Show that the MLE estimate for o? is -E(vi – Bo - Bixi). Li%31

Trigonometry (MindTap Course List)
8th Edition
ISBN:9781305652224
Author:Charles P. McKeague, Mark D. Turner
Publisher:Charles P. McKeague, Mark D. Turner
Chapter4: Graphing And Inverse Functions
Section: Chapter Questions
Problem 6GP: If your graphing calculator is capable of computing a least-squares sinusoidal regression model, use...
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# Ordinary Least Squares (OLS) Derivation

### Key Concepts:
- **Introduction to Regression Line:**
  - The regression model is defined as \( y_i = \beta_0 + \beta_1 x_i + \epsilon_i \).
- **Objective:**
  - Minimize the square error: \( f(\beta_0, \beta_1) = \sum_{i=1}^{n} (y_i - \beta_0 - \beta_1 x_i)^2 \).

### Mathematical Derivation:
- **Quadratic Function:** 
  - Minimized when derivatives are zero.
- **Solving the Derivatives:**
  \[
  \frac{\partial f}{\partial \beta_0} = 2 \sum_{i=1}^{n} (\beta_0 + \beta_1 x_i - y_i) = 0
  \]
  \[
  \frac{\partial f}{\partial \beta_1} = 2 \sum_{i=1}^{n} x_i (\beta_0 + \beta_1 x_i - y_i) = 0
  \]
  - Simplifies to:
  \[
  \hat{\beta_1} = \frac{\sum_{i=1}^{n} x_i y_i - n \bar{X} \bar{Y}}{\sum_{i=1}^{n} x_i^2 - n \bar{X}^2}
  \]
  \[
  \hat{\beta_0} = \bar{Y} - \hat{\beta_1} \bar{X}
  \]

### Prediction:
- Predicted value of \( y_i \): \( \hat{y_i} = \hat{\beta_0} + \hat{\beta_1} x_i \).

# Maximum Likelihood Estimation (MLE) Derivation

### Likelihood Function:
- Definition: Probability that observations arise from a defined probability distribution.
- **Assumption:** \( \epsilon_i \) are normally distributed.

### Probability Distribution:
- Normal distribution function:
  \[
  f(x|\mu, \sigma) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}
  \]

### Likelihood Calculation:
- Aggregated probability:
  \[
  p
Transcribed Image Text:# Ordinary Least Squares (OLS) Derivation ### Key Concepts: - **Introduction to Regression Line:** - The regression model is defined as \( y_i = \beta_0 + \beta_1 x_i + \epsilon_i \). - **Objective:** - Minimize the square error: \( f(\beta_0, \beta_1) = \sum_{i=1}^{n} (y_i - \beta_0 - \beta_1 x_i)^2 \). ### Mathematical Derivation: - **Quadratic Function:** - Minimized when derivatives are zero. - **Solving the Derivatives:** \[ \frac{\partial f}{\partial \beta_0} = 2 \sum_{i=1}^{n} (\beta_0 + \beta_1 x_i - y_i) = 0 \] \[ \frac{\partial f}{\partial \beta_1} = 2 \sum_{i=1}^{n} x_i (\beta_0 + \beta_1 x_i - y_i) = 0 \] - Simplifies to: \[ \hat{\beta_1} = \frac{\sum_{i=1}^{n} x_i y_i - n \bar{X} \bar{Y}}{\sum_{i=1}^{n} x_i^2 - n \bar{X}^2} \] \[ \hat{\beta_0} = \bar{Y} - \hat{\beta_1} \bar{X} \] ### Prediction: - Predicted value of \( y_i \): \( \hat{y_i} = \hat{\beta_0} + \hat{\beta_1} x_i \). # Maximum Likelihood Estimation (MLE) Derivation ### Likelihood Function: - Definition: Probability that observations arise from a defined probability distribution. - **Assumption:** \( \epsilon_i \) are normally distributed. ### Probability Distribution: - Normal distribution function: \[ f(x|\mu, \sigma) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}} \] ### Likelihood Calculation: - Aggregated probability: \[ p
The image describes the setup and assumptions for a linear regression model.

1. **Model Description:**
   - We have a response variable \( y_i \), for \( i = 1, 2, \ldots, n \).
   - One explanatory variable \( x_i \), for \( i = 1, 2, \ldots, n \).
   - The linear relationship is given by:
     \[
     y_i = \beta_0 + \beta_1 x_i + \epsilon_i
     \]
   - Here, \( \epsilon_i \) are normal disturbance terms, often due to measurement error.

2. **Assumptions:**
   - \( \epsilon_i \) is the only randomness of interest. The explanatory variable \( X \) can have any arbitrary distribution or be non-random.
   - Parameters \( \beta_0 \) and \( \beta_1 \) need to be estimated.
   - Each disturbance \( \epsilon_i \) has a mean of 0 and the same variance \(\sigma^2\).
   - Disturbances \( \epsilon_i \) are independent of one another.

3. **Covariance Matrix:**
   - The covariance between different \( \epsilon_i \) and \( \epsilon_j \) is:
     \[
     \text{cov}(\epsilon_i, \epsilon_j) = 
     \begin{cases} 
     0, & \text{if } i = j \\ 
     \sigma^2, & \text{if } i \neq j 
     \end{cases}
     \]

4. **Graph Description:**
   - The graph shows a linear regression line \( E(Y) = \beta_0 + \beta_1 x \).
   - Normal distributions are depicted around the regression line at different \( x \) values (\( x_1, x_2, x_3 \)), symbolizing the variance \(\sigma^2\) of the error terms \( \epsilon_i \).

5. **Maximum Likelihood Estimation (MLE) for Variance:**
   - The goal is to show that the MLE estimate for \(\sigma^2\) is:
     \[
     \hat{\sigma}^2 = \frac{1}{n} \sum_{i=1}^{n} (y_i - \hat{\beta}_0 - \
Transcribed Image Text:The image describes the setup and assumptions for a linear regression model. 1. **Model Description:** - We have a response variable \( y_i \), for \( i = 1, 2, \ldots, n \). - One explanatory variable \( x_i \), for \( i = 1, 2, \ldots, n \). - The linear relationship is given by: \[ y_i = \beta_0 + \beta_1 x_i + \epsilon_i \] - Here, \( \epsilon_i \) are normal disturbance terms, often due to measurement error. 2. **Assumptions:** - \( \epsilon_i \) is the only randomness of interest. The explanatory variable \( X \) can have any arbitrary distribution or be non-random. - Parameters \( \beta_0 \) and \( \beta_1 \) need to be estimated. - Each disturbance \( \epsilon_i \) has a mean of 0 and the same variance \(\sigma^2\). - Disturbances \( \epsilon_i \) are independent of one another. 3. **Covariance Matrix:** - The covariance between different \( \epsilon_i \) and \( \epsilon_j \) is: \[ \text{cov}(\epsilon_i, \epsilon_j) = \begin{cases} 0, & \text{if } i = j \\ \sigma^2, & \text{if } i \neq j \end{cases} \] 4. **Graph Description:** - The graph shows a linear regression line \( E(Y) = \beta_0 + \beta_1 x \). - Normal distributions are depicted around the regression line at different \( x \) values (\( x_1, x_2, x_3 \)), symbolizing the variance \(\sigma^2\) of the error terms \( \epsilon_i \). 5. **Maximum Likelihood Estimation (MLE) for Variance:** - The goal is to show that the MLE estimate for \(\sigma^2\) is: \[ \hat{\sigma}^2 = \frac{1}{n} \sum_{i=1}^{n} (y_i - \hat{\beta}_0 - \
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