I intended to derive the large sample distribution of the intercept of the least squares estimator $\beta_0$, but after many tries I could not obtain the solution that is given in the book of Stock&Watson as is depicted below.
I have been able to derive the large sample distribution of the slope $\beta_1$ myself, so we could treat that as a known random variable.
$\hat{\beta}_1 \sim N(\beta_1,\sigma_{\hat{\beta_1}}^{2})$
$\hat{\beta}_0$ have I earlier defined as $\hat{\beta}_0 = \bar{Y} - \hat{\beta}_1 \bar{X}$
The approach I chose was to treat $\hat{\beta}_0$ as a function of $\hat{\beta}_1$
We know that $\bar{Y}$ and $\bar{X}$ are consistent estimators of their true mean $\mu_X$ and $\mu_Y$ by means of LLN, so the equation becomes the following (given n is large). The assumptions of LLN are assumed to be all statisfied.
$\hat{\beta}_0 = \mu_Y - \hat{\beta}_1 \mu_X$
$\hat{\beta}_0 \sim N(\mu_Y - \beta_1 \mu_X,\frac{\sigma_{\hat{\beta_1}}^{2}}{(n\sigma_x^2)^2})$
Could anyone shed some light to this issue I am facing. I hope someone can help me out!
EDIT
$\bar{X}\sim N(\mu_X,\sigma_x^2/n)$ (assuming the case N is large and $X_i$ is i.i.d., $\bar{X}$ is a consistent estimator, and approximately normally distributed due to CLT)
$\bar{Y}\sim N(\mu_Y,\sigma_y^2/n)$
$\sigma_{\hat{\beta_1}}^2=\frac{1}{n}\frac{var((X_i-\mu_x)u_i)}{(var(x_i))^2}$
$u_i=Y_i-\beta_0-\beta_1 X_i$ (the unobserved error term of the simple linear regression model)
EDIT 2:
$\hat{\beta}_0 = \bar{Y} - \hat{\beta}_1 \bar{X}$
$var(\hat{\beta}_0) = var(\bar{Y} - \hat{\beta}_1 \bar{X})$
$ var(\hat{\beta}_0) = var(Y)/n + \bar{X}^2 var(\hat{\beta}_1)$
Note that X is regarded as nonrandom: $var(Y) = var( \beta_0 + \beta_1X + u) = var(u)$
$var(\hat{\beta}_0) = \frac{var(u)}{n} + \bar{X}^2 var(\hat{\beta}_1)$
$var(\hat{\beta}_0) = \frac{var(u)}{n} + \frac{\bar{X}^2}{n} \frac{var((X-\mu_x)u)}{var(x)^2}$
$var(\hat{\beta}_0) = \frac{1}{n} \Big( var(u)+ \bar{X}^2 \frac{var((X-\mu_x)u)}{var(x)^2} \Big)$
$var(\hat{\beta}_0) = \frac{1}{n} \Big( var(u)+ \bar{X}^2 \frac{var(X)var(u)+\mu_x^2var(u)}{var(x)^2} \Big)$
$var(\hat{\beta}_0) = \frac{1}{n} var(u) \Big( 1+ \bar{X}^2 \frac{var(X)+\mu_x^2}{var(x)^2} \Big)$
Then one might to say that $\bar{X}$ is a consistent estimator of $\mu_x$ given that n is large and $X$ is iid.
$var(\hat{\beta}_0) = \frac{1}{n} var(u) \Big( 1+ \mu_x^2 \frac{var(X)+\mu_x^2}{var(x)^2} \Big)$
$var(\hat{\beta}_0) = \frac{1}{n} var(u) \Big( 1+ E(X)^2 \frac{E(X^2)}{var(x)^2} \Big)$
$var(\hat{\beta}_0) = \frac{1}{n} var(u) \Big( 1+ \frac{E(X)^2E(X^2)}{(E(X^2)-E(X))^2} \Big)$
The final expression looks already very familiar with the $H_i$ in the textbook. But I am not there yet.