# The CDF from the Characteristic Function

• I

## Main Question or Discussion Point

Is there a way to find the CDF of a random variable from its characteristic function directly, without first finding the PDF through inverse Fourier transform, and then integrate the PDF to get the CDFÉ

Related Set Theory, Logic, Probability, Statistics News on Phys.org
mathman
Let F(x) be the desired cdf. You can get $$F(b)-F(a)=\frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{e^{-itb}-e^{-ita}}{-it}\phi(t)dt$$$. Let F(x) be the desired cdf. You can get $$F(b)-F(a)=\frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{e^{-itb}-e^{-ita}}{-it}\phi(t)dt$$$.
Sorry, this isn't clear to me. Could you elaborate more? Where is ##x## in the integration to have ##F(x)##?

mathman
x is a dummy variable. Integrate the expression for f(x) (pdf) from a to b to and then switch the order of integration to get the expession I presented. The important thing is that the same expression holds for the even when you don't have a pdf. If you want F(x), let b=x and see if a going to $$-\infty$$ makes sense.

So, basically

$$F(x)=\int_{-\infty}^{\infty}\frac{e^{-itx}}{-it}\phi(it)\,dt$$

which is the IFT of ##\frac{\phi(it)}{-it}##?

mathman
You must use the expression as I described, since you don't know off hand what the expression (F(x)-F(a)) will look like as a tends to -∞. Also you wrote φ(it) where it should be φ(t).

9t