How to find the moments using the characteristic function?

[Neothilic]

TL;DR

How do you show the Cauchy distribution has no moments, but using the characteristic function?

I have the characteristic function of the Cauchy distribution $C(t)= e^{-(\mid t \mid)}$. Now, how would I show that the Cauchy distribution has no moments using this? I think you have to show it has no Taylor expansion around the origin. I am not sure how to do this.

 

[Office_Shredder]

I can't say much about the first part of your post, but for there to be a Taylor series expansion the function has to be differentiable at the origin. It's not, just draw a graph and take a look at it.

To formally prove it, just calculate what the derivate is to the left of 0, and the right of 0, and observe they are not equal.

 

[Neothilic]

I can't say much about the first part of your post, but for there to be a Taylor series expansion the function has to be differentiable at the origin. It's not, just draw a graph and take a look at it.

To formally prove it, just calculate what the derivate is to the left of 0, and the right of 0, and observe they are not equal.

I have forgotten what you explicitly mean by calculating the derivative is to the left, and to the right. Could you further elaborate? Surely as the function is of modulus t, this wouldn't apply?

 

[Office_Shredder]

Have you tried drawing a plot of the function yet? There's a sharp corner at t=0. On the left of 0 it has one slope, on the right of 0 it has another slope. Similar to how the graph of |t| has a slope of -1 at 0 on the left, and 1 at 0 on the right.

 

[Neothilic]

Have you tried drawing a plot of the function yet? There's a sharp corner at t=0. On the left of 0 it has one slope, on the right of 0 it has another slope. Similar to how the graph of |t| has a slope of -1 at 0 on the left, and 1 at 0 on the right.

I have looked at the plot of the function. What would that do to help? As there is a sharp turn at $t=0$, surely that further enhances my conclusion?

 

[Office_Shredder]

Yeah, I mean that's pretty much it. If you want to prove that the derivative doesn't exist, you can calculate what the derivative at 0 is when restricting to t<0 vs t>0