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Fourier Transform of Probability distribution

  1. Oct 19, 2005 #1
    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.

    Thanks in advance.
  2. jcsd
  3. Oct 19, 2005 #2


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    You need to play [ tex ] ... [ /tex ] around your LaTeX code. (Without the spaces, of course... and note that it's / and not \)
  4. Oct 19, 2005 #3


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