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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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  • $\begingroup$ 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? $\endgroup$ – Kevin May 1 '20 at 23:22

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