Why is the expected value $E(X^2) \neq E(X)^2$?

I wish to use the Computational formula of the variance to calculate the variance of a normal-distributed function. For this, I need the expected value of $X$ as well as the one of $X^2$. Intuitively, I would have assumed that $E(X^2)$ is always equal to $E(X)^2$. In fact, I cannot imagine how they could be different.

Could you explain how this is possible, e.g. with an example?

Solution 1:

Assume $X$ is a random variable that is 0 half the time and 1 half the time. Then $$EX = 0.5 \times 0 + 0.5 \times 1 = 0.5$$ so that $$(EX)^2 = 0.25,$$ whereas on the other hand $$E(X^2) = 0.5 \times 0^2 + 0.5 \times 1^2 = 0.5.$$ By the way, since $Var(X) = E[(X - \mu)^2] = \sum_x (x - \mu)^2 P(x)$, the only way the variance could ever be 0 in the discrete case is when $X$ is constant.

Solution 2:

Let $EX=\mu$ and $E(X-\mu)^2=\sigma^2$, then

$$ EX^2 = E[X-\mu+\mu]^2=\\ =E(X-\mu)^2+2E[(X-\mu)\mu]+E(\mu^2)=\\=\sigma^2+2\mu E(X-\mu)+\mu^2=\\ =\sigma^2+\mu^2 $$

So $EX^2 =\sigma^2+\mu^2$, no matter the distribution, and $EX^2\ne(EX)^2$ unless the variance equals zero.