Powers of Positive Random Variable are monotonic

Question

While reading a paper, I came across the following inequality:

Let $m_1, m_2$ be nonnegative integers, and $X$ a random variable of magnitude at most $1$. Then,

$$E(|X|^{m1})E(|X|^{m2}) \leq E(|X|^{m1+m2})$$

Despite applying Jensen's and Holder's, I have been unable to prove the statement.

Answer 1

You could use the fact that increasing functions of a random variable are positively correlated. Since $x\mapsto x^{m_{1}}$ and $x\mapsto x^{m_{2}}$ are increasing functions on the non-negative reals, we have $$\begin{align*} \mathrm{Cov}\left(|X|^{m_{1}}, |X|^{m_{2}}\right) &\ge 0\\ \Rightarrow \Bbb{E}\left[|X|^{m_{1}}|X|^{m_{2}}\right] - \Bbb{E}\left[|X|^{m_{1}}\right]\Bbb{E}\left[|X|^{m_{2}}\right]&\ge 0 \\ \Rightarrow \Bbb{E}\left[|X|^{m_{1}+m_{2}}\right] &\ge \Bbb{E}\left[|X|^{m_{1}}\right]\Bbb{E}\left[|X|^{m_{2}}\right]. \end{align*} $$

Answer 2

Let $Y$ is an independent copy of $X$.

We have \begin{align*} &\frac12\cdot \mathbb{E}\Big[(|X|^{m_1} - |Y|^{m_1})(|X|^{m_2} - |Y|^{m_2})\Big]\\[6pt] ={}& \frac12 \cdot \mathbb{E}\Big[|X|^{m_1 + m_2} - |Y|^{m_1}\cdot |X|^{m_2} - |X|^{m_1}\cdot |Y|^{m_2} + |Y|^{m_1 + m_2}\Big]\\[6pt] ={}& \frac12\cdot \Big[\mathbb{E}(|X|^{m_1 + m_2}) - \mathbb{E}(|Y|^{m_1})\mathbb{E}(|X|^{m_2}) - \mathbb{E}(|X|^{m_1})\mathbb{E}(|Y|^{m_2}) + \mathbb{E}(|Y|^{m_1 + m_2})\Big]\\[6pt] ={}& \mathbb{E}(|X|^{m_1 + m_2}) - \mathbb{E}(|X|^{m_1})\mathbb{E}(|X|^{m_2}). \end{align*}

On the other hand, using the fact that $(a^{m_1} - b^{m_1}) (a^{m_2} - b^{m_2}) \ge 0$ for all $a, b \ge 0$ (easy to prove), we have $\frac12\cdot \mathbb{E}[(|X|^{m_1} - |Y|^{m_1})(|X|^{m_2} - |Y|^{m_2})] \ge 0$.

We are done.