This may or may not be an answer to Is there an easy proof that the set of $x \in [0,1]$ whose limit of proportion of 1's in binary expansion of $x$ does not exist has measure zero?, depending on how easy a proof has to be to count as easy.

The law of large numbers comes in various versions:

LLN${}_p$ Suppose $X_1,\dots$ are iid with $\Bbb EX_1=0$ and $\Bbb E|X_1|^p<\infty$. Then $\frac1n(X_1+\dots+X_n)\to0$ almost surely.

Proving LLN${}_1$ is considerably harder than proving LLN${}_2$. My impression is that this gives LLN${}_2$ a higher utility/difficulty ratio, since it suffices for many applications. Seems to me that LLN${}_4$ has yet a much higher utility/difficulty ratio. There's an awesomely easy proof of LLN${}_4$. The obligatory question then is this:

Question Are there a lot of standard probability distributions out there, that actually come up, that give $\Bbb E X^4=\infty$ but $\Bbb E X^2<\infty$?

Hoping for an answer to that question is why I'm posting this. The idea that people might be amused by the proof of LLN${}_4$ is absolutely no part of my motivation, I swear; that would not be a legal reason for this post's existence.

Of course you need to see the proof I have in mind in order to evaluate that utility/difficulty ratio:

Say $S_n = X_1+\dots+X_n$. Multiply out the product $S_n^4$: $$S_n^4=\sum X_{j_1}X_{j_2}X_{j_3}X_{j_4}.$$Independence shows that most of those terms have mean $0$; the only terms with non-zero mean are of the form $X_j^4$ or a "permutation" of $X_j^2X_k^2$. A trivial bit of combinatorics shows then that $$\Bbb ES_n^4\le cn^2.$$

Monotone convergence shows that $$\Bbb E\left[\sum\left(\frac 1n S_n\right)^4\right]<\infty.$$Hence $\sum\left(\frac 1n S_n\right)^4<\infty$ almost surely, so $\frac1nS_n\to0$ almost surely.

(Note I'm not claiming to be smart here; I saw this somewhere many years ago.)

  • 1
    $\begingroup$ I wish people would explain the reason for downvotes. The proof is too well known to be interesting? Someone feels that pretending it was a question is inappropriate? $\endgroup$ – David C. Ullrich Aug 28 '15 at 23:46
  • $\begingroup$ Nice question & preamble; I don't get the downvotes. $\endgroup$ – copper.hat Aug 29 '15 at 1:41

The keyword is "fat-tail" distributions. So anything that falls off like $1/x^p$ as $x\rightarrow\infty$ will have all moments up to and including $p-2$. These come up more often than people expect, especially in finance and actuarial sciences and are absolutely catastrophic when it comes to (underestimating) rare events [read something like Black Swan by Taleb], especially if it's something like the Cauchy distribution $\frac{1}{\pi} \frac{1}{x^2+1}$ which has no mean in the first place. Something like $1/x^{2+\epsilon}$ for small $\epsilon$ will also be a headache numerically.

For your benefit, I suggest writing a simple program that samples from the Cauchy distribution to see how the LLN fails. Try a similar exercise with the extra $\epsilon$ as well.

  • $\begingroup$ Hey thanks! Of course when I asked the question that was just a little game, twisting a little essay into the form of a "question". The first version of your answer, just referring to fat-tail distributions, was fine but sort of a "well yes of course" thing. But the current version tells me things I didn't know - turns out it was actually a question after all. heh... $\endgroup$ – David C. Ullrich Aug 28 '15 at 23:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.