Show that $E(|X-Y|^{3/2})\leq 2 E(|X|^{3/2})$ when $X$ and $Y$ are i.i.d.

Question

Let $X$ and $Y$ be i.i.d. random variables such that $E(|X|^{3/2})$ is finite. Prove that $$E(|X-Y|^{3/2})\leq 2 E(|X|^{3/2})$$

This is from a past qualifying exam. I'm really stumped by the question. I tried several things with Jensen, conditional expectations, ..., to no avail.

Answer 1

There exists a constant $c>0$ such that

$$|\xi|^{3/2} = c \int_{\mathbb{R} \backslash \{0\}} (1-\cos(u \xi)) \frac{1}{|u|^{5/2}} \, du \tag{1}$$

for all $\xi \in \mathbb{R}$. (To prove this identity, start with the integral on the right-hand side and perform a change of variables, $z := u \xi$.) Thus,

$$\begin{align*} |X-Y|^{3/2} &= c \int_{u \neq 0} \big(1-\cos(u(X-Y) \big) \frac{1}{|u|^{5/2}} \, du \\ &= c \int_{u \neq 0} \big(1-\cos(uX) \cos(uY)-\sin(uX) \sin(uY) \big) \frac{1}{|u|^{5/2}} \, du. \end{align*}$$

Taking expectation and using the fact that the random variables are independent and identically distributed, we get

$$\begin{align*} \mathbb{E}(|X-Y|^{3/2}) &= c \int_{u \neq 0} (1-\big(\mathbb{E}\cos(uX))^2-(\mathbb{E}\sin(uX))^2 \big) \frac{1}{|u|^{5/2}} \, du. \end{align*}$$

As

$$\big[ \mathbb{E}(1-\cos(uX))\big]^2+ \big[ \mathbb{E}\sin(uX)\big]^2 \geq 0$$

we have

$$-(\mathbb{E}\cos(uX))^2-(\mathbb{E}\sin(uX))^2 \leq 1-2 \mathbb{E}\cos(uX),$$

and so

$$\mathbb{E}(|X-Y|^{3/2}) \leq 2c \mathbb{E} \int_{u \neq 0} (1-\cos(uX)) \frac{1}{|u|^{5/2}} \, du \stackrel{(1)}{=} 2\mathbb{E}(|X|^{3/2}).$$