Variance of a function of a random variable as function of the original variable Suppose $X$ a random variable and $Y=f(X)$ a function of this variable
I know I can write $\mathbb V(Y)$ as $\mathbb E(f(X)^2)-\mathbb E(f(X))^2$
I would like to know if it is possible to write $\mathbb V(Y)$ as a function of $\mathbb V(X)$
e.g : $\mathbb V(Y)=g(\mathbb V(X))$ given the function $f$
In other words, I need to find an expression for the variance of the transformed variable $Y$ in which the variance of the initial variable $X$ appears explicitely, preferably as its parameter.
Details about the problem :

*

*The distribution of $X$ is unknown, and I don't think we need it because we compute variances of the whole population, not a sample of it.

*As it happens, the function $f$ is continuous, strictly decreasing, and positive (somewhat like $f(x)=1/x$ with $x>0$).


Let $X\sim \mathcal N(0,\sigma^2)$ denote a normal random variable and let $f$ be the function $$f(x) = \begin{cases}+1, & x > 0,\\-1, &x \leq 0.\end{cases}$$
Then, $f(X)$ is a random variable taking on values $\pm 1$ with equal probability and so $f(X)$ has variance $1$.  On the other hand, if $X\sim \mathcal N(1,\sigma^2)$, then $f(X)$ takes on values $\pm 1$ with probabilities $\Phi\left(-\frac{1}{\sigma}\right)$ and $1-\Phi\left(-\frac{1}{\sigma}\right)$ and its variance is not $1$ even though the variance of $X$ is unchanged. Thus, it is not necessarily possible that $\mathbb V(f(X))$ can be expressed as a fixed function $g(\cdot)$ of $\mathbb V(X)$, that is, $\mathbb V(f(X))\neq g(\mathbb V(X))$.
Are there any functions $f(\cdot)$ for which $\mathbb V(f(X))$ equals $g(\mathbb V(X))$ for some fixed function $g(\cdot)$? Sure, there are. If
$f(x) = ax+b$, then
$$\mathbb V(f(X)) = \mathbb V(aX+b) = a^2\mathbb V(X) = g(\mathbb V(X))$$ where $g(x) = a^2x$.
What if $X$ is a positive random variable and $f(x) = x^{-1}$ for $x >0$?
Well, one case where the desired relationship might hold is when if $X$ is a Gamma random variable with order parameter $3$ or more. See this answer of mine for some details of how this might be made to work.
