If I have only $\mathrm{Var}(X)$, how can I calculate $\mathrm{Var}(\frac{1}{X})$?
I do not have any information about the distribution of $X$, so I cannot use transformation, or any other methods which use the probability distribution of $X$.
data transformationdistributionsvariance
If I have only $\mathrm{Var}(X)$, how can I calculate $\mathrm{Var}(\frac{1}{X})$?
I do not have any information about the distribution of $X$, so I cannot use transformation, or any other methods which use the probability distribution of $X$.
Best Answer
It is impossible.
Consider a sequence $X_n$ of random variables, where
$$P(X_n=n-1)=P(X_n=n+1)=0.5$$
Then:
$$\newcommand{\Var}{\mathrm{Var}}\Var(X_n)=1 \quad \text{for all $n$}$$
But $\Var\left(\frac{1}{X_n}\right)$ approaches zero as $n$ goes to infinity:
$$\Var\left(\frac{1}{X_n}\right)=\left(0.5\left(\frac{1}{n+1}-\frac{1}{n-1}\right)\right)^2$$
This example uses the fact that $\Var(X)$ is invariant under translations of $X$, but $\Var\left(\frac{1}{X}\right)$ is not.
But even if we assume $\mathrm{E}(X)=0$, we can't compute $\Var\left(\frac{1}{X}\right)$: Let
$$P(X_n=-1)=P(X_n=1)=0.5\left(1-\frac{1}{n}\right)$$
and
$$P(X_n=0)=\frac{1}{n} \quad \text{for $n>0$} $$
Then $\Var(X_n)$ approaches 1 as $n$ goes to infinity, but $\Var\left(\frac{1}{X_n}\right)=\infty$ for all $n$.