Expectation of the Maximum of iid Gumbel Variables

I keep reading in economics journals about a particular result used in random utility models. One version of the result is: if $\epsilon_i \sim_{iid}, $ Gumbel($\mu, 1), \forall i$, then:

$$E[\max_i(\delta_i + \epsilon_i)] = \mu + \gamma + \ln\left(\sum_i \exp\left\{\delta_i \right\} \right), $$

where $\gamma \approx 0.52277$ is the Euler-Mascheroni constant. I’ve checked that this makes sense using R, and it does. The CDF for the Gumbel$(\mu, 1)$ distribution is:

$$G(\epsilon_i) = \exp(-\exp(-(\epsilon_i – \mu)))$$

I’m trying to find a proof of this and I’ve had no success. I’ve tried to prove it myself but I can’t get past a particular step.

Can anyone point me to a proof of this? If not, maybe I can post my attempted proof up to where I get stuck.


I appreciate the work exhibited in your answer: thank you for that contribution. The purpose of this post is to provide a simpler demonstration. The value of simplicity is revelation: we can easily obtain the entire distribution of the maximum, not just its expectation.

Ignore $\mu$ by absorbing it into the $\delta_i$ and assuming the $\epsilon_i$ all have a Gumbel$(0,1)$ distribution. (That is, replace each $\epsilon_i$ by $\epsilon_i-\mu$ and change $\delta_i$ to $\delta_i+\mu$.) This does not change the random variable

$$X = \max_{i}(\delta_i + \epsilon_i) = \max_i((\delta_i+\mu) + (\epsilon_i-\mu)).$$

The independence of the $\epsilon_i$ implies for all real $x$ that $\Pr(X\le x)$ is the product of the individual chances $\Pr(\delta_i+\epsilon_i\le x)$. Taking logs and applying basic properties of exponentials yields

\log \Pr(X\le x) &= \log\prod_{i}\Pr(\delta_i + \epsilon_i \le x) = \sum_i \log\Pr(\epsilon_i \le x – \delta_i)\\
&= -\sum_ie^{\delta_i}\, e^{-x} = -\exp\left(-x + \log\sum_i e^{\delta_i}\right).

This is the logarithm of the CDF of a Gumbel distribution with location parameter $\lambda=\log\sum_i e^{\delta_i}.$ That is,

$X$ has a Gumbel$\left(\log\sum_i e^{\delta_i}, 1\right)$ distribution.

This is much more information than requested. The mean of such a distribution is $\gamma+\lambda,$ entailing

$$\mathbb{E}[X] = \gamma + \log\sum_i e^{\delta_i},$$


Source : Link , Question Author : Jason , Answer Author : whuber

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