Probability Theory – Problem 53: Prove that

Let $\xi_n$ follow the standard normal distribution. Prove that $$\limsup_{n \to \infty} \frac{\xi_n}{\sqrt{2\ln n}} = 1 \\text{ a.s.}$$

Step 1. Prove that $\limsup_{n \to \infty} \frac{\xi_n}{\sqrt{2\ln n}} \le 1$ a.s. By the definition of limsup, this is equivalent to showing that for any given $\epsilon > 0$, the event $\frac{\xi_n}{\sqrt{2\ln n}} > 1 + \epsilon$ can occur only finitely many times. By the first Borel-Cantelli lemma, if the sum of probabilities of a sequence of events $A_n$ converges, i.e., $\sum_{n=1}^\infty P(A_n) < \infty$, then the probability that these events occur infinitely often is 0. Define the event $A_n = \left\{ \frac{\xi_n}{\sqrt{2\ln n}} > 1 + \epsilon \right\}$. We need to estimate the tail probability $P(A_n)$ of the standard normal random variable $\xi_n$. For a standard normal variable $Z$, when $x > 0$, the tail probability has the well-known upper bound: $P(Z > x) = \int_x^\infty \frac{1}{\sqrt{2\pi}} e^{-t^2/2} dt < \frac{1}{\sqrt{2\pi}x} e^{-x^2/2}$ Let $x_n = (1+\epsilon)\sqrt{2\ln n}$, then $P(A_n) = P(\xi_n > x_n)$. Computing the exponential part: $e^{-x_n^2/2} = e^{-((1+\epsilon)\sqrt{2\ln n})^2/2} = e^{-(1+\epsilon)^2 \ln n} = n^{-(1+\epsilon)^2}$ Substituting into the probability upper bound estimate: $P(A_n) < \frac{1}{\sqrt{2\pi}(1+\epsilon)\sqrt{2\ln n}} n^{-(1+\epsilon)^2}$ Consider the series $\sum_{n=1}^\infty P(A_n)$. Since $(1+\epsilon)^2 > 1$, the $p$-series $\sum_{n=1}^\infty n^{-(1+\epsilon)^2}$ converges. The factor $\frac{1}{\sqrt{2\pi}(1+\epsilon)\sqrt{2\ln n}}$ decreases as $n$ increases and does not affect the convergence of the series. Therefore, by the comparison test, $\sum_{n=1}^\infty P(A_n)$ converges. By the first Borel-Cantelli lemma, the probability that $A_n$ occurs infinitely often is 0. This means that almost surely, for any $\epsilon > 0$, $\limsup_{n \to \infty} \frac{\xi_n}{\sqrt{2\ln n}} \le 1 + \epsilon$. Since $\epsilon$ can be arbitrarily small, we conclude $\limsup_{n \to \infty} \frac{\xi_n}{\sqrt{2\ln n}} \le 1$ a.s.

Step 2. Prove that $\limsup_{n \to \infty} \frac{\xi_n}{\sqrt{2\ln n}} \ge 1$ a.s. This is equivalent to showing that for any $0 < \epsilon < 1$, the event $\frac{\xi_n}{\sqrt{2\ln n}} > 1 - \epsilon$ occurs infinitely often. Since the $\xi_n$ are mutually independent random variables, the events $B_n = \left\{ \frac{\xi_n}{\sqrt{2\ln n}} > 1 - \epsilon \right\}$ are also mutually independent. By the second Borel-Cantelli lemma, if the sum of probabilities of independent events $B_n$ diverges, i.e., $\sum_{n=1}^\infty P(B_n) = \infty$, then the probability that these events occur infinitely often is 1. We need a lower bound estimate for the tail probability $P(B_n)$. For a standard normal variable $Z$, as $x \to \infty$, the following lower bound holds: $P(Z > x) > \left(\frac{1}{x} - \frac{1}{x^3}\right) \frac{1}{\sqrt{2\pi}} e^{-x^2/2}$ Let $y_n = (1-\epsilon)\sqrt{2\ln n}$, then $P(B_n) = P(\xi_n > y_n)$. The exponential part is: $e^{-y_n^2/2} = e^{-((1-\epsilon)\sqrt{2\ln n})^2/2} = e^{-(1-\epsilon)^2 \ln n} = n^{-(1-\epsilon)^2}$ As $n \to \infty$, $y_n \to \infty$, so $\frac{1}{y_n} - \frac{1}{y_n^3}$ is of the same order as $\frac{1}{y_n}$. There exists a constant $C > 0$ such that for sufficiently large $n$: $P(B_n) > \frac{C}{y_n} e^{-y_n^2/2} = \frac{C}{(1-\epsilon)\sqrt{2\ln n}} n^{-(1-\epsilon)^2}$ Consider the series $\sum_{n=1}^\infty P(B_n)$. Since $0 < \epsilon < 1$, we have $(1-\epsilon)^2 < 1$. The series $\sum_{n=2}^\infty \frac{1}{n^p \sqrt{\ln n}}$ diverges when $p \le 1$. Here $p = (1-\epsilon)^2 < 1$, so the series $\sum_{n=2}^\infty \frac{n^{-(1-\epsilon)^2}}{\sqrt{\ln n}}$ diverges. By the comparison test, $\sum_{n=1}^\infty P(B_n)$ diverges. By the second Borel-Cantelli lemma, the probability that $B_n$ occurs infinitely often is 1. That is, almost surely, for any $0 < \epsilon < 1$, $\limsup_{n \to \infty} \frac{\xi_n}{\sqrt{2\ln n}} \ge 1 - \epsilon$. Since $\epsilon$ can be arbitrarily close to 0, we conclude $\limsup_{n \to \infty} \frac{\xi_n}{\sqrt{2\ln n}} \ge 1$ a.s.

Step 3. Combining the conclusions of Steps 1 and 2, on the one hand we have almost surely $\limsup_{n \to \infty} \frac{\xi_n}{\sqrt{2\ln n}} \le 1$, and on the other hand almost surely $\limsup_{n \to \infty} \frac{\xi_n}{\sqrt{2\ln n}} \ge 1$. Therefore, both inequalities must hold simultaneously.

QED.