Inequality between $\mathbb{E}[XY]$ and $\mathbb{E}[X^2]$ if $X$ and $Y$ have the same distribution

Question

Does there exist an inequality connecting $\mathbb{E}[XY]$ and $\mathbb{E}[X^2]$ if $X$ and $Y$ have the same distribution, regardless of whether they are independent or not?

Answer 1

The random variables $X$ and $Y$ can be thought of as vectors in a vector space (of infinite dimensions), equipped with an inner product $$\langle X, Y \rangle = \mathbb{E} \left [ XY \right ]$$

Cauchy Schwarz inequality says:

\begin{align*} |\langle X, Y \rangle | &\leq \sqrt{\langle X, X \rangle } \sqrt{\langle Y, Y \rangle }\\ \big | \mathbb{E}[XY] \big | &\leq \sqrt{\mathbb{E}[X^2] } \sqrt{\mathbb{E}[Y^2] } \end{align*}

But $X$ and $Y$ have the same distribution so:

\begin{align*} \big | \mathbb{E}[XY] \big | &\leq \sqrt{\mathbb{E}[X^2] } \sqrt{\mathbb{E}[X^2] }\\ \\ \big | \mathbb{E}[XY] \big | &\leq \mathbb{E}[X^2] \\ \\ \implies \mathbb{E}[XY] &\leq \mathbb{E}[X^2] \end{align*}

Answer 2

Note that $0\le E(X-cY)^2=c^2EY^2-2cEXY+EX^2$ for all $c\in\Bbb R$, so this quadratic in $c$ has discriminant $\le0$, i.e. $4c^2(EXY)^2-4c^2EX^2EY^2\le0$. This simplifies to $|EXY|\le EX^2EY^2$.