Variance and Relation to Mean

Question

I know that $Var(X)=E[(X-E(X))^2]$

if $E(X)=0$,

then $Var(X)=E[X^2]$ do this equals to zero also ?

Answer 1

No.

For every $\epsilon>0$ we have $$\epsilon^21_{\{|X|\geq\epsilon\}}\leq X^2$$ and consequently (take expectation on both sides):$$\epsilon^2P(|X|\geq\epsilon)\leq\mathbb EX^2$$

So $\mathbb EX^2=0$ implies that $P(X=0)=1$.

There are lots of situations where $\mathbb EX=0$ and $P(X=0)<1$.

Answer 2

No, take for example a normal distribution, $X\sim N(0,1)$. The expectation is zero, the variance is known to be $1$, and hence $$1 = \text{Var}(X) = E[(X-E[X])^2] = E[X^2].$$