Formula for the variance of the product of two random variables Let's say I have two random variables $X$ and $Y$.  Can I write that:
$$VAR \left[XY\right] = \left(E\left[X\right]\right)^2 VAR \left[Y\right] + \left(E\left[Y\right]\right)^2 VAR \left[X\right] + 2 \left(E\left[X\right]\right) \left(E\left[Y\right]\right) COV\left[X,Y\right]?$$
If this is not correct, how can I intuitively prove that?
 A: COUNTEREXAMPLE
Let $X=Y\sim N(0,1)$.
Since both have expected value zero, the right-hand side is zero.
However, $XY\sim\chi^2_1$, which has a variance of $2$.
A: The formula you are asserting is not correct (as shown in the counter-example by Dave), and it is notable that it does not include any term for the covariance between powers of the variables.  Some simple moment-algebra yields the following general decomposition rule for the variance of a product of random variables:
$$\begin{align}
\mathbb{V}(XY)
&= \mathbb{E}((XY)^2) - \mathbb{E}(XY)^2 \\[6pt]
&= \mathbb{E}(X^2 Y^2) - \mathbb{E}(XY)^2 \\[6pt]
&= [\mathbb{Cov}(X^2,Y^2) + \mathbb{E}(X^2)\mathbb{E}(Y^2)] - [\mathbb{Cov}(X,Y) + \mathbb{E}(X)\mathbb{E}(Y)]^2 \\[6pt]
&= \mathbb{Cov}(X^2,Y^2) - \mathbb{Cov}(X,Y)^2 - 2 \ \mathbb{E}(X)\mathbb{E}(Y) \mathbb{Cov}(X,Y). \\[6pt]
\end{align}$$
Alternatively, you can get the following decomposition:
$$\begin{align}
\mathbb{V}(XY)
&= \mathbb{E}((XY-\mathbb{E}(XY))^2) \\[6pt]
&= \mathbb{E}((XY - \mathbb{Cov}(X,Y) - \mathbb{E}(X)\mathbb{E}(Y))^2) \\[6pt]
&= \mathbb{E}(([XY - \mathbb{E}(X)\mathbb{E}(Y)] - \mathbb{Cov}(X,Y))^2) \\[6pt]
&= \mathbb{E}([XY - \mathbb{E}(X)\mathbb{E}(Y)]^2) - 2 \ \mathbb{Cov}(X,Y) \mathbb{E}(XY - \mathbb{E}(X)\mathbb{E}(Y)) + \mathbb{Cov}(X,Y)^2 \\[6pt]
&= \mathbb{E}([XY - \mathbb{E}(X)\mathbb{E}(Y)]^2) - 2 \ \mathbb{Cov}(X,Y)^2 + \mathbb{Cov}(X,Y)^2 \\[6pt]
&= \mathbb{E}([XY - \mathbb{E}(X)\mathbb{E}(Y)]^2) - \mathbb{Cov}(X,Y)^2. \\[6pt]
\end{align}$$
