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 $X$.

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

It is impossible.

Consider a sequence $X_n$ of random variables, where



$$\newcommand{\Var}{\mathrm{Var}}\Var(X_n)=1 \quad \text{for all $n$}$$

But $\Var\left(\frac{1}{X_n}\right)$ approaches zero as $n$ goes to infinity:


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



$$P(X_n=0)=\frac{1}{n} \quad \text{for $n>0$} $$

Then $\Var(X_n)$ approaches 1 as $n$ goes to infinity, but $\Var\left(\frac{1}{X_n}\right)=\infty$ for all $n$.

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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

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


\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 $g(X)=\frac{1}{X}$, $\text{Var}[\frac{1}{X}] \approx \frac{1}{\mu^4} \text{Var}(X)$.

Wikipedia: Taylor expansions for the moments of functions of random variables


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
[1] 10.20819  # reasonably close to the population variance

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

[1] 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
[1] 1.069147

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

[1] 0.0001122586

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

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  • 1
    $\begingroup$ 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. $\endgroup$ – cardinal Sep 15 '13 at 13:53
  • $\begingroup$ 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 $x = 0$. $\endgroup$ – cardinal Sep 15 '13 at 13:57
  • $\begingroup$ One must indeed be careful about the behavior of the density near 0. Note that in the above gamma examples, the distribution of the inverse is inverse gamma, for which having a finite mean requires $\alpha>1$ ($\alpha$ being the shape parameter of the gamma we're inverting). The two examples had $\alpha = 10$ and $\alpha = 100$. Even so (with "nice" distributions for inverting) neglect of higher terms can introduce a noticeable bias. $\endgroup$ – Glen_b Nov 6 '19 at 22:41
  • $\begingroup$ this seems in the right direction, of a reciprocal shifted normal distribution instead of a reciprocal standard normal distribution: en.wikipedia.org/wiki/… $\endgroup$ – Felipe G. Nievinski Nov 25 '19 at 15:59

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