# 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?

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.
• 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. Jul 7 at 12:13
• 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! Jul 7 at 14:20