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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?

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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.

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  • $\begingroup$ 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? $\endgroup$
    – user6703592
    Commented Nov 1, 2018 at 14:40
  • $\begingroup$ 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. $\endgroup$
    – krems
    Commented Oct 1, 2020 at 16:00
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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.

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