## Converting Volatility Surfaces from Moneyness to Delta Using an Iterative Method

It often comes up in quantitative finance that you want to convert a vol surface plotted against moneyness, to a vol surface plotted against delta.

See Options, Futures and Other Derivatives by John Hull for a reference on pricing formulas for European options. In the Black-Scholes framework, the delta of a call option is given by

$\Delta = N(d_1),$

Where $$N$$ represents the cumulative normal probability density function, and

$d_1 = \frac{\log(S_0/K) + (r + \sigma^2/2)T}{\sigma \sqrt{T}}.$

(for a put, it is $$\Delta = N(d_1) – 1 )$$ . Rearranging for moneyness, we have

$\frac{S_0}{K} = \exp\left(N^{-1}(\Delta) \sigma \sqrt{T} – (r + \sigma^2/2)T \right).$

Now, our volatility surface would typically be specified using a number of moneyness and volatility pairs $$(m_i,v_i)$$ where the moneyness values would typically be something like

$\{m_i\} = \{0.7, 0.8, 0.9,1,1.1,1.2,1.3\}.$

When calling for a volitility value for a moneyness in between these numbers, the firm would have implemented an interpolation function,

$I: \text{ moneyness} \to \text{ volatility},$

which would typically use a monotonic cubic spline. Inverting this function may be a lot of work, as it would require working out the exact coefficients generated by the cubic spline fitting. Even with an explicit formula, the spline is defined piecewise, which makes inverting it complicated.

Given some delta $$\Delta$$ , we want to find a volatility $$\sigma$$ such that the moneyness corresponding to that volatility according to the cubic spline interpolation is the same as the moneyness from the above formula. This requires solving the following equation for moneyness $$m$$:

$m = \exp\left(N^{-1}(\Delta) I(m) \sqrt{T} – (r + I(m) ^2/2)T \right).$

An equation like this should be solved numerically. This is doubly true due to the complicated definition of the function $$I$$. While inverting $$I$$ would be difficult, evaluating it is easy. This motivates solving using fixed point methods which only require the function to be evaluated.

What we are looking for is a fixed point of the map $$f$$ , i.e. a point $$m$$ such that $$f(m) = m$$. Thus, in the remainder of the article, we’ll look at an iterative fixed point method for solving this equation. The idea is simple. We start with some initial point $$m_0$$, and repeatedly apply the map

$f(m) = \exp\left(N^{-1}(\Delta) I(m) \sqrt{T} – (r + I(m) ^2/2)T \right)$

until the change in $$m$$ is less than some small tolerance.

The critical questions is: under what circumstances does this iterative procedure actually converge?

According to the Banach fixed-point theorem, this process will converge to a unique fixed point if $$f$$ is a contraction mapping, which in the context of a real-valued function means

$|f(m_1) – f(m_2)| \leq L |m_1 – m_2|,$

for some constant $$L \in [0,1)$$. This is also known as the Lipschitz condition, and it is well known that

$L = \sup_m |f'(m)|,$

where the supremum is of course taken over the domain of interest. Thus, our procedure will converge if $$|f'(m)|<1.$$ We calculate,

$f'(m) = f(m) I'(m) \left( N^{-1}(\Delta) \sqrt{T} – I(m)T \right).$

Numerical evidence shows that this derivative does not in general have an absolute value smaller than one, but typically does after just one iteration of our map. Our experience is that this method will almost always converge for all “reasonable” volatility surfaces, and usually within only 2 or 3 iterations!

A possible alternative to searching for a fixed point is to use Newton’s method to search for a zero of the function $$F(m) = m – f(m).$$