Fourier Transform of Probability distribution

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SUMMARY

The discussion focuses on deriving the probability function by inverting its Fourier transform, specifically addressing the equation P(X>x) = 1/2 + (1/π) ∫(0 to ∞) Re[e^(-iθx)f(θ)/(iθ)] dθ, where f(θ) is the characteristic function. The main query revolves around the origin of the 1/2 term in the equation. The user attempts to calculate the Fourier transform of the indicator function I_{X>x} and relates it to the characteristic function, but struggles to understand the contribution of the 1/2 term from the Fourier inversion process.

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  • Basic concepts of probability distributions and indicator functions
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sam2
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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.
 
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Remember that something special happens at the frequency 0 component of the Fourier transform...
 

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