# ATM Implied Volatility and Expected Variance

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?

• 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.
– Ivan
May 15 at 13:44
• @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.
– Alex
May 15 at 15:10

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.