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Centering a Polynomial FX Volatility Smile at the Forward

Article Quant Q&A · Author: Dom
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Summary

The question examines a polynomial interpolation formula for an FX implied-volatility smile and focuses on its at-the-money value. Using the coefficients and equation from a cited textbook example, the author reports that the reproduced value matches only after shifting the function’s moneyness adjustment by one half. The proposed shift centers the quadratic at the forward, where the moneyness ratio is one, and makes the polynomial’s value at zero align with the at-the-money coefficient.

The note also flags a correction to the denominator in the textbook equation, stating that it should use the reference volatility rather than the similarly named alternative parameter. No answer or independent derivation is included, so the proposed adjustment remains a question rather than a verified result. The document offers a specific implementation check but no market data or broader comparison with other smile-fitting methods.

Key ideas

  • The note checks whether an FX polynomial smile should be centered at the forward moneyness.
  • The proposed half-unit shift is intended to make the polynomial’s central value match at-the-money volatility.
  • The author reports an equation erratum concerning the denominator used in the adjustment.
  • No response or derivation confirms whether the proposed shift is correct.

Tags

ForexOptionsVolatilityDerivatives pricing
Full text
# FX Smile Polynomial Fitting


# FX Smile Polynomial Fitting












I am unable to reproduce the example of FX polynomial smile interpolation on page 59 of the book FX Option Pricing by Iain Clark shown below. Consider just the ATM volatility as a specific case. I calculate the following using the parameter values (c0, c1 and c2) given in section 3.9.1:

I can only get agreement for the ATM volatility if I subtract 0.50 from the value of $\delta(x)$ in equation 3.22. This would centre the quadratic around the forward where the moneyness is one. It also ensures that $f(0)=c_0$ so that $\sigma_X(F_{0,T})=\sigma_{ATM}$ if the ATM is forward.

Can anyone confirm that I am correct to subtract 1/2.

PS. Note that there is an erratum (from the book website) that I have fixed where the denominator in (3.22) is $\sigma_0$ rather than $\delta_0$.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.