Let's address each part step by step for the simple linear regression model: \[ Y_i = \beta_0 + \beta_1 X_i + \epsilon_i \] --- ### (a) **Assume that \( X = 0 \) is within the scope of the model. What is the implication for the regression function if \( \beta_0 = 0 \)?** **Answer:** If \( \beta_0 = 0 \), the regression model becomes: \[ Y_i = \beta_1 X_i + \epsilon_i \] So, when \( X = 0 \), \( Y = 0 \) (ignoring the error term). This means the regression line passes through the origin. There is no intercept, so for \( X = 0 \), the expected value of \( Y \) is also 0. --- ### (b) **Under the assumption of \( \beta_0 = 0 \), derive the least squares estimate of \( \beta_1 \).** **Answer:** With \( \beta_0 = 0 \), the model is \( Y_i = \beta_1 X_i + \epsilon_i \). The least squares estimate of \( \beta_1 \) minimizes: \[ S(\beta_1) = \sum_{i=1}^n (Y_i - \beta_1 X_i)^2 \] Setting derivative w.r.t. \( \beta_1 \) to zero: \[ \frac{\partial S}{\partial \beta_1} = -2 \sum_{i=1}^n X_i (Y_i - \beta_1 X_i) = 0 \] \[ \sum X_i Y_i - \beta_1 \sum X_i^2 = 0 \implies \boxed{\hat{\beta}_1 = \frac{\sum X_i Y_i}{\sum X_i^2}} \] --- ### (c) **Find its mean and variance. Is it unbiased?** **Answer:** Assume \( Y_i = \beta_1 X_i + \epsilon_i \), \( E[\epsilon_i] = 0 \), \( Var(\epsilon_i) = \sigma^2 \): - **Expectation:** \[ E[\hat{\beta}_1] = E\left[\frac{\sum X_i Y_i}{\sum X_i^2}\right] = \frac{\sum X_i E[Y_i]}{\sum X_i^2} \] \( E[Y_i] = \beta_1 X_i \), so: \[ E[\hat{\beta}_1] = \frac{\sum X_i (\beta_1 X_i)}{\sum X_i^2} = \frac{\beta_1 \sum X_i^2}{\sum X_i^2} = \beta_1 \] **Unbiased** - **Variance:** \[ Var(\hat{\beta}_1) = Var\left(\frac{\sum X_i Y_i}{\sum X_i^2}\right) = Var\left(\frac{\sum X_i \epsilon_i}{\sum X_i^2}\right) \] Since \( X_i \) are constants, \[ Var(\hat{\beta}_1) = \frac{\sum X_i^2 Var(\epsilon_i)}{(\sum X_i^2)^2} = \frac{\sigma^2 \sum X_i^2}{(\sum X_i^2)^2} = \frac{\sigma^2}{\sum X_i^2} \] --- ### (d) **How do you fit such a model using the lm function?** **Answer:** In R, to fit \( Y \sim X - 1 \) (no intercept) using `lm`: ```r lm(Y ~ X - 1) ``` or ```r lm(Y ~ 0 + X) ``` --- ### (e) **What is the implication for the regression function if \( \beta_1 = 0 \)? How would the regression function plot on a graph?** **Answer:** If \( \beta_1 = 0 \), the model becomes: \[ Y_i = \beta_0 + \epsilon_i \] The regression function is a horizontal line at \( Y = \beta_0 \). On a graph, regardless of \( X \), the regression line is flat at \( \beta_0 \). --- ### (f) **Under the assumption of \( \beta_1 = 0 \), derive the least squares estimate of \( \beta_0 \).** **Answer:** With \( \beta_1 = 0 \), the model is \( Y_i = \beta_0 + \epsilon_i \). Minimize: \[ S(\beta_0) = \sum_{i=1}^n (Y_i - \beta_0)^2 \] Set derivative to zero: \[ \frac{\partial S}{\partial \beta_0} = -2 \sum (Y_i - \beta_0) = 0 \] \[ \sum Y_i - n\beta_0 = 0 \implies \boxed{\hat{\beta}_0 = \frac{1}{n} \sum_{i=1}^n Y_i = \bar{Y}} \] --- ### (g) **Find the Expectation and Variance of the estimate in (f). Is it unbiased?** **Answer:** - **Expectation:** \[ E[\hat{\beta}_0] = E[\bar{Y}] = \frac{1}{n}\sum E[Y_i] = \frac{1}{n} \sum \beta_0 = \beta_0 \] So it is **unbiased**. - **Variance:** \[ Var(\hat{\beta}_0) = Var(\bar{Y}) = \frac{\sigma^2}{n} \] --- Let me know if you need any part explained in more detail!