Does $E(X^2)\ge 0$ always hold?

Question

In probability theory, is always $E(X^2)\geq0$?

I think it is because $X^2\geq0$ and probability $P(x) \in [0,1]$.

Answer 1

It Really depends on where the random variable $X$ outputs. For example if $X:\Omega \rightarrow \mathbb{C}$ and $\Pr{( X=i)}=1$ then $E(X)=-1$ . Anyway if $X$ is, for example, a real r.v. then you are right

Answer 2

As long as $X$ is a real random variable, then yes. Assuming $X$ is a continuous random variable and is distributed over $\mathbb{R}$ according to some valid probability density function $f$, then $$E[X^2]=\int _{\mathbb{R}} x^{2} f( x) dx.$$ Since $x^2\geq0 \ \forall x\in \mathbb{R}$ and by definition $f(x)\geq0 \ \forall x\in \mathbb{R}$ (since $f$ is a PDF), then $E[X^2]\geq0$. This result is easily seen for a discrete case as well.