Absolutely! Let's break down each part in detail, clarifying both the math and the intuition behind each step. --- ## **(a) Setting $\beta_0 = 0$** **Model:** $$ Y_i = \beta_1 X_i + \epsilon_i $$ - **Interpretation**: There is **no intercept**. When $X_i = 0$, $Y_i = \epsilon_i$. - **Expected value at $X=0$**: $E[Y_i | X_i=0] = 0$ (since $E[\epsilon_i] = 0$). - **Graphically**: The regression line **passes through the origin** (0,0). So the model assumes that when $X$ is zero, $Y$ is also (on average) zero. --- ## **(b) Least Squares Estimate of $\beta_1$ (with $\beta_0 = 0$)** You want to find the value of $\beta_1$ that minimizes the sum of squared errors: $$ S(\beta_1) = \sum_{i=1}^n (Y_i - \beta_1 X_i)^2 $$ **Take the derivative w.r.t. $\beta_1$ and set it to zero:** $$ \frac{dS}{d\beta_1} = -2 \sum_{i=1}^n X_i (Y_i - \beta_1 X_i) = 0 $$ **Solve for $\beta_1$:** $$ \sum X_i Y_i - \beta_1 \sum X_i^2 = 0 \implies \hat{\beta}_1 = \frac{\sum X_i Y_i}{\sum X_i^2} $$ - This is the **slope estimate** when the regression is **forced through the origin**. --- ## **(c) Mean and Variance of $\hat{\beta}_1$ (with $\beta_0 = 0$)** We know: $$ \hat{\beta}_1 = \frac{\sum X_i Y_i}{\sum X_i^2} $$ But $Y_i = \beta_1 X_i + \epsilon_i$, so: $$ \hat{\beta}_1 = \frac{\sum X_i (\beta_1 X_i + \epsilon_i)}{\sum X_i^2} = \beta_1 + \frac{\sum X_i \epsilon_i}{\sum X_i^2} $$ ### **Mean** $$ E[\hat{\beta}_1] = \beta_1 + \frac{\sum X_i E[\epsilon_i]}{\sum X_i^2} = \beta_1 + 0 = \beta_1 $$ So, **unbiased**. ### **Variance** Assuming errors are independent and $\text{Var}(\epsilon_i) = \sigma^2$, $$ \text{Var}(\hat{\beta}_1) = \frac{\sum X_i^2 \sigma^2}{(\sum X_i^2)^2} = \frac{\sigma^2}{\sum X_i^2} $$ --- ## **(d) Fitting in R** The formula in R to fit **no intercept** is: ```R lm(Y ~ X + 0) ``` - The `+ 0` (or equivalently, `- 1`) tells R **do not include an intercept**. --- ## **(e) Setting $\beta_1 = 0$** **Model becomes:** $$ Y_i = \beta_0 + \epsilon_i $$ - **Interpretation**: $Y_i$ does **not depend on $X_i$ at all**. All data is modeled as having the same mean, $\beta_0$. - **Graphically**: The regression line is **flat**, parallel to the $X$-axis, at height $Y = \beta_0$. --- ## **(f) Least Squares Estimate of $\beta_0$ (with $\beta_1 = 0$)** Minimize: $$ S(\beta_0) = \sum (Y_i - \beta_0)^2 $$ Take derivative and set to zero: $$ \frac{dS}{d\beta_0} = -2 \sum (Y_i - \beta_0) = 0 $$ $$ \sum Y_i - n\beta_0 = 0 \implies \hat{\beta}_0 = \frac{1}{n} \sum Y_i = \bar{Y} $$ --- ## **(g) Mean and Variance of $\hat{\beta}_0 = \bar{Y}$** ### **Mean** $$ E[\hat{\beta}_0] = E[\bar{Y}] = \beta_0 $$ So, **unbiased**. ### **Variance** Each $Y_i$ has variance $\sigma^2$, and they're independent: $$ \text{Var}(\bar{Y}) = \frac{\sigma^2}{n} $$ --- ## **Summary Table** | Model | Estimate | Mean | Variance | R syntax | |--------------------------|----------------------------|--------------|----------------------|-----------------------| | $\beta_0 = 0$ | $\hat{\beta}_1 = \dfrac{\sum X_i Y_i}{\sum X_i^2}$ | $\beta_1$ | $\dfrac{\sigma^2}{\sum X_i^2}$ | `lm(Y ~ X + 0)` | | $\beta_1 = 0$ | $\hat{\beta}_0 = \bar{Y}$ | $\beta_0$ | $\dfrac{\sigma^2}{n}$ | `mean(Y)` or `lm(Y ~ 1)` | --- **Let me know if you'd like to see worked examples with numbers, or have questions about any step!**