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Let

$E[X]=0$

them How is Expected value of $E[X^2]$ ? in my opinion is $0$ But I need confirmation and explanation

In general, if $ (\Omega,\Sigma,P) $ is a probability space and $ X: (\Omega,\Sigma) \to (\mathbb{R},\mathcal{B}(\mathbb{R})) $ is a real-valued random variable, then $$ \text{E}[X^{2}] = \int_{\Omega} X^{2} ~ d{P}. $$

see for more about this in this quetion Computing the Expectation of the Square of a Random Variable: $E[X^2]$.

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    $\begingroup$ If X is 1 half the time and -1 half the time, E(X) =0, bur X^2=1 all the time, so E(X^2)=1. Hope this example helps see whats going on $\endgroup$ Sep 10, 2021 at 16:15

2 Answers 2

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No way. Since $X^2\geq 0$, $\mathbb{E}[X^2]=0$ iff $X$ is zero almost surely.

Otherwise, if $\mathbb{E}[X^2]$ is well defined, $\mathbb{E}[X^2]\color{red}{>}0.$

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  • $\begingroup$ So, if the expected value of x is zero What I can say about expected value of x squared? $\endgroup$ Feb 2, 2015 at 17:04
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    $\begingroup$ You can't say anything about $E(X^2)=V(X)$. For example, suppose $X\sim N(0,a)$ where $a>0$ is fixed and arbitrary. Then $E(X)=0$ whereas $E(X^2)=a$. $\endgroup$
    – Matthew H.
    Jan 7, 2021 at 1:41
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A really easy example to disabuse yourself of this notion is to look up standard normal and $\chi^2$ distributions.

If $Z \sim N(0, 1)$, then $Z^2 \sim \chi_1^2$. We have $\mathrm E[Z] = 0$, but $\mathrm E[Z^2] = 1$.

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