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Suppose that

$V= max_i${$V_i+\epsilon_i $}

where $i=1...N$ and each $epsilon_i$ has a a type 1 extreme value distribution with $F(x)=e^{e^{-x}}$

I need to show that


where u is eulers constant, i.e. $E[\epsilon_i]$

I have so far that P(option i is picked)=$\frac{e^{V_i}}{\sum_je^{V_j}}$ (This seems to be a standard result, so I'm fairly sure of it)

Using this formula, I get that:

$E[V] = u +\sum_i\frac{e^{V_i}}{\sum_je^{V_j}}V_i $

and then I dont know how to proceed.

EDIT: Ok I realize that my formula for E[V] is incorrect, it should be

$E[V] = \sum_i\frac{e^{V_i}}{\sum_je^{V_j}}(V_i+E[\epsilon_i|V_i+\epsilon_i>V_j+\epsilon_j ] $ for all $j\neq i$

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up vote 2 down vote accepted

Let us first correct the value of $F$ given in the post and use $F(x)=\exp(-\mathrm e^{-x})$, for every $x$, as the common CDF of the random variables $\epsilon_i$.

Then, $V=\max\limits_iv_i+\epsilon_i$ hence, for every $x$, $[V\leqslant x]=\bigcap\limits_i[\epsilon_i\leqslant x-v_i]$. By independence, $$ \mathbb P(V\leqslant x)=\prod\limits_iF(x-v_i)=\prod\limits_i\exp(-\mathrm e^{-x+v_i})=\exp(-\mathrm e^{-x}\mathrm e^w)=F(x-w), $$ with $\mathrm e^w=\sum\limits_i\mathrm e^{v_i}$. Thus, $\mathbb P(V\leqslant x)=\mathbb P(\epsilon_1\leqslant x-w)$ for every $x$, that is, $V$ is distributed as $\epsilon_1+w$. In particular, $$ \mathbb E(V)=\mathbb E(\epsilon_1)+w=\gamma+\log\left(\sum_i\mathrm e^{v_i}\right), $$ where $\gamma=\mathbb E(\epsilon_1)$ is indeed Euler's constant.

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