Error in derivation of variance of $\beta_1$ in SLR [duplicate]

Question

I'm trying to derive the variance of the slope parameter for a simple linear regression in the following way, however I'm running into an issue I don't know how to resolve. Define $y_i=\beta_0+\beta_1\cdot x_i+\epsilon_i$ where $\epsilon_i\sim N(0,\sigma^2)$. Let $\hat\beta_0$ and $\hat\beta_1$ be the least-square estimators for the intercept and slope parameters. My derivation is as follows:

$ \begin{gather*} \\ Var(\hat\beta_1) = Var\left(\frac{\sum(x_i-\overline{x})(y_i-\overline{y})}{\sum(x_i-\overline{x})^2} \right) \\ \text{* since x's are fixed, they act as constants in this derivation } \\ = \frac{1}{(\sum(x_i-\overline{x})^2)^2}\sum(x_i-\overline{x})^2Var(y_i-\overline{y}) \\ = \frac{1}{(\sum(x_i-\overline{x})^2)^2}\sum(x_i-\overline{x})^2[Var(y_i)-Var(\overline{y})] \\ = \frac{1}{(\sum(x_i-\overline{x})^2)^2}\sum(x_i-\overline{x})^2\left[\sigma^2-\frac{\sigma^2}{n}\right] \\ = \frac{1}{(\sum(x_i-\overline{x})^2)^2}\left[\sigma^2-\frac{\sigma^2}{n}\right]\cdot\sum(x_i-\overline{x})^2 \\ = \frac{\left[\sigma^2-\frac{\sigma^2}{n}\right]}{\sum(x_i-\overline{x})}\ne\frac{\sigma^2}{\sum(x_i-\overline{x})} \end{gather*} $

My only thought is that I'm forgetting to account for the covariance between $y_i$ and $\overline{y}$, but I thought that this was simply 0. Any insight would be much appreciated!

Edit : I know there is a way with defining $\hat\beta_1$ as the linear combination of some constants, but I was curious if there is a simple fix to my steps

Edit 2 : I thought the covariance is accounted for by the following:

$ \begin{gather*} Var(y_i-\overline{y}) =Var\left(y_i-\frac{\sum{y_i}}{n}\right)\\ =Var\left(y_i-\frac{y_1}{n}-\frac{y_2}{n}-\ldots-\frac{y_i}{n}-\ldots-\frac{y_n}{n}\right)\\ =Var\left(\left(1-\frac{1}{n}\right)\cdot y_i-\frac{y_1}{n}-\ldots-\frac{y_n}{n}\right)\\ =\left(1-\frac{1}{n}\right)^2\cdot Var(y_i)+\frac{1}{n^2}\cdot Var(y_1)+\ldots+\frac{1}{n^2}\cdot Var(y_n)-\frac{2}{n^2}\mathop{\sum{}\sum{}}_{j\ne k}Cov(y_j,y_k)\\ \text{**I'm probably missing some rigour with the notation of the double sum, but you get the point}\\ \text{The covariances between different response values are equal to 0 by definition (since $\epsilon_i$ are i.i.d). Thus:}\\ =\left(\frac{n-1}{n}\right)^2\cdot \sigma^2+\frac{n-1}{n^2}\sigma^2\\ =\frac{\sigma^2}{n^2}((n-1)^2+(n-1))\\ =\frac{\sigma^2}{n^2}((n-1)(n-1+1)\\ =\frac{(n-1)\sigma^2}{n} \end{gather*} $

Answer 1

You could have made the calculations easier by noting that

\begin{align}\frac{\sum (x_i-\bar x) (y_i-\bar y) }{\sum (x_i-\bar x) ^2}&=\frac{\sum(x_i-\bar x) y_i}{\sum(x_i-\bar x) ^2}-\frac{\bar y\sum (x_i-\bar x) }{\sum(x_i-\bar x) ^2},\end{align} where the second term is zero as $\sum (x_i-\bar x) =0.$ Then you could have expressed $\hat \beta_1 =\sum c_i y_i$ where $c_i:=(x_i-\bar x) /\sum(x_i-\bar x) ^2.$ As $\sum c_i^2 =1/\sum(x_i-\bar x) ^2, $ the variance of $\hat \beta_1$ is $\sigma^2/\sum(x_i-\bar x) ^2.$

Also, a remainder that $\operatorname{Var}(X\pm Y) =\operatorname{Var}(X) + \operatorname{Var}(Y) \pm 2\operatorname{Cov}(X,Y);$ check whether you used that correctly in your calculations. And is $y$ and $\bar y$ independent?