Does $\mathbb{E}(\max\{\epsilon_1,...,\epsilon_J\})<\infty$ imply $\mathbb{E}(\epsilon_j)<\infty$ for each $j=1,...,J$?

Solution 1:

Yes, this is an application of Jensen's inequality.

Jensen's Inequality: Given a random vector $X$ and a convex function $f$, we have that $f(\mathbb{E}[X]) \leq \mathbb{E}[ f(X) ]$.

You'll need the fact that the function $f(x_1, x_2, \dots x_n) = \max(x_1, x_2, \dots x_n)$ is convex, which is described here.

Using this, we can show that

\begin{align} \max \left( \mathbb{E}[\epsilon_1], \mathbb{E}[\epsilon_2] \dots \mathbb{E}[\epsilon_n] \right) = f(\mathbb{E}[\epsilon]) \leq \left( \mathbb{E}[f(\epsilon)] \right) = \mathbb{E} \left[ \max \left( \epsilon_1, \epsilon_2, \dots \epsilon_n \right) \right] \enspace. \end{align}

And thus, if $\mathbb{E} \left[ \max \left( \epsilon_1, \epsilon_2, \dots \epsilon_n \right) \right] < \infty$, then this implies $\max \left( \mathbb{E}[\epsilon_1], \mathbb{E}[\epsilon_2] \dots \mathbb{E}[\epsilon_n] \right) < \infty$.