The CDF from the characteristic function

[mathman]

Look at post #27. \int_0^{\infty}\frac{sin(tx)}{t}dt=0,\ for\ x=0

 

[EngWiPy]

Ooops! I forgot about that. My mistake. Thanks!

So, this is an indication that the derivation is correct, right? The only difference in my equations is that I use the formula $\text{ci}(x)\pm j\,\text{si}(x) = \text{Ei}(\pm j\,x)$, where $\text{Ei}(x) = -\int_{-x}^{\infty}\frac{e^{-t}}{t}\,dt = -E_1(-x)$. The question is: why the numerical integration doesn't evaluate as a CDF for the general case using MATLAB and Mathematica? Is it a precision issue in the software tools?

 

[mathman]

I don't know how MATLAB or Mathematica were set up. There may be some limitations to their applicability. Since I have never used them I can't help here. When I need complicated integrals I look up the table I previously mentioned.

 

[EngWiPy]

I don't know how MATLAB or Mathematica were set up. There may be some limitations to their applicability. Since I have never used them I can't help here. When I need complicated integrals I look up the table I previously mentioned.

But theoretically, the derivation should be correct, right? Because I just raise the CF of a single X to the power K, and then solve for the CDF as we did for a single X.

 

[mathman]

The basic theory (invert the nth power of the char. function gives the cdf of the sum of n terms) is correct. But as we have seen with the case n=1, the calculation can be very tricky. The Gil-Pelaez theorem had to be used and getting the imaginary part can be messy.

 

[EngWiPy]

Yes, but supposedly, the software tools take care of the imaginary part using their functions to extract the imaginary part. We don't have to concern ourselves with doing that manually.

 

[mathman]

I can't comment on these tools, since I've never used them. If you are sure they should work, I suggest you check the input if the answer doesn't look right.