1. Let \( \left\{X_{n}\right\} \) be a sequence of independent identically distributed random variables following \( R(0, \theta) \). Show whether \( X_{n}=\max _{i=1,2, \ldots, n}\left\{X_{i}\right\} \) convergences in law. 2. Show the order statistic \( X_{(r)}, X_{(S)}, r

Ask by Howell Schwartz. in Kenya

Mar 09,2025

Let's break down the problem into its components and solve each part step by step. ### 1. Convergence in Law of \( X_n = \max_{i=1,2,\ldots,n} \{X_i\} \) Given that \( \{X_n\} \) is a sequence of independent identically distributed random variables following \( R(0, \theta) \), we want to show whether \( X_n \) converges in law. **Step 1: Distribution of \( X_n \)** The cumulative distribution function (CDF) of \( X_n \) can be expressed as: \[ P(X_n \leq x) = P(\max_{i=1,2,\ldots,n} X_i \leq x) = P(X_1 \leq x)^n \] Let \( F(x) = P(X_1 \leq x) \). Then, \[ P(X_n \leq x) = [F(x)]^n \] **Step 2: Behavior as \( n \to \infty \)** As \( n \to \infty \), if \( F(x) < 1 \), then \( [F(x)]^n \to 0 \). If \( F(x) = 1 \), then \( [F(x)]^n \to 1 \). Therefore, we need to find the limit of \( X_n \) as \( n \to \infty \). **Step 3: Convergence in Law** The limit distribution of \( X_n \) will depend on the behavior of \( F(x) \). If \( \theta \) is finite, \( X_n \) will converge to the distribution of the maximum of the distribution, which is a Gumbel distribution. Thus, \( X_n \) converges in law to a Gumbel distribution. ### 2. Correlation Coefficient of Order Statistics Let \( X_{(r)} \) and \( X_{(s)} \) be the \( r \)-th and \( s \)-th order statistics from a sample of size \( n \) drawn from \( U(0,1) \). **Step 1: Covariance Calculation** The covariance between \( X_{(r)} \) and \( X_{(s)} \) can be calculated using the properties of order statistics. The expected values and variances can be derived from the uniform distribution. **Step 2: Correlation Coefficient Formula** The correlation coefficient \( \rho \) is given by: \[ \rho = \frac{Cov(X_{(r)}, X_{(s)})}{\sqrt{Var(X_{(r)}) Var(X_{(s)})}} \] Using the known results for order statistics from a uniform distribution, we can derive: \[ Cov(X_{(r)}, X_{(s)}) = \frac{r(n-s+1)}{n+1} \] and \[ Var(X_{(r)}) = \frac{r(n-r+1)}{(n+1)^2(n+2)} \] \[ Var(X_{(s)}) = \frac{s(n-s+1)}{(n+1)^2(n+2)} \] **Step 3: Final Correlation Coefficient** Substituting these into the correlation coefficient formula gives: \[ \rho = \left[\frac{r(n-s+1)}{s(n-r+1)}\right]^{1/2} \] ### 3. Convergence of \( \sqrt{2 \chi_n^2} \) We want to show that \( \sqrt{2 \chi_n^2} \stackrel{a}{\rightarrow} N(\sqrt{2n-1}, 1) \) as \( n \to \infty \). **Step 1: Chi-Squared Distribution** The chi-squared distribution \( \chi_n^2 \) with \( n \) degrees of freedom can be approximated by a normal distribution as \( n \) becomes large. **Step 2: Central Limit Theorem Application** By the Central Limit Theorem, we have: \[ \frac{\chi_n^2 - n}{\sqrt{2n}} \xrightarrow{d} N(0, 1) \] Thus, \[ \chi_n^2 \approx n + \sqrt{2n}Z \quad \text{for } Z \sim N(0, 1) \] **Step 3: Transformation** Taking the square root: \[ \sqrt{2 \chi_n^2} \approx \sqrt{2(n + \sqrt{2n}Z)} = \sqrt{2n} + Z \] As \( n \to \infty \), this converges in distribution to \( N(\sqrt{2n-1}, 1) \). ### 4. Weak Law of Large Numbers **Statement:** The Weak Law of Large Numbers states that if \( X_1, X_2, \ldots, X_n \) are independent and identically distributed random variables with finite mean \( \mu \) and variance \( \sigma^2 \), then: \[ P\left(\left|\frac{1}{n}\sum_{i=1}^{n} X_i - \mu\right| \geq \epsilon\right) \to 0 \quad \text{as } n \to \infty \] **Proof:** Using Chebyshev's inequality: \[ P\left(\left|\frac{1}{n}\sum_{i=1}^{n} X_i - \mu\right| \geq \epsilon\right) \leq \frac{Var\left(\frac{1}{n}\sum_{i=1}^{n} X_i\right)}{\epsilon^2} = \frac{\sigma^2}{n\epsilon^2} \] As \( n \to \infty \), the right-hand side approaches 0, proving the weak law. ### 5. Joint Distribution and Conditional Expectation Given \( X \sim N_4(\mu, \Sigma) \) with specified \( \mu \) and \( \Sigma \): **(i) Joint Marginal Distribution of \( Z = 4y_1 - 2y_2 + y_3 - 3y_4 \)** To find the distribution of \( Z \), we compute: \[ Z \sim N(\mu_Z, \sigma_Z^2) \] where \( \mu_Z = 4\mu_1 - 2\mu_2 + \mu_3 - 3\mu_4 \) and \( \sigma_Z^2 \) is derived from the covariance matrix \( \Sigma \). **(ii) Joint Distribution of \( Z_1 = y_1 + y_2 + y_3 + y_4 \) and \( Z_2 = -2y_1 + 3y_2 - 2y_4 \)** The joint