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In the following article: https://link.springer.com/content/pdf/10.1023%2FA%3A1018054314350.pdf

below equation (4.1) there is a statement that: $$EZ^2 \geq (EZ)^2$$, where $Z$ is a random variable.

I am not sure where does it come from and whether is true. How to prove that inequality?

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  • $\begingroup$ Notice the difference is $\operatorname{Var}Z$. $\endgroup$ – J.G. Apr 13 at 12:41
  • $\begingroup$ You are right ! Good point ! $\endgroup$ – micholeodon Apr 17 at 18:24
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$0\leq E(Z-EZ)^2 = E(Z^2-2ZEZ+(EZ)^2)=EZ^2-2(EZ)^2+(EZ)^2=EZ^2-(EZ)^2$

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Just write $Z$ as $(Z)(1)$ and apply Cauchy -Schwarz inequality.

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I have also found that this is a special case of a Jensen's inequality for convex functions, that is explained and proven here: http://www.stat.cmu.edu/~larry/=stat705/Lecture2.pdf

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