Expected Value – Expectation of the Maximum of iid Gumbel Variables

expected valuegumbel distribution

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.

Best Answer

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

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

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

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

QED.

Related Solutions

Solved – How to find the Inverse Transform of the Gumbel distribution

Finding the inverse transform follows a typical pattern. Start with the CDF, $F_X(x)$ and set equal to $U$.

Solve $F_X(x) = U$ for $x$.

$$\begin{align} \text{e}^{-\text{e}^{-(X-\mu)/\beta}} &= U\\ -\text{e}^{-\left(\frac{X-\mu}{\beta} \right)} &= \text{ln}(U) \\ -\left(\frac{X-\mu}{\beta} \right) &= \text{ln}(-\text{ln}(U)) \\ X-\mu &= -\beta \,\text{ln}(-\text{ln}(U)) \\ X &= \mu -\beta \,\text{ln}(-\text{ln}(U)) \quad \quad \square \end{align}$$

When sampling with this method, $U\sim \text{Uniform}(0,1)$ and $X\sim \text{Gumbel}(\mu,\beta)$.

More on the Inverse Transform here and here.

% MATLAB 2017a
% Code to generate X ~ Gumbel(mu,beta) with inverse transform method
% Parameters
mu = 1;
beta = 2;
n = 1000;                    % number of samples to generate

% Generation
U = rand(n,1);               % U ~ Uniform(0,1)
X = mu - beta*log(-log(U));  % X ~ Gumbel(mu,beta)

Solved – Expectation of inverse of sum of positive iid variables

You cannot bound that expectation in $\sigma, n$. That's because there is the distinct possibility that the expectation do not exist at all (or, is $\infty$.) See I've heard that ratios or inverses of random variables often are problematic, in not having expectations. Why is that?. If the conditions given there is fulfilled for the density of $X_1$, it will so be for the density of $\bar{X}_n$. If densities do not exist, but probability mass functions do, it is simpler, since your assumptions prohibit a probability atom at zero, but a probability density can still be positive at zero even if $P(X >0)=1$.

For a useful bound you will at least need to restrict the common distribution of $X_1, \dotsc, X_n$ much more.

EDIT

After your new information, and with $v_1>0$, the expectation of $1/\bar{X}_n$ certainly will exist (irrespective if $K$ is finite or not.) And, since the function $x\mapsto 1/x$ is convex for $x>0$, we can use the Jensen Inequality to conclude that $\DeclareMathOperator{\E}{\mathbb{E}}\E 1/\bar{X}_n \ge 1/\E \bar{X}_n$.

Best Answer

Related Solutions

Solved – How to find the Inverse Transform of the Gumbel distribution

Solved – Expectation of inverse of sum of positive iid variables

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