Belgrade 2004 paper basically proposes that inflation year-on-year volatilities (and hence yoy options) are basically the spread vols between the Zero-coupon vols from (t0 to T) minus the zero-coupon vols from (t0 to T-1).

Is it possible to express the payoff and hence a year-on-year option unto zero-coupon options? I am trying to map the yoy vega's of my options so that they can be netted off against the vega's of the zero-coupon options.


You can't readily map YY options payoffs into ZC options payoffs.

To go from ZC to YY requires:

  • a convexity adjustment for transforming the CPI forwards ratio into a YY forward
  • CPI correlations for transforming the CPI volatilities into a YY volatility

With flat (in strike) volatilities you can in principle compute CPI implied correlations from ZC implied volatilities and YY implied volatilities. However since there is a smile for both volatilities things would be more complicated as you would need to arbitrarily map YY strikes into ZC strikes.

  • $\begingroup$ hi @Antoine Conze, yes.. you are right there are convexity adjustments for YoY forwards and an assumed CPI(T)/CPI(T-1) correlations. ... However, I believe one can infer the CPI(T)/CPI(T-1) correlations from the implied volcube of ZC-options against YoY-options using any of the assumed volatility function in Belgrade's paper (either Hull-White or simplistic ZC-call-spreads). Given that we *** can *** create these two volcubes, would it not be possible? For example, ignoring the convexity adjustments, we do have the marginal densities for YoY and ZC volatilities. $\endgroup$ – Kiann Apr 23 '19 at 13:51
  • $\begingroup$ given these marginal densities, we can, for example, bump the ZC 2y tenor, K-strike. Convert this into the equivalent actual realized bumps in the YoY volcube. This would then be the vega for YoY capfloor, given a bump in the ZC-cube, converted into the YoY volcube. What I am unsure about, is given the K-strike in ZC versus the K-strike in the vol-cube, how to convert these marginal densities so that they are equivalent. $\endgroup$ – Kiann Apr 23 '19 at 13:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.