According to the accepted answer to a question in this site on the interpolation in the term structure of volatility surface:

A simple linear interpolation on implied variance along iso-moneyness lines is enough to guarantee that there is no arbitrage between maturities as long as the inputted market data is arbitrage free.

In my case, however, I have implied volatilities sampled at fixed delta points for a set of maturities $\{ T_i \}$. If I linearly interpolate the implied variance at a time $T$, where $T_i \leq T \leq T_{i+1}$, along iso-delta lines, will the time-interpolated results at $T$ for all delta points be arbitrage free?

My guess is yes, but I'm hoping someone can confirm. In the Black-Scholes model, the delta of a call option is $\Delta = N(d_1)$ where $N()$ represents the cumulative normal probability density function with $$d_1 = \frac{\log(S_0/K) + (r + \sigma^2/2)T}{\sigma \sqrt{T}}.$$

Because $N()$ is monotonic and non-decreasing, I'm hoping the arbitrage-free result would still be maintained. I would much appreciate if anyone can confirm or refute this.


1 Answer 1


The iso-moneyness approach guarantees no arbitrage in terms of calendar spreads, but it is not proven it does not introduce some butterfly spreads at some interpolated time. See Arbitrages in the Volatility Surface Interpolation and Extrapolation.

Using iso-delta is sometimes done, not for arbitrage concerns, but because it may make more sense from a financial perspective. It does not guarantee the absence of calendar spread arbitrage. It would be interesting to find a counter example. $N()$ is monotonic but $\sigma$ is not. The no-arbitrage condition is derived in Arbitrage-free conditions for implied volatility surface by Delta does not look nice

Arbitrages with iso-delta interpolation is usually not a concern even if it may happen in not-so-realistic/manufactured examples.


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