There is a famous formula for the variance swap strike that reads $$ K_{var}^2 = \int_{-\infty}^\infty dz\, n(z) I^2(z) $$ where $I(z)$ is the Black-Scholes implied volatility function, $$ n(z) = \frac{1}{\sqrt{2\pi}} e^{-\frac12 z^2} $$ and $z$ is the Black-Scholes `$d_2$' function $$ z = \frac{\log S_t/K}{I\sqrt\tau} - \frac{I\sqrt\tau}{2} $$ See for example slide 7 in this presentation by J. Gatheral (2006).

I want to show heuristically first that $K_{var}^2 \geq I^2(z=0)$ if the second derivative of $I^2(z)$ wrt $z$ is $\geq 0$ for all $z$.

First, write $$ I^2(z) = I^2(0) + z \frac{dI^2}{dz}(0) + \frac{z^2}{2!}\frac{d^2 I^2}{dz^2}(a) $$ for some $a\in (0,z)$. This is just Taylor's remainder theorem and is exact.

Substituting this into the integral expression for the variance swap strike, \begin{align*} \int_{-\infty}^\infty dz\, n(z) I^2(z) &= I^2(0) \int_{-\infty}^\infty dz\, n(z) + \frac{dI^2}{dz}(0) \int_{-\infty}^\infty dz\, zn(z) \\ &\quad + \frac{1}{2!}\int_{-\infty}^\infty dz\, z^2 n(z) \frac{d^2 I^2}{dz^2}(a) \quad (a \in (0,z)) \\ &= I^2(0) + \frac{1}{2!}\int_{-\infty}^\infty dz\, z^2 n(z) \frac{d^2 I^2}{dz^2}(a) \quad (a \in (0,z)) \\ &\geq I^2(0) \end{align*} where the second equality is because $\int_{-\infty}^\infty dz\, zn(z) = 0$ because $z$ is uneven and $n(z)$ is even, and the last inequality follows from the assumption that $\frac{d^2 I^2}{dz}(z) \geq 0$ for all $z$.

Does this make sense?


1 Answer 1


It is indeed perfectly correct under your working assumptions.

This is actually what Gatheral also notes in his book 'The Volatility Surface: A Practioner's Guide' (Chapter 11 on Variance Swaps, pages 140 and following). Specifically he writes:

Now consider the following simple paramterization of the BS implied variance skew: $$ \sigma^2_{BS}(z) = \sigma^2_0 + \alpha z + \beta z^2 $$ Substituting into equation (11.5) and integrating, we obtain $$ E[W_T] = \sigma_0^2 T + \beta T $$ We see that skew makes no contribution to this expression, only the curvature contributes.

Caveat: Skew should here be interpreted in the context in which it is defined in the first place. It corresponds to the order one coefficient in a second order representation of the BS implied variance around $z=0$ (which is neither ATM, nor ATMF). It is therefore not the 'usual' implied volatility skew $\partial \sigma_{BS}/\partial K$ that most practitioners are used to think about and to which the fair strike of a variance swap is sensitive, see this related question.

  • $\begingroup$ Ty for looking at my question Quantuple. Yes, now I remember above example from Gatheral, which is one example satisfying the condition I imposed on the curvature (wrt $z$). I'll still need to check though if Gatheral's parameterisation is arbitrage free so that at least there is one non-trivial case that satisfies the curvature condition imposed and is arbitrage free. $\endgroup$
    – Frido
    Jul 7 at 12:13
  • $\begingroup$ I guess that if you work with a real market smile, then it should be arbitrage-free by definition (if the market is efficient). So your question boils down to how good you can locally approximate that smile using a second order polynomial (and you have the answer since as you note it's a Taylor expansion, so you have well-known results to size the remainder). If so you'll have the result as long as $\beta > 0$. Maybe rewriting everything in terms of prices, then deriving those prices to get pdfs will be enlightening as far as arb goes. Good luck! $\endgroup$
    – Quantuple
    Jul 7 at 14:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.