Let $X$ be a discrete random variable and suppose $\mathrm{Var}(X)>1$. How do $\mathrm{Var}(X)$ and $\mathrm{Var}(X^2)$ compare?"
Is this enough information to solve?
I think that $\mathrm{Var}(X^2)$ is larger than $\mathrm{Var}(X)$, but I only arrived at the answer by creating a hypothetical situation and solving it, but not sure if that holds in all cases or if there was a more obvious or intuitive way to come to this conclusion.
If $X$ takes values from {$-1,1$}, then $X$ has a positive variance but $X^2=1$ has zero variance.
On the other hand for a Bernoulli variable $I$, take $Y=aI$ then $$ \text{E}(Y) = ap $$
$$ \text{E}(Y^2) = a^2p $$
$$ \text{E}(Y^4) = a^4p $$
So $$ \text{Var}(Y) = a^2pq $$ $$ \text{Var}(Y^2) = a^4 pq $$ So depending on $a$ we will get any relation.