# I've heard that ratios or inverses of random variables often are problematic, in not having expectations. Why is that?

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?

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 $$0$$: 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. 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.

• 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 Aug 26 '17 at 20:00
• @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.
– whuber
Aug 27 '17 at 20:35

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!

• 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 $Y.$
– whuber
Mar 17 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

EXAMPLES for RATIOS

• 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)
etc

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

• 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.
– whuber
Mar 17 at 22:03
• @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. Mar 18 at 1:51
• What, then, do you mean by "examples"? What are these intended to be examples of?
– whuber
Mar 18 at 13:28
• @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. Mar 18 at 13:37