When is the expectation of the maximum bounded Let $\epsilon\equiv(\epsilon_1,\dots, \epsilon_J)$ be a random vector. Let $F$ be the probability distribution of $\epsilon$.
Assumption 1: $F$ has full support on $\mathbb{R}^J$.
Assumption 2: $F$ has marginals that are identical and symmetric around zero.
Take a vector of reals $U\equiv (U_1,\dots, U_J)\in [a,b]^J$ with $|a|<\infty, |b|< \infty$.
Consider the quantity
$$
\mathbb{E}(\epsilon_{X^*})
$$
where
$$
X^*\equiv argmax_{j\in \{1,\dots,J\}}\Big(U_j+ \epsilon_{j}\Big)
$$
Note that under Assumption 1, $X^*$ is unique with probability 1. Assumption 1 wil be maintained throughout.
Claim 0: Under Assumptions 1 and 2 (and without further assumptions on $F$), $|\mathbb{E}(\epsilon_{X^*})|$ is not bounded away from $\infty$  (i.e., it might be equal to $\infty$).

QUESTION: I want to show Claim 0.    Could you help?

SUB-QUESTION: Suppose we replace Assumption 2 with Assumption 3.
Assumption 3: $\epsilon_1,\dots, \epsilon_J$ are mutually independent and identically distributed. In particular, for $j=1,...,J$,  $\epsilon_j$ has a Gumbel distribution with scale 1 and location 0.
Under Assumption  3 (which   implies also Assumption 1), we know that $
|\mathbb{E}(\epsilon_{X^*})|<\infty$.
Therefore, I wonder which assumption is actually needed to ensure that $|\mathbb{E}(\epsilon_{X^*})|<\infty$ (while maintaining the full support condition).
My prior is that we need some restrictions on the shape  of $F$ in the tails. Is this correct? Could you elaborate?
 A: Suppose (for symplicity) that $\mathsf{P}(U_i+\epsilon_i=U_j+\epsilon_j)=0$ for any $i\ne j$. Then
\begin{align}
\mathsf{E}|\epsilon_{X^*}|&=\sum_{j=1}^J \mathsf{E}[|\epsilon_j|1\{X^*=j\}] \\
&\le J\max_{1\le j\le J}\mathsf{E}|\epsilon_j|. 
\end{align}
Therefore, $\mathsf{E}|\epsilon_{X^*}|<\infty$ as long as each $\epsilon_j$ is integrable. On the other hand, if $\mathsf{E}\min_{1\le j\le J}|\epsilon_j|=\infty$, then
$$
\mathsf{E}|\epsilon_{X^*}|\ge \mathsf{E}\min_{1\le j\le J}|\epsilon_j|=\infty.
$$
A: Question 0 very much depends on the distribution $F$. For example, if $F$ were to be multivariate normal with mean vector 0 and covariance matrix $\Sigma$, then
\begin{align*}
|\mathbb{E} \epsilon_{X^*}| \le \sum_{j=1}^{J}\mathbb{E}|\epsilon_{j}| = \sqrt{\frac{2}{\pi}}\text{trace}(\Sigma) < \infty
\end{align*}
A: simplest example that shows this is: let's have $X \sim N(0, 1)$ (whatever with support $R$ with bounded mean) and for $k \in \mathbb{R} $, $Y$ with distribution $F_{N(k, 0.01)}/2 + F_{N(-k, 0.01)}/2$ to have continuous distribution and behavior close to $P(Y=k)=P(Y=-k)=0.5$, now $\epsilon=(X,Y)$ it's easy to see that $lim_{k \to \infty} E[\epsilon_{X^*}] \to \infty$, in general we can have this value arbitrary big, if we allowed to have $k=\infty$ then expected value can be $\infty$
