How to prove the Cauchy distribution has no moments?

In summary, the Cauchy distribution has no moments because the integral used to calculate the moments blows up for n > 2 and does not go to zero even for n = 2. This means that the distribution has infinite moments and cannot be calculated using the characteristic function. Additionally, the inequality of the geometric mean and arithmetic mean can be used to prove that the expectation of the absolute value of X is infinite, leading to the conclusion that all moments of the Cauchy distribution are infinite.
  • #1
Neothilic
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TL;DR Summary
Proving how the Cauchy distribution has no moments for all n.
How can I prove the Cauchy distribution has no moments?

##E(X^n)=\int_{-\infty}^\infty\frac{x^n}{\pi(1+x^2)}\ dx.##

I can prove myself, letting ##n=1## or ##n=2## that it does not have any moment. However, how would I prove for ALL ##n##, that the Cauchy distribution has no moments?
 
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  • #2
Sorry for the dumb I'm not a mathematician question but why isn't

## E_A(X^n) = \int_{-A}^A \frac{x^n}{\pi(1+x^2)}dx = 0 ##

for all ##n## odd and all ##A##?
 
  • #3
so you've proven
##E\Big[\big \vert X\big \vert\Big] = \infty##

in general
##\big \vert X\big \vert= \Big(\big \vert X\big \vert^n \cdot 1\Big)^\frac{1}{n}= \Big(\big \vert X\big \vert^n \cdot \prod_{k=1}^{n-1}1\Big)^\frac{1}{n} \leq \frac{1}{n} \Big(\big \vert X\big \vert^n + \sum_{k=1}^{n-1}1 \Big) =\frac{1}{n} \big \vert X\big \vert^n + \frac{n-1}{n}##
by ##\text{GM}\leq \text{AM}##

Now take expectations of each side (and use LOTUS)
##E\Big[\big \vert X\big \vert\Big]\leq \frac{1}{n} E\Big[\big \vert X\big \vert^n\Big] + \frac{n-1}{n}##

thus
##E\Big[\big \vert X\big \vert\Big] = \infty \longrightarrow E\Big[\big \vert X\big \vert^n\Big] = \infty##
 
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  • #4
StoneTemplePython said:
so you've proven
##E\Big[\big \vert X\big \vert\Big] = \infty##

in general
##\big \vert X\big \vert= \Big(\big \vert X\big \vert^n \cdot 1\Big)^\frac{1}{n}= \Big(\big \vert X\big \vert^n \cdot \prod_{k=1}^{n-1}1\Big)^\frac{1}{n} \leq \frac{1}{n} \Big(\big \vert X\big \vert^n + \sum_{k=1}^{n-1}1 \Big) =\frac{1}{n} \big \vert X\big \vert^n + \frac{n-1}{n}##
by ##\text{GM}\leq \text{AM}##

Now take expectations of each side (and use LOTUS)
##E\Big[\big \vert X\big \vert\Big]\leq \frac{1}{n} E\Big[\big \vert X\big \vert^n\Big] + \frac{n-1}{n}##

thus
##E\Big[\big \vert X\big \vert\Big] = \infty \longrightarrow E\Big[\big \vert X\big \vert^n\Big] = \infty##

Wow. I have been trying to prove this for a couple weeks now. I have tried doing series test with the integrands etc. However, this is so unique and short and brilliant! How did you come up with this? Btw, what is meant by GM and AM?
 
  • #5
##\frac{|x^n|}{1+x^2}\to \infty## as ##|x|\to \infty##, for ##n\gt 2##. So the integral blows up.
 
  • #6
mathman said:
##\frac{|x^n|}{1+x^2}\to \infty## as ##|x|\to \infty##, for ##n\gt 2##. So the integral blows up.
And even for n=2 the integrand does not go to zero, that means all these cases can be solved without calculating anything. Only n=1 needs some thought.
Neothilic said:
Btw, what is meant by GM and AM?
Geometric mean, arithmetic mean

You opened three threads for basically the same topic - it would have been better to keep the discussion in one thread. But now merging them would create a mess so I keep them separate.

https://www.physicsforums.com/threa...nts-using-the-characteristic-function.992462/
https://www.physicsforums.com/threads/what-is-the-nth-differential-of-this-equation.992492/
 
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1. How do you define the Cauchy distribution?

The Cauchy distribution is a probability distribution that describes the probability of a random variable taking on a certain value within a given range. It is also known as the Lorentz distribution and is characterized by its heavy tails, meaning that it has a higher probability of extreme values compared to other distributions.

2. What are moments in probability theory?

Moments are statistical measures that describe the shape and characteristics of a probability distribution. They are calculated by taking the expected value of a function of the random variable, where the function is raised to a certain power. The first moment is the mean, the second moment is the variance, and so on.

3. Why is it important to prove that the Cauchy distribution has no moments?

The Cauchy distribution has no moments, meaning that its moments do not exist or are undefined. This is important because it makes it difficult to use traditional statistical methods that rely on moments, such as calculating the mean or variance. It also means that the distribution does not follow the law of large numbers, making it challenging to make accurate predictions based on sample data.

4. How do you prove that the Cauchy distribution has no moments?

The Cauchy distribution has no moments because its probability density function does not converge. This can be proven mathematically by showing that the integral of the function does not converge, using techniques such as integration by parts or the Cauchy-Schwarz inequality. Additionally, it can be shown that the Cauchy distribution does not satisfy the conditions for the existence of moments, such as finite mean or variance.

5. Are there any real-life applications of the Cauchy distribution despite not having moments?

Yes, the Cauchy distribution has several real-life applications, especially in physics and engineering. It is commonly used to model the distribution of errors or disturbances in physical processes, such as the movement of particles or the flow of fluids. It is also used in signal processing and image processing to model noise. Additionally, the Cauchy distribution has been used in the field of finance to model extreme events, such as stock market crashes.

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