Variance of a random variable

Question

How do you get the variance of a random variable $X$ where $X = \frac{1}{6}(A \cdot B)$ and where $A$ and $B$ are two independent random variables with variances $\sigma_A^2$ and $\sigma_B^2$, respectively?

Answer 1

Let's start off with the definitions to build some intuition.

First off, if $A$ and $B$ are independent, then $E[AB]=E[A]E[B]$

$Var(X)=E(X^2)-[E(X)]^2=E[(AB/6)^2]-[E(AB/6)]^2$

$\quad\quad\quad\quad\!\!=\dfrac{1}{36}E[A^2]E[B^2]-\dfrac{1}{36}E[A]^2E[B]^2$

Note that $Var(A)=\sigma_A^2=E(A^2)-E(A)^2$ and likewise for $Var(B)$.

It should not be too difficult to go on from here.