Young's inequality in probability theory

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The following is the standard version, in two equivalent statements:

If $a \geq 0$ and $b \geq 0$ are nonnegative real numbers and if $p > 1$ and $q > 1$ are real numbers such that $\frac{1}{p} + \frac{1}{q} = 1,$ then $$\tag{1} a b \leq \frac{a^p}{p} + \frac{b^q}{q}$$

$$\tag{2} a^\alpha b^\beta \leq \alpha a + \beta b, \qquad\, 0 \leq \alpha, \beta \leq 1, \quad\ \alpha + \beta = 1$$

Question. Is there an equivalent statement in probability theory? Say $X$, $Y$ are two random variables. Then is it possible to have, for instance, $$ \mathbb{E}[|XY|]\leq\frac{\mathbb{E}(|X|^p)}{p}+\frac{\mathbb{E}(|Y|^q)}{q} $$

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Of course, because expectation is linear. So you can apply Young's inequality in a pointwise-fashion to say that the random variable $|XY|$ is bounded by $|X|^p/p + |Y|^q/q$, and then take expectation.