Levy processes, vanilla option and Fourier Transform

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The context to this problem is mathematical finance, although the answer does not need specific knowledge of the area. I am trying to work out the expression for the price of a call option using Levy Processes.

Without getting into much details, I am now stuck trying to solve the following Fourier transform:

$$ \textbf{F}f^*(z) = \int_{-\infty}^{+\infty}e^{iyz}(e^{y+rT} - e^k)^+dy $$

such that $(e^{y+rT} - e^k)^+ = (e^{y+rT} - e^k) \mathbb{1}_{\{e^{y+rT} > e^k\}}$, where $r$, $T$ and $k$ are constants.

I have found the solution to this problem, but I am unable to see how one can get there:

$$ \textbf{F}f^*(z) = \frac{e^{k+iz(k-rT)}}{iz(iz+1)} $$

It might ultimately only have to do with classical calculus. Could somebody please show me how one can get there?

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There's a few ways to solve it, but here's one:

  1. Change of variables: Work with asset prices $Y=\exp(y)$ and exercise prices $K=\exp(k)$ rather than log prices $y$ and log exercise prices $k$.

  2. Restrict range of integration: Option prices are bounded. Hint: $(Y-K)^+$

  3. Apply the expression:

$$ \int_u^\infty x^{-v} (x-u)^{w-1} dx = u^{w-v}\beta(v-w, w)$$

The solution can be found here phrased in terms of Mellin transforms. It should contain everything you're looking for, including the relationship between Fourier and Mellin transforms.