How to calculate $\operatorname{Var}(b_0)$ where $b_0$ is estimator for $\beta_0$ in $y_i=\beta_0+\beta_1x_i+\epsilon_i$

Question

Consider the equation $y_i=\beta_0+\beta_1x_i+\epsilon_i$ for $i=1, \dotsc, n$.

We have unbiased estimators $b_0$ and $b_1$ for $\beta_0$ and $\beta_1$ respectively, where $b_0=\bar{y}-b_1\bar{x}$ and $b_1= S_{xy} / S_{xx}$. One can easily show that $$ \operatorname{Var}(b_1) = \operatorname{Var}\left( \frac{S_{xy}}{S_{xx}} \right) = \frac{1}{S_{xx}^2} \operatorname{Var}(S_{xy}) = \frac{\sigma^2}{S_{xx}}. $$

Now, I'm trying to find $\operatorname{Var}(b_0)$ as follows: $$ \operatorname{Var}(b_0) = \operatorname{Var}(\bar{y}) - \operatorname{Var}(b_1\bar{x}) = \operatorname{Var}(\bar{y})-\bar{x}^2 \operatorname{Var}(b_1) = \operatorname{Var}(\bar{y})-\frac{\bar{x}^2\sigma^2}{S_{xx}}. $$ Now, $$ \operatorname{Var}(\bar{y}) = \operatorname{Var}\left( \frac{1}{n}\sum y_i \right) = \frac{1}{n^2} \sum \operatorname{Var}(y_i) = \frac{\sigma^2n}{n^2} = \frac{\sigma^2}{n}. $$ Therefore, $$ \operatorname{Var}(b_0) = \frac{\sigma^2}{n}-\frac{\bar{x}^2\sigma^2}{S_{xx}}. $$

But according to the answer given in my book, it says $$ \operatorname{Var}(b_0) = \frac{\sigma^2}{n}+\frac{\bar{x}\sigma^2}{S_{xx}}. $$

Note: $$ S_{xy} = \sum{x_i(y_i-\bar{y})}, \qquad S_{xx} = \sum{x_i(x_i-\bar{x})}. $$

Answer 1

Hint: The problematic part is

$\operatorname{Var}(b_0) = \operatorname{Var}(\bar{y}) - \operatorname{Var}(b_1\bar{x})$

Suppose $U$ and $V$ are random variables with finite variances then the variance of the difference is

$Var(U-V)=Var(U)\color{red}+Var(V)-2Cov(U,V)$

If $U$ and $V$ are additionally independent then $Cov(U,V)=0$.

Answer 2

$$ var(b_0) = var( \bar{y} ) + var(b_1 \bar{x} ) - 2cov(\bar{y} , b_1 \bar{x} ), $$ where $$ cov(\bar{y} , b_1 \bar{x} ) = \bar{x}cov\left( \bar{y}, b_1\right) = 0, $$ as \begin{align} cov(\bar{y}, b_1) &= \frac{1}{nS_{xx}}cov\left(\sum y_i , \sum ( x_j - \bar{x})y_j \right)\\ & = \frac{1}{nS_{xx}}\sum ( x_j - \bar{x})\sigma ^ 2 \\ & = \frac{\sigma ^ 2 ( \bar{x}- \bar{x})}{nS_{xx}}\\ & = 0. \end{align} Hence, $$ var(b_0) = var( \bar{y} ) + var(b_1 \bar{x} ) = \frac{\sigma ^ 2}{n} + \frac{\sigma^2 \bar{x} ^ 2}{ S_{xx}}, $$