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.


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.

  • 1
    $\begingroup$ How do you define the VS variance ? the variance of a VS fair strike in the future ? $\endgroup$ Feb 10, 2019 at 11:54
  • $\begingroup$ Hi @AntoineConze, I've edited the question to make that clear. My final objective would be to obtain a relationship that would allow me to compare future ATMF volatility in a LV vs. SV framework calibrated to the same smile (hence implied forward start vols), knowing that I can easily compare future ATMF skew and curvature under both models (and since they are both calibrated to the vanilla market, forward variances will coincide). $\endgroup$
    – Quantuple
    Feb 10, 2019 at 12:14
  • $\begingroup$ @Quantuple have you tried to expand the realized variance of the forward price?When you say "ATMF volatility in a LV" you mean the Dupire formula? $\endgroup$
    – FunnyBuzer
    Feb 11, 2019 at 12:31
  • 1
    $\begingroup$ No I didn't, expand it around what? What I meant is that for an ATM forward start option $$ V_0 = \Bbb{E}_0^\Bbb{Q}\left[ \left(\frac{S_{T_2}}{S_{T_1}}-1 \right)^+ \right]$$ $$\iff BS(1,\Delta,\sigma_{1T_1T_2}) = \Bbb{E}_0^\Bbb{Q} \left[ BS(1,\Delta,\sigma_{1\Delta}) \right] $$ where $\sigma_{1T_1T_2}$ represents the IV of an ATM forward start option of forward starting date $T_1$ and expiry $T_2$ as of today and $\sigma_{1\Delta}$ a random variable representing the future IV (prevailing at $T_1$) of an ATM vanilla of residual maturity $\Delta$ under the model at hand: be it LV or SV. $\endgroup$
    – Quantuple
    Feb 11, 2019 at 12:44
  • $\begingroup$ If you have a relationship such as the one I am looking for, then you can compare the distributions of $\sigma_{1\Delta}$ under the two models knowing that the forward variances (calendar spread of VS) will be the same, just by comparing their future skew and curvature. I'm sure most of you have heard the: future skew and convexity are underestimated under LV hence the price of forward starts is higher, I'm trying to show that. But this was only to provide context to my question, which is independent of what I'm trying to prove. $\endgroup$
    – Quantuple
    Feb 11, 2019 at 12:48

1 Answer 1


Assume that the implied volatility for log-forward moneyness $k$ has an expansion $$ \sigma(k,T)=\sigma_0+\alpha k $$ Then up to terms of the first order, $$ f(k)=\frac{k}{\sigma_0 \sqrt{T}}+\frac{\sqrt{T}}{2}(\sigma_0+\alpha k)=\frac{(\sigma_0+\alpha k)-\sigma_0}{\alpha\sigma_0 \sqrt{T}}+\frac{\sqrt{T}}{2}(\sigma_0+\alpha k)= $$ $$ =\bigg(\frac{1}{\alpha\sigma_0 \sqrt{T}}+\frac{\sqrt{T}}{2}\bigg)(\sigma_0+\alpha k)-\frac{1}{\alpha\sqrt{T}} $$ Then $$ \sigma_0+\alpha k=\frac{f(k)+\frac{1}{\alpha\sqrt{T}}}{\frac{1}{\alpha\sigma_0 \sqrt{T}}+\frac{\sqrt{T}}{2}}=\frac{2\sigma_0}{2+\alpha\sigma_0T}(\alpha\sqrt{T}f(k)+1) $$ Changing variables $z=f(k)$ in the integral we get $$ \sigma^2_T=\int^\infty_{-\infty}(\sigma_0+\alpha k)^2\phi(f(k))f'(k)dk= $$ $$ =\int^\infty_{-\infty} \bigg(\frac{2\sigma_0}{2+\alpha\sigma_0T}(\alpha\sqrt{T}z+1)\bigg)^2\phi(z)dz= $$ since $\int^\infty_{-\infty}\phi(z)dz=1,$ $\int^\infty_{-\infty}z\phi(z)dz=0,$ $\int^\infty_{-\infty}z^2\phi(z)dz=1,$ $$ =\frac{4\sigma^2_0(\alpha^2T+1)}{(2+\alpha\sigma_0 T)^2} $$ Resulting approximation is $$ \sigma^2_T\approx \frac{4\sigma^2_0(\alpha^2T+1)}{(2+\alpha\sigma_0 T)^2}, $$ where $\sigma(k,T)=\sigma_0+\alpha k$

  • $\begingroup$ Hi @FunnyBuzer and thanks for your answer. However, I am not totally convinced by it: (i) we look for an expansion at first order at $\alpha$ and it appears at second order in the result (ii) the result is actually not the one I'm looking for. I've edited my question to show the steps I make in my version of the "demonstration". But some questions remain. $\endgroup$
    – Quantuple
    Mar 21, 2019 at 9:36
  • $\begingroup$ @Quantuple it looks like your proof is now complete. What are the open questions? $\endgroup$
    – FunnyBuzer
    Mar 21, 2019 at 16:07
  • $\begingroup$ This part "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)" is not clear to me. Also, I've never used perturbation theory so I just give it my best try but maybe some steps are not how they should be. $\endgroup$
    – Quantuple
    Mar 22, 2019 at 8:55

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