- #1
sam2
- 22
- 0
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.
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.