Does $ \mathbb{E}\Big(\min_{1\leq j\leq J} |\epsilon_j|\Big)=\infty $ imply heavy tails? Consider $J$ continuous random variables
$$
\epsilon_1,\dots, \epsilon_J
$$
Suppose
$$
\mathbb{E}\Big(\min_{1\leq j\leq J} |\epsilon_j|\Big)=\infty
$$
Could you help me to "graphically" interpret this condition? In particular, does it mean that there is an issue of heavy tails? If not, is there any other intuitive interpretation?
 A: Since $(x_1,...,x_J) \mapsto \min_{1 \le j \le J} x_j$ is concave on $\mathbb{R}^J$, Jensen's inequality gives $$\min_{1 \le j \le J} \mathbb{E}[|\epsilon_j|] \ge \mathbb{E}\left[ \min_{1 \le j \le J} |\epsilon_j|\right] = \infty,$$ so $\mathbb{E}[|\epsilon_j|] = \infty$ for all $1 \le j \le J$.  Therefore, the condition $\mathbb{E}\left[ \min_{1 \le j \le J} |\epsilon_j|\right] = \infty$ says that none of the random variables $\epsilon_1,...,\epsilon_J$ are integrable.  You can interpret that as saying that all of those random variables have heavy tails.
A: Here's another way to look at it. I'll assume that the $\epsilon_j$ are iid and let $M = \min_{j\leq J} |\epsilon_j|$. I'll show by contrapositive that the $|\epsilon_j|$ being light tailed means $M$ has a finite mean.
Suppose that the $|\epsilon_j|$ are not heavy tailed, which I'll take to mean that the moment generating function (MGF) $M_{\epsilon}(t) < \infty$ for some $t$ in a neighborhood of zero.
The MGF being finite near zero means that for $t$ sufficiently small we have
$$
\text E[e^{t|\epsilon|}] = \int_0^\infty P(e^{t|\epsilon|} > x)\,\text dx = \int_0^\infty P\left(|\epsilon| > \frac{\log x}t\right)\,\text dx < \infty
$$
so it must be that $P\left(|\epsilon| > \frac{\log x}t\right) \to 0$ as a rate faster than $1/x$ as $x\to\infty$. This means $P(|\epsilon| > x)$ decays at an exponential rate at least. We then have
$$
\int_0^\infty P(M > x)\,\text dx = \int_0^\infty P(|\epsilon|> x)^J\,\text dx 
$$
and the exponential decay of $P(|\epsilon| > x)$ guarantees that this is finite, so $\text E[M] = \int_0^\infty P(M > x) < \infty$.
