# a not directly application of strong/weak law of large numbers

Let $X_n$ be iid random real variables with $P(X_n > x)= e^{-x}$ let $M_n = \displaystyle \max_{1\le m\le n}X_m$.

Show that $\displaystyle \limsup_{n\to \infty} \frac{X_n}{\log n}=1$ and $\frac{M_n}{\log n}\to 1$ a.s

I'm studying borel-Cantelli Lemmas and the weak and strong law of numbers. I don't see how can I use those theorems here :/ please help me

## 1 Answer

Let $\varepsilon>0$. By assumption, $X_n$ is exponentially distributed with parameter $1$; therefore we obtain by applying Markov's (exponential) inquality

$$\mathbb{P} \left( \frac{X_n}{\log n} \geq (1+\varepsilon) \right) \leq e^{- (1+\varepsilon) \cdot \lambda \cdot \log n} \cdot \mathbb{E}e^{\lambda \cdot X_n} = \frac{1}{n^{(1+\varepsilon) \cdot \lambda}} \cdot \frac{1}{1-\lambda}$$

for any $0<\lambda<1$ where we used that the exponential moments can be expressed explicitely. In particular, if we choose $\lambda<1$ close to $1$, this shows

$$\sum_{n \in \mathbb{N}} \mathbb{P} \left( \frac{X_n}{\log n} \geq (1+\varepsilon) \right) < \infty$$

Hence, by Borel-Cantelli's lemma

$$\limsup_{n \to \infty} \frac{X_n}{\log n} \leq 1$$

since $\varepsilon>0$ is arbritrary. Similarly, we can show that

$$\sum_{n \in \mathbb{N}} \mathbb{P} \left( \frac{X_n}{\log n} \geq (1-\varepsilon) \right) = \infty$$

By the independence of the random variables, this implies, again by Borel cantelli's lemma,

$$\limsup_{n \to \infty} \frac{X_n}{\log n} \geq 1$$

This proves the first claim. From $\limsup_{n \to \infty} \frac{X_n}{\log n} =1$, we see that

$$\frac{X_n}{\log n} \leq 1+\varepsilon$$

for $n \geq n_0(\omega)$ sufficiently large. Hence

$$\limsup_{n \to \infty} \frac{M_n}{\log n} \leq 1$$

On the other hand, we can choose a subsequence such that $\lim_{k \to \infty} \frac{X_{n(k)}}{\log n(k)} =1$. Without loss of generality, we may assume $n(k) \uparrow \infty$ as $k \to \infty$. For any $n \in \mathbb{N}$ there exists $k \in \mathbb{N}$ such that $n(k) \leq n \leq n(k+1)$. Thus,

$$\frac{M_n}{\log n} \geq \frac{X_{n(k)}}{\log n(k)} \frac{\log n(k)}{\log n(k+1)}$$

Hence

$$\liminf_{n \to \infty} \frac{M_n}{\log n} \geq 1$$