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I have to solve the following homework problem:

Let $X_{1}, \ldots ,X_{n}$ be i.i.d. random variables with $X_{1}\sim \operatorname{Exp}(1)$ and let $X$ have distribution function $F(x)=1-e^{-e^{-x}}$. Show that $\max_{1 \leq k \leq n} X_{k} - \log(n)$ converges to $X$ in distribution.

What I have done so far: We can compute that $$\mathbb{P}\Big(\max_{1 \leq k \leq n} X_{k} \leq x\Big)=\mathbb{P}(X_{1}\leq x)^{n}=(1-e^{-x})^{n}.$$ Substituting $x+\log(n)$ for $x$, it follows that

$$\mathbb{P}\Big(\max_{1 \leq k \leq n} X_{k} \leq x + \log(n)\Big)= \big(1-e^{-x-\log(n)}\big)^{n}.$$

This does not converge to $1-e^{-e^{-x}}$ as $n$ goes to infinity. Where is my mistake?

Sangchul Lee
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Giuliano Cantina
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1 Answers1

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Clearly $1-\exp(-e^{-x})$ as is not a distribution function (check the limit at $\infty$ for example). But $F(x)=\exp(-e^{-x})$ is a valid distribution function (of the Gumbel/Extreme-value distribution).

You should say that $\mathbb P\left(\max_{1\le k\le n} X_k -\log n\le x\right)\longrightarrow \exp(-e^{-x})$ for all real $x$ as $n\to \infty$. Hence the convergence in distribution.

StubbornAtom
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