When Variance Swap Strike Exceeds Center Implied Variance

# A lower bound for variance swap strike


# A lower bound for variance swap strike












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?

## Answer by Quantuple (score 3, accepted)

https://quant.stackexchange.com/a/71524

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.