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.
 A: If $X$ and $Y$ are independent then $\mathbb{E}[g(X)h(Y)]=\mathbb{E}[g(X)]\mathbb{E}[h(Y)]$ for all continuous functions $g,h$. 

Therefore, using your first expression for the variance of $XY$ we have
\begin{align}
\mathbb{V}ar(XY)=&\mathbb{E}[(XY)^2]-(\mathbb{E}[XY])^2\\
=&\mathbb{E}[X^2 Y^2]-(\mathbb{E}[X]\mathbb{E}[Y]])^2\\
=&\mathbb{E}[X^2]\mathbb{E}[Y^2]- (\mathbb{E}[X])^2(\mathbb{E}[Y])^2
\end{align}
Seems like the solution given to you expanded the first term without expanding the second term. Without knowing more, it is hard to say why the solution recommends this particular form. But it could be that the resulting expressions are particularly easy to compute given a particular form of $X$ and $Y$.
A: 
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.
