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Consider the linear regression $Y_i=\alpha_0+\beta_0 X_i+\epsilon_0$, where $i=1,2,...,n$, and $\epsilon \sim N(0, \sigma^2)$.

$(\hat{\alpha},\hat{\beta})$ estimate $(\alpha_0,\beta_0)$ and are found by minimizing $\sum_{i=1}^n (Y_i-\alpha-\beta X_i)^2.$

We get that $\hat{\alpha}=\bar{Y}-\hat{\beta} \bar{X}$ and $\hat{\beta}=\frac{\sum_{i=1}^n(X_i-\bar{X})Y_i}{\sum_{i=1}^n(X_i-\bar{X})^2}$.

Let $X_1,X_2,...,X_n$ be $i.i.d$ where $X_i \sim N(0, \tau^2)$.

I have to find the limiting distribution of $\sqrt{n}(\hat\beta - {\beta_0})$ as $n$ goes to infinity.

I think I have to show that $\sqrt{n^{-1}\sum_{i=1}^n(X_i-\bar{X})^2}\sqrt{n}(\hat\beta - {\beta_0})$ is independent of $X_i$'s first, but I am not sure.

I was thinking of plugging in for $(\hat\beta - {\beta_0})=\frac{1}{\sum_{i=1}^n(X_i-\bar{X})^2}$ into $\sqrt{n}(\hat\beta - {\beta_0})$, but I am not sure that is true.

Any help is appreciated, thank you.

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Assuming that $X_i \perp Y_j$ for all $i$ and $j$, and you can show that $\bar{X}_n$ is independent of $\sum_{i=1}^n (X_i - \bar{X}_n)^2$. Quick sketch of the proof: Show that $(\bar{X}_n, X_2 -\bar{X}_n, ...., X_n -\bar{X}_n)$ is multivariate normal with diagonal co variance matrix, hence the elements are jointly independent. Then, $$ \hat{\beta}_1 = \sum_{i=1}^n w_iY_i, $$ is linear combination of normal i.i.d r.vs $\{Y_i\}_{i=1}^n$, hence it is normaly distributed and unbiased estimator of $\beta_1$, hence just find its variance $Var(\hat{\beta}_1) =\sigma^2$ and conclude that $$ \sqrt{n}(\hat{\beta}_1 - \beta_1)\xrightarrow{D}\mathcal{N}(0, \sigma^2)\, . $$

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