This answer claims that $$\sigma^2_{ATM}\approx E^Q\left(\frac{1}{T}\int_0^T\sigma^2_t dt\right)$$ ie implied ATM vol = risk-neutral expectation of integrated variance.

Is there some proof available? Where does this approximation come from? Is it model independent (probably relying on no jumps though)? And I guess it applies to the implied vol of European options?

A second question: What happens if the market is incomplete and there are several risk-neutral measures?

  • $\begingroup$ I don’t think that’s true in general otherwise all options would have the same implied i.e. no skew. I believe Dupire and others have shown that it needs to be weighted by gamma. $\endgroup$
    – Ivan
    May 15, 2022 at 13:44
  • $\begingroup$ @Ivan The formula only applies to ATM options? ITM and OTM options could still have a different implied vol ($\Rightarrow$ skew). Also, the formula changes with option maturity, ie might capture term structure effects. $\endgroup$
    – Alex
    May 15, 2022 at 15:10

1 Answer 1


I am not so sure about the ATM approximation from the other answer (i.e. I don't think it's a great approximation). I think it comes from the following for $T \ll 1$: \begin{align} E \left[ \frac{1}{T} \int_0^T \sigma^2_u \, du \right] &\approx E \left[ \frac{1}{T} \int_0^T \left(\sigma_0 + d\sigma_0 \right)^2\, du \right] \\ &\approx \frac{1}{T}\int_0^T \sigma_0^2 \, du \\ &\approx I^2_{ATM} \end{align} since it can be shown rigorously that $$\lim_{u \rightarrow 0} \sigma_u = I_{ATM}$$

You might be better off to use the rather famous expression for the variance swap strike due to Matytsin. It is (under the pricing measure): $$ E \left[ \frac{1}{T} \int_0^T \sigma^2_u \, du \right] = \int_\mathbb{R} I^2(z) N'(z) dz $$ where $z$ is the Black-Scholes `$d_2$' moneyness measure, $$ d_2 := \frac{\log(S_t/K)}{I\sqrt T} - \frac{I\sqrt T}{2} $$ and $N'(z)$ is the standard normal density.

What you can then do is expand the implied volatility in the integrand around $z=0$, $$ I^2(z) = I^2(0) + z(I^2)'(0) + \frac{z^2}{2} (I^2)''(0) + \cdots $$ and substitute this term back in the integral. The lowest order term is then $$ E \left[ \frac{1}{T} \int_0^T \sigma^2_u \, du \right] \approx I^2(0) $$ and I can already tell you that this is not a good enough approximation since $I(0)$ is approximately the volatility swap strike, so the lowest order approximation ignores the convexity correction. Hence you need to go to second order (you can ignore terms with $z^n$ where $n$ is odd): $$ E \left[ \frac{1}{T} \int_0^T \sigma^2_u \, du \right] \approx I^2(0) + \frac{ (I^2)''(0)}{2} \int_{\mathbb R} z^2 N'(z) dz $$ This is an OK approximation and can be used for non trivial $T$ values such as 1 week or 1 month and perhaps even larger $T$. Notice also that this approximation automatically gives you an expression for the convexity correction.


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