Probability Theory – Problem 75: Prove: (1)

Let $X_{1},\cdots, X_{n}$ be i.i.d. random variables satisfying $\mathbb{P}(X_{i}\;>\;x)\;=\;e^{-x}$, and denote $M_{n}\;=\;\operatorname*{max}_{1\leq m\leq n}X_{m}$. Prove: (1) $\operatorname*{lim\,sup}_{n\to\infty}\frac{X_{n}}{\log n}=1\quad\mathrm{a.s.}$ (2) $\operatorname*{lim}_{n\to\infty}\frac{M_{n}}{\log n}=1\quad\mathrm{a.s.}$

(1) Proof that $\limsup_{n \to \infty} \frac{X_n}{\log n} = 1 \quad \text{a.s.}$ Step 1. Prove $\limsup_{n \to \infty} \frac{X_n}{\log n} \le 1$ a.s. By the definition of the limit superior, it suffices to show that for any given $\epsilon > 0$, the event $\left\{ \frac{X_n}{\log n} > 1 + \epsilon \right\}$ can occur only finitely many times (almost surely). By the first Borel--Cantelli lemma, it suffices to show that the series $\sum_{n=1}^\infty P\left(\frac{X_n}{\log n} > 1 + \epsilon\right)$ converges. Using the given tail probability $P(X_i > x) = e^{-x}$: $P\left(\frac{X_n}{\log n} > 1 + \epsilon\right) = P(X_n > (1+\epsilon)\log n)$ Setting $x = (1+\epsilon)\log n$: $P(X_n > (1+\epsilon)\log n) = e^{-(1+\epsilon)\log n} = (e^{\log n})^{-(1+\epsilon)} = n^{-(1+\epsilon)}$ Since $\epsilon > 0$, we have $1+\epsilon > 1$. This is a $p$-series with $p = 1+\epsilon > 1$, so $\sum_{n=1}^\infty n^{-(1+\epsilon)}$ converges. By the first Borel--Cantelli lemma, $P\left( \frac{X_n}{\log n} > 1 + \epsilon \text{ i.o.}\right) = 0$. Since $\epsilon$ is arbitrary, $\limsup_{n \to \infty} \frac{X_n}{\log n} \le 1$ a.s.

Step 2. Prove $\limsup_{n \to \infty} \frac{X_n}{\log n} \ge 1$ a.s. For any $0 < \epsilon < 1$, the events $A_n = \left\{ \frac{X_n}{\log n} > 1 - \epsilon \right\}$ are mutually independent since the $X_n$ are i.i.d. $P(A_n) = n^{-(1-\epsilon)}$. Since $1-\epsilon < 1$, the series $\sum_{n=1}^\infty n^{-(1-\epsilon)}$ diverges. By the second Borel--Cantelli lemma, $P(A_n \text{ i.o.}) = 1$. Since $\epsilon$ is arbitrary, $\limsup_{n \to \infty} \frac{X_n}{\log n} \ge 1$ a.s.

Step 3. Combining Steps 1 and 2 yields $\limsup_{n \to \infty} \frac{X_n}{\log n} = 1$ a.s.

(2) Proof that $\lim_{n \to \infty} \frac{M_n}{\log n} = 1 \quad \text{a.s.}$

Step 4. Prove $\limsup_{n \to \infty} \frac{M_n}{\log n} \le 1$ a.s. From Part (1), for any $\delta > 0$, there exists $N$ such that $X_n < (1+\delta)\log n$ for all $n > N$ a.s. Let $C = \max(X_1, \ldots, X_N)$. Then $M_n \le \max(C, (1+\delta)\log n)$, so $\frac{M_n}{\log n} \le \max\left(\frac{C}{\log n}, 1+\delta\right) \to 1+\delta$. Since $\delta$ is arbitrary, $\limsup \frac{M_n}{\log n} \le 1$ a.s.

Step 5. Prove $\liminf_{n \to \infty} \frac{M_n}{\log n} \ge 1$ a.s. For $0 < \epsilon < 1$: $P(M_n < (1-\epsilon)\log n) = [1 - n^{-(1-\epsilon)}]^n \le e^{-n^\epsilon}$. Since $\sum e^{-n^\epsilon} < \infty$, by the first Borel--Cantelli lemma, $\liminf \frac{M_n}{\log n} \ge 1$ a.s.

Step 6. Combining Steps 4 and 5 completes the proof.

QED.