How to find the moments using the characteristic function?

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Discussion Overview

The discussion revolves around the characteristic function of the Cauchy distribution and the question of whether it has moments. Participants explore the relationship between the differentiability of the characteristic function and the existence of a Taylor series expansion around the origin.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents the characteristic function of the Cauchy distribution and questions how to demonstrate that it has no moments, suggesting the need to show the absence of a Taylor expansion around the origin.
  • Another participant argues that for a Taylor series expansion to exist, the function must be differentiable at the origin, which they claim the characteristic function is not, citing a graphical observation.
  • A further reply emphasizes the need to calculate the derivative from both sides of zero to show they are not equal, supporting the claim of non-differentiability.
  • One participant seeks clarification on how to calculate the derivative from the left and right of zero, questioning the applicability of this method due to the modulus in the function.
  • Another participant reiterates the presence of a sharp corner at t=0, indicating differing slopes on either side, drawing a parallel to the graph of |t|.
  • A later reply confirms that calculating the derivative at zero for t<0 and t>0 can help prove the non-existence of the derivative.

Areas of Agreement / Disagreement

Participants generally agree on the non-differentiability of the characteristic function at the origin, but there is no consensus on the implications of this observation regarding the existence of moments.

Contextual Notes

Participants reference graphical analysis and derivative calculations, but the discussion does not resolve the mathematical steps required to definitively prove the absence of moments.

Neothilic
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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.
 
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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.
 
Office_Shredder said:
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?
 
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.
 
Office_Shredder said:
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?
 
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
 

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