Variance calculation bivariate probability distribution I understand the following statement is true
$$V(XY)=E((XY)^2)-(E(XY))^2$$
In my question X and Y are independent.
The solution recommends
$$ V(XY)=E((X)^2)E((Y)^2)-(E(XY))^2 $$
and I curious to understand why this approach fails
$$V(XY)=E(XY)E(XY)-(E(XY))^2$$
Seems that there are arbitrary rules applying here, I am certain there are not and so would appreciate it if someone could unpick the issue.


The solution recommends
V(XY)=E((X)^2)E((Y)^2)-(E(XY))^2

Because $X$ and $Y$ are independent, so too are $X^2$ and $Y^2$. The expectation of a product of independent random variables is the product of their expectations.
You can further say $\mathsf V(XY)=\mathsf E(X^2)~\mathsf E(Y^2)-(\mathsf E(X))^2~(\mathsf E(Y))^2$

and I curious to understand why this approach fails
V(XY)=E(XY)E(XY)-(E(XY))^2

The product rule only applies for independent random variables.  $XY$ is not independent from $XY$; that product is actually very dependent on itself.
