$$\lim_{n\to\infty} P(X_1 + X_2 + \cdots + X_n > n \log n)$$
Properties of Random Variables
We are given independent random variables $X_i$ ($i \ge 1$) that are symmetric about 0.
- Symmetry about 0 implies that the expected value $E[X_i] = 0$.
- The variance is given as $\text{Var}(X_i) = 2i-1$.
Variance of the Sum
Let $S_n = X_1 + X_2 + \cdots + X_n$.
Since the $X_i$'s are independent, the variance of the sum $S_n$ is the sum of their variances:
$ \text{Var}(S_n) = \sum_{i=1}^n \text{Var}(X_i) $
Substitute the given variance formula:
$ \text{Var}(S_n) = \sum_{i=1}^n (2i-1) $
This is the sum of the first $n$ odd positive integers, which equals $n^2$.
$ \text{Var}(S_n) = n^2 $
The standard deviation of $S_n$ is $\sigma_{S_n} = \sqrt{\text{Var}(S_n)} = \sqrt{n^2} = n$.
The expected value of the sum is $E[S_n] = \sum_{i=1}^n E[X_i] = \sum_{i=1}^n 0 = 0$.
Applying Chebyshev's Inequality
We want to evaluate the limit $\lim_{n\to\infty} P(S_n > n \log n)$.
Chebyshev's inequality provides a bound for the probability that a random variable deviates from its mean. It states:
$ P(|Y - E[Y]| \ge k \sigma_Y) \le \frac{1}{k^2} $ where $Y$ is a random variable, $E[Y]$ is its mean, $\sigma_Y$ is its standard deviation, and $k > 0$.
Apply this to $Y = S_n$, with $E[S_n] = 0$ and $\sigma_{S_n} = n$. We are interested in the event $S_n > n \log n$.
Since $X_i$ are symmetric about 0, $S_n$ is also symmetric about 0. This means $P(S_n > a) = P(S_n < -a)$ for any $a$. Therefore, $P(|S_n| > a) = 2 P(S_n > a)$.
Specifically, $P(S_n > n \log n) = \frac{1}{2} P(|S_n| > n \log n)$.
Using Chebyshev's inequality with $Y=S_n$, $E[Y]=0$, $\sigma_Y=n$, and the threshold $a = n \log n$. We need $k$ such that $k \sigma_{S_n} = n \log n$, so $k \cdot n = n \log n$, which gives $k = \log n$. (Note: $k = \log n > 0$ for $n > 1$).
$ P(|S_n| \ge n \log n) \le \frac{\text{Var}(S_n)}{(n \log n)^2} = \frac{n^2}{(n \log n)^2} $
$ P(|S_n| \ge n \log n) \le \frac{n^2}{n^2 (\log n)^2} = \frac{1}{(\log n)^2} $
Evaluating the Limit
From the relationship $P(S_n > n \log n) = \frac{1}{2} P(|S_n| > n \log n)$ and the Chebyshev bound:
$ 0 \le P(S_n > n \log n) \le \frac{1}{2} P(|S_n| \ge n \log n) \le \frac{1}{2} \frac{1}{(\log n)^2} $
Now, consider the limit as $n \to \infty$:
$ \lim_{n\to\infty} 0 = 0 $
$ \lim_{n\to\infty} \frac{1}{2 (\log n)^2} = 0 $ since $\log n \to \infty$ as $n \to \infty$.
By the Squeeze Theorem, since $P(S_n > n \log n)$ is bounded between two functions that both approach 0 as $n \to \infty$, the limit must also be 0.
$ \lim_{n\to\infty} P(S_n > n \log n) = 0 $
Important Questions from Central Limit Theorems
- Let $X_1, X_2, . . .$ be a sequence of independent and identically distributed random variables with $E(X_1) = 0, E(X_1^2) = 1, E(X_1^3) = 0, E(X_1^4) = 3$. Let $S_n = \sum_{i=1}^n X_i, T_n = \sum_{i=1}^n X_i^2, U_n = \sum_{i=1}^n X_i^3$ and $V_n = \sum_{i=1}^n X_i^4$. Then, which of the following statements are true?
- Let $\{X_i; i \ge 1\}$ be a sequence of independent random variables each having a normal distribution with mean 2 and variance 5. Then which of the following are true
- For $n \ge 1$, let $X_n$ be a Poisson random variable with mean $n^2$. Which of the following are equal to $\frac{1}{\sqrt{2\pi}} \int_2^\infty e^{-x^2/2} dx$?
Let $\{X_n : n \ge 1\}$ be a sequence of independent and identically distributed random variables and the probability mass function of $X_1$ is the following;
$P(X_1 = 1) = P(X_1 = 3) = \frac{1}{2}.$ If $Y_n = X_1 + \cdots + X_n$,
then which of the following statements are correct?
Suppose $X_1, X_2, \dots$ are independent random variables. Assume that $X_1, X_3, \dots$ are identically distributed with mean $\mu_1$ and variance $\sigma_1^2$, while $X_2, X_4, \dots$ are identically distributed with mean $\mu_2$ variance $\sigma_2^2$. Let $S_n = X_1 + X_2 + \dots + X_n$. Then $\frac{S_n - a_n}{b_n}$ converges in distribution to $N(0,1)$ if
