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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?

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    $\begingroup$ The variance of a random variable is a constant, so you have a constant on the left and a random variable on the right. Therefore the identity is basically always false for any non trivial random variables $X$ and $Y$ $\endgroup$
    – Stratos supports the strike
    Commented Mar 22, 2022 at 11:49
  • $\begingroup$ @StratosFair apologies it should be Expectation of the rv. I corrected this in my post $\endgroup$
    – Brian Smith
    Commented Mar 22, 2022 at 11:55
  • $\begingroup$ Suppose $E[X]=E[Y]=0:$ your formula would have you conclude the variance of $XY$ is zero, which clearly is not implied by those conditions on the expectations. $\endgroup$
    – whuber ♦
    Commented Mar 22, 2022 at 14:15

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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:

$$\begin{align} \mathbb{V}(XY) &= \mathbb{E}((XY)^2) - \mathbb{E}(XY)^2 \\[6pt] &= \mathbb{E}(X^2 Y^2) - \mathbb{E}(XY)^2 \\[6pt] &= [\mathbb{Cov}(X^2,Y^2) + \mathbb{E}(X^2)\mathbb{E}(Y^2)] - [\mathbb{Cov}(X,Y) + \mathbb{E}(X)\mathbb{E}(Y)]^2 \\[6pt] &= \mathbb{Cov}(X^2,Y^2) - \mathbb{Cov}(X,Y)^2 - 2 \ \mathbb{E}(X)\mathbb{E}(Y) \mathbb{Cov}(X,Y) \\[6pt] &\quad \quad + \mathbb{E}(X^2)\mathbb{E}(Y^2)-[\mathbb{E}(X)\mathbb{E}(Y)]^2 \\[6pt] &= \mathbb{Cov}(X^2,Y^2) - \mathbb{Cov}(X,Y)^2 - 2 \ \mathbb{E}(X)\mathbb{E}(Y) \mathbb{Cov}(X,Y) \\[6pt] &\quad \quad + \mathbb{Var}(X)\mathbb{Var}(Y)+\mathbb{Var}(X)\mathbb{E}(Y)^2+\mathbb{Var}(Y)\mathbb{E}(X)^2.\\[6pt] \end{align}$$

Alternatively, you can get the following decomposition:

$$\begin{align} \mathbb{V}(XY) &= \mathbb{E}((XY-\mathbb{E}(XY))^2) \\[6pt] &= \mathbb{E}((XY - \mathbb{Cov}(X,Y) - \mathbb{E}(X)\mathbb{E}(Y))^2) \\[6pt] &= \mathbb{E}(([XY - \mathbb{E}(X)\mathbb{E}(Y)] - \mathbb{Cov}(X,Y))^2) \\[6pt] &= \mathbb{E}([XY - \mathbb{E}(X)\mathbb{E}(Y)]^2) - 2 \ \mathbb{Cov}(X,Y) \mathbb{E}(XY - \mathbb{E}(X)\mathbb{E}(Y)) + \mathbb{Cov}(X,Y)^2 \\[6pt] &= \mathbb{E}([XY - \mathbb{E}(X)\mathbb{E}(Y)]^2) - 2 \ \mathbb{Cov}(X,Y)^2 + \mathbb{Cov}(X,Y)^2 \\[6pt] &= \mathbb{E}([XY - \mathbb{E}(X)\mathbb{E}(Y)]^2) - \mathbb{Cov}(X,Y)^2. \\[6pt] \end{align}$$

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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 $2$.

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