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.

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    $\begingroup$ 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
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    $\begingroup$ 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

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 $C$.


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