The title is the question. I am told that ratios and inverses of random variables often are problematic. What is meant is that expectation often do not exist. Is there a simple, general explication of that?


3 Answers 3


I would like to offer a very simple, intuitive explanation. It amounts to looking at a picture: the rest of this post explains the picture and draws conclusions from it.

Here is what it comes down to: when there is a "probability mass" concentrated near $X=0$, there will be too much probability near $1/X\approx \pm \infty$, causing its expectation to be undefined.

Instead of being fully general, let's focus on random variables $X$ that have continuous densities $f_X$ in a neighborhood of $0$. Suppose $f_X(0)\ne 0$. Visually, these conditions mean the graph of $f$ lies above the axis around $0$:

Figure showing the graph of a density and the area below it.

The continuity of $f_X$ around $0$ implies that for any positive height $p$ less than $f_X(0)$ and sufficiently small $\epsilon$, we may carve out a rectangle beneath this graph which is centered around $x=0$, has width $2\epsilon$, and height $p$, as shown. This corresponds to expressing the original distribution as a mixture of a uniform distribution (with weight $p\times 2\epsilon=2p\epsilon$) and whatever remains.

Figure showing the graph as a mixture.

In other words, we may think of $X$ as arising in the following way:

  1. With probability $2p\epsilon$, draw a value from a Uniform$(-\epsilon,\epsilon)$ distribution.

  2. Otherwise, draw a value from the distribution whose density is proportional to $f_X - p I_{(-\epsilon,\epsilon)}$. (This is the function drawn in yellow at the right.)

($I$ is the indicator function.)

Step $(1)$ shows that for any $0 \lt u \lt \epsilon$, the chance that $X$ is between $0$ and $u$ exceeds $p u / 2$. Equivalently, this is the chance that $1/X$ exceeds $1/u$. To put it another way: writing $S$ for the survivor function of $1/X$

$$S(x) = \Pr(1/X \gt x),$$

the picture shows $S(x) \gt p / (2x)$ for all $x \gt 1/\epsilon$.

We're done now, because this fact about $S$ implies the expectation is undefined. Compare the integrals involved in computing the expectation of the positive part of $1/X$, $(1/X)_{+} = \max(0, 1/X)$:

$$E[(1/X)_{+}] = \int_0^\infty S(x)dx \gt \int_{1/\epsilon}^x S(x)dx \gt \int_{1/\epsilon}^x \frac{p}{2x}dx = \frac{p}{2} \log(x\epsilon).$$

(This is a purely geometric argument: every integral represents an identifiable two-dimensional region and all the inequalities arise from strict inclusions within those regions. Indeed, we don't even need to know the final integral is a logarithm: there are simple geometric arguments showing this integral diverges.)

Since the right side diverges as $x\to\infty$, $E[(1/X)_{+}]$ diverges, too. The situation with the negative part of $1/X$ is the same (because the rectangle is centered around $0$), and the same argument shows the expectation of the negative part of $1/X$ diverges. Consequently the expectation of $1/X$ itself is undefined.

Incidentally, the same argument shows that when $X$ has probability concentrated on one side of $0$, such as any Exponential or Gamma distribution (with shape parameter less than $1$), then still the positive expectation diverges, but the negative expectation is zero. In this case the expectation is defined, but is infinite.

  • 3
    $\begingroup$ Am I right in suspecting that the assumption $f_X(0)\neq 0$ is crucial for the result? I mean, we have cases where $1/X$ has moments at least for some range of involved parameters, and it appears that it is in cases where $f_X(0) = 0$, like Gamma/Inverse-Gamma $\endgroup$ Aug 26, 2017 at 20:00
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    $\begingroup$ @Alecos No, that assumption is not crucial. That and the continuity of $f$ at $0$ make the argument simple, but neither is essential. Consider an $X$ with density $f_X$ proportional to $-1/\log(x)$ for $0 \lt x \lt 1/e$ and $f_X(0)=0$. This is continuous at $0$ but $1/X$ has no expectation. $\endgroup$
    – whuber
    Aug 27, 2017 at 20:35
  • $\begingroup$ For a continuous distribution, $f_X(0) = 0$ is not a sufficient condition for the integral $\int_0^b 1/x f_X(x) dx$ to converge. But when the slope has a finite limit $\lim_{x\to 0} f_X^\prime(0) = a$ then you do get convergence. You can imagine the zooming in on the piece from $-\epsilon$ to $-\epsilon$ being a wedge shaped function $a |x|$. $\endgroup$ Jun 16, 2022 at 10:08

Ratios and inverses are mostly meaningful with nonnegative random variables, so I will assume $X \ge 0$ almost surely. Then, if $X$ is a discrete variable which take on the value zero with positive probability, we will be dividing with zero with a positive probability, which explains why the expectation of $1/X$ will not exist.

Now look at the continuous distribution case, with $X \ge 0$ a random variable with density function $f(x)$. We will assume that $f(0)>0$ and that $f$ is continuous (at least at zero). Then there is an $\epsilon > 0$ such that $f(x) > \epsilon $ for $0 \le x < \epsilon$. The expected value of $1/X$ is given by $$ \DeclareMathOperator{\E}{\mathbb{E}} \E \frac1{X} = \int_0^\infty \frac1{x} f(x)\; dx $$ Now let us change variable of integration to $u=1/x$, we have $du = -\frac1{x^2} \; dx$, obtaining $$ \E \frac1{X} = -\int_{\infty}^0 u f(\frac1{u}) (\frac1{u})^2 \; du = \\ \int_0^\infty \frac1{u} f(\frac1{u}) \; du $$ Now, by assumption $f(u) > \epsilon$ on $[0,\epsilon)$ so $f(\frac1{u}) > 1/\epsilon$ on $(1/\epsilon, \infty)$, using this we have $$ \E \frac1{X} > \epsilon \int_{1/\epsilon}^\infty \frac1{u}\; du =\infty $$ showing that the expectation does not exist. An example fulfilling this assumption is the exponential distribution with rate 1.

We have given an answer for inverses, what about ratios? Let $Z=Y/X$ be the ratio of two nonnegative random variables. If they are independent, we can write $$ \E Z = \E\frac{Y}{X}=\E Y \cdot \E\frac1{x} $$ so this pretty much reduces to the first case and there is not much new to say. What if they are dependent, with joint density factoring as $$ f(x,y) = f(x \mid y) g(y) $$ Then we get (using same substitution as above) $$ \E \frac{Y}{X} = \int_0^\infty y \int_0^\infty \frac1{x} f(x\mid y) \; dx \;g(y)\; dy = \\ \int_0^\infty y \int_0^\infty \frac1{u} f(\frac1{u}\mid y) \; du \; g(y) \; dy $$ and we can reason as above on the inner integral. The result will be that if the conditional density (given $y$) is positive and continuous at zero, for a set of $y$'s with positive marginal probability, the expectation will be infinite. I guess it will not be easy to find examples where the marginal expectation of $1/X$ is infinite, but the expectation of the ratio $Y/X$ is finite, unless there is a perfect correlation. It would be nice to see some such examples!

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    $\begingroup$ Re your last remark: such examples abound. Take $U$ to be any variable with infinite expectation and $V$ be an independent variable with finite expectation. Set $X=1/U$ and $Y=V/U,$ so that $(1/X,Y/X)=(U,V).$ Want the correlation to be small? Then let $B$ be an independent Rademacher variable and set $Y=BV/U$ instead; now, any measure of "correlation" must be zero due to the symmetry of $Y.$ $\endgroup$
    – whuber
    Mar 17, 2021 at 22:08

Let's offer a "dissenting" view:
Ratios and inverses of random variables can be fine in the following sense:

  • It may be the case that in many cases they do not possess moments
  • But it is also the case that in many cases they result in recognizable, "named" and exhaustively studied distributions.
  • ...and there is distribution-life beyond moments, like probabilities and quantiles


  • Student's t-distribution is the ratio of a Normal and a Chi distribution
  • F-distribution is the ratio of two Chi-squares
  • Ratio of two Normals is Cauchy
  • Ratio of two Exponentials is Lomax (shifted Pareto)

On the contrary, it is products of random variables that appear to not lead to recognizable distributions that often.

  • $\begingroup$ Many of your examples are non-examples: for instance, the F distribution has infinite mean when its second df parameter is $2$ or smaller; the Cauchy distribution (same as Student's t with 1 df) has no mean; Pareto distributions with sufficiently long tails have no mean. $\endgroup$
    – whuber
    Mar 17, 2021 at 22:03
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    $\begingroup$ @whuber I don't understand your point. The first thing I state in my answer is that ratios often don't have moments, and then I explain why they are more "friendly" than products, regardless. $\endgroup$ Mar 18, 2021 at 1:51
  • $\begingroup$ What, then, do you mean by "examples"? What are these intended to be examples of? $\endgroup$
    – whuber
    Mar 18, 2021 at 13:28
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    $\begingroup$ @whuber Examples of ratios of random variables that map to well-known and well-studied distributions, and so not so "problematic" as regards their usability, in reference to the word that the OP uses in the title of the post. $\endgroup$ Mar 18, 2021 at 13:37

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