Statistical metric to measure how well does the volatility surface fit the market

Question

Suppose that I have a model for implied volatility surface and want to figure out required recalibration frequency based on historical quotes. Since I have a large range of strikes and tenors over a long period of time I need to somehow automate this process, i.e. I need a computable metric rather than "ahh it seems pretty close to market".

What kind of statistical metric will make the most sense? I'm thinking about the mean of percentage differences between market and model quotes, i.e. the mean value of $$100\cdot\frac{\sigma^{market}-\sigma^{model}}{\sigma^{market}}$$ over the entire volatility surface, however the mean over the entire surface can be quite misleading as it will not capture large single outliers on a big enough surface and will cancel out differences with similar magnitude but opposite signs. Nevertheless I can't see a better single metric to assess an overall surface fit.

How much sense does an average percentage difference over the entire surface make to assess the quality of a fit? Is there a better metric? Any help will be appreciated.

UPD: Does it make more sense to pick a squared sum of differences across all tenors and strikes $$\sum_{K,T}(\sigma^{market}-\sigma^{model})^2$$ as a metric?

Answer 1

I suspect you want to use a weighted norm: https://math.stackexchange.com/questions/394237/understanding-weighted-inner-product-and-weighted-norms

Generally, your volatility surface (or volatility cube if you include skews), can be considered an element of the set of all possible volatility surface parametrisations. Each individual volatility or parameter of that element would represent a measured dimension.

You are seeking some function which represents the distance between two such elements of the set.

Here are two suggestions:

• l2 norm, i.e. the usual sqrt of vector inner product

• indicator function, where if a dimensional difference is greater than some tolerance it is set to 1, and summed across all dimensions, for example:

  the elements (1,2,3) and (1,9,9) have a distance of 2 (0+1+1) if the tolerance is 1, and a distance of 0 (0+0+0) if the tolerance is 10.

Now which of these distance measures is better? I would have to go with the l2 norm, because that creates and satisfies the axioms of a metric space, whereas the second does not. It does not satisfy the triangle inequality axiom that for two elements, A and B, the distance between them is smaller than or equal to the sum of the distance between A and C and C and B. Now, practically, why might this matter?

If you are coding some algorithm that recalibrates when the distance between two elements differs by a set value then the second distance function might create different results across days.

Suppose you set a tolerance of 2 and a recalibrate trigger when the distance is greater than 1, then:

On day 0 your surface is: (1,2,3) and moves to (2,3,4) which has an indicator distance of 0, so no trigger. On day 1 your surface is: (2,3,4) and moves to (3.5, 4.5, 5.5) which has an indicator distance of 0, so no trigger. The distance sum over two days is 0, and generates no trigger.

However, if instead: (1,2,3) moved to (3.5, 4.5, 5.5) on the first day the indicator distance would be +3 and would trigger a recalibration.

This inconsistency here is due to the triangle inequality axiom and the fact you are not operating inside a metric space.