The CDF from the Characteristic Function

  • I
  • Thread starter EngWiPy
  • Start date
  • #1
1,367
61

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
 

Answers and Replies

  • #2
mathman
Science Advisor
7,800
430
Let F(x) be the desired cdf. You can get [tex]F(b)-F(a)=\frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{e^{-itb}-e^{-ita}}{-it}\phi(t)dt[/tex]$.
 
  • #3
1,367
61
Let F(x) be the desired cdf. You can get [tex]F(b)-F(a)=\frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{e^{-itb}-e^{-ita}}{-it}\phi(t)dt[/tex]$.
Sorry, this isn't clear to me. Could you elaborate more? Where is ##x## in the integration to have ##F(x)##?
 
  • #4
mathman
Science Advisor
7,800
430
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 [tex]-\infty[/tex] makes sense.
 
  • #5
1,367
61
So, basically

[tex]F(x)=\int_{-\infty}^{\infty}\frac{e^{-itx}}{-it}\phi(it)\,dt[/tex]

which is the IFT of ##\frac{\phi(it)}{-it}##?
 
  • #6
mathman
Science Advisor
7,800
430
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
 

Related Threads on The CDF from the Characteristic Function

Replies
36
Views
2K
  • Last Post
Replies
6
Views
519
Replies
3
Views
511
Replies
9
Views
1K
  • Last Post
Replies
4
Views
4K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
1
Views
3K
Replies
2
Views
5K
  • Last Post
Replies
3
Views
3K
Top