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.

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  • $\begingroup$ 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

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