Bounded Random Variables with mean $0$ have $ E(X^2)=0$?

Question

I know the statement in the title is wrong (take X to be a signed bernulli(1/2) ). But I have a fake proof of it, and I am not sure where I went wrong. Any help would be appreciated :

Fake Proof:

Say $|X| <K$ then $0 \leq E(X^2) = E(XX) \leq E(KX) = KE(X) = 0$

Here the only thing I use is monotonicity in the second inequality ($ E(XX) \leq E(KX) $)

Where is the mistake ?

Answer 1

$X^2 \le KX$ is only true when $$X(X-K) \le 0$$ i.e., $0 \le X \le K$

However, $X$ can be negative and your inequality is not true in such a scenario, as demonstrated by your counterexample.

Answer 2

$E[XX] \leq E[KX]$ isn't justified, and in fact isn't valid. It would need to be $$E[XX] = E[|X||X|] \leq E[K|X|]$$

Answer 3

The inequality $E(XX)\leq E(KX)$ doesn't hold. For example, $X$ takes values $-1$ or $0$ with probability $1/2$ each and $K = 4/3$.

Then $|X| < K$ but $E(X^2) = \frac{1}{4}$ and $E(KX) = KE(X) = -\frac{2}{3}$