Finite-Difference Approximations for Third-Order Option Greeks

Summary

The document gives two finite-difference formulas for approximating a third derivative from function values sampled around a point. In options work, third derivatives of value with respect to relevant inputs correspond to higher-order sensitivities such as Speed, Zomma, Color, and Ultima, depending on the variable being differentiated. The formulas use equally spaced observations with step size h and combine values at offsets on either side of the evaluation point.

One approximation has first-order accuracy and uses an asymmetric stencil extending two steps in one direction and one in the other. The second has second-order accuracy and uses a symmetric stencil extending two steps on both sides. The answer supplies the coefficients and scaling factors, but does not explain how to map each Greek to a particular underlying derivative or input. It also gives no numerical error analysis or guidance on choosing h, so implementation should account for truncation error and numerical precision.

Key ideas

• Third derivatives can be estimated from function values at regularly spaced points around the target value.
• The first formula has first-order accuracy and uses an asymmetric set of sample points.
• The second formula has second-order accuracy and uses a symmetric stencil spanning two steps on each side.
• The step size affects approximation error, but the document does not provide selection guidance.
• Which third-order Greek is estimated depends on the input variable being differentiated.

Tags

# What's the formula to compute the divided difference approximation for the third order greeks?


# What's the formula to compute the divided difference approximation for the third order greeks?












I can't seem to find the quotient required to approximate the third order greeks Speed, Zomma, Color and Ultima

## Answer by Magic is in the chain (score 3, accepted)

https://quant.stackexchange.com/a/46681

For first order accuracy you can use:

$f^3(x)=\frac{1}{h^3}\left(f(x+2h)-3f(x+h)+3f(x)-f(x-h) \right)$

For the second order accuracy:

$f^3(x)=\frac{1}{2h^3}\left(f(x+2h)-2f(x+h)+2f(x-h)-f(x-2h) \right)$

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.