# Solved – 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 + lnleft(sum_i expleft{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.

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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(Xle x)\$ is the product of the individual chances \$Pr(delta_i+epsilon_ile x)\$. Taking logs and applying basic properties of exponentials yields

\$\$eqalign{ log Pr(Xle x) &= logprod_{i}Pr(delta_i + epsilon_i le x) = sum_i logPr(epsilon_i le x – delta_i)\ &= -sum_ie^{delta_i}, e^{-x} = -expleft(-x + logsum_i e^{delta_i}right). }\$\$

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

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

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

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

QED.

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