Fourier Pricing of a Truncated Exponential Payoff

Summary

The document considers Fourier-domain pricing for a payoff that equals a decaying exponential above a threshold and zero below it. It gives the Fourier transform of that truncated exponential and the transform of a Gaussian density. It then states a solution-manual relationship expressing the time-zero transform as the discounted maturity transform multiplied by the transform of a Gaussian kernel, and asks why that Gaussian factor appears.

The relation points to a standard pricing interpretation: under a diffusion model, the distribution of log-price changes over the remaining time acts as a Gaussian smoothing kernel, while discounting accounts for the risk-free rate. The passage does not derive this connection, specify all model conventions, or complete the algebra leading to the displayed target identity. Thus it offers useful transform and propagation pieces, but the underlying pricing assumptions and derivation remain incomplete.

Key ideas

• The payoff is an exponential above a threshold and zero below it.
• Its Fourier transform has a factor involving the threshold and the sum of decay and frequency terms.
• The document states that time-zero value is obtained by discounting and applying a Gaussian transform kernel.
• The Gaussian factor represents diffusion across the time to expiry, though the document does not derive that step.

Tags

# Fourier transform of price function


# Fourier transform of price function












If the expiry value is given by $f(x,T) = e^{-c x}$ for $x \ge a$ and 0 otherwise and c is a +ve constant, prove that in the Fourier domain:

$$ (c + j \omega) F(\omega, 0) = e^{-rT} e^{-a(c+j\omega)}e^{-r T \omega^2} $$

### Solution

First, the Fourier transform of $e^{-cx}$ is $\frac{e^{-a(c+j \omega)}}{c + j \omega}$.

Then, the Fourier transform (FT) of the gaussian pdf $g_{\sigma^2}(x) = \frac{1}{\sqrt{2 \pi \sigma^2}}e^{-\frac{x^2}{2 \sigma^2}}$ is:

$$ G_{\sigma^2}(\omega) = e^{-\sigma^2 \frac{\omega^2}{2}} $$

Here, is where I don't understand how to proceed.

The solution manual says that the following is true:

$$ F(\omega, 0) = e^{-r T} G_{2 r T}(\omega) F(\omega, T) $$

Where does the $e^{-r T} G_{2 r T}(\omega)$ come from? I can understand $e^{-r T}$ as discounting from T to 0 but what about $G_{2 r T}(\omega)$?

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