Fourier Transform of a VIX Call Payoff on Squared Volatility

Summary

The document considers a European call option on the VIX, expressing its payoff as the positive part of the square root of a variable representing squared VIX, minus the strike. It states a Fourier transform involving the error function and asks how that expression follows from the payoff.

The answer sets up the transform as an integral over the payoff’s positive region, which begins at the squared strike, and suggests substituting a square-root expression involving the transform variable and integration variable. It does not show the remaining integration steps or derive the stated closed form. The material is therefore a starting point for working through the transform rather than a complete derivation. It provides no numerical validation, pricing comparison, or discussion of transform conventions and convergence conditions.

Key ideas

• The call payoff is zero below the squared strike and positive above it.
• The Fourier transform can be written as an integral beginning at the squared strike.
• A square-root change of variable is proposed to evaluate the integral.
• The document states an error-function transform but does not provide its full derivation.

Tags

# Transform of payoff function $w_c=(\sqrt{y}-K)^+$


# Transform of payoff function $w_c=(\sqrt{y}-K)^+$












I am working on a project where I price EU call options written on the VIX index.

The payoff function of interest looks like

$w_c=(\sqrt{y}-K)^+$

where K is the strike price and y is the value of $VIX^2$

The Fourier transform of this function takes the form:

$\hat{w_c}=\frac{\sqrt{\pi}}{2}\frac{1-\text{erf}(K\sqrt{-\phi})}{(\sqrt{-\phi})^3}$

Here $\phi$ is the transform variable.

I would like to know more about how to get from $w_c$ to $\hat{w_c}$.

I have tried to look around in different transform tables, but with no luck.

Any help is much appreciated. Thanks.

## Answer by Freelunch (score 6)

https://quant.stackexchange.com/a/43945

The fourier transform is \begin{equation} \hat{w}_c= \int_{-\infty}^\infty (\sqrt{y}-K)^+ e^{-i\phi y}dy = \int_{K^2}^\infty (\sqrt{y}-K) e^{-i\phi y}dy \end{equation}

Now do a change of variable with $t=\sqrt{i\phi y}$ and solve the resulting integral.

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