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Cauchy-Schwarz inequality has been applied to various subjects such as probability theory. I wonder how to prove the following version of the Cauchy-Schwarz inequality for random variables:

$$\bigr|\mathbb E[XY]\bigr| \leq \mathbb E(|XY|) \leq \sqrt{\mathbb E X^2 } \sqrt{\mathbb E Y^2 }$$

I can see why the first inequality is true. Basically, it is rephrase of the triangular inequality, i.e. the fact that the absolute value of an integral is less than or equal to the integral of the absolute value of the integrand. In addition, I understand how to show $\bigr|\mathbb E[XY]\bigr| \leq \sqrt{\mathbb E X^2 } \sqrt{\mathbb E Y^2 }$ by using the fact that $\mathbb E\left[ (X - \alpha Y)^2 \right] \geq 0$ for any real number $\alpha$. However, I could not figure out how to show the second inequality. Could anyone provide any help, please? Thank you!

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    $\begingroup$ Apply $\bigr|\mathbb E[XY]\bigr| \leq \sqrt{\mathbb E X^2 } \sqrt{\mathbb E Y^2 }$ to the random variables $|X|$ and $|Y|$. $\endgroup$
    – Did
    May 14, 2014 at 8:47
  • $\begingroup$ @Did Oh, that is cool. Thank you! $\endgroup$
    – LaTeXFan
    May 14, 2014 at 8:49

1 Answer 1

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Apply the inequality $\bigr|\mathbb E[XY]\bigr| \leq \sqrt{\mathbb E X^2 } \sqrt{\mathbb E Y^2 }$ to the random variables $|X|$ and $|Y|$.

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