There have been some posts on this topic, but not what I am looking for, so a new post on an old topic..

I think some/most of us here are familiar with the following formula expressing implied volatility as the break-even constant BlackScholes hedge volatility to make the expected final P/L equal to zero. After some re-arranging we get the familiar formula, and restricting now to a pure stochastic volatility model as the true dynamics:

$$ \Sigma^2(S_t,t,K,T) = \frac{E_t^Q \int_t^T \sigma_u^2 S_u^2 \Gamma^{BS}(S_u,K,\Sigma(S_t,t,K,T)) du}{E_t^Q \int_t^T S_u^2 \Gamma^{BS}(S_u,K,\Sigma(S_t,t,K,T)) du} $$

where $\sigma_u$ is the stochastic volatility, and $Q$ is the risk-neutral measure for this SV model. Note that the implied volatility in the Black-Scholes gamma terms is always the initial implied volatility (this formula is afterall obtained under the assumption of hedging at constant initial implied volatility).

Now this is where I think I am either making an error, or where things get interesting:

Assume that the correlation between the spot and vol is zero always. Then we can apply conditioning. So, for a given path of volatility we can take the expectation of the dollar gamma terms, and then take expectation over all volatility paths.

But for a given path of realized variance, we know that the dollar gamma is a martingale, and so the resulting initial gamma appears in both denominator and numerator and can therefore be cancelled out. What is left is then the following formula (when $dSd\sigma = 0$)

$$ \Sigma^2(S_t,t,K,T) = \frac{1}{T-t} E_t^Q \int_t^T \sigma_u^2 du $$

This cannot be true of course since that would mean for all strikes the break even implied vol is the variance strike and the smile would be flat. Not what we observe in a SV model with zero correlation.

The formula for implied volatility as break even constant delta hedge volatility is used by many (e.g. Bergomi in his book, Reghai in his book, Gatheral, and others) as a starting point for expansions and the like. These are very knowledgeable guys so I am sure they won't use something that doesn't make sense logically.

My question is, in which step did I make the brain-fart?


  • 1
    $\begingroup$ It seems to me that in a stoch vol world, OTM options have two sources of gamma : one with respect to the underlying , and the other with respect to movements in vol. Is the second part properly captured in your thinking? $\endgroup$
    – dm63
    Commented May 12, 2019 at 10:50
  • $\begingroup$ You have a good point. Going to mull over it and post when I have something. $\endgroup$
    – user34971
    Commented May 12, 2019 at 13:01

1 Answer 1


Just answering and closing my own question for the sake of completeness (no need to receive points for this).

Re-reading my question above it is clear where I made the mistake: $S^2_u \Gamma^{BS} (S_u,K,\Sigma(S_t,t,K,T))$ is only a martingale under the `Black-Scholes risk-neutral measure' with constant vol $\Sigma(S_t,t,K,T)$, not under the SV model risk-neutral measure. Hence, and as expected, the implied vol will be symmetric but not flat in stoch vol models when correlation is zero.


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