$Var(X_1 X_2)=Var(X_1) Var(X_2)$

Question

Let : $X_1$ and $X_2$ two independent variables, square-integrable et non constant.

At which necessary and sufficient condition, do we have $ Var(X_1 X_2)= Var(X_1) Var(X_2)$ ?

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Attempt : $V(Z)= V( E(Z|X_1) ) + E(V(Z|X_1))$ if $Z$ is a random variable.

Answer 1

$$var(X_1 X_2) = E(var(X_1 X_2|X_2))+var(E(X_1 X_2 | X_2)\\ =E(X_2^2 var(X_1)) + var(X_2 E(X_1))\\ = var(X_1) (E(X_2)^2 + var(X_2)) + E(X_1)^2 var(X_2)\\ = var(X_1) var(X_2) + E(X_2)^2 var(X_1) + E(X_1)^2 var(X_2)$$

Since $X_1, X_2$ are not constants then we must have $E(X_1)=E(X_2)=0$ which is necessary and sufficient.

Answer 2

With $Z=X_1 X_2$ your attempt leads to \begin{align} V(X_1 X_2) &= V(X_1 E[X_2 \mid X_1]) + E[X_1^2 V(X_2 \mid X_1)] \\ &= E[X_2]^2 V(X_1) + V(X_2) E[X_1^2] \\ &= V(X_1)V(X_2) + V(X_1)E[X_2]^2 + V(X_2) E[X_1]^2 \end{align}