Let random variables $X, Y$ be independent, then
$\text{Var}(XY) \geq \text{Var}(X)\text{Var}(Y)$
I thought $\text{Var}(XY) = [E(X)]^2 \text{Var}(Y) + [E(Y)]^2 \text{Var}(X) + \text{Var}(X)\text{Var}(Y)$
but I don't know how prove the rest of this.
Could someone explain me that thing step by step how to get the answer to the task?
Sorry for bad format, but it's my first post on this website.
$Var(X) = E[X^2] - E[X]^2$, similarly $Var(Y) = E[Y^2] - E[Y]^2$
We have: $Var(XY) = E[X^2Y^2] - E[XY]^2$.
Since $X,Y$ are independent, we have independence of $X^2,Y^2$, too. In particular $E[XY] = E[X]E[Y]$ and $E[X^2Y^2]=E[X^2]E[Y^2]$, so:
$Var(XY) = \mathbb E[X^2] \mathbb E[Y^2] - (\mathbb E[X]\mathbb E[Y])^2 = \mathbb E[X^2]\mathbb E[Y^2] - (\mathbb E[X]\mathbb E[Y])^2 + \mathbb E[X^2] \mathbb E[Y]^2 - \mathbb E[X^2]\mathbb E[Y]^2 = \mathbb E[X^2](\mathbb E[Y^2] - \mathbb E[Y]^2) + \mathbb E[Y]^2(\mathbb E[X^2] - \mathbb E[X]^2) = \mathbb E[X^2]Var(Y) + \mathbb E[Y]^2Var(X) = \mathbb E[X^2]Var(Y) + \mathbb E[Y]^2Var(X) + \mathbb E[X]^2Var(Y) - \mathbb E[X]^2Var(Y) = Var(X)Var(Y) + \mathbb E[X]^2Var(Y) + \mathbb E[Y]^2Var(X)$
Since both $Var$ and $\mathbb E^2$ operators are $\ge 0$, then $Var(XY) \ge Var(X)Var(Y)$