0
$\begingroup$

Let $X$ be a random variable and suppose that we have $E(|X|) <+\infty$ and $E(|X|^2)<+\infty$. In particulary, I can find a constant C such that $$ E(|X|^2) < C\quad\quad (1) $$ I consider the higher moments of $X$ as $E(|X|^{2p})$ where $p>1$. I have two questions:

1) Can we always conclude that $E(|X|^{2p}) < + \infty$ by using (1) (maybe not!) or it depends on the properties of $X$?

2) Suppose that $E(|X|^{2p}) < + \infty$ and if I want to find an upper bound for $E(|X|^{2p})$. Is there any inequality that allow us to find the upper bound by using (1)?

Thank you in advance for any answer.

$\endgroup$
  • 2
    $\begingroup$ Answer is no to both questions when p>1. But the answer is yes when p <1 $\endgroup$ – Amr Mar 24 '16 at 11:42
  • 1
    $\begingroup$ For example, knowing $E(|X|^4)$ puts a natural upper bound on $E(|X|^2)$ (the square root) but not the other way round. $\endgroup$ – Henry Mar 24 '16 at 15:38
0
$\begingroup$

To see the answer is no to both questions, you can consider the following counter examples

1. We can have $\mathbb E(X^{2p})=\infty$

Let us assume $P(X)\propto\frac1{|X|^\nu+1}$. If $\nu>1$, it is a well defined probability distribution, which has a finite $\mathbb{E}(|X|)$ for $\nu>2$ and a finite $\mathbb{E}(|X|^2)$ for $\nu>3$.

Actually we have \begin{align} \mathbb{E}(|X|^\mu)&\text{ finite} & \text{if }&\mu<\nu\\ \mathbb{E}(|X|^\mu)&=+\infty & \text{if }&\mu\ge\nu, \end{align} therefore, a good choice of $\nu$ allows set the rank after which all moments are infinite, even if all the lower moments are bounded

2. We can have really high finite $\mathbb E(X^{2p})$, even when $\mathbb E(X^{2p})\le C$

Just look at the following simple probability distribution for $X$, defined for $\varepsilon>0$: \begin{align} P(X=0)&=1-\varepsilon; & P\left(X=\sqrt{\frac C\varepsilon}\right)&=\varepsilon. \end{align} We have then, assuming $\varepsilon\ll 1$ and $p>1$ \begin{align} \mathbb E(|X|)&=\sqrt{\varepsilon C}&\ll \sqrt{C}\\ \mathbb E(|X|^2)&=\varepsilon C&\\ \mathbb E(|X|^{2p})&=\frac{C^p}{\varepsilon^{p-1}}& \gg C^p \end{align} Chosing a small enough $\varepsilon$ allows to choose a $E(|X|^{2p})$ arbitrary large, but finite, while keeping the first two moments below $C$.

$\endgroup$

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.