Responder
1. The sequence \( X_n = \max\{X_1, X_2, \ldots, X_n\} \) converges in law to a Gumbel distribution as \( n \to \infty \).
2. The correlation coefficient between the \( r \)-th and \( s \)-th order statistics \( X_{(r)} \) and \( X_{(s)} \) from a uniform distribution is:
\[
\rho = \left[\frac{r(n-s+1)}{s(n-r+1)}\right]^{1/2}
\]
3. As \( n \to \infty \), \( \sqrt{2 \chi_n^2} \) converges in distribution to \( N(\sqrt{2n-1}, 1) \).
4. **Weak Law of Large Numbers (WLLN):**
- *Statement:* If \( X_1, X_2, \ldots, X_n \) are independent and identically distributed 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:* Apply Chebyshev's inequality to show that the probability of the sample mean deviating from the population mean by more than \( \epsilon \) becomes negligible as \( n \) increases.
5. **Joint Distributions and Conditional Expectation:**
- *(i)* The joint marginal distribution of \( Z = 4y_1 - 2y_2 + y_3 - 3y_4 \) is a normal distribution with mean and variance determined by \( \mu \) and \( \Sigma \).
- *(ii)* The joint distribution of \( Z_1 = y_1 + y_2 + y_3 + y_4 \) and \( Z_2 = -2y_1 + 3y_2 - 2y_4 \) is also a bivariate normal distribution with means, variances, and covariance derived from \( \mu \) and \( \Sigma \).
- *(iii)* The conditional expectation \( E(X_2 \mid X_1) \) can be found using the properties of multivariate normal distributions, resulting in a linear function of \( X_1 \).
Solución
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
Respondido por UpStudy AI y revisado por un tutor profesional

Explicar

Simplifique esta solución