# What is that inequality - Cauchy-Schwarz for a single random variable?

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?

• Notice the difference is $\operatorname{Var}Z$. – J.G. Apr 13 at 12:41
• You are right ! Good point ! – micholeodon Apr 17 at 18:24

$$0\leq E(Z-EZ)^2 = E(Z^2-2ZEZ+(EZ)^2)=EZ^2-2(EZ)^2+(EZ)^2=EZ^2-(EZ)^2$$
Just write $$Z$$ as $$(Z)(1)$$ and apply Cauchy -Schwarz inequality.