Probability Theory – Problem 40: prove that for , .\\

Let $X = X_1, X_2, \dots$ be independent and identically distributed random variables.\\ (1) If $\mu = E X > 0$, prove that for $\alpha < 1$, $\lim\limits_{n \to \infty} P(X_n \ge n^{\alpha}) = 1$.\\ (2) If $\mu = E X = 0$ and $E X^2 < +\infty$, compute $\lim\limits_{n \to \infty} P(X_n \ge n^{\alpha})$.

Step 1. Establish the asymptotic behavior of partial sums via the Law of Large Numbers. Since $X_1,X_2,\dots$ are i.i.d. with $\mathbb{E}X = \mu > 0$, by the Law of Large Numbers, $\lim\limits_{n \to \infty} \frac{S_n}{n} = \mu$ holds almost surely. This means that for any $\varepsilon \in (0, \mu)$, there exists a positive integer $N_0$ such that for $n > N_0$, $\frac{S_n}{n} > \mu - \varepsilon$ almost surely, i.e., $S_n > n(\mu - \varepsilon)$ almost surely.

Step 2. Analyze the order-of-magnitude relationship between $n^\alpha$ and $n(\mu - \varepsilon)$. Since $\alpha < 1$, we have $1 - \alpha > 0$, so $n^{1 - \alpha} = \frac{n}{n^\alpha} \to \infty$ as $n \to \infty$. For a fixed $\varepsilon \in (0, \mu)$, when $n$ is sufficiently large, $n^{1 - \alpha} > \frac{1}{\mu - \varepsilon}$; multiplying both sides by $n^\alpha$ yields $n(\mu - \varepsilon) > n^\alpha$.

Step 3. Derive the probability limit. By Steps 1 and 2, for sufficiently large $n$, $S_n > n(\mu - \varepsilon) > n^\alpha$ almost surely, so $\mathbb{P}(S_n \geq n^\alpha) \geq \mathbb{P}(S_n > n(\mu - \varepsilon))$. Moreover, the Strong Law of Large Numbers implies convergence in probability, i.e., $\lim\limits_{n \to \infty} \mathbb{P}\left( \left| \frac{S_n}{n} - \mu \right| < \varepsilon \right) = 1$, and therefore $\lim\limits_{n \to \infty} \mathbb{P}(S_n > n(\mu - \varepsilon)) = 1$. Since probability is bounded above by 1, we conclude $\lim\limits_{n \to \infty} \mathbb{P}(S_n \geq n^\alpha) = 1$.

We continue with the partial sum $S_n = X_1 + \cdots + X_n$ and analyze the probability limit using the Central Limit Theorem. Given $\mathbb{E}X = 0$ and $\mathbb{E}X^2 = \sigma^2 < \infty$, by the i.i.d. Central Limit Theorem, $\frac{S_n}{\sqrt{n}} \stackrel{d}{\to} N(0, \sigma^2)$ (convergence in distribution), where $N(0, \sigma^2)$ denotes the normal distribution with mean 0 and variance $\sigma^2$, and $\Phi(x) = \mathbb{P}(N(0,1) \leq x)$ is the standard normal distribution function.

Rewriting the target probability as $\mathbb{P}(S_n \geq n^\alpha) = \mathbb{P}\left( \frac{S_n}{\sqrt{n}} \geq n^{\alpha - \frac{1}{2}} \right)$, let $t_n = n^{\alpha - \frac{1}{2}}$ and consider three cases according to the limit of $t_n$:

- When $\alpha > \frac{1}{2}$, we have $\alpha - \frac{1}{2} > 0$, so $t_n \to \infty$ as $n \to \infty$. By the tail behavior of the normal distribution, $\Phi(x) \to 1$ as $x \to \infty$, and therefore $\mathbb{P}\left( \frac{S_n}{\sqrt{n}} \geq t_n \right) = 1 - \mathbb{P}\left( \frac{S_n}{\sqrt{n}} \leq t_n \right) \to 1 - \Phi(\infty) = 0$.

- When $\alpha = \frac{1}{2}$, we have $t_n = n^0 = 1$. By the definition of convergence in distribution, $\lim\limits_{n \to \infty} \mathbb{P}\left( \frac{S_n}{\sqrt{n}} \leq 1 \right) = \Phi\left( \frac{1}{\sigma} \right)$, so $\lim\limits_{n \to \infty} \mathbb{P}(S_n \geq \sqrt{n}) = 1 - \Phi\left( \frac{1}{\sigma} \right)$.

- When $\alpha < \frac{1}{2}$, we have $\alpha - \frac{1}{2} < 0$, so $t_n \to 0$ as $n \to \infty$. By the continuity of convergence in distribution, $\lim\limits_{n \to \infty} \mathbb{P}\left( \frac{S_n}{\sqrt{n}} \leq 0 \right) = \Phi(0) = \frac{1}{2}$, and therefore $\mathbb{P}\left( \frac{S_n}{\sqrt{n}} \geq t_n \right) \to 1 - \Phi(0) = \frac{1}{2}$.

(1) QED. (2) The limiting results are as follows: - If $\alpha > \frac{1}{2}$, the limit is 0; - If $\alpha = \frac{1}{2}$, the limit is $1 - \Phi\left( \frac{1}{\sqrt{\mathbb{E}X^2}} \right)$ (where $\Phi$ denotes the standard normal distribution function); - If $\alpha < \frac{1}{2}$, the limit is $\frac{1}{2}$.