What are examples of DISCRETE probability distributions (a) with a mean but no finite variance, (b) without a mean?

Question

What are examples of DISCRETE probability distributions (a) with a mean but no finite variance, (b) without a mean?

For continuous distributions, good answers are (a) the pareto distribution with alpha parameter between 1 and 2, and (b) the Cauchy distribution. (Actually, does anybody know a better common example than the pareto, since it has have that particular parameter range?)

It should be pretty easy to construct a discrete example with fat tails too, where the infinite series converges for probability but not for higher moments, but my googling didn't find one.

Answer 1

I'm not sure if this is what you're looking for, but for $n \in \mathbb{N} $ $$ p_n = A n^{-(1+m)}$$ With $A$ a normalizing constant, and $m=1,2,...$ being the first moment you want to diverge. This works because the sum of $1/n$ diverges but the sum of $1/n^2$ converges. There are of course many other such distributions.

Answer 2

Regarding (a), the comment here suggests a random variable $X$ with probability mass function $$p_{X}(x) = \dfrac{1}{\zeta(3)x^3}$$ for $x = 1, 2, \dots$.

We have $\sum_{x=1}^{\infty}p_{X}(x) = 1$ by definition.

It is also known that $$\mathbb{E}[X] = \sum_{x=1}^{\infty}xp_{X}(x) = \sum_{x=1}^{\infty}\dfrac{1}{\zeta(3)x^2} = \dfrac{1}{\zeta(3)} \cdot \dfrac{\pi^2}{6}$$ and $$\mathbb{E}[X^2] = \sum_{x=1}^{\infty}x^2p_{X}(x) = \sum_{x=1}^{\infty}\dfrac{1}{\zeta(3)x}\text{.}$$ We have $$\dfrac{1/[\zeta(3)x]}{1/x} = \dfrac{x}{\zeta(3)x} = \dfrac{1}{\zeta(3)} < \infty$$ so since $\sum_{x=1}^{\infty}\dfrac{1}{x}$ diverges, it follows that $\mathbb{E}[X^2]$ is not finite. Thus, $\text{Var}(X)$ does not exist.

Regarding (b), the answer here gives the example of a random variable $X$ with probability mass function $$p_{X}(x) = \dfrac{1}{x} - \dfrac{1}{x + 1}$$ for $x = 1, 2, \dots$.

We observe that $$\sum_{x=1}^{\infty}p_{X}(x) = \sum_{x=1}^{\infty}\left(\dfrac{1}{x}-\dfrac{1}{x+1} \right) = \lim_{n \to \infty}\sum_{x=1}^{n}\left(\dfrac{1}{x}-\dfrac{1}{x+1} \right) = \lim_{n \to \infty}\left(1 - \dfrac{1}{n + 1} \right) = 1$$ but $$\begin{align} \mathbb{E}[X] &= \sum_{x=1}^{\infty}xp_{X}(x) \\ &= \sum_{x=1}^{\infty}x\left(\dfrac{1}{x} - \dfrac{1}{x+1} \right) \\ &= \lim_{n \to \infty}\sum_{x=1}^{n}x\left(\dfrac{1}{x} - \dfrac{1}{x+1} \right) \\ &= \lim_{n \to \infty}\left[1\left(1 - \dfrac{1}{2} \right) + 2\left(\dfrac{1}{2} - \dfrac{1}{3} \right) + 3\left(\dfrac{1}{3} - \dfrac{1}{4} \right) + \cdots + n\left(\dfrac{1}{n} - \dfrac{1}{n+1} \right)\right] \\ &= \lim_{n \to \infty}\left(1 - \dfrac{1}{2} + 1 - \dfrac{2}{3} + 1 - \dfrac{3}{4} + \cdots + 1 - \dfrac{n}{n+1} \right) \\ &= \lim_{n \to \infty}\left(\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \cdots + \dfrac{1}{n+1} \right) \\ &= \lim_{n \to \infty}\sum_{x=1}^{n}\dfrac{1}{x+1} \end{align}$$ which diverges since $$\int_{1}^{\infty}\dfrac{1}{x+1}\text{ d}x $$ is divergent, and then applying the integral test.

Answer 3

For (b)

For $n\geq1,P(X=2^n)=(\frac{1}{2})^n$ and $P(Xa)=0$ otherwise