# The CDF from the Characteristic Function

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EngWiPy
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

## Answers and Replies

Let F(x) be the desired cdf. You can get $$F(b)-F(a)=\frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{e^{-itb}-e^{-ita}}{-it}\phi(t)dt$$$. EngWiPy 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)##?

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.

EngWiPy
So, basically

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

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

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).

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