Under what conditions does the variance-to-mean-ratio decrease when a random variable is multiplied by another? Let $A$ and $B$ be two independent random variables.
Variance to mean ratio is defined as $D=\frac{\sigma^2}{\mu}$. Are there any conditions under which the VMR of $AB$ is less than the VMR of $A$?
 A: The mean and variance of the product of two independent random variables with first two moments $\mu_A, \mu_B, \sigma^2_a, \sigma^2_B$ are:
$$\begin{align}
\mu_{AB} &= \mu_A \mu_B \\ 
\sigma^2_{AB} &= \mu^2_A\sigma^2_B + \mu^2_B\sigma^2_A + \sigma^2_A\sigma^2_B
\end{align}$$
If we take the ratio of the two ratios $D_{AB} = \sigma^2_{AB}/\mu_{AB}$ and $D_A = \sigma^2_A/\mu_A$ we get:
$${D_{AB} \over D_A} = {\sigma^2_{AB} \over \sigma^2_A}{\mu_{A} \over \mu_{AB}}$$
which expands to:
$${D_{AB} \over D_A} = {\mu^2_A\sigma^2_B \over\mu_B\sigma^2_A} + \mu_B +{\sigma^2_B \over \mu_B}$$
which, obviously, can be rearranged some, but not in any particularly useful way.  You can compare this ratio to $1$ to see whether the VMR of $AB$ is less than or equal to the VMR of $A$; obviously sometimes it will be and sometimes it won't.
Similarly, subtracting $D_A$ from $D_{AB}$ results in:
$$D_{AB}-D_A = \sigma^2_{AB} = {\mu_A \over \mu_B\sigma^2_B} + {\mu_B \over \mu_A\sigma^2_A} + {\sigma^2_A\sigma^2_B \over \mu_A\mu_B} - {\sigma^2_A \over \mu_A}$$
which is not especially helpful either; nonetheless, you can compare the difference to $0$  to see whether the VMR of $AB$ is less than or equal to the VMR of $A$; again, sometimes it will be and sometimes it won't.
