# Fourier Transform of Probability distribution

1. Oct 19, 2005

### sam2

Hi,
Sorry about the text, but Latex doesnt 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.

2. Oct 19, 2005

### Hurkyl

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

3. Oct 19, 2005

### Hurkyl

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