# 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 $$G_{2 r T}(\omega)$$?

• I suppose $F(\omega)$ is the Fourier transform of $f(x)=e^{-cx}\mathbf{1}_{\{x\geq a\}}$? How can $F$ depend on $T$ if $f$ does not? – Kevin May 1 '20 at 23:22