Fourier Transform of Probability distribution

[sam2]

Hi,
Sorry about the text, but Latex doesn't work.

Can anyone please give me an outline for the derivation of the probability function by inverting its Fourier transform, i.e.

P(X>x) = \frac{1}{2} + \frac{1}{\pi} \int_{0}^{\infty} Re \bigg[\frac{e^{-i \theta x}f(\theta)}{i \theta} \bigg] d\theta

where f is the characteristic function.
Basically, I do not understand where the 1/2 comes from. My approach was to calculate the Fourier transform of the probabity function:

E \big[ I_{X>x} \big]

and this reduces to being a function of the characteristic function as shown above (f/i theta). I then inverted the Fourier transform and got the integral above. But I don't see where the 1/2 would come from.

Thanks in advance.

 

[Hurkyl]

You need to play [ tex ] ... [ /tex ] around your LaTeX code. (Without the spaces, of course... and note that it's / and not \)

 

[Hurkyl]

Remember that something special happens at the frequency 0 component of the Fourier transform...