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 $X$.

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# Variance Calculation – How to Calculate Var(1/X) from Known Var(X)

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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 $n$.