Let $(X_n)_{n\geq 1}$ be an i.i.d sequence of random variables with $P(X_1>x)=e^{-x}$ and put $M_n=\max_1^n X_i$. Then $M_n/\log n\to 1$ a.s.

My attempt

I have shown that $\limsup X_n/\log n=1$ a.s and $\lim\inf M_n/\log n\geq 1$ a.s.

For the first claim it suffices to note that $$ \sum _{n=1}^\infty P(X_n\geq c \log n)=\sum _n n^{-c} $$ which is finite if $c>1$ and equal to $\infty$ if $c<1$. In particular for any $\epsilon >0$, $X_n/\log n\ge 1+\epsilon $ finitely many times with probability one which implies that $\lim \sup X_n/\log _n\leq 1$. Similarly, for any $\epsilon >0$, $X_n/\log n\geq 1-\epsilon$ infinitely often with probability one, so $\limsup X_n/\log n \geq 1$,

For the other claim it suffices to show that $P(M_n/\log n <1-\epsilon \quad \text{i.o})=0$. This can be shown by an application of Borel Cantelli. Indeed, $$ P(M_n<(1-\epsilon)\log n)=P(X_1<(1-\epsilon)\log n)^n=(1-n^{-(1-\epsilon)})^n\leq \exp(-n\times n^{-(1-\epsilon)}) $$ But $\sum \exp(-n^{\epsilon})<\infty$.

My Problem

I am unable to show that $\limsup M_n/\log n\leq 1$ a.s. I tried to use Borel cantelli by showing that $P(M_n>(1+\epsilon)\log n \quad \text{i.o})=0$. To this end, put $c_n=(1+\epsilon)\log n)$ and $$ P(M_n>c_n)=1-P(M_n\le c_n)=1-(1-e^{-c_n})^n\leq \exp (-(1-e^{-c_n})^n) $$ but I am not sure if this sequence is summable. Any help is appreciated and other methods are welcome.


You don't any probbaility theory to complete that argument. Try to prove the following lemma:

if $a_n$ increases to $\infty$ and $\lim \sup \frac {x_n} {a_n} \leq 1$ then $\lim \sup \frac {y_n} {a_n} \leq 1$ where $y_n=\max \{x_1,x_2,\cdots, x_n\}$.

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