if X and Y are independent Random variable then what is the variance of XY?
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$\begingroup$ Do you have any thoughts about this yourself? $\text{Var}(X)\text{Var}(Y)$ would be wrong - consider an almost surely constant non-zero $X$ $\endgroup$– HenryCommented Mar 16, 2019 at 11:12
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$\begingroup$ 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. $\endgroup$– DevCommented Mar 16, 2019 at 11:22
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$\begingroup$ If $X$ and $Y$ are independent then $X^2$ and $Y^2$ are also independent and $E[X^2Y^2]=E[X^2]E[Y^2]$ $\endgroup$– HenryCommented Mar 16, 2019 at 11:24
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1$\begingroup$ General product case here: stats.stackexchange.com/questions/52646/… (product of 2 is given in the question) $\endgroup$– Glen_bCommented Mar 16, 2019 at 13:18
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1 Answer
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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*}
answered Mar 16, 2019 at 11:48
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3$\begingroup$ $E(Y)^2\, \text{Var}(X) + \text{Var}(Y) E(X^2)$ 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 $\endgroup$– HenryCommented Mar 16, 2019 at 13:29
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3$\begingroup$ @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. $\endgroup$ Commented Mar 16, 2019 at 13:52