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Let $(X_i)_{i \geq 1}$ be a sequence of i.i.d. normal $\mathcal{N}(0,1)$ random variables. Let $M_n = \max_{i=1,\ldots,n} X_i$. Show that $$P[\sqrt{2 \log n} M_n - 2 \log n \leq u ] \rightarrow e^{-e^{-u}} \text{ as } n \rightarrow \infty$$ I thought maybe to use the following property: $$P[\sqrt{2 \log n} M_n - 2 \log n \leq u ] = P(M_n \leq \frac{u+2 \log n}{\sqrt{2 \log n}})= F^n(\frac{u+2 \log n}{\sqrt{2 \log n}})$$ but dont know how to proceed..

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are you sure? I don't get the same result. – mookid Mar 25 '14 at 13:59
I thought maybe using the following property: $P[\sqrt{2 \log n} M_n - 2 \log n \leq u] = P[M_n \leq \frac{u+2 \log n}{\sqrt{2 \log n}}] = F^n ( \frac{u+2 \log n}{\sqrt{2 \log n}})$ but then i dont know how to proceed.. – user137864 Mar 25 '14 at 14:05
This comment is your most interesting contribution: it should be in the question. – Did Mar 25 '14 at 18:52

1 Answer 1

As you mentionned,$$P[a_n M_n - b_n \leq u ] = F\left(\frac{u +b_n}{a_n}\right)^n $$ As $$ \frac 1{\sqrt{2\pi}(x+1/x)}\exp\left(-\frac {x^2}2\right) < 1-F(x) < \frac 1{\sqrt{2\pi}x}\exp\left(-\frac {x^2}2\right),\\ 1-F(x) \sim \frac 1{\sqrt{2\pi}x}\exp\left(-\frac {x^2}2\right). $$

$$ -\log P[a_n M_n - b_n \leq u ] = -n\log \left(1-\left( 1- F\left(\frac{u +b_n}{a_n}\right)\right)\right)\\ \sim n \left( 1- F\left(\frac{u +b_n}{a_n}\right)\right) \sim n \frac {a_n}{\sqrt{2\pi}(u+b_n)} \exp\left(-\frac {(u+b_n)^2}{2a_n^2}\right) \sim n \frac {a_n}{\sqrt{2\pi}b_n} \exp\left(-\frac {(u+b_n)^2}{2a_n^2}\right); \\ \exp\left(-\frac {(u+b_n)^2}{2a_n^2}\right)= \exp\left(-\frac {u^2}{2a_n^2}\right) \exp\left(-\frac {ub_n}{a_n^2}\right) \exp\left(-\frac {b_n^2}{2a_n^2}\right) \sim \frac {\exp\left(-u\right)}n $$ because $a_n\to\infty$, $a_n^2 = b_n$ and $(b_n/a_n)^2 = 2\log n$.

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Are u sure? Because if I want to compute this limit with Mathematica as $n \rightarrow \infty$ with the given values for $a_n$ and $b_n$, I get $\infty$ as result or did i make a typo...? – user137864 Mar 25 '14 at 14:27
I checked it again, the inital question is correct, at least it's stated like this in the exercise.. – user137864 Mar 25 '14 at 18:35
The double inequality is valid for $1-F$, not for $F$. If you correct this and follow the idea in your answer, the result should come... – Did Mar 25 '14 at 18:51
Well... what comes next? – Did Mar 25 '14 at 20:54
Where did the second line with the two inequalities come from? – lightfish Mar 27 '14 at 20:05

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