Fair Strike for Variance Swap with no Skew in IV Surface

I am reading through Derman's 1999 research notes, "More than you ever wanted to know about Volatility Swaps."

In equation B4 of Appendix B, the author takes the Taylor Series of the variance swap replication portfolio with the constant term being equal to implied volatility squared. For me, this implies that the fair strike of a variance swap is equal to implied volatility squared when there is no skew in the IV surface.

I developed a tool to replicate fair variance strike and I always get a value higher than the IV squared when I use constant volatility to price each option in the replication. I get similar results when I use my trader's pricer at my job. I doubt there is a numerical error as I am simply doing a partial sum. Hence, I would expect the error to underprice the fair strike.

Should the fair variance price be equal to IV squared in the absence of skew and curvature in the IVS? If not, why?

Thanks!

• You have no skew, but what about convexity? Check out this related question: quant.stackexchange.com/questions/27539/… Oct 17 '19 at 10:21
• Thanks, Quantuple. In my example, I use a constant IV for every option in my VS replication. The results are still slightly higher than the ATM IV. For me, there is no convexity with this approach. Oct 19 '19 at 19:13
• Can you describe the exact way you replicate it? Is this simply a discretisation error? Oct 19 '19 at 19:15
• I assume rates and dividend are zero and compute $K^2 = \frac{2}{T}\Big[\sum_{i=1}^{20} \frac{P(i*5)}{(i*5)^2} + \sum_{i=21}^{40} \frac{C(i*5)}{(i*5)^2}\Big]$ where I assume the underlying name is trading at 100 and all the calls and puts are evaluated at the same IV. Only strike changes. Oct 19 '19 at 19:24
• Do you miss a times 5 in your expression? What you are observing is a discretisation error. You have a single integral which you replace by two sums. Because the put (resp. call) price as a function of strike is increasing (resp. decreasing) replacing the integral $\int_{K_i}^{K_i+\delta K} f(K) dK$ by $f(K_i) \delta K$ will lead to an underestimatation (resp. overestimatation). Note that for the integral over $K=100$ and $K=105$ you will also have an overestimation since you are using put prices. Just plot the true function and visualise the true area under the curve versus your rectangles. Oct 21 '19 at 11:59