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:
$$
S_{xy} = \sum{x_i(y_i-\bar{y})},
\qquad
S_{xx} = \sum{x_i(x_i-\bar{x})}.
$$
Best Answer
Hint: The problematic part is
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 $Cov(U,V)=0$.