Relationship between probability and expected value Suppose we have a random variable $Y$ which takes on discrete values 1, 2, 3, ... (i.e. Y is a positive random variable). Assume that we know that $\lim_{n\to\infty} P(Y<n)=1$ (or then $\lim_{n\to\infty} P(Y>n)=0$). Is it possible to show that the first moment of Y, $E(Y)$, exists and is finite? And if not in this general setting, which additional assumptions need to be made for $\lim_{n\to\infty} P(Y<n)=1$ to imply $E(Y)<\infty$?
 A: The existence of $E(Y)$ is the condition that the series
$$
\sum_{n \geq 1} n \, P(Y = n) \quad \quad \quad (*)
$$
must converge.  I can cook up an example of divergence: let $S$ be the sum of the convergent series $\sum_{n \geq 1} n^{-2}$, and let
$$
P(Y = n) = {1 \over n^2 S} \quad \mbox{for $n = 1, 2, \ldots$}.
$$
Then series (*), above, becomes
$$
{1 \over S} \sum_{n \geq 1} {1 \over n}
$$
and, therefore, diverges.
For a general criterion of convergence, I would look for a suitable special case of the Cauchy-Hadamard Theorem: https://en.wikipedia.org/wiki/Cauchy%E2%80%93Hadamard_theorem
A: It's possible to have $Y<\infty$ with probability $1$ and $\mathbb{E}[Y]=\infty$. One famous example is to let $Y$ be the first time a symmetric random walk on the integers starting at the origin reaches $1$. An easier example is to let $\mathbb{P}(Y=n)=\frac{6}{\pi^2}\frac{1}{n^2}$ for $n=1,2,3,\dots$.
On the other hand, one can show that
$$ \mathbb{E}[Y]=\sum_{n=1}^{\infty}\mathbb{P}(Y\geq n) $$
(with both sides possibly infinite) for any random variable taking values in the non-negative integers, so this gives a necessary and sufficient condition for the expectation to be finite.
