# Example of a non-negative discrete distribution where the mean (or another moment) does not exist?

community wiki 09/14/2018. 4 answers, 1.513 views

I was doing some work in scipy and a conversation came up w/a member of the core scipy group whether a non-negative discrete random variable can have a undefined moment. I think he is correct but do not have a proof handy. Can anyone show/prove this claim? (or if this claim is not true disprove)

I don't have an example handy if the discrete random variable has support on $\mathbb{Z}$ but it seems that some discretized version of the Cauchy distribution should serve as an example to get an undefined moment. The condition of non-negativity (perhaps including $0$) is what seems to make the problem challenging (at least for me).

whuber 11/16/2018 at 20:52.

Let the CDF $$F$$ equal $$1-1/n$$ at the integers $$n=1,2,\ldots,$$ piecewise constant everywhere else, and subject to all criteria to be a CDF. The expectation is

$$\int_{0}^\infty (1-F(x))\mathrm{d}x = 1/2 + 1/3 + 1/4 + \cdots$$

which diverges. In this sense the first moment (and therefore all higher moments) is infinite. (See remarks at the end for further elaboration.)

If you're uncomfortable with this notation, note that for $$n=1,2,3,\ldots,$$

$${\Pr}_{F}(n) = \frac{1}{n} - \frac{1}{n+1.}$$

This defines a probability distribution since each term is positive and $$\sum_{n=1}^\infty {\Pr}_{F}(n) = \sum_{n=1}^\infty \left(\frac{1}{n} - \frac{1}{n+1}\right) = \lim_{n\to \infty} 1 - \frac{1}{n+1} = 1.$$

The expectation is

$$\sum_{n=1}^\infty n\,{\Pr}_{F}(n) = \sum_{n=1}^\infty n\left(\frac{1}{n} - \frac{1}{n+1}\right) =\sum_{n=1}^\infty \frac{1}{n+1} = 1/2 + 1/3 + 1/4 + \cdots$$

which diverges.

This way of expressing the answer it makes it clear that all solutions are obtained by such divergent series. Indeed, if you would like the distribution to be supported on some subset of the positive values $$x_1, x_2, \ldots, x_n, \ldots,$$ with probabilities $$p_1, p_2, \ldots$$ summing to unity, then for the expectation to diverge the series which expresses it, namely

$$(a_n) = (x_n p_n),$$

must have divergent partial sums.

Conversely, every divergent series $$(a_n)$$ of non-negative numbers is associated with many discrete positive distributions having divergent expectation. For instance, given $$(a_n)$$ you could apply the following algorithm to determine sequences $$(x_n)$$ and $$(p_n)$$. Begin by setting $$q_n = 2^{-n}$$ and $$y_n = 2^n a_n$$ for $$n=1, 2, \ldots.$$ Define $$\Omega$$ to be the set of all $$y_n$$ that arise in this way, index its elements as $$\Omega=\{\omega_1, \omega_2, \ldots, \omega_i, \ldots\},$$ and define a probability distribution on $$\Omega$$ by

$$\Pr(\omega_i) = \sum_{n \mid y_n = \omega_i}q_n.$$

This works because the sum of the $$p_n$$ equals the sum of the $$q_n,$$ which is $$1,$$ and $$\Omega$$ has at most a countable number of positive elements.

As an example, the series $$(a_n) = (1, 1/2, 1, 1/2, \ldots)$$ obviously diverges. The algorithm gives

$$y_1 = 2a_1 = 2;\ y_2 = 2^2 a_2 = 2;\ y_3 = 2^3 a_3 = 8; \ldots$$

Thus $$\Omega = \{2, 8, 32, 128, \ldots, 2^{2n+1},\ldots\}$$

is the set of odd positive powers of $$2$$ and $$p_1 = q_1 + q_2 = 3/4;\ p_2 = q_3 + q_4 = 3/16;\ p_3 = q_5 + q_6 = 3/64; \ldots$$

### About infinite and non-existent moments

When all the values are positive, there is no such thing as an "undefined" moment: moments all exist, but they can be infinite in the sense of a divergent sum (or integral), as shown at the outset of this answer.

Generally, all moments are defined for positive random variables, because the sum or integral that expresses them either converges absolutely or it diverges (is "infinite.") In contrast to that, moments can become undefined for variables that take on positive and negative values, because--by definition of the Lebesgue integral--the moment is the difference between a moment of the positive part and a moment of the absolute value of the negative part. If both those are infinite, convergence is not absolute and you face the problem of subtracting an infinity from an infinity: that does not exist.

community wiki 09/14/2018.

Here's a famous example: Let $X$ take value $2^k$ with probability $2^{-k}$, for each integer $k\ge1$. Then $X$ takes values in (a subset of) the positive integers; the total mass is $\sum_{k=1}^\infty 2^{-k}=1$, but its expectation is $$E(X) = \sum_{k=1}^\infty 2^k P(X=2^k) = \sum_{k=1}^\infty 1 = \infty.$$ This random variable $X$ arises in the St. Petersburg paradox.

community wiki 09/14/2018.
1. The zeta distribution is a fairly well-known discrete distribution on the positive integers that doesn't have finite mean (for $1<\theta\leq 2$) .

$P(X=x|\theta)={{\frac {1}{\zeta (\theta)}}x^{-\theta}}\,,\: x=1,2,...,\:\theta>1$

where the normalizing constant involves $\zeta(\cdot)$, the Riemann zeta function

(edit: The case $\theta=2$ is very similar to whuber's answer)

Another distribution with similar tail behaviour is the Yule-Simon distribution.

2. Another example would be the beta-negative binomial distribution with $0<\alpha\leq 1$:

$P(X=x|\alpha ,\beta ,r)={\frac {\Gamma (r+x)}{x!\;\Gamma (r)}}{\frac {\mathrm{B} (\alpha +r,\beta +x)}{\mathrm{B} (\alpha ,\beta )}}\,,\:x=0,1,2...\:\alpha,\beta,r > 0$

community wiki 09/14/2018.

some discretized version of the Cauchy distribution

Yes, if you take $p(n)$ as being the average value of the Cauchy distribution in the interval around $n$, then clearly its zeroth moment is the same as that of the Cauchy distribution, and its first moment asymptotically approaches the first moment of the Cauchy distribution. As far as "the interval around $n$", it doesn't really matter how you define that; take $(n-1,n]$, $[n,n+1)$, $[n-.5,n+.5)$, vel cetera, and it will work. For positive integers, you can also take $p(n) =\frac6{(n\pi)^2}$. The zeroth moment sums to one, and the first moment is the sum of $\frac6{n\pi^2}$, which diverges.

And in fact for any polynomial $p(n)$, there is some $c$ such that $\frac c {p(n)}$ sums to 1. If we then take the $k$th moment, where $k$ is the order of $p(n)$, that will diverge.