# Is this formula correct if we are given that $X_1$ and $X_2$ are independent

If I am given that $X_1$ and $X_2$ are independent, can the formula $Var(X_1X_2)$ be "manipulated" anymore than it already is? I think this is as far as I can go but I'm not sure. The formula is supposed to be in terms of $\mu_1=E(X_1)$ , $\sigma_1^2=Var(X_1)$ , $\mu_2=E(X_2)$, and $\sigma_2^2=Var(X_2)$. So what I did was this: $$Var(X_1X_2)=Var(X_1)Var(X_2)=\sigma_1^2\sigma_2^2$$ $$=(E(X_1^2)-(E(X_1))^2)(E(X_2^2)-(E(X_2))^2)$$ and since $E(X_1)$ and $E(X_2)$ equal $\mu_1$ and $\mu_2$, $$(E(X_1^2)-\mu_1)(E(X_2^2)-\mu_2)$$ Can this be manipulated any further? I feel like it can but I'm not sure how.

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It's not generally true that $V(X_1X_2) = V(X_1)V(X_2)$. –  Alan Guo Oct 23 '12 at 2:49
This looks a lot like homework, and if so, please add the homework tag. That being said, I suggest that you consider using the formula $$\text{var}(Z) = E[Z^2] - \left(E[Z]\right)^2$$ with $Z = X_1X_2$ and use the result that if $X_1$ and $X_2$ are independent random variables, then $g(X_1)$ and $h(X_2)$ are also independent random variables for (measurable) functions $g$ and $h$. –  Dilip Sarwate Oct 23 '12 at 2:54