# Var(XY), if X and Y are independent random variables [duplicate]

if X and Y are independent Random variable then what is the variance of XY?

• Do you have any thoughts about this yourself? $\text{Var}(X)\text{Var}(Y)$ would be wrong - consider an almost surely constant non-zero $X$ – Henry Mar 16 '19 at 11:12
• No sir. I know Var(XY)= E(X^2 Y^2)- (E(XY))^2 and E(XY)= E(X) E(Y) as X, Y are independent but no idea about X^2 and Y^2 are independent or not. – Dev Mar 16 '19 at 11:22
• If $X$ and $Y$ are independent then $X^2$ and $Y^2$ are also independent and $E[X^2Y^2]=E[X^2]E[Y^2]$ – Henry Mar 16 '19 at 11:24
• General product case here: stats.stackexchange.com/questions/52646/… (product of 2 is given in the question) – Glen_b Mar 16 '19 at 13:18
• Thank u so much Glen_b – Dev Mar 24 '19 at 12:16

## 1 Answer

You can follow Henry's comments to arrive at the answer. However, another way to come to the answer is to use the fact that if $$X$$ and $$Y$$ are independent, then $$Y | X = Y$$ and $$X |Y = X$$.

By iterated expectations and variance expressions

\begin{align*} \text{Var}(XY) & = \text{Var}[\,\text{E}(XY|X)\,] + \text{E}[\,\text{Var}(XY|X) \,]\\ & = \text{Var}[\,X\, \text{E}(Y|X)\,] + E[\,X^2\, \text{Var}(Y|X)\,]\\ & = \text{Var}[\,X\, \text{E}(Y)\,] + E[\,X^2\, \text{Var}(Y)\,]\\ & = E(Y)^2\, \text{Var}(X) + \text{Var}(Y) E(X^2)\,. \end{align*}

• $E(Y)^2\, \text{Var}(X) + \text{Var}(Y) E(X^2)$ may be correct, but it is strangely non-symmetric as $E(Y^2)\, \text{Var}(X) + \text{Var}(Y) E(X)^2$ would be. I would have thought $\text{Var}(X)E(Y)^2 + \text{Var}(Y) E(X)^2 +\text{Var}(X)\text{Var}(Y)$ would be more natural while $\text{Var}(X)E(Y^2) + \text{Var}(Y) E(X^2) -\text{Var}(X)\text{Var}(Y)$ would also be true – Henry Mar 16 '19 at 13:29
• @Henry Well, using $E(X^2) = Var(X) + E(X)^2$, we get $Var(XY) = E(Y)^2Var(X) + Var(Y)Var(X) + Var(Y) E(X)^2$. That's symmetric. – Greenparker Mar 16 '19 at 13:52