Consider a set of random variables $Z_1, Z_2, \ldots, Z_m$ drawn independently from a standard normal distribution, i.e., $Z_i \sim \mathcal{N}(0,1)$ for $1 \leq i \leq m$. Given these $m$ samples, index $l$ is selected based on the Gibbs distribution with inverse temperature $\beta$, i.e., the probability of $l$ being equal to $i$ is proportional to $e^{\beta Z_i}$. Are there good approximate upper bounds on the expectation $E[Z_l]$?
Note that when $\beta = 0$, $Z_l$ is the mean of $Z_1, \cdots,Z_m$ and when $\beta \rightarrow \infty$, $Z_l$ is $\max \{Z_1, \cdots, Z_m\}$, for both of which we have good approximate solutions for the expected value of $Z_l$:
For the case when $\beta = 0$, we know that $l$ is uniformly distributed between $1$ and $m$, resulting in $$\mathbb{E}[Z_l] = E\left[\frac{1}{m} \sum_{i=1}^m Z_i\right] = 0.$$
For the case when $\lim \beta \rightarrow \infty$, we know that $Z_l \rightarrow \max\{Z_1, Z_2, \cdots, Z_l\}$, in which case this answer provides a tight upper bound: $$\mathbb{E}[Z_l] = \frac{1}{t}\log e^{\mathbb{E}[tZ_l]} \leq \frac{1}{t}\log\mathbb{E}[e^{tZ_l}] = \frac{1}{t}\log\mathbb{E}[\max\{ e^{tZ_1}, \cdots, e^{tZ_m}\}] \leq \frac{1}{t}\log\mathbb{E}[\sum_{i=1}^m e^{tZ_i}] = \frac{\log m}{t} + \frac{t}{2}.$$ Since the above holds for all $t$, by setting $t = \sqrt{2\log m}$, we get $\mathbb{E}[Z_l] \leq \sqrt{2\log m}$.
I'm interested in finding an upper bound for the expectation $\mathbb{E}[Z_l]$ in terms of the inverse temperature $\beta$ that interpolates the two extreme cases.
