# Why linear interpolation not appropriate for volatility surface construction?

We know linear interpolation is not appropriate for constructing a surface, but why?

In the book, "Foreign Exchange Option Pricing: A Practitioners Guide", the author writes:

native linear interpolation with regard to time can lead to unrealistic forward volatility dynamics... this implies a negative forward variance between ...

I am not sure I understand the reasoning. Why does linear interpolation imply negative forward volatility ? Can anyone provide a better explanation? Is there any other reason that the simple linear interpolation should not be used?

## 2 Answers

Note that total implied variance defined as $$V(T,K) = T\Sigma(T,K)^2$$ should be an increasing function of $T$. Otherwise you have a calendar arbitrage (sell the call with shorter expiry and buy the cheap longer one).

If you interpolate linearly your implied volatility is $$\Sigma(T,K) = w\Sigma(T_i,K) + (1-w)\Sigma(T_{i+1},K)$$ with weight $w = \frac{T_{i+1}-T}{T_{i+1}-T_i}$. This can also be written as $$\Sigma(T,K) = \Sigma(T_i,K) + s(T-T_i)$$ with slope $s = (\Sigma(T_{i+1},K)-\Sigma(T_{i},K))/(T_{i+1}-T_i)$. Note that $s$ can be negative, i.e. $\Sigma(T_{i+1},K) < \Sigma(T_{i},K)$ even in an arbitrage-free situation: $V(T_{i+1},K) \ge V(T_{i},K)$.

Now all you have to do is check for calendar arbitrage: $$\partial_T V(T,K) \geq 0$$ A simple computation will show you that the lhs is a 2nd order polynomial in $T$ and that it can turn negative.

• how about linear interpolation of a single smile, i mean fix the time T then linear interpolate the implied vol repect to the strike K. is it no arbitrage? Commented Nov 1, 2018 at 14:40
• Note that K must be a delta, not strike! You can't use the same strike for different maturities due to e.g. discounting. That's why FX options are quoted in terms of delta. Commented Oct 1, 2020 at 16:00

It implies negative forward variance. I have the book, and went through the section following your quote. In math terms, he is making a proof by contradiction. He first assumes that you can interpolate Iinearly, and comes to the conclusion that it is not a good assumption. The argument does involve some calculus. I don't think I have a better explanation, so let me know if you have any questions about his argument.