In a pure diffusion setting, it is a well known result that the volatility $\sigma_T$ of a fresh-start variance swap of maturity $T$ as seen of $t=0$ verifies \begin{align} \sigma_T^2 &= \Bbb{E}_0^\Bbb{Q} \left[ \frac{1}{T}\int_0^T d\langle \ln S \rangle_t \right] \\ &= \int_{-\infty}^{+\infty} \tilde{\sigma}^2(z,T) \phi(z) dz \\ &= \int_{-\infty}^{+\infty} \sigma^2(f^{-1}(z),T) \phi(z) dz\\ &= \int_{-\infty}^{+\infty} \sigma^2(k,T) \phi(f(k)) f'(k) dk \\ &= \color{blue}{\Bbb{E}[\sigma^2(k,T)]} \end{align} where $\phi(\cdot)$ is the pdf of a standard Gaussian, $\sigma(k,T)$ is the implied volatility smile for log-forward moneyness $k$ and time to expiry $T$ and $$ f : k \to \frac{k}{\sigma(k,T)\sqrt{T}} + \frac{1}{2} \sigma(k,T) \sqrt{T} $$
As a result, if the smile in the modified moneyness space $z$ admits the following parameterisation $$ \tilde{\sigma}^2 = \tilde{\sigma}_0^2 + \alpha z + \beta z^2 $$ then $$ \sigma_T^2 = \tilde{\sigma}_0^2 + \beta $$ by the properties of the standard Gaussian and it is often said that only the "ATM" level and "ATM" convexity contribute to the price of the VS. Of course "ATM" is here to be understood as at the money in the modified moneyness space, i.e. $z=0$ not $k=0$.
I've been looking to obtain a similar expression relating the VS variance, ATMF variance, ATMF skew and curvature under the standard moneyness space $k$ but am surprised by how difficult this is. Certainly missing something, if you have pointers, I'd gladly take them.
I am aware that such relationships exist for generic stochastic volatility models as given by the Guyon-Bergomi expansion, but I was (secretly) hoping for a model-free result.
[Edits]
Since this question seems harder than it seems. I'm also willing to take answers to a lighter version of it.
Indeed,a model-free result can be obtained when considering a smile linear in log-forward moneyness $$ \sigma(k, T) = \sigma_{F_T T} + \mathcal{S}_T k $$ It is not an exact result but rather an approximation based on a perturbation analysis. More specifically, $\sigma(k,T)$ is perturbed around $\sigma_0 = \sigma_{F_T T}$ at order 1 in the ATMF skew $\mathcal{S}_T \ll 1$ and the resulting order 1 expansion of VS volatility then reads $$ \sigma_T = \sigma_{F_T T} - \frac{1}{2} \sigma_{F_T T}^2 T \mathcal{S}_T $$ which is already one step towards the relationship I'm looking for.
Starting from the integral formulation of VS volatility above, I have worked out a (sloppy) demonstration of this result as I'm not an expert of perturbation theory.
So if anyone is willing to provide a rigourous demonstration of how perturbation theory can be used here, or comment the one I have below, I would be a happy man.
Consider that the IV smile prevailing at $T$ is linear in log-forward moneyness $$ \sigma_{k T} = \sigma_{F_T T} + \mathcal{S}_T k $$
Wwe look for an expansion of the VS volatility $\sigma_T$ after considering a perturbation of the implied volatility smile around $\sigma_0 := \sigma_{F_T T}$ at order 1 in $\mathcal{S}_T$. In other words, we look for \begin{align} \sigma_T &= \underbrace{\sigma_{T,0}}_{\text{Order 0 (no skew)}} + \underbrace{\delta \sigma_T}_{\text{Order 1 contrib. of skew}} \end{align} given the perturbed smile \begin{gather*} \sigma_{kT} = \sigma_0 + \delta{\sigma_{kT}} \\ \sigma_0 := \sigma_{F_T T},\, \delta{\sigma_{kT}} := \mathcal{S}_T k \end{gather*} In the non-perturbed case, i.e. flat smile $\sigma_0$, the VS volatility is trivially equal to $\sigma_{T,0} = \sigma_0$. In the perturbed case, we first note that at order $1$ in $\mathcal{S}_T$ $$ \sigma^2_{kT} = \sigma_0^2 + 2 \sigma_0 \delta\sigma_{kT} $$ Then, from the results of the previous paragraph we get \begin{align} \require{cancel} \sigma_T^{2} + \delta \sigma_T^2 &= \color{blue}{\Bbb{E}_{\sigma_{kT}} \left[ \sigma^2_{kT} \right]} \\ &= \Bbb{E}_{\sigma_0 + \cancel{\delta \sigma_{kT}}} \left[ \sigma_0^2 + 2 \sigma_0 \delta\sigma_{kT} \right] \\ &= \int_{-\infty}^{\infty} \left( \sigma_0^2 + 2 \sigma_0 \mathcal{S}_T k \right) \phi_{X,\sigma_0} (k) dk \\ &= \sigma_0^2 + 2 \sigma_0 \mathcal{S}_T \int_{-\infty}^{+\infty} k \phi_{X,\sigma_0}(k) dk \\ &= \sigma_0^2 + 2 \sigma_0 \mathcal{S}_T \int_{-\infty}^{+\infty} k \phi( f_{\sigma_0}(k) ) f_{\sigma_0}'(k) dk \\ &= \sigma_0^2 + 2 \sigma_0 \mathcal{S}_T \int_{-\infty}^{+\infty} f_{\sigma_0}^{-1}(z) \phi( z ) dz \\ &= \sigma_0^2 + 2 \sigma_0 \mathcal{S}_T \int_{-\infty}^{+\infty} \left( z - \frac{1}{2}\sigma_0 \sqrt{T}\right) \sigma_0 \sqrt{T} \phi(z) dz \\ &= \sigma_0^2 - \mathcal{S}_T \sigma_0^3 T \end{align} with the following remarks:
- In the second equation, notice how the perturbed smile appears explicitly in the integrand as well as implicitly in the density used for computing the expectation. For the sake of computing the order-one perturbation in $\delta \sigma_{kT}$, however, the contribution generated by the density vanishes (same reasoning as in Bergomi's "Stochastic Volatility Modelin" p.42), but I'm not sure why?
- The rest of the demonstration is straightforward as soon as one notices that the mapping $f$ becomes linear when considering a flat smile, which is the case in the non-perturbed problem.
At the end of the day, at order 1 in $\mathcal{S}_T$, we get that $$ \delta \sigma_T^2 = - \mathcal{S}_T T \sigma_0^3 $$ and since at that order $\delta \sigma_T = \frac{1}{2 \sigma_0} \delta \sigma_T^2$, from $\sigma_0 = \sigma_{F_T T}$ we finally obtain \begin{align} \sigma_T &= \sigma_0 + \delta\sigma_T \\ &= \sigma_{F_T T} - \frac{1}{2} \mathcal{S}_T \sigma_{F_T T}^2 T \end{align} But equivalently since $\sigma_0 = \sigma_T$ we can write that, at order 1 in the skew \begin{align} \sigma_{F_T T} &= \sigma_T + \frac{1}{2} \mathcal{S}_T \sigma_{T}^2 T \end{align} which is equation (8.26) in Bergomi's book. It tells us that for equity smiles that are typically negatively skewed ($\mathcal{S}_T < 0$) the ATM volatility will lie lower than the VS volatility.
