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Let $\epsilon\equiv(\epsilon_1,\dots, \epsilon_J)$ be a continuous random vector. Consider the quantity $$ \mathbb{E}(\max\{\epsilon_1,...,\epsilon_J\}) $$

Question: Is $\mathbb{E}(\epsilon_j)<\infty$ for each $j=1,...,J$ necessary for $\mathbb{E}(\max\{\epsilon_1,...,\epsilon_J\})$ to be finite?

StubbornAtom
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Star
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    Well, not with the absolute value there. We could imagine that $\epsilon_1$ is always positive and $\epsilon_2, \dots, \epsilon_J$ are always negative; in that case, the negative ones could have expected value $-\infty$ if they wanted to, as long as $\mathbb{E}(\max{\epsilon_1,\dots,\epsilon_J}) = \mathbb E(\epsilon_1)$ is finite. But it's worth asking whether we need $\mathbb E(\epsilon_j) < \infty$ for all $j$. – Misha Lavrov Dec 08 '21 at 16:09
  • Thanks. Let me remove the absolute value. – Star Dec 08 '21 at 16:14

1 Answers1

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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 $\max \left( \mathbb{E}[\epsilon_1], \mathbb{E}[\epsilon_2] \dots \mathbb{E}[\epsilon_n] \right) < \infty$.

Chris Harshaw
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