0

Fix a random variable $X$ on $R$. I want to compare two following conditions for $X$:

(1) Assume that $\lim_{x\to \infty}x^4P(|X|\ge x)=0$

(2) Assume that $E[|X|^4]<\infty$.

Which one is stronger?

It seems that there is a counterexample: the distribution whose density $f(x) $ decays as $|x|^{−5}\log|x|$,then it does not have finite fourth moment though condition (1) holds for it.

I know that $$ E[X]=\int_R xP(X\ge x).$$ But how about the fourth moment?

Hermi
  • 1,488
  • $x^4 P(|X| \geq x) = E(x^4I(|X|^4 \geq x^4)) \leq E(|X|^4I(|X|^4 \geq x^4))$. By the way, the problem is essentially equivalent if you replace 4 with 1. – Mason Oct 31 '22 at 20:18
  • (2) implies (1). There are examples where (1) holds but not (2). – Mittens Oct 31 '22 at 20:25
  • @OliverDíaz Can you explain why? It seems that there is a counterexample: the distribution whose density $f(x) $ decays as $|x|^{−5}\log|x|$,then it does not have finite fourth moment though condition (1) holds for it. – Hermi Nov 01 '22 at 20:13
  • @Hermi: I think you would be able to cook-up some examples from this posting. But you are on the right track, the logarithm function which grows slowly is helpful. – Mittens Nov 01 '22 at 20:15
  • @OliverDíaz Thanks! How to prove that (2) implies (1)? – Hermi Nov 01 '22 at 20:18
  • 1
    @Hermi: The Markov-Chebyshev inequality yields $\mu(|X|>\lambda)\leq \frac{1}{\lambda^4}\int_{{|X|>\lambda}}|X|^4,d\mu$. Dominated convergence arguments show that $\lim_{\lambda\rightarrow\infty}\int_{{|X|>\lambda}}|X|^4,d\mu=0$. – Mittens Nov 01 '22 at 20:21

0 Answers0