How do E[X] and E[|X|] relate?

[MidgetDwarf]

TL;DR

For expectations of random variables. Relationship between E[X] n E[|X|]?

Greetings,

I am studying probability theory [non-measure theory] from a textbook.

I stumbled to the topic stating that Cauchy Distribution has no moments.

It was not proved, and I tried working it via direct calculation of the improper integral of E[X^n] for the case n=1.

Anyhow, I wanted to generalize this without success. I stumbled upon this thread here:

Undergrad Thread 'How to prove the Cauchy distribution has no moments?'

Aug 14, 2020

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?

• Neothilic
• Cauchy Distribution Expectation Moments
• Replies: 5
• Forum: Set Theory, Logic, Probability, Statistics

I really enjoyed the proof given StoneTemplePython. However, I am unsure of the following:

Why does E[|x|] diverges implies E[x] diverges?I believe this is the crux of the poof. Since it assumes, in my understanding, that having shown E[|X|] diverges, then applying the comparison test for improper integrals in the last line of that thread, then E[|X|^n] diverges.

How does this relate to E[X^n]?

. My hunch tells me that it has something to do with f is integrable iff |f| is integrable, and how we can decompose |f| into the positive and negative parts of f?

My knowledge of Measure Theory is very poor. So hopefully someone here can why.

 

[anuttarasammyak]

Integral for even n diverges clearly to plus infinite. Integral for odd n is zero for integrand [-L,L]. However we do not have such a constraint thus the integral diverges plus or minus.

 

[MidgetDwarf]

Let $X_-$ and $X_+$ denote the negative and positive parts of $X$. Then $E[X]=E[X_+]+E[X_-]$ and $E[|X|]=E[X_+]-E[X_-]$. If $E[|X|]$ is infinite or undefined, then one or both of $E[X_+]$ and $E[X_-]$ are infinite, so $E[X]=E[X_+]+E[X_-]$ also diverges.
I'm not sure about the higher orders, $E[X^n]$ for $n\gt 1$. If $X$ is restricted to $|X| \lt 1$, then those moments may exist.

Thank you fact checker!

I now understand that the proof given was using the measure theory [lebesgue integral] formulation of probability.

the decomposition of E into its positive and negative parts comfirmed my hunch.

from my understanding, E[|X|] exists iff E[X] does using lebesgue integral. Since it comes from the fact that f is integrable iff |f| is integrable.

This is not necessarily true under the Reimann Integral. Ie., E[X] finite does not imply E[|X]|] is finite.

 

[Filip Larsen]

Let $X_-$ and $X_+$ denote the negative and positive parts of $X$. Then $E[X]=E[X_+]+E[X_-]$ and $E[|X|]=E[X_+]-E[X_-]$.

For this to hold true it seems to me there also needs to be a requirement along the lines that $X_+$ has same "cardinality" or "density" as $X_-$?

E.g. for a finite sequence of $N = P + M$ numbers, $P$ non-negative and $M$ negative, I would expect the relation to be $N E[X] = P E[X_+] + M E[X_-]$.

 

[FactChecker]

Thank you fact checker!

I now understand that the proof given was using the measure theory [lebesgue integral] formulation of probability.

I wouldn't say that necessarily depends on the type of integral. Lebesgue integration allows you to talk about integration in a more general context, but it is easy to use this with either integral method. That being said, I do not know what treacherous details would be required in a formal proof.

the decomposition of E into its positive and negative parts comfirmed my hunch.

Yes, it is a simple step in that direction.

This is not necessarily true under the Reimann Integral. Ie., E[X] finite does not imply E[|X]|] is finite.

That may be true. I am too rusty on this to say. It would depend on how the limits are defined. I would say that if $E[|X|]$ is infinite, with both positive and negative parts infinite, then the value of $E[X]$ depends on how limits are taken.

I will have to leave this discussion for those who are more knowledgeable.

 

[anuttarasammyak]

The contraposition: "E[ X ] converges thus E[ |X| ] converges" seems easier to handle.

$$ E[|x|]:=\int_{-\infty}^{+\infty} f(x)|x| dx = -\int_{-\infty}^{0} f(x)x dx + \int_{0}^{+\infty} f(x)x dx = E[x]-2 \int_{-\infty}^{0} f(x)x dx $$

As a finite value cannot be divided into two infinites, $\int_{-\infty}^{0} f(x)x dx$ is finite thus E[|x|] is finite.

 

[statdad]

Notice that the integral $$\int_0^1 f(x)dx$$ converges for the Cauchy density. Now, for x larger than 1 $$1+x^2 < 2x^2$$ so $$\frac 1 {1+x^2} > \frac 1 {2x^2}$$

This means that for $n \ge 2$ you have $$\frac{x^n}{1+x^2} > \frac{x^n}{2x^2} = \frac{x^{n-2}}{2}$$, so $n - 2 \ge 0$, and since $$\int_1^{\infty} x^k dx$$ diverges when $k \ge 0$ you've shown $$\int_0^{\infty} x^n f(x) dx$$ diverges for $n \ge 2$.

For the integral over the negative portion of the number line apply that sort of argument to $|x|^n$.

Of course since $$\int_{-\infty}^{\infty} f(x) dx$$ is defined as $$\int_{-\infty}^{0} f(x) dx + \int_{0}^{\infty} f(x) dx$$ (I could have used any number other than $0$), since the integral over the positive portion diverges the entire integral diverges.