Var(X) is known, how to calculate Var(1/X)?

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 $X$.

• I think that this might help you. – Christoph_J Nov 5 '12 at 9:27

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 $n$.

You can use Taylor series to get an approximation of the low order moments of a transformed random variable. If the distribution is fairly 'tight' around the mean (in a particular sense), the approximation can be pretty good.

So for example

$$g(X) = g(\mu) + (X-\mu) g'(\mu) + \frac{(X-\mu)^2}{2} g''(\mu) + \ldots$$

so

\begin{eqnarray} \text{Var}[g(X)] &=& \text{Var}[g(\mu) + (X-\mu) g'(\mu) + \frac{(X-\mu)^2}{2} g''(\mu) + \ldots]\\ &=& \text{Var}[(X-\mu) g'(\mu) + \frac{(X-\mu)^2}{2} g''(\mu) + \ldots]\\ &=& g'(\mu)^2 \text{Var}[(X-\mu)] + 2g'(\mu)\text{Cov}[(X-\mu),\frac{(X-\mu)^2}{2} g''(\mu) + \ldots] \\& &\quad+ \text{Var}[\frac{(X-\mu)^2}{2} g''(\mu) + \ldots]\\ \end{eqnarray}

often only the first term is taken

$$\text{Var}[g(X)] \approx g'(\mu)^2 \text{Var}(X)$$

In this case (assuming I didn't make a mistake), with $g(X)=\frac{1}{X}$, $\text{Var}[\frac{1}{X}] \approx \frac{1}{\mu^4} \text{Var}(X)$.

---

Some examples to illustrate this. I'll generate two (gamma-distributed) samples in R, one with a 'not-so-tight' distribution about the mean and one a bit tighter.

a <- rgamma(1000,10,1)  # mean and variance 10; the mean is not many sds from 0
var(a)
 10.20819  # reasonably close to the population variance

The approximation suggests the variance of $1/a$ should be close to $(1/10)^4 \times 10 = 0.001$

var(1/a)
 0.00147171

Algebraic calculation has that the actual population variance is $1/648 \approx 0.00154$

Now for the tighter one:

a <- rgamma(1000,100,10) # should have mean 10 and variance 1
var(a)
 1.069147

The approximation suggests the variance of $1/a$ should be close to $(1/10)^4 \times 1 = 0.0001$

var(1/a)
 0.0001122586

Algebraic calculation shows that the population variance of the reciprocal is $\frac{10^2}{99^2\times 98} \approx 0.000104$.

• Note that in this case, a quite weak hypothesis leads to the conclusion that no mean (whence variance) for $1/X$ will exist, i.e., that the approximation in the answer will be rather misleading. :-) An example hypothesis is that $X$ has a density $f$ that is continuous in an interval around zero and such that $f(0) \neq 0$. The result then follows because the density will be bounded away from zero on some interval $[-\epsilon,\epsilon]$. The hypothesis just given is not the weakest possible, of course. – cardinal Sep 15 '13 at 13:53
• The reason the Taylor series argument then fails is because $\approx$ hides the remainder (error) term, which in this case is $$R(x,\mu) = \frac{(x+\mu)(x-\mu)^2}{x\mu} \>,$$ and this behaves badly around $x = 0$. – cardinal Sep 15 '13 at 13:57