Let $\{X_i; i \ge 1\}$ be a sequence of independent random variables, where each $X_i$ follows a normal distribution with mean $\mu = 2$ and variance $\sigma^2 = 5$. We analyze the convergence in probability of the given expressions.
Normal Variables Convergence Analysis
Option A: Sample Mean Convergence
We examine the expression $\frac{1}{n}\sum_{i=1}^n X_i$. Let $\bar{X}_n = \frac{1}{n}\sum_{i=1}^n X_i$. The Weak Law of Large Numbers (WLLN) states that for i.i.d. random variables with a finite mean, the sample mean converges in probability to the expected value.
Given $E[X_i] = \mu = 2$. Therefore, $\bar{X}_n \xrightarrow{p} 2$. This statement is true.
Option B: Convergence of Sample Mean of Squares
We examine the expression $\frac{1}{n}\sum_{i=1}^n X_i^2$. Let $Y_i = X_i^2$. The $Y_i$ variables are i.i.d.
The expected value of $Y_i$ is calculated using $E[X_i^2] = Var(X_i) + (E[X_i])^2$. Using the given variance $\sigma^2 = 5$ and mean $\mu = 2$: $E[X_i^2] = 5 + (2)^2 = 5 + 4 = 9$.
By WLLN, the sample mean of $Y_i$ converges in probability to $E[Y_i]$: $ \frac{1}{n}\sum_{i=1}^n X_i^2 \xrightarrow{p} 9 $ This statement is true.
Option C: Convergence of Squared Sample Mean
We examine the expression $\left(\frac{1}{n}\sum_{i=1}^n X_i\right)^2$. From Option A, we know $\frac{1}{n}\sum_{i=1}^n X_i \xrightarrow{p} 2$. Let $g(x) = x^2$. Since $g(x)$ is a continuous function, the Continuous Mapping Theorem applies.
Applying this theorem: $ \left(\frac{1}{n}\sum_{i=1}^n X_i\right)^2 \xrightarrow{p} (2)^2 = 4 $ This statement is true.
Option D: Convergence of Sum of Squared Scaled Variables
We examine the expression $\sum_{i=1}^n \left(\frac{X_i}{n}\right)^2$, which equals $\frac{1}{n^2} \sum_{i=1}^n X_i^2$.
Let $W_n = \frac{1}{n^2} \sum_{i=1}^n X_i^2 = \frac{1}{n} \left(\frac{1}{n} \sum_{i=1}^n X_i^2\right)$.
From Option B, $\frac{1}{n} \sum_{i=1}^n X_i^2 \xrightarrow{p} 9$. Let $Z_n = \frac{1}{n} \sum_{i=1}^n X_i^2$. Thus, $Z_n \xrightarrow{p} 9$, implying $Z_n$ is bounded in probability.
Since $W_n = \frac{1}{n} Z_n$, and $\frac{1}{n} \to 0$ as $n \to \infty$, while $Z_n$ is bounded in probability, the product $W_n$ converges in probability to $0 \times 9 = 0$. This statement is true.
Conclusion
All four statements (Options A, B, C, D) are true based on the Weak Law of Large Numbers and properties of continuous functions applied to convergent sequences.
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?
- 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
- Let $X_i$'s be independent random variables such that $X_i$'s are symmetric about 0 and $\text{Var}(X_i) = 2i-1$, for $i \ge 1$. Then,
$$\lim_{n\to\infty} P(X_1 + X_2 + \cdots + X_n > n \log n)$$
