Inequalities of expectations Given two random variables $X$ and $Y$, integrable and everything. Is it true that,
$E(|XY|^2) \leq E(|X|^2) E(|Y|^2) $ ?
If not, can I use something so that I get,
$E|XY|^2 \leq (E|X|^p)^{1/p} E|Y|^2 \ $ ?
Thank you very much guys!
 A: The inequality 
$$
E[|XY|^2] \leq E[|X|^2] E[|Y|^2]
$$
cannot be obtained in general. Indeed we can find an example of a pair of square-integrable random variables $X,Y$ for which the product is not square integrable, breaking the above inequality in a very fundamental way.
Proof: Let $\Omega = (0,1]$ and let $P$ be Lebesgue measure. Then let
$$
X(x) = Y(x) = \frac{1}{x^{1/4}}
$$
Note now that both $X,Y$ are $P$-square integrable (just integrate and see), yet $(XY)(x) = 1/\sqrt{x}$ is not square integrable.
A: I can't make more comments so I write here. Sorry.
Taking indicator functions, f.ex: $X=1_A$ and $Y=1_B$. I obtain:
$P(A\cap B)\leq P(A)P(B)$
but is this true for any events $A$,$B$? (equality is true if they are independent)
In any case, is the inequality true?
I found out that, using Hölder one gets,
$E[|XY|^2] \leq \displaystyle \sup_{\omega \in \Omega}|X(\omega)|^2 E[|Y|^2]$
So $X\in L^{\infty}(\Omega)$. Is this the only way?
A: Due to Hölder's Inequality, for $p,q \in (1,\infty)$ with $1/p +1/q=1$, 
$$ 
E(|XY|) \leq [E(|X|^p)]^{1/p} [E(|Y|^q)]^{1/q}.
$$ 
Taking $p=q=2$ you get $E(|XY|)^2 = E(|X|^2) E(|Y|^2). $
Also note Lyapunov’s Inequality: If $0<s<t$, then
$$
[E(|X|^s)]^{1/s} \leq [E(|X|^t)]^{1/t}.
$$
