If X and Y are independent and $E[(XY)^2] = 0$, then $P(X = 0) = 1$ or $P(Y = 0) = 1$ Let $(X, Y)$ be a discrete random vector. Prove that: If $X$ and $Y$ are independent and $E[(XY)^2] = 0$, then $P(X = 0) = 1$ or $P(Y = 0) = 1$
Since $X$ and $Y$ are independent, covariance, i.e., $E(XY)-E(X)E(Y)=0 \implies E(XY)=E(X)E(Y), V((XY))=E[(XY)^2] - E[XY]^2=0-E(X)^2E(Y)^2 \leq 0$, since variance $\geq 0, V(XY) =0$. Since the variance is $0, P(XY=c)=1$ but the value of $c$ is not known, so how exactly do we conclude it? Feels like I'm missing something, can someone provide hints?

HINT 1. By independence and the zero assumption, $E((XY)^2) = E(X^2 Y^2) = E(X^2)E(Y^2)=0$. Thus, either $E(X^2)=0$ or $E(Y^2)=0$ or both. But if $E(X^2)=0$, this means
$$
E(X^2)  = \sum_{x\in \mathcal{X}} x^2 P(X=x) = 0.
$$
Can you see now why the result follows?
HINT 2. For all $x$, $0\leq x^2$ and $0< P(X=x)$, thus $0\leq x^2P(X=x)$; without loss of generality, you can think $\mathcal{X}$ includes only those $x$ for which $0< P(X=x)$.
