Standard Error of Coefficients in simple Linear Regression

Question

In the book "Introduction to Statistical Learning" page 66, there are formulas of the standard errors of the coefficient estimates $\hat{\beta}_0$ and $\hat{\beta}_1$. I know the proof of $SE(\hat{\beta}_1)$ but I am confused about how to derive the formula for $$SE(\hat{\beta}_0)^2 = \sigma^2\left[\frac{1}{n} + \frac{\bar{x}^2}{{\sum_{i=1}^n}(x_i - \bar{x})^2}\right]$$ since $\sigma^2 = Var(\epsilon)$, not the variance of $y_i's$.

My calculation so far is as follows: $$Var(\hat{\beta}_0) = Var(\bar{y} - \hat{\beta}_1\bar{x}) = Var(\bar{y}) + \bar{x}^2\frac{\sigma^2}{{\sum_{i=1}^n}(x_i - \bar{x})^2} - 2\bar{x} Cov(\bar{y}, \hat{\beta}_1) $$in which $\sigma^2 = Var(\epsilon)$.

$Cov(\bar{y}, \hat{\beta}_1) = 0$ since $\bar{y}$ and $\hat{\beta}_1$ are uncorrelated. $Var(\bar{y}) = \frac{\sigma^2}{n}$ in which $\sigma^2 = Var(y_i)$.

So how can we have the formula for $SE(\hat{\beta}_0)^2$ as above since the 2 $\sigma's$ are different from each other?

Many thanks in advance for your help!

Answer 1

You have found out that

$$Var(\hat{\beta}_0) = Var(\bar{y} - \hat{\beta}_1\bar{x}) = Var(\bar{y}) + \bar{x}^2\frac{\sigma^2}{{\sum_{i=1}^n}(x_i - \bar{x})^2} - 2\bar{x} \underbrace{Cov(\bar{y}, \hat{\beta}_1)}_{=0}$$ in which $Var(\bar{y}) =\frac{\sigma^2}{n}$.

Inserting $\frac{\sigma^2}{n}$ for $Var(\bar{y}) $

$$Var(\hat{\beta}_0) = \frac{\sigma^2}{n} + \bar{x}^2\frac{\sigma^2}{{\sum_{i=1}^n}(x_i - \bar{x})^2}=\sigma^2 \cdot \left(\frac{1}{n} + \bar{x}^2\frac{1}{{\sum_{i=1}^n}(x_i - \bar{x})^2} \right)$$