive (negative) values of $f^{''}(K)$ depending on the options price.
My question is - if we choose to express the curvature of the entire volatility skew as:
$$Curvature = \sum_i{f^{''}(K_i)}$$
How would $Curvature$ be affected if we reduced $h$ by fitting a volatility skew model to the market-implied volatility skew (making the volatility skew "more continuous")?
Some of my thoughts would be:
- Smoothing of the volatility skew removes concave parts of the market-implied volatility skew, increasing the overall curvature measure.
- As we reduce $h$, the number of second derivative components $f^{''}(K_i)$ would increase as there are more points on the volatility skew. However, the individual contributions of $f^{''}(K_i)$ are ambiguous as we are unsure how they would turn out as the $f(K_{ATM}+h)-2f(K_{ATM})+f(K_{ATM}-h)$ decreases as well.
-- EDIT --
- One answer from a user in another post "Moreover, as h becomes small, small errors in the measurement of the sigma of the calls and puts are magnified. Choosing a large h reduces this noise in the calculations at the expense of approximating the average slope over a wider interval."
