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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!

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The first property $E[XY]^2 \leq E[X]^2 E[Y]^2$ is the well known Cauchy-Schwarz inequality. –  jay-sun Jan 21 '13 at 17:47
You need some parentheses. I don't know whether the powers are inside or outside the expected values. –  Robert Israel Jan 21 '13 at 17:48
I mean $E[|XY|^2] \leq E[|X|^2]E[|Y|^2]$. I thought the Cauchy-Schwarz inequality was without the square on the left-hand side. I want it with the square. –  Dan Jan 21 '13 at 17:51
Hint: Consider indicator functions. –  cardinal Jan 21 '13 at 18:30
You may also want to take a look at Holder's Inequality associated with Probability measures. –  jay-sun Jan 21 '13 at 18:45

3 Answers 3

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.

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This is also an explicit counterexample to the second inequality you posited, for all $p < 4$. –  A Blumenthal May 8 '13 at 22:59

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?

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In a certain sense, yes! See my above answer. Practically speaking, the Holder inequality one derives is usually the optimal inequality one can obtain. –  A Blumenthal May 8 '13 at 23:01

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 $$ [E(|X|^s)]^{1/s} \leq [E(|X|^t)]^{1/t}. $$

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I see nothing wrong in this answer, but it doesn't address the question at hand. –  cardinal Jan 22 '13 at 0:44
The question may ask $E(|X|^2) \leq E(|X|^p)^{2/p}$ for $p>2$, taking $Y$ to be the indicator function, that's why I noted Lyapunov’s Inequality. –  беркай Jan 22 '13 at 0:51

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