Proof for the asymptotic correlation coefficient $\lim_{x\to\infty} \rho_{{\hat{\beta_0}},{\hat{\beta_1}}} = -\frac{\sqrt{3}}{2}$

Question

Be my regression model: $$Y_i = 1 + X_i + \epsilon_i, \hspace{5mm} \epsilon_i \sim N(0,1), \hspace{5mm} X_i = \frac{i-1}{n-1}, \;\;\; i = 1, 2, ..., n$$ I want to prove that the asymptotic correlation coefficient equals: $$\lim_{x\to\infty} \rho_{{\hat{\beta_0}},{\hat{\beta_1}}} = -\frac{\sqrt{3}}{2}$$

Hints: $\begin{cases} S_{xx} = \frac{n(n+1)}{12(n-1)} \\ \sum\limits_{i=1}^n i = \frac{n(n+1)}{2} \\ \sum\limits_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6} \end{cases}$

Answer 1

Use the formulas below to obtain $\rho_{{\hat{\beta_0}},{\hat{\beta_1}}}=\frac{\operatorname{cov}\left(\hat{\beta_0},\hat{\beta_1}\right)}{\sqrt{\operatorname{Var}\left(\hat{\beta_0}\right)\cdot \operatorname{Var}\left(\hat{\beta_1}\right)}}$. Since $\epsilon_i \sim N(0,1)$ it follows that $\sigma^2=1$

$$\operatorname{Var}\left(\hat{\beta_0}\right)=\frac{\sigma^2\sum_{i=1}^nx_i^2}{n\sum_{i=1}^n\left(x_i-\bar{x}\right)^2}$$

$$\operatorname{Var}\left(\hat{\beta_1}\right)=\frac{\sigma^2}{\sum_{i=1}^n\left(x_i-\bar{x}\right)^2}$$

and

$$\operatorname{cov}\left(\hat{\beta_0},\hat{\beta_1}\right)=\frac{-\sigma^2\sum_{i=1}^nx_i}{n\sum_{i=1}^n\left(x_i-\bar{x}\right)^2}$$

If we look at the formula of $\rho_{{\hat{\beta_0}},{\hat{\beta_1}}}$ we see that $\sum_{i=1}^n\left(x_i-\bar{x}\right)^2$ are cancelling out. With the formulas of the hints it is straigtforward to calculate the limit of $\rho_{{\hat{\beta_0}},{\hat{\beta_1}}}$. Feel free to ask if something is still unclear.