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